math-5-ev

Elementary Mathematics Class Five

100 National Curriculum And Textbook Board, Dhaka

Prescribed by the National Curriculum and Textbook Board as the textbook for class five from the academic year-2013

ELEMEMTARY MATHEMATICS CLASS FIVE

Written by

Illustration

Shamsul Haque Mollah A.M.M. Ahsan Ullah Dr. Amal Halder Shawpon Kumar Dhali

Kazi Saifuddin Abbas Art Editing

Hashem Khan Kazi Saifuddin Abbas

Translated & Edited by

Dr. Munibur Rahman Chowdury Sheikh Kutubuddin

NATIONAL CURRICULUM AND TEXTBOOK BOARD, DHAKA

Published by

National Curriculum and Textbook Board 69-70, Motijheel Commercial Area, Dhaka-1000

(All rights reserved by the publisher)

Trial Edition First Print : December, 2012

Co-ordinator Mohd. Manirul Islam

Graphics Md. Abul Hossain

Design NATIONAL CURRICULUM AND TEXTBOOK BOARD, DHAKA

To be distributed free of cost by the Government of the People’s Republic of Bangladesh under the Third Primary Education Development Program

Printing: Mousumi Of-set Press, 38 Banglabazar, Dhaka.

Preface The children themselves are boundless wonder of the world. Exploration of children’s world of wonder has been the subject of thinking of all scholars from primitive age to the present age of information and technology. These outcomes of thinking of the scholars have been the basis of children education, and have been placed in Education policy of 2010. In the light of adapting Education policy of 2010 and to help the children for their normal and gradual development, aims and objectives of primary education have been re-fixed and placed in curriculum document. The terminal competencies for primary education, subject wise terminal competencies and class wise competencies for all the subjects in the form of essential learning continua have been developed in the light of re-fixed aims and objectives, giving proper importance and emphasis on the total development of the children. Against this back-drop, all the steps of the curriculum development have been reflected in the developed textbooks. This book is the English version of the Bangla textbook prepared for the students of Class Five. The book has been written on the basis of the revised curriculum to be effective from 2013.Throughout the early years of life, children notice and explore mathematical dimensions of their world. They compare quantities, find patterns, navigate in space, and grapple with real problems. Mathematics helps children making sense of their world-outside of school and helps them to construct a solid foundation for success in school.

The revised curriculum takes note of it, and the textbook ensures that the children get adequate mathematical understanding and skills. Special attention has been given to make the contents of the book, as far as possible, plain and lucid to our tender aged learners. The salient feature of the new book is that it is learner-centered and activity-based. To make the book attractive to the young learners, it is printed in four colours and varieties of pictures are included in it. The book encourages that the students will learn through games.

Despite careful efforts of all concerned, some errors might have remained in the book. Such errors, if any, will be removed in the next edition. Finally, I would like to express my heartfelt thanks and gratitude to those who have provided their valuable contributions in writing, editing, evaluating and translating this book. I sincerely hope that the book will be useful to those for whom it has been prepared.

Profe Professor f ssor Md. Mostafa fe Mostofa f Kamaluddin fa Chairman National Curriculum and Textbook Board, Dhaka

Content Chapter

Subject

Page

One

Multiplication

1

Two

Division

6

Three

Problems Involving the Four Rules

12

Four

Average

22

Five

G.C.D and L.C.M

27

Six

Mathematical Symbols and Statements

36

Seven

Common Fractions

42

Eight

Decimal Fractions

75

Nine

Percentage

86

Ten

Measurement

92

Eleven

Time

108

Twelve

Arrangements of Data

118

Thirteen

Geometry

130

Fourteen

Calculator and Computer

134

Answers

141

Chapter One

Multiplication Product Multiplicand Multiplier Multiplier Multiplicand Product Product Multiplicand Multiplier

We know

Example 1. Multiply 437 by 235.

In the second step 437 Solution : Multiplicand multiplication by 3 tens has Multiplier 235 been written down one place 2185 437 5 to the left; in the third step multiplication by 2 hundreds 437 1311 3 tens has been written down 437 874 2 hundreds two places to the left.

Product 102695

Example 2. Multiply 2456 by 293.

Solution : 2456 Multiplicand Multiplier 293 7368 2456 3 2456 90 221040 491200 2456 200 719608 Product

In the second step multiplication by 9 tens has been written down putting 0 in ones place; in the third step multiplication by 2 hundreds has been written down putting zero in ones and in tens places.

Fill in the blanks : (a) 357

29

10353 Here, multiplicand

multiplier

product

(b) 723

64

46272 Here, multiplicand

multiplier

product

(c) 287

375

(d) 823

356

(e) 8527

672

(f) 8452

795

Elementary Mathematics

Example 3. Multiply 6582 by 308.

The tens place of the Solution : 6 5 8 2 multiplier is 0 ; so multiplication by tens is not shown. In the 308 second step multiplication by 52656 6582 8 3 hundreds has been written 1974600 6 5 8 2 3 hundreds down putting zero in ones and in tens places.

Product 2027256

Example 4. Multiply 7396 by 600.

The ones as well as tens place of the multiplier is 0; so multiplication by ones and tens is not shown. Multiplication by 7 3 9 6 6 hundreds 6 hundreds has been written Product down putting zero in ones and in tens places.

Solution :

7396 6 00 44376 00

Example 5. Multiply 3271 by 420.

Solution : 3 2 7 1 420 65420 1308400 1373820

The ones place of the

Multiplicand multiplier is 0; so multiplication by ones is not shown. In the first Multiplier step multiplication by 2 tens has been written down by putting zero in 3 2 7 1 2 tens ones place; in the second step 3 2 7 1 4 hundreds multiplication by 4 hundreds has been written down by putting zero in ones and in tens places.

Product

Fill in the blanks :

(a)

(b)

4586 4 9 4 274 1834400 18756 2

9207 64 36 280 55242 0 58924 0

Elementary Mathematics Multiplication by Easy Method Example 6. Multiply by easy method (a) 9999 425 10000 1 425 425 1 10000 425 4250000 4249575 (c) 3010 893 3000 10 893 3000 893 10 2679000 8930 2687930

425

(b) 8254 990 10000 10 8254 1000 8254 8254 8254000 82540 8171460

893

(d) 9099 857 9100 1 857 9100 857 1 7798700 857 7797843

10

857

Multiply by Easy Method (a) 9990 457 10000 10000 4570000

(b) 5010 358 5000 5000 1790000

457 457

358 358

Fill in the blanks : 1 9 366 54 0 5856

5 1 0

2 1 8 1 8 4 2 2 8 1 8 2 4 0 0 1 0 4

3

4 5 9 0 0 0 5 7 0 0 3 6

1 5

0

Elementary Mathematics

Example 7. A person's daily income is 216 taka.

What is her income in one year ?

1 year = 365 days Solution :

1 year = 365 days In one day she earns 216 taka in 365 days she earns 216

365

taka.

365 216 2190 3650 73000 78840 In one year the income of the person is 78840 taka.

Example 8. In a plantation there are 2605 betelnut trees. How many betelnut trees are there in 316 such plantations ?

Solution :

In 1 plantation there are 2605 betelnut trees in 316 plantations there are 2605

316 betelnut trees

2605 316 15630 26050 781500 823180 In 316 plantations there are 823180 betelnut trees.

4

Elementary Mathematics

Exercise 1 1.

Find the products : (a)

(b)

589 162

(e) 7 6 9 5 620 2.

3.

(f) 5 0 3 6 387

2 4 70 359

(h)

7692 609

(b) 746 by 635

(c) 2674 by 628

(d) 7091 by 890

(e) 9236 by 903

(f)

6759 by 900

Fill in the blanks : (b)

9

1

9.

10.

11.

12.

(g) 8 6 3 4 700

(d)

(a) 537 by 560

(c)

3

8

1 4 2

3 1

0

0 1 7 2 0 0 4 3 0 0 0

0 0 1 9 1 0 0

5.

6.

7.

8.

703 249

Multiply :

(a)

4.

(c)

427 307

6 2

6

0 2 1 5 0 3 7 0 0 8 4

1

9 6 2

Find the products by the easy method : (a) 567 99 (b) 99 990 (c) 6273 (d) 8593 990 (e) 9999 607 (f) 9999

999 400

Multiplicand is 6381 and multiplier is 215; what is the product ?

The price of a bicycle is 5825 taka. What is the price of 165 such bicycles ?

A ream of paper has 500 sheets. How many sheets of paper are there in 298 reams ?

A person's daily income is 275 taka. What will be her income in one year ?

1 year = 365 days The price of a chair is 800 taka. What is the price of 2614 such chairs ?

A book has 439 pages. How many pages are there in 2038 such books ?

A nursery has 5834 saplings. How many saplings are there in 486 such nurseries ?

A godown has 8326 kg of rice. How many kg of rice are there in 602 such godowns ?

5

Chapter Two

Division We know, in case of division without remainder : Dividend

Divisor

Quotient

Dividend

Quotient

Divisor

Divisor

Quotient

Dividend

In case of division with remainder : Dividend

Divisor

Quotient

Remainder

Divisor

Dividend Remainder

Quotient

Quotient

Dividend Remainder

Divisor

Example 1. Divide 8368 by 16.

Example 2. Divide 6489 by 47.

Solution :

Solution :

16 8368 523

47 6 4 8 9 1 3 8

80 36 32 48 48 0 Quotient is 523

47 178 141 379 376 3 Quotient is 138, remainder is 3

Elementary Mathematics

Example 3. Divide 9415 by 23 Solution :

23 9 4 1 5 4 0 9

23 9 4 1 5 4 0 9

92 215

92 21

207

00 215

8 Here one step has been suppressed.

207 8

Quotient is 409, remainder is 8

Example 4. Divide 79513 by 25

Example 5. Divide 49735 by 245

Solution :

Solution :

25 7 9 5 1 3 3180 7 5

245 4 9 7 3 5 203 4 9 0 73

4 5 2 5 2 0 1

00 73 5

2 0 0 1 3 0 0 1 3

73 5 0

Quotient is 203

Quotient is 3180, remainder is 13

Which digit or digits, when placed in the blank space, will make the quotient less than 10 ?

84 8 5

7

Elementary Mathematics

Division by 10 or 100 To divide by 10 or by 100 by the easy method, we put a comma before one digit, or two digits, from the right of the dividend, respectively. The number to the left of the comma is the quotient; the number to the right of the comma is the remainder.

Example 6. Divide 2465 by 10.

There is one zero (0) to the right

Solution : 10 2 4 6 5 246 2 0 46 40 65 60 5

of 1 in the divisor. Putting a comma before one digit from the right of the dividend we get 246, 5. The number 246 to the left of the comma is the quotient; the number 5 to the right of the comma is the remainder.

Quotient is 246, remainder is 5.

Example 7. Divide 30845 by 100.

There are two zeros (00) to the right of 1 in the divisor. Putting a Solution : comma before two digits from the right of the 100 3 0 8 4 5 308 dividend we get 308, 45. The number 308 to the 3 0 0 left of the comma is the quotient; the number 8 45 45 to the right of the comma is the remainder.

8 00 Similarly, the quotient and remainder of a number upon division by 1000, or by 45 10000, are easily written down.

Quotient is 308, remainder is 45.

8

Elementary Mathematics

Find out the mistake and rectify : (a) 4 7

5 6 4 1 20

(b)

(c)

3 2 9 6 6 4 32

68 7 5 1 0 10 9

4 7

6 8

9 6

9 4 9 4

6 4 6 4

0

0

71 0 61 2 9 8

Example 9. Divide 63524 by 50.

Example 8. Divide 42578 by 100

Solution : Here the divisor is 50. We multiply both dividend and divisor by 2 : 63524 2=127048; 50 2 = 100.

Putting a comma before two digits form

Solution : In the divisor there are two zeros (00) to the right of 1.

Putting a comma before two digits from the right of the dividend, we get 425, 78.

So, the quotient is 425 and the remainder is 78.

the right of the dividend we get 1270, 48.

So the required quotient is 1270, and the required remainder is 48

2 = 24.

Explanation: Multiplying Dividend and divisor by 2 does not change the required quotient,but the required remainder is multiplied by 2. So we have to divide 48 by 2 to get the required quotient.

Fill in the blanks : (a)

433

12

quotient

remainder

(b)

4050

19

quotient

remainder

(c)

2700

30

quotient

remainder

(d)

56789

100

quotient

remainder

(e)

3795

50

quotient

remainder

9

Elementary Mathematics

Example 11. In a division problem, dividend is 28087, the divisor is 264 and the remainder is 103. What is the quotient ?

Example 10. In a division problem, the dividend is 37037, the quotient is 89 and the remainder is 13. What is the divisor ?

Solution : We know,

Solution : We know, divisor = (dividend

remainder)

Here, dividend = 37037 divisor

quotient = (dividend remainder)

quotient

Here, dividend

remainder 13 = 37024

37024 416

= 28087 quotient

89

remainder 103 = 27984

27984 106

264

quotient is 106

divisor is 416

Example 12. The price of 325 kg of rice is 7800 taka. What is the price of 1 kg of rice ?

Solution : Price of 325 kg = 7800 price of 1 kg = 7800 taka 325

divisor

Example 13. The price of 1 quintal of rice is 4200 taka. What is the price of 1 kg of rice ?

[1 quintal = 100 kilogram] Solution : Price of 100 kg = 4200 taka price of 1 kg = 4200 taka 100

325

7 8 0 0 24 Here the divisor is 100.

Putting a comma before two digits of the dividend from the right, we get 42,00.

65 0 1 3 00 1 3 00 0

quotient is 42

The price of 1 kg of rice is 24 taka.

The price of 1 kg of rice is 42 taka.

10

Elementary Mathematics

Exercise 2 1. Fill in the blanks : (a) 76965 (b) 21547 (c) 7358

15 29

(e) 3652 2.

Divide (a) 25748 (d) 75089 (g) 14970 (j) 43560 (m) 39768

98 325 365 100 100

divisor 5131 Here dividend divisor 743 Here dividend (d) 52895 419406 149 (f) 52185 213 317724 (b) (e) (h) (k) (n)

54871 52889 24135 65900 83090

37 289 10 100 100

(c) (f) (i) (l) (o)

42835 43702 87520 67500 93500

quotient quotient

197 342 10 100 100

3.

(a) Dividend is 37920, quotient is 12 and remainder is 0. What is the divisor ? (b) Dividend is 73635, quotient is 152 and remainder is 67. What is the divisor ? (c) Dividend is 35792, divisor is 47 and remainder is 25. What is the quotient ? (d) Divisor is 325, quotient is 72 and remainder is 9. What is the dividend ?

4.

How many days are there in 10008 hours ?

5.

Divide by 87 the largest number of five digits.

6.

A basket can hold 355 mangoes. How many such baskets will be needed to hold 25560 mangoes ?

7.

Dividing a number by 54, the quotient is18 and the remainder is 50. What will be the quotient when the number is divided by 73 ?

8.

Divide the largest number of 5 digits that can be formed by using the digits 8, 3, 5, 0, 7 once, by the largest number of three digits.

9.

10500 taka is needed if every person is given 140 taka. What is the number of persons ?

10.

The product of two numbers is 43290. One number is 555; what is the other number ?

11.

Divide by 165 the smallest number that can be formed by using the digits 9, 4, 0, 2, 6 once.

12.

The product of two numbers is 89262. One number is 342; what is the other number ?

11

Elementary Mathematics Example 1. Mina and Rina together have 7532 taka. Mina has 560 taka more than Rina. What amount of money does Mina and Rina each have ?

Solution : Subtracting Mina's excess amount from the total amount, will render their amounts of money equal.

7532 taka 560 taka = 6972 taka Rina has 6972 2 taka = 3486 taka Mina has 3486 560 taka = 4046 taka Mina has 4046 taka and Rina has 3486 taka.

Example 2. Mr Altaf's monthly pay is 9870 taka. Every month he spends 3800 taka on house rent and 5650 taka on household expenses. The remaining money he saves in a bank. What amount of money does Mr. Altaf save in a year ?

Solution : Every month he spends (3800 + 5650) taka = 9450 taka on house rent and household expenses.

Every month he saves ( 9870 9450 ) taka = 420 taka In 1 year he saves 420

12

taka = 5040 taka

Example 3. The sum of ages of a father and his daughter is 80 years. Father's age is four times the age of the daughter. What are their ages ?

Solution : Daughter's age = 1 time daughter's age Father's age = 4 times daughter's age Sum of father's and daughter's ages = 5 times daughter's age 5 times daughter's age =80 years daughter's age = 80 years 5 = 16 years father's age = 16 years 4 = 64 years father`s age is 64 years and daughter`s age is 16 years.

[Alternatively, father's age = total age - daughter's age = (80-16) years = 64 years]

13

Elementary Mathematics Example 1. Mina and Rina together have 7532 taka. Mina has 560 taka more than Rina. What amount of money does Mina and Rina each have ?

Solution : Subtracting Mina's excess amount from the total amount, will render their amounts of money equal.

7532 taka 560 taka = 6972 taka Rina has 6972 2 taka = 3486 taka Mina has 3486 560 taka = 4046 taka Mina has 4046 taka and Rina has 3486 taka.

Example 2. Mr Altaf's monthly pay is 9870 taka. Every month he spends 3800 taka on house rent and 5650 taka on household expenses. The remaining money he saves in a bank. What amount of money does Mr. Altaf save in a year ?

Solution : Every month he spends (3800 + 5650) taka = 9450 taka on house rent and household expenses.

Every month he saves ( 9870 9450 ) taka = 420 taka In 1 year he saves 420

12

taka = 5040 taka

Example 3. The sum of ages of a father and his daughter is 80 years. Father's age is four times the age of the daughter. What are their ages ?

Solution : Daughter's age = 1 time daughter's age Father's age = 4 times daughter's age Sum of father's and daughter's ages = 5 times daughter's age 5 times daughter's age =80 years daughter's age = 80 years 5 = 16 years father's age = 16 years 4 = 64 years father`s age is 64 years and daughter`s age is 16 years.

[Alternatively, father's age = total age - daughter's age = (80-16) years = 64 years]

13

Elementary Mathematics Unitary Method In day-to-day life we need to do calculations. As an example, let us find the price of 4 pencils, given that the price of 10 pencils is 60 taka. First we divide the price of 10 pencils by 10, and thus find the price of 1 pencil. Then we multiply the price of 1 pencil by 4, which gives us the price of 4 pencils, This process of solving problems is called the unitary method. The quantity to be determined has to be placed towards the end of every working line; see the following examples.

Example 4. The price of 9 books is 216 taka, what is the price of 12 books ?

Price of 9 books is 216 taka Solution : price of 1 book is 216 9 taka = 24 taka price of 12 books is 12 24 taka = 288 taka Example 5. The price of 12 dozen writing pads is 2304 taka, what is the price of 8 writing pads ?

Solution :

12 dozen = 12

12

144

1 dozen = 12

Price of 144 writing pads is 2304 taka price of 1 writing pad is 2304 144 taka = 16 taka price of 8 writing pads is 8 16 taka = 128 taka Example 6. The price of one hundred lychees is 300 taka. What is the price

of 25 lychees ?

Solution : Price of 100 lychees is 300 taka price of 1 lychee is 300 100 taka = 3 taka price of 25 lychees is 25 3 taka = 75 taka Example 7. A worker earns 1575 taka per week. What amount of money will she earn in 12 days ?

Solution :

In 7 days she earns 1575 taka In 1 day she earns 1575 7 In 12 days she earns 12 225 14

taka = 225 taka taka = 2700 taka

Elementary Mathematics Example 8. In a hostel there is food for 40 students for 20 days. If 10 new students arrive, how long will the food last ?

Solution : With the arrival of 10 new students, the number of students will be 40+10=50.

For 40 students the food will last 20 days for 1 student the food will last 20 40 days = 800 days for 50 students the food will last 800 50 days = 16 days

We observe : For consuming a given amount of food, the less the number of consumers the more the number of days; the more the number of consumers the less the number of days. So, in the second step we have multiplied by 20 and in the third step we have divided by 50

Example 9. 6 persons can harvest the crop of a piece of land in 21 days.

In how many days will 18 persons harvest that crop ?

Solution : 6 persons can harvest the crop in 21 days 1 21 6 days = 126 days 18 126 18 days = 7 days

Example 10. 200 persons need 15 days to excavate a pond. How many additional persons must be employed if the pond is to be excavated in 10 days ?

Solution : 15 days are needed to excavate the pond by 200 persons 1 day is needed to excavate the pond by

200 10 days are needed to excavate the pond by 3000

300 200 = 100 100 additional persons must be employed.

15

15 persons = 3000 persons 10 persons = 300 persons

Elementary Mathematics

Exercise 3 (A) 1. Fill in the blanks : (a) If 1 dozen bananas cost 30 taka, 3 dozen bananas will cost (b) If 10 eggs cost 40 taka, 1 egg will cost taka. taka. (c) If 6 pencils cost 24 taka, 1 pencil will cost (d) 3 baskets hold 48 mangoes. 1 basket will hold mangoes. (e) 1 basket holds 8 mangoes. 9 baskets will hold mangoes.

taka.

2. 20 persons can do a piece of work in 15 days. In how many days can 15 persons do that work ?

3. 16 persons can do a piece of work in 5 days. In how many days can 20 persons do that work ?

4. 8 persons can harvest the crop of a land in 21 days. In how many days will 14 persons harvest that crop ?

5. A certain quantity of food can be consumed by 200 persons in 20 days. How many persons will consume that food in 40 days ?

6. A worker earns 490 taka per week (working 7 days a week), How many days will she need to earn 1050 taka ?

7. A certain amount of food serves 45 persons for 20 days. How many persons will that food serve for 25 days?

8. In a girls' hostel 16 students have food for 25 days.. Some new students came in; as a result the food was consumed in 20 days. What is the number of new students ?

9. In a hostel 500 students have food for 50 days. After 10 days 300 more students joined the hostel. How many days will the ramaining food last ?

10. A family of 8 persons has food for 26 days. After 5 days 1 person went away. How many days will the remaining food last ?

11. 200 persons are needed to excavate a pond in 25 days. How many additional persons are needed if the pond is to be excavated in 20 days ?

12 30 persons can complete a piece of work in 18 days. How many more persons are needed if the work is to be completed in 12 days ?

13. 16 persons need 56 kg of rice in 1 week. How much rice will be needed by 24 persons in four weeks ?

16

Elementary Mathematics

Use of Brackets We use brackets to form one mathematical statement from two mathematical statements.

Apurba's father gave him 50 taka to purchase writing pads and pencils. Apurba bought a writing pad for 16 taka and two pencils for 12 taka. After buying writing pad and pencils Apurba was left with 22 taka.

Mathematical statement 50 50

Mathematical statement

16 12 28

50

22

16

12

22

Observing the two methematical statements we find that brackets ( ) have been used to combine the two numbers to be subtracted from 50; the numbers inside the brackets have been added; finally their sum has been subtracted from 50. We observe that 50 (16 + 12) = 50 28 = 22 and 50 16 12 = 34 12 = 22 Thus 50 (16 + 12) = 50 16 12.

In other words, if there is a minus sign before a bracket, then the signs of the numbers inside the brackets must be changed upon removal of the brackets. In a mathematical statement, work involving muliplication and division has to be carried out before addition or subtraction. If there are brackets, then calculations inside the brackets has to be carried out first. For example : 15 42 (11 + 3) = 15 42 14 = 15 3 = 12 Let us remember the Rules of Simplification : Calculations are to be done from left to right.

First we do work involving division, muliplication then we do work involving addition and subtraction.

If there are brackets, calculations inside the brackets has to be carried out first.

First we do work on first brackets ( ), then we do work on second brackets { }; finally we do work on third brackets [ ].

17

Elementary Mathematics Example 1. Simplify : 25 Solution : 25 5 5 25 5 5 5 25 5 40 25 5 40 5 200

5 24 24 8

Example 2. Simplify : 78 56 Solution : 78 56 165 56 165 78 56 165 78 56 165 78 56 21 78 78 77 1 Example 3. Simplify : 48 Solution : 48 4 28 4 28 48 4 28 48 4 28 48 4 48 4 48 8 6

4 4 4 4 7

5

24 18 18 15

15

3

165 48 6 8 9 72 2 144

28 12 12

48 9 2

4 7 4

6 2

12 3

9

7

3

Example 4. Simplify : 36 3 4 5 4 3 4 5 4 8 1 Solution : 36 12 4 5 4 1 8 2 12 4 12 8 96 18

8

1

2

3

Elementary Mathematics

Example 5. The price of 6 chairs and 4 tables totals 9570 taka. The price of one chair is 625 taka; what is the price of one table ?

Brief solution : Price of 1 table Solution : Price of 1 chair is 625 taka Price of 6 chairs is 675

6 taka

4050 taka Total price of 6 chairs and 4 tables is 9570 taka.

9570

675 6

4 taka 4 taka

9570 4050 5520 4 taka 1380 taka Price of 1 table is 1380 taka.

Price of 6 chairs is 4050 taka.

Price of 4 tables is

9570

4050 taka

5520 taka Price of 1 table is 5520

4 taka = 1380 taka.

Example 6. In a division problem, the divisor is 78, the quotient is 25 and the remainder is one-third of the divisor. What is the dividend ?

Solution : We know Dividend Dividend

Divisor 78 1950

Quotient

Remainder

25

3

78 26

1976 The required dividend is 1976.

19

Elementary Mathematics

Exercise 3 (B) 1. Minuend is 985214 and subtrahend is 97465; what is the difference ?

2. Subtracting 68975 from a number, the difference is 794768. What is the number ?

3. From the smallest number of six digits, subtract the largest number of five digits.

4. What number when added to the largest number of four digits will make the sum 9 lac ?

5. What is the difference between the largest number and the smallest number of six digits that can be formed by using each of the digits 6, 8, 9, 5, 0, 4 once?

6. Raju's father sold paddy for 25830 taka, wheat for 30645 taka and lentil for 9786 taka; he then bought jute costing 45927 taka. What amount of money remained with him ?

7. A school had 785 students. At the beginning of the year 142 students left the school and 250 students were newly admitted. What became the number of students of the school ?

8. Proma, Rimi and Monisha made 70 flags to decorate the school on Victory Day. It was found that Proma had made 5 more flags than Rimi, again Monisha had 6 more flags than Proma. How many flags did each one of them make ?

9. Mina has 45987 taka, Raju has 8250 taka less than Mina. Rony has 985 taka more than Raju. What is the total amount of money of the trio ?

10. The sum of four numbers is 468520. The first two numbers are 73584 and 64209. The third number is less than the first number by 9485. What is the fourth number ?

11.

9 dozen pencils cost 1620 taka. What is the cost of 1 pencil ?

12. 38 hali of bananas cost 1216 taka. What is the cost of 1 banana ? [ 1 hali is any collection of 4 objects of the same kind ] 13. The sum of the ages of a father and his son is 96 years. The age of the father is 3 times that of the son. What are their ages ?

14. A basket holds 168 mangoes. From the mangoes of 15 such baskets, Mina was given 780, and Rani was given 750 mangoes. The rest of the mangoes was given to Raju; how many mangoes did Raju get ?

20

Elementary Mathematics 15. In a division problem, the dividend is 8903, the divisor is 87 and the remainder is 29.

What is the quotient ?

16. In a division problem the divisor is 12 times the remainder and the dividend is 9896.

The remainder is 8; what is the quotient ?

17. The product of two numbers is 6272, 4 times of one of the numbers is 256; what is the other number ?

18

The price of 2 cows and 3 goats together is 25080 taka. The price of 1 goat is 3560 taka; what is the price of 1 cow ?

19. The price of 14 chairs and 6 tables together is 17650 taka. The price of one table is 1250 taka; what is the price of one chair ?

20. Raju and Rony together have 690 lychees. Rony has 86 lychees less than Raju. How many lychees does Raju, as well as Rony, have ?

21. The pays of Farida and Fatema total 19950 taka. Fatema's pay is 2450 taka more than that of Farida. What are the pays of Farida and Fatema each ?

22. Mr. Jalal's monthly salary is 8765 taka. Every month he spends 3225 taka on house rent and 4850 taka on other items; the rest of the money he deposits in a bank. What will be his saving in 8 months ?

23. From a departmental store Zahidul Hasan bought 40 kg of rice, Soyabean oil for 265 taka and fish for 588 taka. Each kg of rice costs 18 taka. He gave 2000 taka to the cashier.

What amount will the cashier refund him?

21

Chapter Four

Average

In the first row of the picture above, there are 7 mangoes in the first tray, 9 mangoes in the second tray, 11 mangoes in the third tray, 9 mangoes in the fourth tray. The mangoes in these four trays are put together in one large tray (middle row of the picture); so there are in all 7 + 9 + 11 + 9 = 36 mangoes in the large tray. Dividing by 4, which is the number of trays in the first row, we get 9. If each of the original four trays had contained 9 mangoes, then also the total number of mangoes in the four trays would have been the same, because 9 + 9 + 9 + 9 = 36 (see last row of the picture).

We express this fact by saying that the average number of mangoes in the four trays is 9.

Here, 9

36

4

Total number of mangoes in the four trays

number of trays

In general : Average of several quantities of the same kind = Sum of the quantities

Number of quantities

In the example above, there were in all 36 mangoes in the 4 trays; so the average number of mangoes in the trays was 9. Observe that 9 4 = 36 In general : Average

Number of quantities = Sum of the quantities

Elementary Mathematics Finding Average Example 1. The ages of Mina, Mithu Runu, Rony and Nila are respectively 10, 12, 13, 11 and 14 years. What is their average age ?

Solution : Sum of their ages 10 12 13 11 14 years 60 years Number of persons = 5 Average age 60 years 5 60 5 years 12 years So the required average age is 12 years.

Example 2. The runs scored by a cricketer in the eight matches of a one-day series are 47, 19, 23, 12, 37, 3, 14, 13. How many runs did he score on average ?

Solution : Total runs 47 19 23 12 37 3 14 13 168 Number of matches = 8 Average run 168 8 21 So the criketer scored 21 runs on average in that series.

Example 3. The amount of rainfall at a certain place in Bangladesh during the twelve months of the year 2011 were as follows. What is the average amount of monthly rainfall at the place ?

Month

Amount of rainfall

Month

Amount of rainfall

January

23 mm

July

296 mm

February

06 mm

August

271 mm

March

18 mm

September

104 mm

April

57 mm

October

33 mm

May

31 mm

November

21 mm

June

108 mm

December

04 mm

Solution : Total amount of rainfall at the place in twelve months of 2011 23 06 18 57 31 108 296 271 104 33 21 04 mm 972 mm. Number of months = 12 Average monthly rainfall ( 972 mm) 12 972 12 mm 81 mm Average monthly rainfall at that place in the year 2011 was 81 mm.

23

Elementary Mathematics

Example 4. In a partnership business Helen invested 15,000 taka, Kamal 20,500 taka, Halim 18,000 taka, Titas 13,500 taka and Shishir 17,000 taka. How much money did they invest on average ?

Solution : Total investment 15,000 20,500 84, 000 taka Number of investors Average investment

18,000

13,500

5 84,000 taka

17,000 taka

5

84,000 5 taka 16,800 taka

Example 5. In a test series of five cricket matches, the average of the runs made by six batsmen of the visiting team was 76; the average of the runs made by four bowlers was 21. What average run did those players make in that series ?

Solution : Runs made by six batsmen in five matches 6 76 5 76 30 2280 Runs made by four bowlers in five matches 4 21 5 21 20 420 Total runs 2280 420 2700 Total number of matches in the series = 5 The total number of quantities to be considered in finding average run is Number of players 10 5 Average run

2700

50

Number of matches

50

54.

Players of the visiting team made 54 runs on average in that series.

24

Elementary Mathematics

Example 6. The sum of thirteen numbers is 1924. The average of seven of those numbers is 172. What is the average of the other six numbers ? What is the average of all the numbers ?

Solution : Average of 7 numbers is 172. So sum of those 7 numbers Sum of the other 6 numbers

1924 720 720 120 1924 148

So, average of those 6 numbers Average of all 13 numbers

172 7 1204

1204 6 13

Average of the other 6 numbers is 120. Average of all 13 numbers is 148.

Example 7. In the concluding examination of Class Five, Dolon got 82 marks in Bangla, 75 in English, 92 in Mathematics, 78 in Social Studies, 93 in Science and 96 in Religion. What marks did she get on average ?

Solution : Total marks obtained by Dolon is = 82

75 92

516

78 93 96

Number of subjects = 6 Average marks

516

6

86

So Dolon got 86 marks on average in those subjects.

25

Elementary Mathematics

Exercise 4 1.

Find the average : (a) 23

37

(c) 364 taka (d) 47 cm (e) 42 kg 2.

3.

47

(b) 22

61

541 taka 49 cm 32 kg

46

60

72

775 taka.

54 cm

37 kg

52 cm 29 kg

53 cm.

41 kg

35 kg

Lily bought eight laces of varying prices. The average price of the laces is 5 taka 75 paisa; how much did Lily spend on laces ?

In the annual examination Molly obtained 68 marks in Bangla, 96 in Mathematics, 81 in English, 77 in Environmental Science and 73 in Religion. What is her average marks in those subjects ?

4.

The heights of Koli, Dolly, Poly, Molly and Lily are respectively 123 cm, 131cm, 135 cm, 126 cm and 130 cm. What is their average height ?

5.

The average age of Apu and Dipu is 22 years; the average age of Dipu and Tipu is 24 years. Apu's age is 21 years; what are the ages of Dipu and Tipu ?

6.

The sum of seven numbers is 401. The average of the first three numbers is 56; the average of the last three numbers is 58. What is the fourth number ?

7.

Out of eleven numbers the average of first six numbers is 87 and the average of the last five numbers is 131. What is the average of all the numbers ?

8.

The population of the five villages of a union is respectively 1327, 1872, 2187, 2516 and 2943. What is the average population of those villages ?

9.

The marks obtained by ten students in Mathematics in the annual examination are as follows : 76, 61, 87, 56, 42, 64, 73, 68, 50, 73. What is their average marks in Mathematics ?

10.

The average age of three children and their father is 17 years. The average age of these children and their mother is 13 years. Mother's age is 22 years; what is the father's age?

11.

During the month of Ashar, the average daily rainfall in Dhaka during the first ten days was 23mm, during the second ten days daily average was 27 mm; during the third ten days it was 29 mm; 16 mm rain fell on the last day of the month. What was the average daily rainfall in Dhaka during that month ?

26

Chapter Five G.C.D. and L.C.M. G.C.D. stands for Greatest Common Divisor. It is the largest number which divides each of two (or more) given numbers. Let us find the G.C.D. of 24 and 36.

First Method We write down all factors (divisors) of the two numbers.

Write down the factors of 24

Write down the factors of 36

All factors of 24 are : 1 , 2 , 3 , 4 , 6 , 8 , 12 , 24 All factors of 36 are : 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36 Now we identify the common factors of these numbers. They are : 1 , 2 , 3 , 4 , 6 , 12 Among these numbers 12 is the largest. So the G.C.D. of 24 and 36 is 12.

Second Method We write down all factors (divisors) of the first number. We then determine which of these factors divide the other number. The largest of these common factors is the desired G.C.D.

Write down the factors of 24.

These are 1 , 2 , 3 , 4 , 6 , 8 , 12 , 24.

12 is the largest factor of 24 which divides 36.

All factors of 24 are : 1 , 2 , 3 , 4 , 6 , 8 , 12 , 24 Factors of 24 which divide 36 are : 1 , 2 , 3 , 4 , 6 , 12 Among the common factors of 24 and 36, 12 is the largest.

So the G.C.D of 24 and 36 is 12.

Elementary Mathematics Third Method We find all prime factors (with repetitions, if any) of the two numbers.

Find the prime factors of 24 and 36.

The product of their common prime factors (with repetitions, if any) is the G.C.D.

All prime factors of 24 are : 2, 2, 2, 3 because 24 2 2 2 2 All prime factors of 36 are : 2, 2, 3, 3 because 36 The common prime factors are : 2, 2, 3; their product is 2 2 3 So The G.C.D. of 24 and 36 is 12.

2 3 3 3 12.

Next we find the G.C.D. of the three numbers 18, 24, 30, by the first and third methods.

All factors of 18 are : 1 2 3 6 9 18 All factors of 24 are : 1 2 3 4 6 8 12 24 All factors of 30 are : 1 2 3 5 6 10 15 30 The common factors of the three numbers are : 1, 2, 3, 6. Among them 6 is the largest. So the G.C.D. of 18, 24, 36 is 6.

Find out the prime factors of 18, 24, 30. They are : 2, 2, 2, 3; 2, 2, 3, 3; 2, 3, 5 The common prime factors are 2 and 3.

Their product is 6.

The common prime factors of the numbers 18, 24, 36 are 2 and 3.

Their product, 6, is the G.C.D. of the three numbers 18, 24, 36.

The largest number among the common factors (divisors) of two or more numbers is their G.C.D. G.C.D. of two or more numbers is the product of their common prime factors.

Two or more numbers have G.C.D. 1 if they have no common prime factor.

Another term for divisor is factor.

28

Elementary Mathematics Finding G.C.D. with the help of prime factors Example 1. Find the G.C.D. of 125 and 225.

25 Solution : 125 5 5 5 5 225 5 45 5 5 9 5 5 3 3 The common prime factors of 125 and 225 are 5, 5.

Required G.C.D 5 5 25

Example 3. Find the G.C.D. of 48, 72 and 168.

Example 2. Find the G.C.D. of 40, 60 and 75.

Solution : 40 60 75

3 20 3 25

2 2 3 3

20 2 2 10 2 2 5 2 2 5 5 5

Solution: 48 72 168

It is seen that the only common prime factor of the three numbers 40, 60, 75 is 5.

Their G.C.D. is 5.

6 8 2 2 2 2 8 9 2 8 21

2 2 2 2

3 2 3 2

2 2 2 3 3 3 7

It is seen that the common prime factors of the three numbers 48, 72 and 168 are 2, 2, 2, 3, So the G.C.D of the three numbers is 2 2 2 3 24

Example 4. Find the G.C.D. of 24, 30 and 77.

3 2 2 2 2 2 2 3 Solution : 24 3 8 5 6 5 2 3 2 3 5 30 77

7 11

It is seen that the three numbers 24, 30, 77 have no common prime factor.

But 1 is a factor of any number. So 1 is the only common factor of the given numbers. So their G.C.D. is 1.

If two or more numbers have no common prime factor, then their G.C.D. is 1.

29

Elementary Mathematics

Least Common Multiple (L.C.M.) L.C.M. stands for Least Common Multiple. It is the smallest number which is divisible by two or more given numbers.

Let us find the L.C.M. of 24 and 36.

First Method We write down the multiples of the numbers.

Write down the multiples of 36.

Write down the multiples of 24.

Multiples of 24 are : 24 , 48 , 72 , 96 , 120 , 144 , 168 , 192 , 216 , 240, etc.

Multiples of 36 are : 36 , 72 , 108 , 144 , 180 , 216 , 252 , 288, etc.

Among the common multiples of the two numbers, 72 is the smallest.

So the L.C.M. of 24 and 36 is 72.

Second Method We write down the multiples of the first number. We determine which of these multiples is divisible by the other number.

Write down the multiples of 24.

These are 24, 48, 72, 96, 120, 144, 168, ... .The first of these numbers which is divisible by 36 is 72. So the L.C.M. of 24 and 36 is 72.

Multiples of 24 are : 24, 48, 72, 96, 120, 144, 168, 192, 216, 240, etc.

Divisible by 36 are : 72, 144, 216.

Among the common multiples of the two numbers, 72 is the smallest.

So the L.C.M. of 24 and 36 is 72.

30

Elementary Mathematics Third Method We find out the prime factors of the two numbers.

Find out the prime factors of 24 and 36; these are 2 and 3. Among them, 2 appears most often 3 times in 24 and 3 appears most often 2 times in 36. So their 3 = 72.

L.C.M. is 2 2 2 3

All prime factors of 24 are : 2 All prime factors of 36 are : 2

2

2

3

2

3

3

because 24 = 2 because 36 = 2

2

2

3

2

3

3

Among the common prime factors, 2 appears most often 3 times in 24; 3 appears most often 2 times in 36.

So the L.C.M. of 24 and 36 is 2

2

2

3

3 = 72.

In the same manner we can find the L.C.M. of three or more numbers.

As an exmple, we find the L.C.M. of 18, 24 and 27, by the first method.

We write down the multiples of 18, 24 and 27 : Multiples of 18 are : 18

216

36 54 72 90 108

126

144 162 180 198

234, etc.

Multiples of 24 are : 24 48 72 96 120 Multiples of 27 are : 27 54 81 108

144 168 192

135 162 189 216

216

240, etc.

243, etc.

In all three rows of multiples of the three given numbers, the number 216 appears.

So 216 is a common multiple of 18, 24 and 27. None of the rows has a number smaller than 216 which appears in all three rows.

So 216 is the least common multiple (L.C.M.) of the numbers 18, 24 and 27.

31

Elementary Mathematics

Example 5. Find the L.C.M. of 18, 24 and 30.

2 3 3 Solution : 18 24 2 2 2 3 30 2 3 5 Among the prime factors of the three numbers, 2 appears most often 3 times (in 24); 3 appears most often 2 times (in 18). In addition, 5 appears once (in 30).

So the L.C.M of 18, 24, 30 is 2 2 2 3 3 5 4 9 10 360 The L.C.M. of the three numbers is 360.

Determination of L.C.M. by Short Method In this method we determine all prime factors of the given numbers in succession.

The product of all these prime factors is the L.C.M. of the given numbers.

Example 6 . Find the L.C.M of 18, 24 and 40.

Solution : 2 18 24 40 2 9 12 20 2 9 6 10 3 9 3 5 3 1 5 Required L.C.M.

2 2 2 3 3 5 360

The prime number 2 has divided 24 thrice; the last quotient being 3 which is a prime number.

So 24 2 2 2 3.

2 has divided 18 once; the quotient being 9. The prime number 3 has divided 9 twice. So 18 2 3 3.

2 has divided 40 thrice; the last quotient being 5. which is a prime number.

So 40 2 2 2 5.

So the L.C.M of 18, 24 and 40 is 2 2 2 3 3 5 360.

The short method of finding L.C.M. is a consolidated form of finding L.C.M. by finding all prime factors of the given numbers. In this method, at every step we choose as divisor some prime number which divides at least two of the numbers appearing in that step. The quotients of the divisible numbers are written down in the next step; any number not divisible by the chosen prime number is carried unchanged to the next step. The last step in this method is reached when the numbers appearing in some step have no common prime factor (that is, no common factor except 1).

32

Elementary Mathematics

Example 7. Find the L.C.M. of 16, 24, 30, 42, 45 Solution : 2 16 24 30 42 45 28

12 15 21 45 2 4 6 15 21 45 3 2 3 15 21 45 5 2 1 5 7 15 2

1 1

7

3

L.C.M. of the given numbers

2

2 2

3 5

2

7 3

5040.

Simple Problems Involving G.C.D. and L.C.M. Example 8. Dividing 138, 215, 457 by which largest number leaves respectively the remainder 3,5,7 ?

Example 9. Four bells having tolled together, began tolling after every 6, 9, 12 and 15 minutes. After what minimum time will the bells toll together again ?

Solution: If we subtract the remainder from the respective dividend then the required number will be a divisor of all these differences.

138

3

135

215

5

210

Solution : The required minimum time is the L.C.M. of 6, 9, 12 and 15 minutes.

3 6 9 12 15 2 2 3 4 5 1 3 2 5

457 7 450.

So the required number is the G.C.D. of 135, 210 and 450.

135

5 27

5 3 3 3

210

3 70

3 7 2 5

So, L.C.M. of the numbers 6, 9, 12, 15 3 2 3 2 5 18 10 180.

So, the four bells will toll together for the first time again after 180 minutes or 3 hours.

450 9 50 3 3 5 2 2 So G.C.D. of 135, 210 and 450 3 5 15 So the required largest number is 15.

33

Elementary Mathematics

Example 10. Find the smallest number which when divided by 6, 10, 15 and 21 leaves in each case the remainder 4.

Example 11. Which smallest number when divided by 4, 6, 10 leaves respectively the remainder 2, 4, 8 ?

Solution : The required number is 4 more than the L.C.M. of 6, 10, 15 and 21.

3 6 10 15 21 2 2 10 5 7 5 1 5 5 7 1 1 1 7

Solution : 4 2 2 6 4 2 10 8 2 So, in each case divisor remainder 2 So the required number is 2 less than the L.C.M. of 4, 6 and 10.

2 4 6 10 2 3 5 L.C.M. of 4, 6, 10 2 2 3 5 60 The required smallest number is 60 2 58

L.C.M. of 6, 10, 15, 21 3 2 5 7 21 10 210 The required smallest number is 210 4 214

Example 12. The length of a rectangular room is 7.20 metre and breadth is 4.40 metre. What is the size of largest marble tiles of square shape which can be used to pave the floor of the room so that no tile needs to be broken ?

Solution : Length of the room 7.20 metre = 72 decimetre Breadth of the room 4.40 metre = 44 decimetre The square shape of the tiles and the condition that no tile can be broken, imply that the length ( = breadth) of the tiles must a be common divisor of the length and breadth of the room. So the length of any one side of the square tiles is the G.C.D. of 72 and 44 decimetre.

72 8 9 2 2 2 3 3 2 2 11 44 4 11 So, G.C.D. of 72 and 44 is 2 2 4 4 decimetre 4 10cm 40 centimetre.

Length of one side of the square tiles of largest size is 40 centimetre.

34

Elementary Mathematics

Exercise 5 1. Find the G.C.D. using prime factors : (a) 144 192 (b) 160 275 (c) 112 343 (d) 24 60 72 (e) 45 75 120 (f) 22 77 110 (g) 48 72 160 240 (h) 35 56 84 119 (i) 30 50 90 2. Find the L.C.M. using prime factors :

140

(a) 51 85 (b) 76 95 (c) 42 112 (d) 32 48 80 (e) 35 55 75 (f) 28 42 70 3. Find the L.C.M. by the condensed method : (a) 24 36 40 (b) 15 33 45 (c) 25 45 85 (f) 18 27 45 54 (d) 21 35 49 56 (e) 24 36 48 72 (g) 26 44 77 99 (h) 12 18 24 30 42 (i) 15 25 35 45 55 4. What is the largest number which divides 76, 114 and 228 without remainder ?

5. What is the largest number of children among whom 60 mangoes and 150 lychees can be divided exactly ? How many mangoes and how many lychees will each of them get ?

6. Two drums have capacity 228 litres and 348 litres respectively. A bucket of what largest capacity can be used to fill up the two drums with water using the bucket (to its) full capacity an integral number of times ? Which drum will hold how many buckets of water ?

7. Which number divides 137, 212 and 452 leaving the remainder 2 in each case ?

8. Which largest number divides 129, 236 and 364 leaving respectively the remainder 3, 5 and 7 ?

9. The length of a retangular hallroom is 12 metre and breadth is7 metre. What is the largest size of tiles of square shape which can be used to pave the floor of the room, so that no tile needs to be broken ?

10. Which smallest number is exactly divisible by 18, 24, 30, 36 ?

11. Which smallest number leaves in each case the reminder 6 when it is divided by 16, 24, 32, 40 ?

12. Which smallest number leaves respectively the remainder 6, 12, 24 when it is divided by 12, 18, 30 ?

13. Four bells having tolled together, began to toll after every 5, 7, 12 and 15 minutes. After what minimum time will the bells toll together again ?

14. A number of saplings is such that when 3, 5, 6, 8, 10 or 15 saplings are planted in each row, every time two saplings are left out. What is the minimum number of saplings ?

15. 7 added to which smallest number will make the sum exactly divisible by 15, 18, 20, 24 and 32 ?

35

Chapter Six

Mathematical Symbols and Statements Symbols Mina said, "I am thinking of a number. Adding two to the number and multiplying the sum by three makes the product greater than thirty".

Let us see how we can express this statement with the help of symbols.

Raju said, "Denoting the unknown number by the symbol x, the above statement becomes : (x

2)

3

30."

Here we have used several kinds of symbols.

In Mathematics various kinds of symbols are used. These are Symbols

Kind of symbols Numerical symbols Operational symbols

1, 2, 3, 4, 5, 6, 7, 8, 9, 0 addition,

subtraction,

greater than

equal to

multiplication, less than

division

not equal to

Relational symbols less than or equal to

Bracketing symbols Literal symbols

First brackets

greater than or equal to

Second brackets

a, b, c, . . ., x, y, z.

Third brackets

Elementary Mathematics

Mathematical Statement

Raju and Mina are playing with marbles. Mina has seven more marbles than Raju. In all they have 45 marbles. How many marbles does each of them have ?

Suppose Raju has marbles numbering x ; then Mina has (x + 7) marbles.

Together Raju and Mina have x + ( x + 7 ) marbles.

According to the question, x + ( x + 7) = 45. or 2x + 7 = 45, because x + ( x + 7) = (x + x) + 7 = 2x + 7.

Here, 2x + 7 = 45 is an example of an open statement, An open statement is a mathematical statement containing some special symbol indicating some unknown quantity. A mathematical statement is a statement containing numbers, symbols, expressions or mathematical concept, about which it can be said without ambiguity whether it is true or false. For example : 48

6

5

48

6

false statement

5

Every triangle has three sides and three angles

true statement

17 is a prime number

true statement

The G.C.D. of 12 and 15 is 6

false statement

Thus, a mathematical statement is any statement of a mathematical nature which is either definitely true, or definitely false.

37

Elementary Mathematics

Open Statement On her birthday, out of 30 chocolates Sumi gave 2 chocolates to each of her friends; 6 chocolates remained.

Here, the number of Sumi's friends is not known. Suppose the number of friends is x.

Every one got 2 chocolates. So friends numbering x got x 2 chocolates. Since out of 30 chocolates 6 remained, we have the relation x 2 6 30 Here, x 2 6 30 is an open statement.

An open statement is a mathematical statement which contains a literal symobl indicating an unknown number or quantity. For example : (x 5) 3 45.

Example 1. Express the problem with the help of a symbol, and determine the unknown number : Which number yields the difference 33 when 17 is subtracted from it ?

Example 2. Express the problem with the help of a symbol, and determine the unknown number : Which number yields the quotient 9 when divided by 15 ?

Solution : Denoting the unknown number by the literal symbol x the problem x 17 33. becomes Addition being the reverse process of subtraction, the number x will be the sum of 33 and 17.

Solution : Denoting the unkown number by the symbol x, the problem becomes x 15 9.

x

33

17

Multiplication being the reverse process of division, the number x will be the product of 9 with 15. x

50

38

9

15 15

9

135

Elementary Mathematics Example 3. Use a literal symbol to form a mathematical statement from the information given in the sentence below : From 20 mangoes every one of a number of children was given 3 mangoes, and 5 mangoes remained.

Solution : We use the literal symbol x to denote the number of children . Every one got 3 mangoes; so x children got x 3 mangoes.

According to the given information : x

3

The desired open statement is : x

5 3

20 5

20.

Example 4. Use bracketing symbols to form the indicated numerical expression and evaluate the expression.

Solution : Seventyfive is divided by five and the quotient is multiplied by 2.

The desired expression is 75 5 2 15 2 30

Example 5. Find the value of the symbol x from the following open statement : x 5 4 80 5 4 Solution :

80 x

We have 1

5 5 4

4 4

80 4

20

4

4. So we must have. x

5

1.

x

1

5

5

Example 6. Find the value of the symbol x from the following open statement : x 6 4 80 5 4 Solution : 80 x We have 1

5 6 4

80 20 4 4 4 4 4 . So we must have, x

39

6

1,

x

1

6

6

Elementary Mathematics

Exercise 6 1.

Use symbols to express the following statements : (a) To subtract one thousand five hundred twentyfive from two thousand three hundred seventyfive and multiply the difference with the sum of one hundred thirty and seventyeight (b) The difference of five hundred eightyseven and the sum of two hundred twentyfive and one hundred thirtyeight is equal to the successive difference of five hundred eightyseven, two hundred twentyfive and one hundred thirtyeight. (c) Three hundred sixty multiplied by the sum of fortyfive and fifteen is equal to the sum of the products of three hundred sixty with fortyfive and with fifteen. (d) The result of dividing fifty by five and multiplying the quotient with two, is not equal to the result of dividing fifty by the product of five and two.

2.

Express each problem with the help of a symbol and find the unknown number : (a) Which number when added to 8 will yield the difference 33 when 7 is subtracted from the sum ? (b) Which number when 12 is subtracted from it will yield the sum 28 when 10 is added to the difference ? (c) Which number multiplied by 7, and the product divided by 6 will yield the quotient 21? (d) Which number when divided by 12 will yield the product 56 when the quotient is multiplied by 7 ?

3.

Verify the truth (or otherwise) of each mathematical statement below and write down true or false (as the case may be) against each: (a)

639

(b)

375

15

5

(c)

850

17

10

850

(d)

230

75

15

5

227

85

639 375

227 15

5

10 235

40

85

17 5

75

5

15

5

Elementary Mathematics 4. Find the value of the literal symbol from each open statement below : (a)

x

60

3

x

30

(c)

x

8

9

15

(d)

x

5

5

75

3

7

3 4

91

5 9

For what value of the literal symbol will each open statement below be true ? (a)

x

(b) 3

6.

10

(b) 15

(e) x 5.

5

10

5

x

20

100

(c)

x

4

2

14

(d)

x

6

3

12

Form an open statement using a literal symbol and find the unknown number : (a) Anu had some chocolates. He gave five chocolates to Nilu. As a result Anu was left with seven chocolates. (b) Mou had some money . Her mother gave her one hundred taka more. As a result her amount of money became fifteen taka more than twice her previous amount. (c) Doubling the age of Sion and adding 5 years to it makes the sum equal to 25 years. (d) Peau's father gave her some money to buy books, writing pads and pencils. Her mother gave her seventyfive taka more. Peau bought one book for 80 taka, two writing pads for 40 taka and two pencils for 20 taka. After purchasing book ,writing pads and pencils, Peau was left with thirtyfive taka.

41

Chapter Seven

Common Fractions Fractions with Common Denominator Observe the following two pictures. What portion of each picture is coloured ?

2 5

1 5

Picture - A Picture - B 1 In Picture - A , is a fraction, whose numerator is 1 and denominator is 5.

5 2 In Picture - B, is a fraction, whose numerator is 2 and denominator is 5.

5 The fractions have the same denominator, 5. They are fractions with a common denominator.

Fractions with the same denominator are fractions with a common denominator.

Observe again the following two pictures :

1 4

1 3 Picture - C

Picture - D

1 1 and are different. They are not 3 4 fractions with a common denominator. They can be expressed as fractions with Here the denominators of the fractions a common denominator ; 1 1 4 4 3 3 4 12 1 4

1 3 4 3

3 12

1 1 and are not fractions with a common 3 4 4 denominator. They are respectively equal to 12 3 and . So these fractions have been expressed 12 as fractions with a common denominator.

Elementary Mathematics

Fractions with Common Numerator

2

2

3

5

Picture - B Picture - A 2 In Picture -A, is a fraction, whose numerator is 2 and denominator is 3.

3 2 In Picture -B, is a fraction, whose numerator is 2 and denominator is 5.

5 The fractions have the same numerator 2. They are fractions with a common numerator.

2 3 and are fractions with different numerators. They can be expressed as fractions with a 3 8 common numerator: 2 3 and are not fractions with a common 2 2 3 6 3 8 6 numerator.

They are respectively equal to 3 3 3 9 9 6 and .

So these fractions have been expressed 6 3 3 2 16 8 8 2 16 as fractions with a common numerator.

1 1 5 and as fractions with common denominator 24.

3 4 6 Solution : We observe that 24 is a common multiple of 3, 4 and 6, the denominators of the given fractions.

2 2 8 16 24 3 8 3 3 8 24 Example 1. Express

24

4

6

1 4

1 6 4 6

6 24

24

6

4

5 6

5 4 6 4

20 24

43

Elementary Mathematics 3 5 1 and as fractions with common numerator 15.

8 12 16 Solution : We observe that 15 is a common multiple of the numerators 3, 5, 1 of the given fractions.

3 3 5 15 15 3 5 8 8 5 40

Example 2. Express

15

5

3

5 12

5 3 12 3

15

1

15

1 16

1 15 16 15

15 36 15 240

Reduced Form of a Fraction Observe the following two pictures:

The pictures show that 4 6

4 2 ; in other words, 6 3

and

2 3

are equivalent fractions.

2 4 is obtained from by cancelling the common factor 2, of the numerator and denominator.

3 6 2 2 The numerator and denominator of have no common factor, except 1. So is the 3 3 4 reduced form of the fraction 6 The reduced form of a fraction is an equivalent fraction whose numerator and denominator have no common factor except 1. It is obtained by dividing the numerator and denominator by their G.C.D. A fraction expressed in its reduced from is said to be in its lowest terms.

44

Elementary Mathematics 12 in reduced form.

36

Example 3. Express Solution :

12 36

1 2 2 1

1 2 2 1

1 3 3 3 1

1 3

1 3 4 7 and as fractions having the least 5 10 15 20 common denominator

Example 4. Express

Solution : Here the denominators of the fractions are 5, 10, 15 and 20.

5 5 10 2 1 2 1

1

15 20 3 4 3

2

L.C.M. of 5, 10, 15 and 20 is

60

5

60

10

6

60

15

4

60

20

3

1 5

12

12 60

3 10

5

1 5

1 12 5 12

12 60

3 10 4 15 7 20

3 10 4 15 7 20

18 60 16 60 21 60

18 60

6 6 4 4 3 3

4 15

16 60

2 3

7 20

2

60

21 60

To express two or more fractions as fractions with the least common denominator, we find the L.C.M. of the denominators and express the given fractions as fractions with the L.C.M as their common denominator.

45

Elementary Mathematics Comparison of Fractions In the two pictures below, each of two equal rectangular regions has been divided into five equal parts by the vertical lines. In Picture-A, two of these parts have been coloured; in Picture-B, three of the parts have been coloured.

Picture - A

Picture - B

What portion of each figure has been coloured ? Which is the greater portion ?

2 2 In Picture-A,the coloured portion is ; in Picture - B, the coloured portion is , 5 5 2 3 and are fractions with the same denominator 5. Their numerators are 5 5 3 3 2 2 and 3. Since 3 > 2, we clearly have > . So is the greater portion.

5 5 5 Among the fractions 1 , 2 and 3 which is the greatest ?

5 5 5 Observe the following pictures.

3 1 2 > > .

5 5 5 3 1 2 Each of the fractions , and has denominator 5; in the decreasing 5 5 5 3 1 2 order the numerators are 3 > 2 > 1. So > > .

5 5 5 From the picture it is clear that

Among fractions with the same denominator, the fraction with the greater numerator is the greater fraction.

46

Elementary Mathematics

18 30 3 Therefore 5

Since 18 < 20 < 25 it follows that

Arrangement in ascending order of magnitude is

20 30 2 3

25 30 5 6

3 5

2 3

Arranging fractions successively from the smallest to the largest, is called arrangement in ascending order of magnitude.

5 6

Example 8. Arrange the fractions 5 7 and 3 in descending order of magnitude.

8 16 4 Solution : The denominators of the given fractions are 8, 16 and 4.

Now 8 divides 16, as does 4. So the L.C.M. of 8, 16 and 4 is 16.

We express the given fractions as fractions with the common denominator 16.

16

8

16

16

16

4

2 1 4

Since 12

10

Therefore

3 4

5 8 7 16

5 2 8 2 7 1 16 1

3 4

3 4 4 4

7, we have 5 8

10 16 7 16 12 16

12 16 7 .

16

Arranging fractions 10 16

7 16

successively from the largest to the smallest is called arrangement in descending order of magnitude.

Arrangement of the given fractions in descending order of magnitude is 3 4

5 8

7 .

16

48

Elementary Mathematics

18 30 3 Therefore 5

Since 18 < 20 < 25 it follows that

Arrangement in ascending order of magnitude is

20 30 2 3

25 30 5 6

3 5

2 3

Arranging fractions successively from the smallest to the largest, is called arrangement in ascending order of magnitude.

5 6

Example 8. Arrange the fractions 5 7 and 3 in descending order of magnitude.

8 16 4 Solution : The denominators of the given fractions are 8, 16 and 4.

Now 8 divides 16, as does 4. So the L.C.M. of 8, 16 and 4 is 16.

We express the given fractions as fractions with the common denominator 16.

16

8

16

16

16

4

2 1 4

Since 12

10

Therefore

3 4

5 8 7 16

5 2 8 2 7 1 16 1

3 4

3 4 4 4

7, we have 5 8

10 16 7 16 12 16

12 16 7 .

16

Arranging fractions 10 16

7 16

successively from the largest to the smallest is called arrangement in descending order of magnitude.

Arrangement of the given fractions in descending order of magnitude is 3 4

5 8

7 .

16

48

Elementary Mathematics Now observe the following picture: 2 3

2 5 2 2 portion is greater than portion.

It is clear from the picture that 5 3 2 and 2 have the same numerator 2.

The fractions 5 3 2 2 has denominator 3 and has denominator 5.

Among fractions with 3 5 the same numerator, the 2 2 Since 3 < 5, we have fraction with the smaller 5 3 denominator is the larger fraction.

Example 9 Show, using symbol, which of the fractions Solution :

Since 15 < 30, we have

5 Example 10. Arrange the fractions 7 magnitude.

7 15 ,

>

7 7 and is the greater ?

15 30

7 30

5 5 and in ascending order of 12 18

Solution : The fractions have the same numerator 5, and their denominators are 7, 12 and 18.

5 5 5 Since 7 12 18. we have .

18 12 7 Arrangement of the given fractions in ascending order of magnitude is : 5 18

5 12

5 .

7 49

Elementary Mathematics

Exercise 7 (A) 1. Transform into fractions with a common denominator : (a)

1 2

1 3

(b)

(d)

3 4

5 6

4 9

1 2

1 4

7 18

1 6

(c) 5 7

(e)

3 14

5 6

3 4

2 21

1 42

3 4

5 12

5 12

7 18

5 8

2. Transform into fractions with denominator 40 : 1 4

2 5

7 8

3 10

9 20

3. Transform into fractions with a common numerator : (a)

1 2

2 3

(b)

2 3

3 4

1 6

(c)

3 8

4. Transform into fractions with numerator 12 : 1 6

4 5

2 3

3 8

6 7

5. Transform into fractions with a least common denominator : (a)

3 4

5 8

(b)

5 6

4 9

(c)

2 3

6. Divide each of the equal rectangular regions into seven equal parts (by drawing vertical lines); and colour green two, three and five parts of Picture A, B, C respectively.

Picture A

Picture B

Picture C

What portion of each picture is coloured green ? Use symbol to express which portion is greater.

50

Elementary Mathematics 7.

8.

Arrange the following fractions in increasing order of magnitude; use symbols to indicate the result : (a)

4 5

1 5

3 5

(d)

7 36

11 36

23 36

13 36

9 17

7 17

15 17

(e)

19 48

23 48

(c) 31 48

7 24

17 24

11 24

5 48

Arrange the following fractions in decreasing order of magnitude; use symbols to indicate the result : (a)

9.

(b)

2 7

6 7

5 7

(b)

3 16

9 16

11 16

(c)

9 25

12 25

23 25

Compare the values of the following fractions; use symbols to indicate the result : (a)

2 3

2 5

2 9

(b)

7 10

7 15

7 20

(c)

3 5

3 7

3 11

1 3 portion of it red, and portion of it green.

5 5 Which colour covered the greater part of the region ?

10. Arghya drew a square region and painted

2 11. Himu was given portion of a bottle of cold drinks and his younger sister was given 7 3 portion. Who got more cold drinks ?

7 7 6 8 hour, Shafiq spends hour and Saima spends hour reading 15 15 15 newspapers. Who spends the least time reading newspapers ?

12. Everyday Rafiq spends

13. Shuvo painted

2 portion of the wall of their room. Shuvo's younger sister Shetu painted 25

2 portion. Who painted the greater portion ?

17

5 5 5 portion, portion and portion of a 18 9 36 basket of oranges. Who got the largest share of the oranges ?

14. Javed, Shoumik and Mouli got respectively

51

Elementary Mathematics

Proper Fractions

1

3

5

3

4

8

Observe the above pictures.

5 1 , 3 and are fractions whose numerators are less than their denominators. These 8 4 3 are proper fractions.

Fraction whose numerators are less than their denominators, are proper fractions. The value of any proper fraction is less than 1.

Improper Fractions Four friends Mahi, Momo, Esha and Urmi were having tiffin together. The four of them had five apples. How can they share the five apples equally ? They can divide each apple into four equal pieces; each of the four friends then takes one piece from each of the five apples. In this way every one of the four friends gets an equal share of the four apples.

4 = 5 apples. Here 5 is a fraction whose numerator 4 4 5 is greater than its denominator 4. Such fractions are improper fractions.

3 , 12 , 18 , are examples of improper fractions 2 7 5 In symbol, each of them gets 5

Fractions whose numerators are gereater than their denominators, are improper fractions. The value of any improper fraction is greater than 1.

Elementary Mathematics Mixed Fractions Let us return to the problem of sharing five apples equally by the four friends Mahi, Momo, Esha and Urmi. Alternatively, each of them could take one whole apple; then divide the remaining fifth apple into four equal pieces and each of them take one of these pieces.

1 portion of an apple. Expressed in 4 1 1 apples.

1 1 symbols, each friend would get 1 + or 1 1+ is written as 1 .

4 4 4 4 1 is read as one and one-fourth. Such fractions are mixed fractions.

1 4 3 .

Some other examples of mixed fractions are 1 1 2 1 5 2 3 4 In this way, each of them would get 1 apple and

Fractions having an integer part and a proper fractional part, are mixed fractions.

Conversion of Improper Fractions into Mixed Fractions

Picture-A

Picture-B

Picture-C

Observe the pictures above. What portion of Picture-A, Picture-B and Picture-C has been coloured ?

53

Elementary Mathematics In Picture - A, three one-third, in Picture-B, also three one-third and in Picture-C, one-third portion has been coloured. So, in the three pictures taken together, ( 3 + 3 + 1) = 7 one-third portion has been coloured. In other words , 7 portion has been coloured.

3 7 is an improper fraction. It is clear from our discussion that the improper fraction 7 3 3 1 1 is equal to the mixed fraction 2 + 2 3 3 We have 7 3 2 + 1.

So

7 3

3

3

2+1 3

2 3

+

1 3

1 3

2+

2

1 3

Observe that 2 is the quotient and 1 is the remainder of 7 upon division by 3. Thus 7 the improper fraction 3 has been converted into a mixed fraction.

We observe : * To convert an improper fraction into a mixed fraction, we divide the numerator by the denominator. The quotient is the integer part of the desired mixed fraction. * The remainder is the numerator of the fractional part of the mixed fraction. * The denominator of the fractional part of the mixed fraction corresponding to a given improper fraction is the same as that of the improper fraction.

Conversion of Mixed Fractions into Improper Fractions In a mixed fraction, an integer is added to a proper fraction. Any integer can be written as a fraction with denominator 1. For example : 3

Let us convert the mixed fraction 2 2

3 4

2

3 4

2 1

3 4

2 1

4 4

3 4

3 4

3 1

5

5 1

into an improper fraction: 2 4 4

54

3 4

2 4 3 4

11 4

11

11 .

1

Elementary Mathematics This working is briefly written as follows : 2

2

3

4 4

4

11 4

Denominator Denominator

Integer part

Mixed fraction

3

Numerator

Improper fraction

Example 1. Convert the following improper fractions into mixed fractions : (a)

11 5

Solution :

(b)

55 16

11 5 11 = 2 11 2 5 55 (b) 16 55 = 3

5 11 10 1

(a)

5+ 1 1 5

16 55 3 48 7

16 + 7

55 16

3

2

7 16

Example 2. Convert the following mixed fractions into improper fractions :

Solution :

(a)

3

1 5

(a)

3

1 5

3 5 1 15 1 5 5

16 5

(b)

4

5 7

4 7 5 28 5 7 7

33 7

(b)

4

55

5 7

Elementary Mathematics

Exercise 7 (B) 1. Identify the proper fractions from among the following fractions : (a)

2

9

(b)

3

1

(c) 2

5

4

(d)

3

35

(e)

27

17

2. Separate the improper and mixed fractions from among the following fractions : 2 3 7 (f) 20

1 4 1 (g) 10 5 (b) 1

(a)

1 3

(c) 25 (h)

5 18 75 (i) 18

9 11 8 (j) 25

(d)

40 33

(e)

3. Convert the following improper fractions into mixed fractions : (a)

8 3

(b)

39 18

(c)

47 19

(d)

227 25

(e)

211 30

(f)

315 37

(g)

407 40

(h)

521 45

(i)

750 119

(j)

1416 203

4. Convert the following mixed fractions into improper fractions : 3

3 4

(b) 12

5 9

(c) 16

2 7

(d) 25

9 16

(e) 30

7 30

(f) 55

3 10

(g) 68

2 15

(h) 75

17 25

(i)

11 23

(j) 110

13 29

(a)

80

5. Fill in the blanks with correct numbers : (a)

85 12

(d) 16

37 43

7

1

43

(b) 15 (e)

7 30

(c)

30

1485 247 56

3 247

27 13

1 13

Elementary Mathematics 6. Jaya had 5 pens. She gave her younger brother 2 pens. What portion of her total pens did Jaya give to her younger brother ?

7. Mamun had 200 taka. He bought a book for 50 taka. What portion of his total money did Mamun spend on the book ?

8. Anik had 20 chocolates. He took out 10 chocolates and gave half of these to his friends.

What portion of his total chocolates did Anik give to his friends ?

9. Mr. Rafiq out of his 3 bighas of land cultivated marigold on 2 bighas and tuberose on half of the remaining land. On how much bigha of land did he cultivate tuberose ?

10. In a class test in Mathematics Rashed got 8 marks out of 10. What portion of total marks in Mathematics did he get in that class test ?

57

Elementary Mathematics

Addition of Fractions Observe the Picture-A to the right. How many segments in this picture are coloured ? Two segments out of four equal segments are coluored. Each of these segments constitute one-fourth of the picture; the two coloured portions constitute one-half of the picture. So Picture-A 1 4

1 4

2 4

1 1 4

1 2

Observe the Picture-B. From the picture it is clear that 1 9

1 9

Example 1. Add : Solution :

1 11

Example 2. Add : Solution :

1

1 9

9 7

1 1 9

1 11

3 11

3 9 5 11

10 7 9 10 7 27 7 6 3 7

10 7

Picture-B

The sum of two or more fractions with a common denominator is a fraction with the same denominator. Its numerator is the sum of the numerators of the given fractions.

3 5 11 11 1 3 5 11 9 11 9 7

1 3

8 7

8 7

If the sum is an improper fraction, it is generally expressed as a mixed fraction.

8

58

Elementary Mathematics

Example 3. Add : 2

Solution :

2

1

5

13

3

3

13

1 13

5

27

68

44

13

13

13

27

68 13

3

3

13

13

5 13 The mixed fractions have been converted into improper fractions and then added.

Finally, the sum has been expressed as a mixed fraction.

44

139 13 10

5

9 13

Now observe the following pictures.

1 2

1 3

Picture-A

2 6

Picture-B

3 6

5 6

1 2 1 3 or portion of Picture-A and or portion of Picture-B is coloured. In total, 3 6 2 6 1 2 3 1 5 and are fractions portion of the two pictures is coloured. Here 2 6 6 3 6 with different denominators, 3 and 2.

59

Elementary Mathematics To add fractions with different denominators, we first convert them into fractions with a common denominator. In the second row of the pictures above, 1 and 1 have 3 2 been expressed as fractions with the common denominator 6; 1 and 2 as well as 3 6 1 3 and , are equivalent fractions. As such 2 6 1 1 2 3 2 3 5 6 3 2 6 6 6

1 3

Example 4. Add :

7 10

2 5

Solution : Denominators of the given fractions are 3, 5 and 10 5 3 5 3

10 1 2

G.C.D. of 3, 5 and 10 is

30

3

10

30

5

6

30

10

3

1 3

2 5

7 10

5

3 2

1 3 2 5

1 3 2 5

10 10 6 6

10 30 12 30

7 10

7 3 10 3

21 30

10 30

12 30

21 30

10

12 21 30

43 30

1

13 30

60

30

Before adding fractions with different denominators, they have to be converted into fractions with a common denominator.

Elementary Mathematics

Subtraction of Fractions 2 3

Observe the following picures. is coloured. We observe that how can we do that ?

2 3

2 3

portion of Picture-A is coloured ; 1 portion of Picture-B 2

is the greater fraction. If we wish to subtract

1 2

from

2 3

1 2

are not fractions with a common denominator. We need to 2 1 express them as fractions with a common denominator; the expressions of and 3 2 4 3 with a common denominator are (Picture-C) and (Picture-D) as fractions and

6

6

To subtract a smaller fraction from a larger fraction with the same denominator, we subtract the numertor of the smaller fraction from the numerator of the larger fraction; the denominator of the difference fraction is the common denominator of the minuend and subtrahend fractions.

1

2 3

2 express them

Picture-A

Picture-B

3 6

4 6

Picture-D

Picture-C So

1 6

2 3

1 2

4 6

Example 5. Subtract : Solution :

13 25

8 25

13 8 25 1 5

3 6

4

3 6

8 25

13 25

1 6 The denominator of the difference of two fractions with the same denominator is their common denominator.

The numerator of the difference fraction is the difference of the numerators.

1 5

The sum or difference of

25

fractions should always be expressed in the reduced form.

5

61

Elementary Mathematics 31 6

Example 6. Subtract : Solution :

31 6

23 6

Example 7. Subtract : Solution : 2 3 4

23 6

6 8 6 1 13

4 3

3 4

1

1 4

1 4

11 5 4 4 11 5 4 6 4 3 2 1 12

23

31

1

2

3 4

Example 8. Subtract :

2 5

2 3 5 4 The denominators of the given fractions are 4 and 5. Their L.C.M. = 20 Solution :

20 20

4 5

5

3 4

3 4

5 5

4

2 5

2 5

4 4

3 4

2 5

3 4

2 5

15 20 8 20

15 8 20 20 15 8 7 20 20

The two fractions do not have the same denomiator.

They have been expressed as fractions with a common denominator

Expressed briefly, 15 20

15 8 20

8 20

62

7 20

Elementary Mathematics

Example 9. Subtract :

5

1 3

4

2

Example 10. Simplify : 1 1 2 2 3 6 1 1 2 2 Solution : 3 6 7 13

9

Solution : 5

1 3

4

2 9

16 3

38 9

48

38 9

9

Mixed fractions have been converted into improper fractions before subtraction.

9 is the L.C.M of 3 and 9

3 14 6 14

38

48

1

2 1

2 3

6 2 13 9 6 6 13 9 6

27

9

1

1

9 6

10 9

3 18 6

1 1 9

1

3

1 1 metre of a 10 metre long white tape was coloured blue and 4 3 12 metre was coloured red. What length of the tape remained white ?

1 1 4 metre of the tape is coloured blue and Solution : 5 3 metre is coloured red.

12 length of the segment sum of the lengths of the coloured segments 1 1 which remained white is of the tape is = 5 metre 4 12 3 113 metre 10 61 13 12 metre 12 3 120 113 metre 61 52 12 metre 12 7 metre 113 metre 12 12 Example 11. 5

63

Elementary Mathematics

Exercise 7 (C) 1. Add : (a)

1

3

2

7

7

7

28 (d)

31

37

39

25

25

25

25

1 5

1

2 5

3

(g) 2

(j)

10

1 2

4

3 38

(b)

(e)

1 1 (h) 2 5 3

3 5

2

5 38

7 38

8 38

1

2

5

18

9

6

1 2

3 4

(f)

5 4

1

1 4

4 (k) 5

3 4

4 5

1

(b)

31 9

11 9

11 9

(c)

7 8

(i) 7

1

3 4

10

7

1 2

13 9

10 9

2

7

11

5

15

25

5 16

17 6

7 12

3 3 8

5

(d) 5

7 12

5

3 10

4 3 7

1 5

2. Subtract : (a)

27 32

(e) 6

9 32

1 9 3 4

(i) 2

7 9

4

(f) 8 3 16 4 5

1

(c)

7

1 16

13 12

(g) 1

9 10

1

(k) 4

2 3

2

5 8

(j) 3

43 12

4 5

(h) 2

3 4

(d) 7

3 4

5

4

1 5

4

5 12

4 5

3. Simplify : (a)

3 4

7 8

11 12

(e) 7

2 5

3

1 2

(h) 5

3 4

1

1 2

(j) 1

1 3

5

1 4

(b)

1

1 4

2

7 15

(f) 8

1 1 3

2

5 16

5 6

2 3 3 4

3 7

9 20 2

3 5

(i) 5

(k) 2

(c)

4 1 4

1 2

2 7

64

(g) 1

1

4

5 32

1 7

3 5

11 12

7 8

1 2

3

5 14

1 2

3 8

5

1

7 15

3 4

4 5

6

3 10

Elementary Mathematics 1 1 4. Shomu spends portion of a day studying at school, portion playing out, 4 12 1 portion studying at home, in the remaining time he does other work and takes rest.

3 What portion of the day does he spend studying and playing ?

1 portion 3 1 and younger brother Mamun got portion. What portion of the total property did they get ?

6 1 1 portion of a book in the first day, portion in the second day and 6. Mimi read 3 4 1 portion in the third day. In three days in all, what portion of the book did she read ?

12 1 1 kilometre in the first hour, 3 7. Rahim walked 4 kilometre in the second hour 3 6 1 and 1 kilometre in the third hour. In three hours, in all how many kilometres did 3 he walk ?

8. 1 portion of a bamboo is in mud, 2 portion is in water and the remaining portion is above 5 5 water. What portion is above water ?

5. Shafiq got

1 4

portion of his father's property, while his younger sister Minu got

9. Apu has 50 3 taka. He bought a writing pad for 25 1 taka and a pencil for 10 1 taka.

4 2 2 How much taka does he have now ?

10. Mr. Hanif gave away his property to his wife, one son and two daughters. The son got 3 portion and each daughter got 2 portion. What portion did the his wife get ?

8 8 3 3 11. The difference of two fractions is 8 . The smaller number is 3 ;what is the larger number ?

8 8 12. Fill in the blanks : (a) (d)

(f) 1

1 4

5

1

8

4

(b)

7

3

10

10 2

1 2

2

5 6

(e) (g) 3

1 4 65

1

1

5

2

(c)

5

12

27

27

3

1 2

4

3 4

1

1

2

6

Elementary Mathematics Multiplicaton of Fractions Multiplication of a Fraction by an Integer Picture-A

Picture-B

Picture-C

Observe the following pictures.

In Picture-A, the lower left 1 portion 4 of the circular region has been coloured.

1 In Picture-B, the lower right portion 1 1 2 4 portion colour portion colour portion colour 4 4 of the circular region (of the same size) 4 1 has been coloured. In Picture-C, both lower portions have been coloured. If we superimpose 4 1 Picture-A on Picture-B, we get Picture-C, in Picture-C, portion of the circular region is 2 coloured.

1 4

So,

1 4

1 4

2

1 2 4 2 4

Fraction

Integer

Numerator of the fraction Integer Denominator of the fraction

1 2

To multiply a fraction by an integer is to multiply the numerator by the integer; the denomintor is not changed. The product is to be expressed in reduced form.

1 4 1 4

In Picture-D, a circular region has been divided into four equal parts. Each

1 4 1 4

part is 1 portion of the circular region. Taking 3 of these parts amounts to

4 multiplying 1 by 3. Again, taking 3 of these parts amounts to writing down 1 4

Picture-D

4

three times and then adding them.

1 Multiplication is a condensed form of addition. So multiplying by 3 is the same as writing 4 1 down three times and then adding the three fractions.

4

1 4

1 4

1 4

1 1 1 4

1 3 4

3 4 66

Elementary Mathematics 1

kilometre from his home.

3 On Independence Day Apu went to school very early and returned home after hositing of the national flag. In the afternoon he went to the school cultural function and returned home in the evening. What distance did Apu cover on that day walking to and from school ?

1 Solution : On Independence Day Apu walked the distance kilometre 4 times.

3 1 km by 4.

So the required distance is the product of 3 Example 1 . Apu's school is at a distance of

1

1 4 3

4

3

4

1

3

3

Apu walked 1

1

1

kilometre on that day.

3 Now let us observe the following pictures. They illustrate the result :

1 3

4

Example 2.

1

1

1

1

3

3

3

3

15 7

Solution : 15 7

5

15 5 7

5 75 7

what ?

10

Example 3. 8

Solution : 8

1 4 231 462 4 2

5 7

1 4 14

14

3

what ?

33 4

14

231 2

115

To multiply a mixed fraction by an integer, the mixed fraction is first converted into an improper fraction 67

1

1

33 14 4

1 2

Elementary Mathematics Multiplication of a Fraction by a Fraction The adjoining picture shows a square region, each of whose sides has length 1 metre. The area of the square region is 1 square metre. The square region has been divided into 15 rectangular regions of the same size,

1

as shown. So area of each of these rectangular regions m 1 e square metre. is t 15 r Now observe the coloured region in the picture. e It consists of 8 of those rectangular regions adjoining one another. As these regions do not overlap, the area of the 1 8 1 metre coloured region is 8 sq metre sq metre.

15 15 4 metre and its breadth is On the other hand, the length of the coloured rectangular region is 5 2 metre. So the area of the coloured region is, 4 metre 2 metre 2 ) sq metre.

(4 3 5 3 5 3 8 4 2 .

Equating the two expressions for the same area, we get 15 5 3 8 2 Observe that 4 15 5 3 Therefore

4 5

2 3

4 5

2 3

The product of two fractions is a fraction whose numerator is the product of the two numerators and whose denominator is the product of the two denominators.

Example 4. 5 6 Solution : 5 6

3 8

3 5 8 6 2 5 16 .

Example 5.

what ?

1 3 8

Solution :

12

1

5

6 12

1

2 12

5

6

5

2 5

68

what ?

.

1 6 1

Elementary Mathematics A rectangular region has been divided into 15 equal and smaller rectangular regions, as shown.

The four coloured smaller rectangular regions to the lower left side of Picture-D constitute 4 15

portion of the whole region. The length of the coloured region

the whole rectangle, and its breadth is

2

2 5

portion of the length of

portion of the breadth that rectangle.

2 2 2 2 4 portion of the area So the area of the coloured rectangular region is 5 3 5 3 15 of the whole rectangular region.

3

Therefore, the coloured region of picture-D can be treated as 2 3

of

2 5

. So

2 3

of

2

2

2

5

3

5

.

Thus,

2 5

of

2

2

2

3

3

5

'of ' is tantamount to multiplication.

Example 8. 34 portion of one metre long piece of cloth is coloured red. A handkerchief is made from

1 2

portion of the red part. What portion of the one metre long cloth

has been used to make the handkerchief ?

Solution : (

1 of 3 ) portion of the one metre long cloth has been used in making the 2 4

handkerchief.

Now

1 3 of 2 4

In making the handkerchief,

1 3 2 4

3 8

3 portion of the one metre long cloth has been used.

8

Reciprocal Fraction Every fraction has a denominator and a numerator. The fraction obtained by interchanging the denominator and numerator of a given fraction, is called the reciprocal fraction of that fraction.

3 is a fraction whose denominator is 4 and numerator is 3. Interchanging their positions 4 3 4 4 3 we get the fraction . So is the reciprocal fraction of ; likewise is the reciprocal 4 3 3 4 4 fraction of . If the product of two fractions is equal to 1, then they are mutually reciprocal 3

fractions.

Any fraction

Reciprocal of that fraction = 1 70

Elementary Mathematics A rectangular region has been divided into 15 equal and smaller rectangular regions, as shown.

The four coloured smaller rectangular regions to the lower left side of Picture-D constitute 4 15

portion of the whole region. The length of the coloured region

the whole rectangle, and its breadth is

2

2 5

portion of the length of

portion of the breadth that rectangle.

2 2 2 2 4 portion of the area So the area of the coloured rectangular region is 5 3 5 3 15 of the whole rectangular region.

3

Therefore, the coloured region of picture-D can be treated as 2 3

of

2 5

. So

2 3

of

2

2

2

5

3

5

.

Thus,

2 5

of

2

2

2

3

3

5

'of ' is tantamount to multiplication.

Example 8. 34 portion of one metre long piece of cloth is coloured red. A handkerchief is made from

1 2

portion of the red part. What portion of the one metre long cloth

has been used to make the handkerchief ?

Solution : (

1 of 3 ) portion of the one metre long cloth has been used in making the 2 4

handkerchief.

Now

1 3 of 2 4

In making the handkerchief,

1 3 2 4

3 8

3 portion of the one metre long cloth has been used.

8

Reciprocal Fraction Every fraction has a denominator and a numerator. The fraction obtained by interchanging the denominator and numerator of a given fraction, is called the reciprocal fraction of that fraction.

3 is a fraction whose denominator is 4 and numerator is 3. Interchanging their positions 4 3 4 4 3 we get the fraction . So is the reciprocal fraction of ; likewise is the reciprocal 4 3 3 4 4 fraction of . If the product of two fractions is equal to 1, then they are mutually reciprocal 3

fractions.

Any fraction

Reciprocal of that fraction = 1 70

Elementary Mathematics Division of Fractions On the occasion of cultural week of her school, Chaiti bought 1 meter long red tape and used half it to make three equal badges. What portion of the 1 metre long tape was needed to make each badge ?

1 metre tape.

2

Half of 1 metre tape

1 metre tape into three equal parts and made a badge from each part.

2 1 1 For making each badge she needed portion of metre.

3 2 Chaiti divided the

1 metre 1 1 of 3 2

1 1 of 3 2

1 1 of 3 2

1 6

1 6

1 6

1 metre 2

1 2

3 metre 6

1 metre 2

1 From the picture it is clear that 1 portion of metre is equal to 1 portion of 1 metre.

2 3 6 In other words.

1 1 3 1 1 1 3 2 2 1 2 3 6 So.

1 metre portion of the 1 metre tape was needed to make each badge.

6

To divide a fraction by a fraction, we need to multiply the first fraction (dividend) by the reciprocal of the second fraction (divisor).

71

Elementary Mathematics 3

Example 9.

Solution :

Example 10. 1

3 5

8

3

1

1

5

8

8

what ?

8

5

3

8

5

1

Solution : 1

is the reciprocal of

8

5 7

1

7

Example 12.

1 1

2

what ?

23 75 Soluton : 46 35

Solution : 1 15

1

15

2

5 15

2

1

3

3 2

10 1

what ?

12

12

1 12

3 40

Example 11. 15

5 7

12

12

7

1

1

1

12 1

7

23 75 46 35

what ?

23 75

46 35

1 23 75 15

7 35 46 2

7 30

1 Example 13. Mr. Habib kept portion of his property for himself and divided the 4 the rest of the property equally between his two children. What portion of the propety did each child get?

1 Solution: Mr Habib kept portion of the property for himself.

4 4 1 3 remaining property 1 1 portion portion portion.

4 4 4 This portion he divided equally between his two children 3 1 2 3 3 3 portion.

So each child got 2 8 2 1 4 4 4

72

Elementary Mathematics

Exercise 7 ( D ) 1. Find the products : (a) (e)

6 7

14

16 3

17 12

(b) (f)

7

18

8

(c)

5

68

13

18

13

34

2

9

3

16

(g) 3

7 12

(d)

1

5 21

17

12

18

17

(h) 8

3 14

3

1 35

2. Find the quotients : 3 4

5

(e) 15

2

(a)

1 2

(b) 18

24 11

(c) 1

1 20

7

17 30

34 5

(g) 2

3 4

5

(f)

1 2

(d) 2

7 10

9

(h) 5

3 5

1

1 7

3. Evaluate : (a)

8 3 of 9 4

(b)

9 5 of 8 12

(c)

27 8 of 32 9

of 15 (d) 24 7

19 3 (e) 38

1 2 (f) 1 5 4 3

8.

1 In a hostel every day 1 quintal of rice in needed. In that hostel how many quintals 4 of rice is needed in 8 days ?

4 Sister's age is portion of brother's age. The brother's age is 15 years; what is the sister's 5 age ?

7 The divisor is 8 times the quotient. The quotient is ; what is the dividend ?

8 1 8 times of what number is equal to 10 times ?

3 2 2 The product of two numbers is 10 7 . One number is 2 7 ; what is the other number ?

9.

A number when divided by 5

4.

5.

6.

7.

1 1 yields the quotient 20 . What is the number ?

2 2 73

Elementary Mathematics 10.

Which number when divided by 12 will yield a quotient equal to the sum of 1 1 and ?

3 4 3 11. The value of portion of Mr. Mizan's property is 60,000 taka. What is the value of 20 1 portion of that property ?

10 1 1 portion of a pole is in mud, portion is in water and the remaining portion is above 12.

6 2 water. The length of the portion above water is 2 metre; what is the length of the portion in water ?

3 dozen of bananas were divided equally among 21 persons; how many bananas did 13. 8 4 every one get ?

14.

15.

16.

17.

18.

19.

Molly bought paper with 1 portion of her money, pens with 1 portion and writing 5 3 pad with 1 portion. She found that she still had 52 taka. What amount of money did she 4 have to begin with ?

5 3 portion of his money to an orphanage, Mr. Sajjad had 24000 taka. He donated 12 8 portion to an eductional institution. What amount of money was he left with ?

1 1 portion of his property for himself and gave protion to his wife.

8 8 The remaining property he divided equally among his four sons. Each son got property worth 15000 taka. What is the value of the total propety ?

3 1 Mr. Kamal spends portion of his monthly income on house rent , portion on 8 5 3 portion for other purposes. The remaining educational expenses for his children and 10 money he deposits in a bank. At the end of the year the bank deposit amounted to 14400 taka. What is his monthly income ?

3 2 portion of In the annual examination Rony and Panna respectively obtained and 4 3 the total marks. Rony obtained 50 marks more than Panna. What were the total marks and who obtained what marks ?

1 4 Mr. Matin bought a bicycle with portion of his money. He then bought a radio at 2 5 portion of the cost of the bicycle and distributed the remaining money equally between Mr. Nagen kept

his two daughters. Each daughter got 300 taka. How much money did Mr. Matin have ?

74

Chapter Eight

Decimal Fractions Multiplication of Decimal Fractions Mina and Raju went to the market and bought 1 5 kg of vegetables at the rate of 15 50 taka per kg. What amount of money will they owe the seller ? Clearly, the cost of the vegetables will be ( 15 50 1 5 ) taka . But the 15 50 and 1 5 are decimal fractions. To find the cost of the vegetables Mina and Raju must know how to multiply decimal fractions. Let us try to determine the product 0 1 0 1.

We know 0 1

1 . So 0 1 0 1 10

1 10

1 10

1 1 10 10

1 100

Now observe the adjoining picture. The coloured strip is one-tenth of the entire square region; so the 1 can be used to represent that strip.

10 1 1 1 1 We know, means of .

10 10 10 10

fraction

So if we divide the coloured strip into 10 equal parts, 1 1 then each of these smaller parts will represent of =0 1 10 10 One such part is shown coloured in the adjoining picture. Can we represent the small coloured square region in another manner ?

The square region has been divided into 100 small square regions as shown, So the coloured small square is

1 1 portion of the square region. is 100 100

written as 0 01.

1 1 Therefore of =0 1 10 10

0 1 = 0 01.

0 1.

0 01.

Elementary Mathematics Example 2. Multiply : 15 06 Solution :

25 Explanation :

1506 25

Hundreds

Tens

Ones

1

5

0

6 25

7

6

5

0

7530

One-tenths One-hundredths

30120 37650

3

Omitting the decimal point, the product 1506

25 has been found. In the

multiplicand there are two digits to the right of the decimal point. So the desired product is obtained by putting the decimal point before two digits from the right of the product 37650.

15 06

25

376 50

Multiplication of an Integer by a Decimal Fraction Example 3. Multiply : 12 Solution :

3 125 3125 12 6250 31250

Explanation : Hundreds

37500 Tens

3

Ones

One-tenths One-hundredths One-thousandths

3

1

2

7

5

0

5 12 0

In the multiplier there are three digits to the right of the decimal point. So the desired product is obtained by putting the decimal point before three digits from the right of the product 37500.

12 3 125 37 500 37 5 The last two zeros to the right of 5 have been omitted as being after the decimal point.

77

Elementary Mathematics Example 2. Multiply : 15 06 Solution :

25 Explanation :

1506 25

Hundreds

Tens

Ones

1

5

0

6 25

7

6

5

0

7530

One-tenths One-hundredths

30120 37650

3

Omitting the decimal point, the product 1506

25 has been found. In the

multiplicand there are two digits to the right of the decimal point. So the desired product is obtained by putting the decimal point before two digits from the right of the product 37650.

15 06

25

376 50

Multiplication of an Integer by a Decimal Fraction Example 3. Multiply : 12 Solution :

3 125 3125 12 6250 31250

Explanation : Hundreds

37500 Tens

3

Ones

One-tenths One-hundredths One-thousandths

3

1

2

7

5

0

5 12 0

In the multiplier there are three digits to the right of the decimal point. So the desired product is obtained by putting the decimal point before three digits from the right of the product 37500.

12 3 125 37 500 37 5 The last two zeros to the right of 5 have been omitted as being after the decimal point.

77

Elementary Mathematics

Multiplication of Decimal Fractions by 10, 100 Explanation : Example 4. Multiply : 5 237

10

5 237

Solution :

5 237

10

5237 thousandths 10 52370 thousandths 5237 hundredths

52 37

52 37

In the multiplier there is one zero to the right of 1. The desired product is found by shifting the decimal point one place to the right in the multiplicand.

Example 5. Find the product : 9 853 Solution :

Explanation : 9 853

100

9853

9 853 100 985 3

1000 9853

100 100

10 985 3

In the multiplier there are two zeros to the right of 1. The desired product is found by shifting the decimal point two places to the right in the multiplicand.

78

Elementary Mathematics

Multiplication of Decimal Fractions by Decimal Fractions Example 6. Multiply :0 78 Solution :

0 25

0 78

0 25

78

25

Explanation :

78

100 100 78 25

25 390

10000 1950

1560 1950

0 1950

10000

The required product is 0 1950

To the right of the decimal point there are 2 digits in the multiplicand and 2 digits in the multiplier in all 4 digits. So the desired product is found by placing the decimal point four places to the left of the product 78 25 = 1950. So 0 78

0 25 = 0 1950

Observe that the decimal points have been omitted and the numbers 78 and 25 have been multiplied as integers. Then we have placed the decimal point before as many digits from the right of that product as there are total digits to the right of the decimal points in the multiplicand and in the multiplier.

Example 7. Multiply : 0 23 Solution :

0 23

0 178

To the right of the decimal point in the multiplicand there are 2 digits, in the multiplier there are 3 digits, in all 5 digits. But the product 23 178 = 4094 has 4 digits; so we need to put one zero to the left of 4094 before putting the decimal point so as to have 5 digits.

23 178 184 1610 2300 4094 0 178 0 04094

Example 8. A train travels 42 5 kilometre in an hour. How many kilometre does it travel in 5 4 hours ?

425 Solution : In 1 hour the train travels 42 5 kilometre. in 5 4 hour the train travels 42 5

54 kilometre

229 50 kilometre.

79

54 1700 21250 22950

Elementary Mathematics

Exercise 8 (A) 1.

Find the products : (a) 0 24 12 (d) 39 3 679 (g) 0 00125 100 (j) 7 0 7 0 007 (k) 0 3 (l) 1 8

(b) 6 265 (e) 87 67 (h) 11

(c) 25 6 05 (f) 123 585 100

18 10

0 11

1 1

(i) 0 25

0 25

2 5

7000

0 03

0 003

30

0 18

0 8

0 01

2. 1 metre equals 39 37 inches; how many inches are there in 12 metres ?

3. 1 inch equals 2 54 centimetre; how many centimetres are there in 8 5 inches ?

4. One pencil costs 15 75 taka; what is the cost of 1 dozen pencils ?

5. A car travels 42 5 kilometres in one hour. How many kilometres will it travel in 15 3 hours ?

6. Urmi and Moumi respectively took 0 5 and 0 4 portion of one kuri of lychees. How many lychees did each get ?

[1 kuri is any collection of 20 objects of the same kind ]

7. One litre of milk costs 45 75 taka . Shamim bought 2 litres of milk and gave the milkman 100 taka. What refund will Shamim get ?

8. One kuri of eggs cost 160 taka. Molly bought 0 8 kuri and Rumi bought 0 5 kuri of eggs.

Who bought how many eggs and how much did each of them pay ?

9. Mr. Sayeed pays 0 15 portion of his monthly pay as income tax. He spends 0 8 portion of the remaining money in family maintenance; he saves the residual money. He saves 1020 taka per month; what is his monthy pay ?

10. 0 15 portion of a bamboo is in mud, 0 65 portion is in water. The length of the bamboo above water is 4 metre; what is the length of the bamboo as a whole ?

80

Elementary Mathematics Division of Decimal Fractions by Integers Example 9. Find the quotient : 34 8 Solution :

Putting decimal point in the quotient

6

6 348 58 30 48 48

6 34 8 5 8 same

30 48 48

0

Explanation :

0

34 8

348 tenths 6 348 tenths 58 tenths 30 48 48

quotient

0

58 tenths 5 8

We observe, in finding the quotient as soon as the digit after the decimal point has been taken down, the decimal point has been placed in the quotient.

Example 10. Find the quotient : 38 5 Solution :

16 38 5 2 40625 32 65 64 100 96 40 32 80 80

16

Division has been carried out till the remainder zero ( 0 ) is reached. After the second step a zero (0) has been put after the remainder 10, and after every subsequent remainder (4 and 8), because 38 5 38 5000 So

0 81

38 5

16

38 5000

16

Elementary Mathematics Division of Integers by Decimal Fractions Example 11. Find the quotient :

32

2 5

2 5 Solution : 32 Here the dividend is 32 and divisor is 2 5. Shifting the decimal point in the divisor one place to the right we get 25; this amounts to multiplying by 10. Multiplying the dividend by 10 we get 320.

So, 32 2 5 320 25 Explanation : 25 320 12 8 32 25 32 2 5 2 5 70 32 10 50

32

2 5

2 5

200 200

320

0

25

10

320

25

12 8

We observe : The decimal point in the divisor has been shifted to the right as far as is necessary to make the divisor an integer. To the right of the dividend as many zeros have been placed as the number of places the decimal point in the divisor was shifted. The desired quotient is the quotient upon division of the changed dividend by the changed divisor.

Example 12. The price of 3 5 kg of potatoes is 63 taka . What is the price of 1 kg of potatoes ?

Solution : The price of 1 kg of potatoes is 63 dividend and the divisor by 10 we get 63 35

630 35 280 280 0

35

3 5. Multiplying the 630

35

18

price of 1 kg of potatoes is 18 taka.

82

Elementary Mathematics Division of Decimal Fractions by 10, 100

Explanation : Exmaple 13. Find the quotient : 57 48 Solution : 57 48

10 57 48

10

1 10 5748 thousandths 5 748 5748 hundredths

5 748 57 48

10

10

5 748

We observe : In the divisor there is one zero (0) to the right of 1. The quotient has been obtained by shifting the decimal point in the dividend one place to the left.

The digits in the quotient are the same (and in the same order) as in the dividend.

Only the decimal point has changed its position.

Example 14. Find the quotient : 247 8 Solution : 247 8

100

Explanation :

2 478

247 8

100

100

2 478

247 8 100 1 2478 tenths 100 1 1 2478 10 100 1 2478 1000 2478 thousandths 2 478

We observe : * In the divisor there are 2 zeros (00) to the right of 1. * The quotient has been obtained by shifting the decimal point in the dividend 2 places to the left. * If in the divisor there are 3 zeros (000) to the right of 1, that is, if the divisor is 1000, then the quotient is obtained by shifting the decimal point 3 places to the left. * Required number of zeros (0, 00, 000, .....) have to be inserted as and when needed.

83

Elementary Mathematics Division of Decimal Fractions by Decimal Fractions Example 16. Find the quotient : 3 36 32

Example 15. Find the quotient : 33 6 3 2 Solution: 33 6 3 2 Here dividend is 33 6 and divisor is 3 2.

Shifting the decimal point in the divisor one place to the right we get 32; shifting the decimal point one place to the right in the dividend 33 6 we get 336 = 336 00 32 336 00 10 5 32

Solution: 3 36 32 Here dividend is 3 36 and divisor is3 2.

Shifting the decimal point in the divisor one place to the right we get 32; shifting the decimal point one place to the right in the dividend 3 36 we get 33 6.

32 33 6 1 05 32 160 160

160 160 33 6

3 2

0

0 10 5

3 36

3 2

1 05

We observe : * To free the divisor from decimals, it has been multiplied by 10; that means, the decimal point has been shifted one place to the right. * To balance (that is, to keep the quotient unchanged), the decimal point in the dividend has been shifted one place to the right. * Divison to the right of the decimal in the quotient has been carried out till the remainder 0 is reached by putting one zero (0) to the right of every remainder.

Example 17. A car travels 43 2 kilometre on 2 4 litre of petrol. How far will it travel on 1 litre of petrol ?

Solution: On 2 4 litre of petrol the car travels 43 2 kilometre on 1 litre of petrol the car travels 43 2 43 2 43 2 10 432 Now 2 4 2 4 10 24 On 1 litre of petrol the car travels 18 kilometre.

84

2 4 kilometre.

24 432 18 24 192 192 0

Elementary Mathematics

Exercise 8 (B) 1. Find the quotient : (a) 6 13 14 6

(b) 14 49

7

(d) 0 80 16 (g) 15 0 04 (j) 12 38 10

(e) 65 4 6 (h) 42 7 5 (k) 28 5 10

(m) 83 65 (p) 0 027

(n) 57 5 0 23 (q) 0 0007 0 035

100 18

(c) 46 5

5

(f) 0 005 25 (i) 18 3 60 (l) 78 3 100 (o) 6 4 0 05 (r) 65 61 0 0027

2. One dozen ballpens cost 51 60 taka, what is the price of one ballpen ?

3. One quintal of rice costs 3875 00 taka, what is the price of 1 kg of rice ?

4. 3 5 metres equal 137 795 inches; how many inches are there in 1 metre ?

5. A car travels 46 8 kilometre in one hour. What time will the car need to travel 257 4 kilometres ?

6. A 3-minute call from a mobile phone costs 2 04 taka; what is the per minute cost of calling from that phone ?

7. The product of two numbers is 10 5. One number is 2 8; what is the other number ?

8. The product of two numbers is 12 775. One number is 0 7; what is the other number ?

9. 0 000625 divided by which number will yield the quotient 0 125 ?

10. Everyone of 40 students contributed 12 75 taka. The money thus raised was divided equally among 8 poor persons. Wht money did each poor person get ?

11. A teacher bought oranges for 393 90 taka, costing 60 60 taka per dozen, and divided them equally among 13 students. How many oranges did each student get ?

12. The sum of two numbers is 70 60. The larger number is 4 50 more than the smaller number. Find the two numbers.

85

Chapter Nine

Percentage The square region to the right has been divided into 100 small square regions, as shown. 15 of these small square regions to the upper left corner have been coloured. So the coloured 15 (rectangular) region is portion of the whole 100 (square) region. We express this by saying that the coloured region is 15% of the whole region.

The denominator of the above fraction is 100; 15 % is read as ' fifteen per cent ' meaning fifteen per hundred.

15 .

So 15% = 100 Full marks Siam Sirat Shihab Partha Subject Mathematics

100

82

75

85

70

Mita 73

We see that out of 100 marks, Siam has got 82 marks; that is he has got 82% marks.

Similarly, Sirat has got 75%, Shihab has got 85%, Partha has got 70% and Mita has got 73% marks.

We observe : A percentage is a fraction whose denominator is 100.

4 into 5 percentage.

Example 1. Convert

Solution :

4 5

4 5 80 100

20 20

13 into 25 percentage.

Example 2. Convert

Solution : 13 25 80%

13 4 25 4 52 52 % 100

We observe : To convert a fraction into percentage, the denominator needs to be 100.

Elementary Mathematics

Example 3. Express 45% as a

Example 4. Express 18% as a

fraction Solution : 45 %

9

45 100 20

fraction.

Solution : 18 %

9 20

18 9 100 50

9 50

The fraction representing a percentage should be expressed in its lowest terms.

Example 5. The population of Sakhipur village is 1280. 40% of the population of that village is educated. Find the number of educated persons.

Solution : 40 % of the inhabitants are educated.

So, out of 100 persons the number of educated persons is 40 40 100

1 1280

2 40

256 1280 100 1 5

512 persons.

Example 6. A pen is bought for 50 taka and sold for 55 taka. What is the percentage profit ?

Solution : Selling price Cost price

55 taka 50 taka

Profit 5 taka 55 taka 50 taka On cost price 50 taka profit is 5 taka 5 1 50 2 5 100 taka 100 50 1

The percentage profit is 10 taka.

87

10 taka.

Elementary Mathematics * Profit is made if selling price is more than the cost price * Profit = Selling price

Cost price

* Percentage profit is calculated on the basis of the cost price.

Example 7. A pen is bought for 50 taka and sold for 46 taka. What is the percentage loss ?

Solution : Cost price = 50 taka Selling price = 46 taka Loss

50 taka

46 taka

4 taka

On cost price 50 taka loss is 4 taka 4 1 50 2 4 100 8 taka.

100 50 1 Percentage loss is 8 taka.

Loss is made if selling price is less than the cost price Loss = Cost price

Selling price.

Percentage loss is calculated on the basis of the cost price.

Example 8. The rate of interest is 7 % per annum. What is the interest on 800 taka for 5 years ?

Solution: On 100 taka deposit interest for 1 year is 7 taka.

1 800

1 1

800

5

Interest

Rate of interest

7 100 7 7

800 100 8 800 5 taka 100 1

Principal 100 88

280 taka.

Time (in years)

Elementary Mathematics

Example 9. Mahfuza deposited 500 taka in a bank and in 3 years got 105 taka as interest.What is the percentage rate of interest per annum ?

Solution : On 500 taka interest for 3 years is 105 taka 105 3 105 1 1 500 3 7 21 1 105 100 1 7 taka.

100 500 3 1 51 So the percentage rate of interest is 7 taka per annum.

500

1

Percentage rate of interest

Interest 100 Principal Time (in years)

Money deposited in a bank is the 'principal' Money (additional to the principal) credited to an account is 'interest' Interest earned by depositing 100 taka for 1 year is the percentage rate of interest.

Example 10. The rate of interest being 8% what amount of money will earn 240 taka as interest in 5 years ?

Solution: 8 taka is the interest in 1 year when principal is 100 taka 100 1 1 8 100 240 240 1 8 6 100 240 30 600 taka 240 5 8 5 1 1 So, 600 taka will earn 240 taka as interest in 5 years at 8% per annum.

Principal

Interest Rate of interest

100 Time ( in years )

For the same amount of interest , the more the time the less the principal.

So in the last step, one has to divide by 5.

89

Elementary Mathematics

Exercise 9 1. Express as percentage : (a) 3 4

(b) 5 8

(c) 4 15

(d) 18 15

(e) 15 24

(f) 5 12

(g) 28 50

2. Express as fraction : (a) 6%

(b) 15%

(c) 30%

(d) 45%

(e) 50%

(f) 65%

(g) 75%

(g) 80%

3. Evaluate : (a) 8% of 25 taka (b) 5% of 75 taka

(c) 25% of 60 taka

(d) 20% of 40 kg (e) 30% of 60 gram (f) 15% of 80 km 4.

The population of Alipur village is 1800. 45% of the people of the village is educated. What is the number of educated persons in Alipur ?

5.

Increasing by 5% the population of Birampur village became 2100.

What was the previous population of the village ?

6.

In an examination out of 40 students 4 failed,, What percentage of students passed in that examination ?

7.

In Bangladesh every year 25 persons are born and 15 persons die, for every one thousand of the population. What is the percentage rate of increase of polulation in a year ?

8.

Increasing by 4 percent the population of Chandanpur village became 1040 persons. What was the previous population of that village ?

9.

In 1981 the population of Bangladesh was 9 crore and in 1982 it was 9 crore 27 lac. What was the percentage rate of increase of population in 1982 ?

10.

A pen was bought for 50 taka and sold for 56 taka. What is the percentage profit ?

90

Elementary Mathematics 11.

A pen was bought for 60 taka and sold for 54 taka.

What is the percentage loss ?

12.

The cost price of a trouser is 250 taka. At what price must it be sold to make 20% profit ?

5% loss was incurred by sellling a radio for 1140 taka. What was the purchase price of the radio ?

A person sold a goat at 6% loss. The selling price was 3196 taka; what was the cost price?

13.

14.

15.

Apples bought for 800 taka were sold at 4% loss. What was the total amount of loss in taka ?

16.

Apples bought for 800 taka were sold at 4% profit. What was the total amount of profit in taka ?

The profit for 3 years on 600 taka is 144 taka. Find the annual rate of interest.

17.

18.

Depositing 900 taka in a bank Mahfuza got 81 taka as interest for 3 years.

What is the annual rate of interest ?

19.

The annual rate of interest is 7 taka. What is the interest on 700 taka for 7 years ?

20.

What is the interest for 5 years on 480 taka at annual 8% interest ?

21.

Rate of interest being 6%, what amount of taka shall earn 180 taka as interest in 5 years ?

22.

Annual percentage rate of interest being 8 taka, which principal will earn 360 taka as interest in 5 years ?

23.

Interest for how many years on 300 taka at 5% per annum will be 60 taka ?

24.

Annual percent rate of interest being 5 taka, what is the interest on 900 taka for 3 years ?

At 5% annual rate of interest in how many years will 425 taka become 510 taka with interest ?

25.

91

Chapter Ten

Measurement Measurement of Length The basic unit for measurement of length is : Metre.

Every bounded object or region has a definite linear measure; this measure is called the length of the object or region.

For convenience various multiples and submultiples of metre are used as secondary units : 1 kilometre (km) = 1000 metre (m) 1 hectometre (hm)

= 100 metre (m)

1 decametre (dm)

= 10 metre (m)

1 metre (m)

= 1metre (m)

1 decimetre (dcm)

=

1 centimetre (cm) 1 millimetre (mm)

1 m or 0.1 m 10 1 = 100 m or 0.01 m =

1 m or 0.001 m 1000

On the increasing side, beginning with 1 metre, these units are such that each unit is10 times the preceding unit. On the decreasing side, 1 beginning with 1 metre they are such that each unit is one-tenth (10 ) of the preceding unit.

10

10

kilo is

thousand times

hecto is

hundred times

deca is

ten times

10 10 10

deci is

one-tenth

1 10

centi is

one-hundredth

milli is

one- thousandth

1 100 1 1000

10

1000 100 10 0.1 0.01 0.001

Elementary Mathematics

Measurement of Weight The basic unit for measurement of weight is : Gram Metric units for Measurement of Weights Replacing 'metre' by 'gram' in the chart on the preceding page, we get the metric units for measurement of weights. On the increasing side these are decagram (dg), hectogram (hg), kilogram (kg); on the decreasing side these are decigram (dcg), centigram (cg), milligram (mg) Two widely used unis for measuring larger weights are : 100 kilogram (kg) = 1 quintal 10 quintal or 1000 kilogram = 1 metric ton

Example 1. Express 8 Kilogram 35 gram 79 centigram 6 millgram in milligram.

Solution : 8 kilogram = 8 1000 gram = 8000 gram = 8000 1000 mg = 8000000 mg 35 gram = 35 1000 mg = 35000 mg 79 centigram = 79 10 mg = 790 mg 6 milligram = 6 mg Sum = 8035796 mg there are 8035796 milligram in 8 kilogram 35 gram 79 centigram 6 milligram.

Example 2. Express 58 kilogram 357 gram 629 milligram in kilogram.

Solution : 58 kg = 58 kg 357 gram = 357 0.001 kg = 0.357 kg 629 milligram = 629 0.001 gram = 0.629 gram = 0.629 0.001kg = 0.000629 kg Sum = 58.357629 kg there are 58.357629 kilogram in 58 kilogram 357 gram 629 milligram.

Example 3. How many qunitals are there in 9876000 grams ? How many metric tons ?

Solution : 9876000 grams = 9876000 kg = 9876 kg = 9876 quintal 1000 100 = 98.76 quintal.

98.76 metric ton = 9.876 metric ton.

10 there are 98.76 quintal or 9.876 metric ton in 9876000 grams.

Again, 9876000 grams = 98.76 quintal =

93

Elementary Mathematics

Example 4. Add : (a)

km

hm

dm

metre

98 76 53 42

7 8 0 9

6 4 8 3

5 9 7 1

hm

dm

metre

7 8 0 9

6 4 8 3

5 9 7 1

6

3

2

Solution : (a) km 98 76 53 42 271

(b)

(b)

271 km 6 hm 3 dm 2 metre

kg

hg

dg

g

39 52 13 48

6 1 5 6

7 0 4 5

8 9 2 3

kg

hg

dg

g

39 52 13 48 153

6 1 5 6

7 0 4 5

8 9 2 3

9

8

2

153 kg 9 hg 8 dg 2 gram

Example 5. Subtract : (a)

km 89 47

hm 5 6

dm 7 4

metre 3 8

(b)

kg 76 53

hg 9 8

dg 2 7

g 4 6

hm dm +10 5 7

metre 3 +10

(b)

kg

hg

dg

g

Solution : (a)

km 89 47 41

+1

6

4

9

2

+1

76

9

2 +10

4+10

8

53

8 +1

7 +1

6

5

23

0

4

8

41 kg 9 hm 2 dm 5 metre

23 kg 4 dg 8 gram

94

Elementary Mathematics

Measurement of Volume of Liquids The basic unit for measurement of volume of liquids is : Litre Replacing 'metre' by 'litre' in the chart on page 92, we get the metric units for measurement of volume of liquids. On the increasing side these are: decalitre (dl), hectolitre (hl), kilolitre (kl), On the decreasing side these are decilitre (decil), centilitre (cl), millilitre (ml).

International Units for Measurement of Volume of Liquid.

1000 millilitre (ml) = 1 litre (l) or 1000 cubic centimetre 1000 litre = 1 cubic metre

The following chart exhibits the basic and secondary units for measurement of length, weight and volume of liquid at a glance; it can be used for ready reference : Thousands

Hundreds

Tens

Ones

One-tenths One-hundredths One-Thousandths

metre kilo

hecto

deca

gram

deci

centi

milli

1 10

1 100

1 1000

litre

1000

100

10

1 96

Elementary Mathematics

Measurement of Volume of Liquids The basic unit for measurement of volume of liquids is : Litre Replacing 'metre' by 'litre' in the chart on page 92, we get the metric units for measurement of volume of liquids. On the increasing side these are: decalitre (dl), hectolitre (hl), kilolitre (kl), On the decreasing side these are decilitre (decil), centilitre (cl), millilitre (ml).

International Units for Measurement of Volume of Liquid.

1000 millilitre (ml) = 1 litre (l) or 1000 cubic centimetre 1000 litre = 1 cubic metre

The following chart exhibits the basic and secondary units for measurement of length, weight and volume of liquid at a glance; it can be used for ready reference : Thousands

Hundreds

Tens

Ones

One-tenths One-hundredths One-Thousandths

metre kilo

hecto

deca

gram

deci

centi

milli

1 10

1 100

1 1000

litre

1000

100

10

1 96

Elementary Mathematics

Exercise 10 (A) 1. Express in millimetre : (a) 29 kilometre 89 centimetre 7 millimetre (b) 18 kilometre 459 metre 98 millimetre (b) 59 kilometre 368 metre 87 centimetre 2. Express in metre : (a) 98 metre 48 centimetre 7 millimetre (b) 76 kilometre 584 centimetre 5 millimetre (c) 58 kilometre 356 centimetre 3. Express in kilometres: (a) 56 kilometre 37 metre 48 centimetre (b) 39 kilometre 84 metre 95 millimetre (c) 87 kilometre 60 metre 51 centimetre 9 millimetre 4. Express in milligrams : (a) 81 gram 52 centigram 7 milligram (b) 95 kilogram 83 gram 76 centigram (c) 67 kilogram 98 gram 67 centigram 8 milligram 5. Express in grams : (a) 75 kilogram 643 gram (b) 28 kilogram 89 gram 45 centigram 8 milligram (c) 76 kilogram 85 gram 43 centigram 6. Express in kilograms : (a) 54 kilogram 52 gram 95 centigram (b) 21 kilogram 89 gram 541milligram (c) 98 kilogram 76 gram 54 centigram 3 milligram 7. How many quintals are there in 796000 gram? How many metric tons ?

8. How many metric tons are there in 97865 kilograms ?

97

Elementary Mathematics 9. Add : (a)

km

hm

dm

metre

12

9

6

7

59

5

0

82

8

17

hg

dg

gram

82

7

5

6

4

43

0

8

4

2

5

70

3

0

8

3

9

8

29

2

7

0

km

hm

dm

hg

dg

gram

89

5

2

1

93

7

3

2

82

9

3

7

45

6

8

9

(b)

kg

10. Subtract : (a)

11.

metre

(b)

kg

(a) How many millilitres are there in 987600 litre ? (b) How many cubic centimetres are there in 237 litre ? (c) How many cubic centimetres are there in 879500 litre ?

12. Multiply : (a) 9 km 7 hm 4 metre by 8 (a) 7 km 10 dm 5 metre by 9 (a) 29 kg 8 hg 19 gram by 7 13. Divide : (a) 82 kilometre 7 hectometre 6 decametre 4 metre by 9 (b) 94 km 8 hm 5 dm 6 metre by 8 (c) 76 kg 7 hg 6 dg 1 gram by 9 14.

Out of 598 kg of potatoes Latif Miah sold 127 kg 256 gram to Malek, 134 kg 125 gram to Khaleq and 89 kg 348 gram to Liton . What amount of potatoes remained with him ?

15. A drum can store 53 kg 9 hg 8 dg 7 gram of flour. How much flour can be stored in 9 such drums ?

98

Elementary Mathematics

Area

The coloured figure to the left is a region. This region occupies a definite portion of this page. As such it has a definite measurement.

That measure is the area of the region.

Any closed region has a definite measurement as to how much space it covers; this measurement is the area of the region.

Triangular region

Quadrangular region Rectangular region

Square region

Each of the above coloured figures is a region.

The space enclosed by a triangle is a triangular region.

The space enclosed by a quadrilateral is a quadrangular region.

The space enclosed by a rectangle is a rectangular region.

The space enclosed by a square is a square region.

Each of these regions, being bounded, has a definite measurement as to how much space it covers. That measurement is the area of the region.

A definite unit is needed for measurement of area. The area of the region enclosed by a square, each of whose sides has length 1 unit, is a natural choice for measurement of area.

Such units are referred to as 'square units'. Thus, 1 square metre is the area of a square region each of whose sides has length 1 metre (this square is referred to as the '1 metre square') 99

Elementary Mathematics

Metric Units for Measurement of Land 1 square metre 1 square decametre 1 square hectometre 1 hectare

= = = =

square decimetre 100 square metre 100 square decametre 100 10,000 square metre

= = = =

10,000 square centimetre 1 are [pronounced: a-or] 10,000 square metre 1 square hectometre

Area of a Square Region

1cm

1cm 1cm

The figure to the left is a square: each of its sides has length 1cm 3 cm. What is the area of the region enclosed by it ?

Beginning from one corner, say the lower left corner, we mark 1cm points on each of the two sides meeting at that corner, whose distances from that corner are 1 cm, 2 cm. From these points we draw lines parallel 1cm to the other side meeting at the corner. These lines divide the square region into 3 3 = 9 square regions: each of these 9 square regions has area 1 square centimetre, because each side of each of these square regions has length 1 cm. 3 3 is written as 32 : so the area of the given square region is 32 = (length) 2 square units.

Area of a Square Region = (length) 2 square units.

Area of a Rectangular Region A

B 1cm

3 cm (breadth)

1cm 1cm

C

1cm

1cm

1cm

1cm

D

4 cm (length)

ABCD is a rectangular region; its length is 4 cm and breadth is 3 cm. Beginning from one corner, say the lower left corner, we mark points on each of the two sides meeting at that corner whose distances from that corner are 1 cm, 2 cm and so on. From these points we draw lines parallel to the other side meeting at that corner. These lines divide the rectangular region into 4 3 = 12 square regions; each of these square regions has area 1 square cm, as explained above.

100

Elementary Mathematics

Therefore, area of the rectangular region ABCD = Sum of the areas of 12 square regions, each having area 1 sq cm = 12 sq cm = (4

3) sq cm = 4 cm

3 cm = length

breadth

We observe : The area of the rectangular region is obtained by multiplying the measeure of its length by the measure of its breadth.

Area of a rectangular region = (length

breadth) square units

Example 1. The length of a rectangular region is 60 cm and its breadth is 45 cm. What is its area ?

Solution :

Area of a rectangular region = (length Area = 60 cm

breadth) square units

45 cm = (60 45) square cm = 2700 square cm

Required area is 2700 square cm.

Example 2. What is the area of a rectangular region of length 7 cm and breadth 3 cm 5 mm?

Solution :

Length = 7 cm = 7 10 mm = 70 mm breadth = 3 cm 5 mm = 35 mm Area of the rectangular region = (length breadth) = 70 mm 35 mm = 2450 square mm

= 24 50 square cm

[100 square mm = 1 square cm]

In finding area, if the length and breadth are expressed in different units, then they must first be expressed in the same unit.

101

Elementary Mathematics

Area of a Triangular Region The triangle ABC enclosing the region can be any one of three types : right angled ; acute angled ; obtuse angled. Suppose ABC is a right angle.

Draw the line through A parallel to the side BC, and the line through C parallel to the side BA. Suppose these lines A D intersect at D. Then ABCD is a rectangle, because it is a parallelogram and ABC is a right angle. Therefore ADC too is a right angled triangle, whose sides AD and DC are respectively B C equal to the sides BC and AB of the triangle ABC. The side AC is common to both triangles. So the triangles ABC and ADC are identical except for position; so they have the same area. Therefore the area of the triangular region ABC is half of the area of the rectangular region ABCD. 1 The latter area is AB BC. So area of the triangular region ABC is = 2 (AB BC) AB is called the height and BC is called the base of the triangle ABC. So, the area 1 of a right angled triangular region = 2 (base height) square units.

Suppose the triangle ABC is acute angled; that means, each of the three angles is less than one right angle. Through A draw the line parallel to the side BC. Through B and C draw the lines perpendicular to the side BC; suppose that these lines intersect the line parallel to BC at E and F respectively.The lines BE and CF, being perpendicular to BC, are parallel. Draw through A the line AD perpendicular to BC.

E

A

F

B

D

C

Picture 1 102

Elementary Mathematics

Each of the figures AEBD and AFCD is a rectangle, and each of the triangles ABD and ACD is right angled.

1 1 Area of the triangluar region ABD = 2 (base height) = 2 (BD AD); 1

1

Area of the triangluar region ACD = 2 (base

height) = 2 (DC AD);

So, Area of the triangular region ABC = Area of the triangular region

ABD 1

1

+ Area of the triangular region ACD = 2 (BD AD) + 2 (DC AD) 1 1 1 = 2 (BD + DC) AD = 2 (BC AD) = 2 (base height) square units.

Lastly, suppose

E

F

A

B

C Picture 2

D

ACB is an obtuse angle. As before, the lines BE and CF are drawn

perpendicular to BC; they meet the line drawn through A and parallel to BC at E and F respectively. But now E and F lie on the same side of the point A ; the perpendicular from A to the side BC now meets the extended side BC at D. Each of the figures AEBD and AFCD is a rectangle, and each of the triangles ABD and ACD is right angled.

So, area of the triangular region ABC = area of the triangular region ABD 1

1 area of the triangular region ACD = 2 (BD AD) (CD AD) 2 1 1 1 = 2 (BD CD) AD = 2 (BC AD) = 2 (base height).suqare units

Therefore, in all cases we get the same result : Area of a triangular region = 1 (base 2

103

height) suqare units

Elementary Mathematics

Example 3. A rectangular region is formed by two adjoining and non over lapping triangular regions of the same size; the length of the rectangle is 16 metre and breadth is 12 metre. What is the area of the rectangular region ?

What is the area of each triangular region ?

Solution : Area of the rectangular region

Length breadth 16 12 square metre 192 square metre 1 2 1 2

Area of the triangular region

Area of the rectangular region

192 square metre 96 square metre Area of the rectangular region is 192 square metre and the area of each triangular region is 96 square metre

Area of a Quadrangular Region In the adjoining figure ABCD is a quadrilateral.

AC is a diagonal. This diagonal divides the quadrangular region ABCD into the two triangular regions ABC and ADC, which do not overlap. So area of the quadrangular region ABCD = Area of the triangular region ABD + Area of the triangular region ACD Draw BE perpendicular from B on AC, and DF perpendicular from D on AC. Then BE is the height of the triangle ABC and DF is the height of the triangleADC; AC is their common base. So, Area of the quadrangular region ABCD

A E D F C

B Picture-3

= Area of the triangular region ABC + Area of the triangular region ADC 1

= 2 (AC

1

BE) + 2 (AC

1

DF) = 2 AC

(BE + DF) suqare units.

104

Elementary Mathematics

Example 3. A rectangular region is formed by two adjoining and non over lapping triangular regions of the same size; the length of the rectangle is 16 metre and breadth is 12 metre. What is the area of the rectangular region ?

What is the area of each triangular region ?

Solution : Area of the rectangular region

Length breadth 16 12 square metre 192 square metre 1 2 1 2

Area of the triangular region

Area of the rectangular region

192 square metre 96 square metre Area of the rectangular region is 192 square metre and the area of each triangular region is 96 square metre

Area of a Quadrangular Region In the adjoining figure ABCD is a quadrilateral.

AC is a diagonal. This diagonal divides the quadrangular region ABCD into the two triangular regions ABC and ADC, which do not overlap. So area of the quadrangular region ABCD = Area of the triangular region ABD + Area of the triangular region ACD Draw BE perpendicular from B on AC, and DF perpendicular from D on AC. Then BE is the height of the triangle ABC and DF is the height of the triangleADC; AC is their common base. So, Area of the quadrangular region ABCD

A E D F C

B Picture-3

= Area of the triangular region ABC + Area of the triangular region ADC 1

= 2 (AC

1

BE) + 2 (AC

1

DF) = 2 AC

(BE + DF) suqare units.

104

Elementary Mathematics

In the adjoining figure ABCD is a quadrilateral of A which the two opposite sides AD and BC are parallel and the angle ABC is a right angle; AD = 8 cm BC = 12 cm DE is the perpendicular from D on BC. Then ABED is a rectangle and DEC is a right angled triangle. DE is the distance B

8 cm

E 12 cm

between the parallel sides; so DE = 6 cm.

EC = BC

BE = BC AD = (12

D

8) cm = 4 cm

Area of the rectangular region ABED = AD AB = 8 cm 6 cm = 48 sq cm Area of the triangular region DEC = 12 (base = 12 (4 cm

height) = 12 (EC 6 cm) = 12

DE)

24 sq cm = 12 sq cm

Area of the quadrangular region ABCD = Area of the rectangular region ABED + Area of the triangular region DEC = 48 sq cm + 12 sq cm = 60 sq cm.

Exercise 10 (B) 1. Find the areas of the following rectangular regions : (a) Length 54 metre and breadth 47 metre. (b) Length 3 metre and breadth 75 centimetre. (b) Length 87 metre and breadth 65 25 metre.

2. The length of one side of a square region is given. Find the area : (a) 67 metre

(a) 7 metre 45 centimetre

106

(c) 9 75 metre

C

Chapter Eleven

Time According to Bangla practice, the time span from one sunrise to the next sunset is called day-time and the time span from one sunset to the next sunrise is called night-time.

The names of twelve Bangla months and the number of their days : Month

Number of days

Month

Number of days

Baishakh

31

Kartik

30

Jaisthya

31

Agrahyon

30

Ashar

31

Paush

30

Shrabon

31

Magh

30

Bhadra

31

Falgun

30

Ashwin

30

Chaitra

30

According to Bangla practice day and date begins with sunrise.

The names of twelve English months and the number of their days : Month

Number of days

Month

Number of days

January

31

July

31

February

28

August

31

March

31

September

30

April

30

October

31

May

31

November

30

June

30

December

31

When the year is a leap year, the month of February has 29 days (see next page).

According to English and international practice, day and date begins just after 12 p.m (midnight)

Elementary Mathematics

Bangla and English Calendar English Year 2012; Bangla Year 1418 - 1419 January Sun

Paush- Magh

Mon Tues

Wed

1

2

3

8

9

10 11

18 25

19 26

20

4

21

26

28

Thurs Fri

5

22

Sat

6

23

Sun

Mon Tues

7

30

Wed

1

24

19

12 13 14 29

Magh-Falgun

February

5

1

23

6

24

7

25

8

26

Thurs Fri

2

3

20

9

27

Sat 21

4

22

10 11 28

29

15 16 17 18 19 20 21

12 13 14 15 16 17 18

22 23 24 25 26 27 28

19 20 21 22 23 24 25

29 30 31

26 27 28 29

2 9

16

3

4

10

5

11

17

6

12

13

21

22

6

23

1

7

Wed

Thurs Fri

7

24

2

8

14

3

9

15

16

10

April

13

Sun

Mon Tues

Wed

Thurs

Fri

Sat

4

5

6

7

1

2

3

8

9

10

8

9

10 11

26

12

6

Sat

3

25

11

5

Chaitra-Baishakh

2

19

4

17

1

18

5

30

15

Falgun-Chaitra Mon Tues

4

14

8

18

March Sun

7

20

18

27

25

19 26

20 27

21

28

22

23

24

12 13 14 29

30

1

11 12 13 14 15 16 17

15 16 17 18 19 20 21

18 19 20 21 22 23 24

22 23 24 25 26 27 28

24 26 27 28 29 30 31

29 30

28 8

11

29

5

12

30

6

13

31 7

14

1

8

15

2

9

16

2

3

10

3

9

17

16

10

4

11

5

12

6

13

7

14

8

15

17

Fill in the blanks : (a) The 6th day of January 2012 is

[ weekday]

(b) The 18th day of April 2012 is

[ weekday] [ weekday]

(c) The 19th day of Magh of Bangla year 1418 is (d) The second Monday in the month of March 2012 is

[ date ]

(e) The first Saturday in the month of Baishakh of Bangla year 1419 is (f) The month of February of 2012 has

days

109

[ date ]

Elementary Mathematics

The last two digits of the number 1988 are not both zero, and the number is divisible by 4. So 1988 was a leap year. So there were 29 days in the month of February,1988.

Example 2. How many days were there in the month of February 1998 ?

Solution : We need to determine whether 1998 was a leap year.

4 ) 1 9 9 8 ( 499 16 39 36 38 36 2 The last two digits of the number 1998 are not both zero and the number is not divisible by 4. So 1998 was not a leap year. So there were 28 days in the month of February 1998.

Example 3. Was 2000 a leap year ?

Solution : 2000 ends with 00, so 2000 was a leap year if 2000 is divisible by 400.

400 ) 2 0 0 0 (5 2000 0

2000 ends with 00, and the number 2000 is divisible by 400; so 2000 was a leap year.

Example 3. Was 1900 a leap year ?

Solution : 1900 ends with 00, so 1900 was a leap year if 1900 is divisible by 400.

400 ) 19 0 0 (4 16 0 0 300

1900 ends with 00, and the number 1900 is not divisible by 400; so 1900 was not a leap year.

111

Elementary Mathematics

The last two digits of the number 1988 are not both zero, and the number is divisible by 4. So 1988 was a leap year. So there were 29 days in the month of February,1988.

Example 2. How many days were there in the month of February 1998 ?

Solution : We need to determine whether 1998 was a leap year.

4 ) 1 9 9 8 ( 499 16 39 36 38 36 2 The last two digits of the number 1998 are not both zero and the number is not divisible by 4. So 1998 was not a leap year. So there were 28 days in the month of February 1998.

Example 3. Was 2000 a leap year ?

Solution : 2000 ends with 00, so 2000 was a leap year if 2000 is divisible by 400.

400 ) 2 0 0 0 (5 2000 0

2000 ends with 00, and the number 2000 is divisible by 400; so 2000 was a leap year.

Example 3. Was 1900 a leap year ?

Solution : 1900 ends with 00, so 1900 was a leap year if 1900 is divisible by 400.

400 ) 19 0 0 (4 16 0 0 300

1900 ends with 00, and the number 1900 is not divisible by 400; so 1900 was not a leap year.

111

Elementary Mathematics

Decade, Yuga, Century A period of 10 consecutive years is a decade.

A period of 12 consecutive years is a yuga. [ There is no term in English for yuga ] A period of 100 consecutive years is a century.

For example : The period from 2001 to 2010 is a decade. The period from 2001 to 2012 is a yuga.

The period from 1901 to 2000 is a century (twentieth century) The period from Bangla years 1201 to 1300 is the thirteenth century.

The period from Bangla years 1301 to 1400 is the fourteenth century.

The period from Bangla years 1401 to 1500 is the fifteenth century.

Example 5. Convert 3 years 2 months 12 days into hours.

Solution :

3 years = 3 2 months = 2

365 days = 1095 days 60 days 30 days = 12 days

Sum

= 1167 days 24

1 year = 365 days 1 month = 30 days

1 day = 24 hours

4668 23340 28008 hours When no specific month is mentioned, a month is assumed to have 30 days.

Example 6. Convert 3 years 5 months 15 days into minutes.

Solution : 3 years = 3 365 days = 1095 days 1 year = 365 days 150 days 1 month = 30 days 5 months = 5 30 days = 15 days = 1260 days 1260 days 1 day = 24 hours

24 5040 25200 30240 hours 60 1814400 minutes 112

1 hour = 60 minutes

Elementary Mathematics Example 7. Convert 1 day into seconds.

Solution: 1 day = 24 hours 24 hours 60 1440 minutes 60 86400 seconds.

1 hour = 60 minutes

1 minute = 60 seconds

1 day = 86400 seconds.

Example 8. Convert 5 months 4 days into seconds.

1 month = 30 days Solution: 5 months = 5 30 days = 150 days 4 days 154 days 5 months 4 days 154 days 24 616 3040 3696 hours 60 221760 minutes 5 months 4 days

60 13305600 seconds.

1 day = 24 hours

1 hour = 60 minutes 1 minute = 60 seconds

Example 9. Convert 89765 hours into year, month and days .

Solution 24 ) 8 9 7 6 5 hours ( 3740 days 72 365 ) 3 7 4 0 day ( 10 years 177 365 168 90 96 00 96 9 0 days 5 30 ) 9 0 days ( 3 months 0 90 5 hours 0 89765 hours = 10 years 3 months 5 hours 113

Elementary Mathematics

We observe : In determining the number of years, 1 year has been assumed to have 365 days.

1 month has been assumed to have 30 days.

Example 10. How many years, months, days and hours are there in 87450 minutes ?

Solution : 60 87450 minutes 6 0 2 7 4 2 4 0 3 4 3 0 4 4

24 5 0 5 0 2 0

1457 hours 1 4 5 7 hour 60 days 1 4 4 1 7 0 0 1 7 hours

3 0 minutes

30

60 60 0

2 months

87450 minutes = 2 months 17 hours 30 minutes.

Example 11. How many hours, minutes and seconds are there in 84648 seconds ?

Solution : 60 8 4 6 4 8 seconds 60 246 240 64 60 48 00

1410 minutes 60 1 4 1 0 minutes 23 hours 120 210 180 3 0 minutes

4 8 seconds 84648 seconds = 23 hours 30 minutes 48 seconds.

114

Elementary Mathematics International Time Table According to international practice, the time span from one midnight to the next midnight is 1 day. In the international time reckoning midnight is 00.00 hour. One hour past midnight is 0100 hour , and so on. When it is 30 minutes past 1 in the night, it is 0130 hour in the international system. When it is 30 minutes past 1at midday, it is 1330 hour in the international system.

Some exmaples of expressing times in the international system are given below. night 25 minutes past 12

11

12

1

11 2

10 9

4

8

1

11 2

9

7

0025 hour

6

11 2

4

8 7

0715 hour

6

12

1 2

10 3

5

night 10 minutes past 9

1

9

4

8

12

10 3

5

6

12

10 3

7

morning 15 minutes past 7 afternoon 45 minutes past 4

9

3 4

8

5

7

1645 hour

6

5

2110 hour

In the international system, there is no morning, noon, afternoon, evening or night.

Express the times shown in the clocks below in the international system, taking into consideration the additional information given below each clock.

11

12

11

1

9

3 4

8 7

6

night

5

1

11 2

10

2

10

12

9

6

11 2

9

4 7

1

10 3

8

12

4 7

noon

6

5

morning 115

1 2

10 3

8

5

12

9

3 4

8 7

6

night

5

Elementary Mathematics Railway Time Table Dhaka - Chittagong Station

702 Distance 704 4 722 from Mahanagar Karno Mahanagar Subarno Dhaka Probhati Phuli Godhuli Express km Express

2 742 Ctg Turna Mail Express

Dhaka Departs

00

0740

0530

1515

1630

2230

2300

Dhaka airport Departs

19

0821

0625

1558

1705

2315

2338

Tongi Departs

23

0747

2330

Ghorashal Departs

47

0828

0003

Narsingdi Departs

58

0850

0028

Bhiarab Bazar Departs

87

Ashuganj Departs

91

1017

1045

0115

1705

0142

1100

Brahmanbaria Departs

104

Akhaura Departs

120

Comilla Departs

168

Laksam Departs

192

1445

Hasanpur Departs

210

1523

Feni Departs

232

1322

1606

2036

Chittagong Arrives

321

1515

1840

2255

1050

1129

0210

1805

0300

1215 1211

1350

0150

1928

2235

0400

0305

0435

0410

0527

0458

0755

0705

Fom the time table one can find out all relevant information about trains from Dhaka to Chittagong; the names of stations where each train stops, their times of departure from each such station, and the time of arrival in Chittagong. All times are given in the international system; we may for convenience convert them to our national system.

For example, Subarno Express leaves Dhaka at 4.30 in the afternoon and arrives at Chittagong at 10.35 in the evening . Consult the above time table and fill in the blanks : (a) Karnaphuli Express leaves Dhaka station at

and leaves Comilla station at

(b) Mahanagar Godhuli leaves Dhaka station at

and leaves Feni station at

(c) Turna Express leaves Brahmanbaria station at (d) Mahanagar Probhati leaves Bhairab Bazar station at 116

and arrives at Chittagong station at and arrives at Chittagong station at

Elementary Mathematics

Exercise 11 1.

2.

Write down the number of days of the following months : (a) Shrabon (b) Bhadro

(c) Agrahayan

(d) Chaitro

(e) April

(g) August

(h) December

(f) July

Determine which of the following years are leap years : (a) 1920

(b) 1922

(c) 1928

(d) 1930

(e) 1940

(f) 1960

(g) 2012

(h) 2000

(i) 2008

(j) 2004

3.

Convert 5 days 8 hours 15 minutes into seconds.

4.

Convert 4 years 3 months 10 days into hours.

5.

Convert 5 years 4 days into minutes.

6.

Convert into years, months, days : (a) 95624 minutes (b) 36290 hours (c) 25850 minutes (d) 39456 seconds

7.

Fill in the blanks : (a) Bangla years from 1101 to 1200 constitute the twelfth (b) Bangla years from 1301 to

constitute the fourteenth century

(c) English years from 1901 to

constitute twentieth century

(d) English years from 1991 to 2002 is

yuga.

8.

Your school ends at 3 o'clock in the afternoon. According to the international system, at what time does your school end ?

9.

According to the international system, Bagladesh Biman leaves for Kolkata at 2000 hour. At what time in the evening does Biman leave ?

10.

A BRTC bus leaves Dhaka at 40 minutes past 1 in the afternoon and reaches Comilla at 1 minite past 4 in the afternoon. At what time does the bus leave Dhaka and reach Comilla according to the international system ?

117

Chapter Twelve

Arrangements of Data Unarranged Data and Arranged Data Unarranged Data : Data which have not been arranged according to some given or chosen characteristic or criterion, are called unarranged data.

Arranged Data : Data which have been arranged according to some given or chosen characteristic or criterion, are called arranged data.

The marks in Mathematics at the annual examination obtained by 25 students are given below : 59

72

70

78

72 68

52 75

69 98

41 74

44 73

85 77

88

76

80

58

89

85

96

84

71

These data are not arranged according to any characteristic; these are unarranged data. To get an overall picture of the performance of the students, it is convenient to group together students who got marks in a certain range; such ranges are called classes and the difference between the highest and lowest marks in a class is called the class difference. The number of students whose marks fall into a particular class is called the frequency of that class. Here the highest and lowest marks obtained are 98 and 41. Their difference

98

41

57.

So a convenient choice for class difference would be 10. The classes are 40 - 49, 50 - 59, . . . , 90 - 99. These classes are called class intervals.

The resulting frequency table is shown below :

Classes of marks 40 50 60 70 80 90

49 59 69 79 89 99

Tally marks

Number of students/ Frequency 2 3 2 10 6 2 Total: 25

Elementary Mathematics Having written down the classes, we have to determine the frequency of each class.

The first number is 59; it belongs to the class 50 - 59; so we put a tally mark in the next column against that class. The next number is 72; it belongs to the class 70 - 79; so we put a tally mark against that class. In this way all the marks are counted by means of tally marks. When we get five tally marks in any one class, as in the class 70 - 79, the fifth tally mark is entered not vertically as the four preceding tally marks, but diagonally across them; this makes the counting of total tally marks in a class transparent and easy.

Example 1. The daily wage (in taka) of 20 workers are given below : 290 246

339 348

326 316

319 276

300 288

247 264 269 316

279 299 295 308 296. Classify the data.

327

Solution : Here the highest wage is 348 taka and the lowest is 246 taka.

Difference 348 246 taka 102 taka 102 5 20 4 So a convenient choice for class difference is 20. Beginning with 240, the class intervals are 240 - 259. 260 - 279, . . . , 340 - 359.

The resulting frequency table is as follows : Classes of wages

Tally marks

Number of workers/ Frequency

240

259

2

260

279

4

280

299

5

300

319

5

320

339

3

340

359

1 Total: 20

119

Elementary Mathematics Example 2. The weights (in kg) of 20 workers are given below : 60

52

54

50

52

54

56

58

60

60

50

54

56

58

60

50

60

60

58

54. Classify the data.

Solution : Weight (in kg)

Tally mark

Number of workers/ Frequency

50

3

52

2

54

4

56

2

58

3

60

6 Total: 20

Example 3. The marks obtained in Bangla by 25 students are given below : 75 68 74 66 80 65 75 67 86 75 68 82 88 84 72 66 85 73 76 78 67 79 86 82 84.

Arrange the data in classes, taking 5 as class difference.

Solution : The lowest among the marks is 65, the highest is 88. The resulting classification is shown in the table below.

Classes of marks

Tally mark

Number of students/ Frequency

65 - 69

7

70 - 74

3

75 - 79

6

80 - 84

5

85 - 89

4 Total: 25

120

Elementary Mathematics

Exercise 12 (A) 1. The number of students of 5 classes of a school are given below. Use tally marks to express the number of students : Class

Tally marks

Number of Students

One

40

Two

45

Three

42

Four

30

Five

28

2. The marks obtained by 20 students in Mathematics are given below : Classify the data.

75

63

75

75

71

75

63

72

72

69

72

70

61

75

60

71

69

63

65 69

3. The amount (in taka) of a day's sale of 20 shops are given below : Classify the data.

125

200

170

225

325

270

180

210

300

315

390

250

260

220

270

375

315

220

250

270

4. The daily wage (in taka) of 15 workers are given below. Classify the data.

325, 300, 325, 350, 300, 325, 325, 300, 350, 350, 325, 400, 350, 325, 325

121

Elementary Mathematics

C 12 11 10 9 8 7 6 5 4 3 2 1 0

A Class 1

Class 2

Class 3

Class 4

B

Class 5

The height of the bar representing the number of students of Class 3, 4, 5 are respectively 6 5 cm, 7 2 cm, 7 0 cm.

Example 2. The number of students present in 6 school days of one week of your class are given below. Draw a bar graph incorporating the data.

Day Present

Sat

Sun

Mon

Tue

Wed

Thu

45

40

30

42

50

35

C 50 40 30 20 10

0 A

Sat

Sun

Mon

Tue

Wed

Thu

B

( The distance between each pair of equidistant points on the line AC represents 10 students )

123

Elementary Mathematics

C 12 11 10 9 8 7 6 5 4 3 2 1 0

A Class 1

Class 2

Class 3

Class 4

B

Class 5

The height of the bar representing the number of students of Class 3, 4, 5 are respectively 6 5 cm, 7 2 cm, 7 0 cm.

Example 2. The number of students present in 6 school days of one week of your class are given below. Draw a bar graph incorporating the data.

Day Present

Sat

Sun

Mon

Tue

Wed

Thu

45

40

30

42

50

35

C 50 40 30 20 10

0 A

Sat

Sun

Mon

Tue

Wed

Thu

B

( The distance between each pair of equidistant points on the line AC represents 10 students )

123

Elementary Mathematics Example 3. The marks obtained by Zafar in various subjects in the annual examination are shown in the bar graph below. Referring to the graph, answer the following questions.

C 10 9 8 7 6 5 4 3 2 1 0

A

B Bangla English Mathematics Science Social Studies ( The distance between each pair of equidistant points on the line AC represents 10 marks)

(a) In which subject has he got the highest mark ? (b) In which subject has he got the lowest mark ? (c) What marks has he got in Mathematics ? (d) What is his total marks ? (e) What is his average mark ?

Solution : (a) In Mathematics he has obtained the highest mark. (b) In Social Studies he has obtained the lowest mark. (c) He has obtained 89 marks in Mathematics. (d) His marks : Bangla 69, English 65, Mathematics 89, Science 78, Social Studies 50. So his total marks = 69 + 65 + 89 + 78 + 50 = 351. (e) There are 5 subjects ; so his average marks

124

351

5

70 2

Elementary Mathematics

Exercise 12 (B) 1. The number of absentee students of a classs on the days of the last week are given below.

Draw a bar graph based on the data.

Day of the week Number of absentee students

Sat

Sun

Mon

2

3

5

Tue 4

Wed

Thu

3

6

2. The number of students of the five classes of Baluchara Primary School are given below.

Draw a bar graph based on the data.

Class

Number of Students

One

80

Two

75

Three

78

Four

65 55

Five

3. The yields last year of five crops in Swarupkathi Upazilla are shown on the table below.

Draw a bar graph based on the data.

Crop

Amount (in ton)

Paddy

450

Jute

260

Wheat

350

Chick-pea

50

Lentil

80

4. The distances by road of five cities from Dhaka (in nearest multiples of ten kilometre) are given below. Draw a bar graph based on the data.

City

Distance

Chittagong

290

Rangpur

440

Khulna

330

Rajshahi

310

Sylhet

400 125

Elementary Mathematics 5. In the annual examination 6 students obtained the following marks in Mathematics.

Draw a bar graph based on the data.

Student

Sumon

Mimi

Marks obtained

80

90

Omar 80

Shapnil

Tania

Ratan

100

75

85

6. In a one-day cricket match between Bangladesh and Australia, Saqib bowled ten overs.

The runs conceded by him in various overs are shown in the bar graph below.

Answer the following questions by referring to the bar graph. (a) In which over did he concede the most runs ? (b) In which over did he concede the least runs ? (c) In all, and on average, how many runs did he concede in ten overs ? (d) Arrange the number of runs in ascending order of magnitude and write down the corresponding ordinal number of the over .

C 13 12

Runs

11 10 9 8 7 6 5 4 3 2 1 0 A

1st

2nd

3rd

4th

5th

6th

7th

8th

9th

10th

Over

126

B

Elementary Mathematics Population Bangladesh is our dear homeland. The population of Bangladesh in the year 2011 is shown in the following chart Total number of males

7,12,55,000

Total number of females

7,10,64,000

Total

14,23,19,000

The number of females and the number of males are nearly equal. These numbers have been found by counting. The counting of population is called census. Through census not only is the population counted, but many essential information and data also are collected.

The numbeer of Shila's family members Member

Shila's father

Shila's mother

Shila

Number

1

1

1

Shila's sister 1

Total 4

The number of Shila's family members is 4. It is a small family; it is an ideal family.

Population of Bangladesh according to the last four censuses are given below * : Year

Population

1981

8 crore 99 lac

1991

11 crore 14 lac

2001

12 crore 39 lac

2011

14 crore 23 lac

Population and Housing Census 2011, Preliminary Result, Bureau of Statistics 127

Elementary Mathematics The total land area of Bangladesh is 1,47,570 square kilometre. In terms of land Bangladesh occupies the ninetieth position among countries of the world; but in terms of population it is ninth.

The population of some countries of Asia are given below * : Country

Population

India

121 crore 45 lac

Pakistan

18 crore 48 lac

Myanmar

5 crore 5 lac

Nepal

2 crore 99 lac

Sri Lanka

2 crore 4 lac

Bhutan

9 lac

* Source : State of the World Population 2010, UNFPA Problems Concerning Population Population Density : The average number of people per square kilometre of a region is called the population density of that region.

Example 1. The population of Patuakhali district is 15 lac 17 thousand (nearly) and its area is 3221 square kilometre. What is the population density of Patuakhali district ?

Solution : 15 lac 17 thousand = 1517000 In 3221 square kilometre there are 1517000 persons 1

1517000 3221

471 persons (nearly)

So, the population density of Patuakhali district is 471 (nearly).

128

Elementary Mathematics

Exercise 12 (C) 1. Wr Write rite down the number of your fa family amily members : Yoour mother Yo our father faather Your Relation Your

You Yoou

Your Yoour brother

Your Yoour sister

Total Tootal

Number Is your fa family amily a small fa family amily ? frrom the year 2001 to the year 2011?

2. What has been the increase in population of Bangladesh from 3. Wr Write family families rite down the number of fa amily members of two of your neighbouring fa amilies : Grand

Family fa father ather 1

Grand Father mother

Mother Brother

Sister

Uncle

Aunt

Others

To otal Total

2 4. What is population density? How can it be determined ?

5. The area of Ratanpur village is 5 square kilometre. 3,000 persons live in that village.

What is population density of the village ?

6. The population of Panchagarh district is 9 lac 81 thousand and area is 1405 sq kilometre, What is the population density of this district ?

7. The area of Rangamati district is 61609 sq kilometre and population is 5 lac 25 thousand.

Find the population density of that district.

8. Fill in the blanks from frrom the census data of 2011 given in the table below : Area in sq km Bangladesh Barisal Division

Population

1,47,570

14,23,19,000

13,297

81,47,000

Chittagong Division Dhaka Division

2,80,79,000 31,120

Density (per sq km)

831

4,67,29,000

Rangpur Division

1,56,65,000 129

960

Chapter Thirteen

Geometry Quadrilateral

D

A quadrilateral is a figure bounded by four line segments. The adjoining figure ABCD is a quadrilateral. The line segments AB, BC, CD, DA, are its four sides; the points A, B, C, D are its four corner points or vertices. The line segments AC and BD are its two diagonals.

C

A

B

Classification of Quadrilateral Parallelogram A parallelogram is a quadrilateral whose both pairs of opposite sides are parallel. The quadrilateral ABCD in the adjoining figure is a parallelogram.

Using a scale we measure the perpendicular distances of the sides at several points on AB and DC. We find that they are all equal, confirming that the sides AB and DC are parallel. Similarly, we find that AD and BC are parallel. Measuring the sides we find that the lengths of any two opposite sides A are equal : AB = DC and AD = BC. Measuring the angles

DAB, DAB

ABC,

BCD,

BCD and

C

D

M B

CDA, we find that ABC

CDA.

DAB and BCD, as well as ABC and CDA, are two pairs of opposite angles of the parallelogram. So, every pair of opposite angles are equal. Now let us draw the two diagonals of the parallelogram; they have intersected each other at M. Measuring, we find that the line segments AM and MC, as well as BM and MD, have equal lengths. So the diagonals of a parallelogram bisect each other at their point of intersection.

The opposite sides of a parallelogram are parallel and equal.

The opposite angles of a parallelogram are equal.

The diagonals of a parallelogram bisect each other.

Elementary Mathematics Square A square is a rectangle whose sides are all

D

C

equal. In other words, a square is a parallelogram whose sides are all equal and whose angles are all right angles. As the opposite sides of a rectangle are equal, a rectangle having any two of its adjacent sides equal, is a square. Put differently, a parallelogram of which two adjacent sides are equal and one angle is a right angle, is a square.

The adjoining figure ABCD is a square. Since A every square is a rectangle , its diagonals have equal lengths. Measuring the four angles made at their point of intersection, we find each of them is a right angle.

B

A rectangle having two equal adjacent sides is a square.

A parallelogram having two equal adjacent sides and one of whose angles is a right angle, is a square.

The diagonals of a square are equal and they bisect each other at right angles.

Do yourself : 1. Draw a quadrilateral. Measure the length of its four sides and two diagonals.

Measure the four angles of the quadrilateral and find the sum of their measures.

2. Draw visually two quadrilaterals, no two of whose sides are equal. (a) In each case measure the four sides and the two diagonals and enter your results in your notebook. (b) In each case measure the four angles and enter your results in your notebook.

Find the sum of the four angles, and state whether you get the same sum in both cases.

132

Elementary Mathematics Square A square is a rectangle whose sides are all

D

C

equal. In other words, a square is a parallelogram whose sides are all equal and whose angles are all right angles. As the opposite sides of a rectangle are equal, a rectangle having any two of its adjacent sides equal, is a square. Put differently, a parallelogram of which two adjacent sides are equal and one angle is a right angle, is a square.

The adjoining figure ABCD is a square. Since A every square is a rectangle , its diagonals have equal lengths. Measuring the four angles made at their point of intersection, we find each of them is a right angle.

B

A rectangle having two equal adjacent sides is a square.

A parallelogram having two equal adjacent sides and one of whose angles is a right angle, is a square.

The diagonals of a square are equal and they bisect each other at right angles.

Do yourself : 1. Draw a quadrilateral. Measure the length of its four sides and two diagonals.

Measure the four angles of the quadrilateral and find the sum of their measures.

2. Draw visually two quadrilaterals, no two of whose sides are equal. (a) In each case measure the four sides and the two diagonals and enter your results in your notebook. (b) In each case measure the four angles and enter your results in your notebook.

Find the sum of the four angles, and state whether you get the same sum in both cases.

132

Elementary Mathematics

Exercise 13 1. Which of the following statement is alway true ? (a) A rhombus is a parallelogram. (b) The diagonals of a parallelogram bisect each other at right angles. (c) A square is a rhombus. (d) A parallelogram is a quadrilateral whose two pairs of opposite sides are parallel and equal. (e) A square is a rectangle.

2. Draw visually a parallelogram, a rhombus and a rectangle. (a) In each case, ascertain by measuring, whether each pair of opposite sides are equal in length. (b) In each case, ascertain by measuring, whether each pair of opposite angles are equal in measure. (c) In each case, ascertain by measuring, whether two diagonals are bisected at their point of intersection. (d) Measure the angles made at the point of intersection of the two diagonals of the rhombus, and ascertain wheather they have intersected at right angles.

3. Draw visually a parallelogram whose two adjacent sides have lengths 4 cm and 3 cm.

Measure the lengths of the opposite sides, and each pair of opposite angles. Draw the two diagonals of the parallelogram and measure the lengths of the four segments of the two diagonals made by their point of intersection.

4. Draw visually a square each of whose sides has length 4 cm. (a) Measure the length of each diagonal and record in your notebook (b) Identify the middle points of the sides. Connect the middle points in succession. What type of a quadrilateral does the resulting quadrilateral appear to be ? Measure its sides and angles. Comment on the accuracy of your drawing.

133

Chapter Fourteen

Calculator and Computer Calculator A calculator is a hand-held electronic device which can carry out calculations following appropricate commands by the person using it. It runs on electricity producted by battery inside it; no outside electirc connection is needed. If electricity runs out, new battery has to be installed. Calculators vary a great deal in their build-up, shape, size and scope. Some calculators have 26 buttons, others have as many as 42 or 47 buttons. To start a calculator one presses the (ON/AC) button; to shut it one presses the (OFF) button.

The adjoining picture is that of a calculator. It has 26 buttons. It can be started and shut off by pressing the (ON) button.

Calculators are widely used in shops and businesses. Calculators play an important role in carrying out extensive calculations needed to solve mathematical problems of all kinds. Only through actual use can one truly appreciate the applicability and usefulness of calculators.

Elementary Mathematics

Computer To computer is to calculate. A computer is an electronic device; it can cerry out calculation on a much larger scale than a calculator. But the function and usefulness of a computer is not limited to calculations. Besides carrying out complicated calculations, it can be used to generate graphs, pictures, correspondence, to receive and send e-mail, for browsing websites, accessing internet and so on. Thanks to the computer, keeping in touch with friends and relations all over the world is now an easy matter. Computers have drastrically changed the life of mankind.

Abacus, a calculating device used in the late Middle Age, was a precursor of a calculator. In the seventeenth century Pascal, as well as Leibniz, had built mechanical calculators. In the nineteenth century, Charles Babbage, professor at the University of Cambridge in England, built an 'Analytical Engine ' which is considered as a precursor of modern computers.

The working process of a computer is straight-forward. A computer has four main components Input, Memory, Processor and Output. The computer works in combination of these four components. The input device is needed to enter numbers and data; keyboard and mouse are input devices.

135

Elementary Mathematics The intended directives are given by pressing the keyboard or klicking the mouse.

Input

Output

Memory

Processor

The directives are stored in the memory of the computer. Next the processor processes the data and directives stored in memory and sends the result to memory. Memory in turn sends it to output devices such as monitor or picture.

Those of you who have used computers would know that a keyboard or mouse is an input device. Data is fed into the computer through them. At the end of the work the computer shows the result in the monitor. The monitor is an output device. The memory and processor of a computer are not visible from outside; they are inside the computer.

So far we have talked about what is called hardware, or the mechanical parts of a computer.

But the computer cannot function on hardware alone. Appropriate commands are needed to make it function. Such directives constitute the software. What task the computer processor will perform, how it will perform, what directives will be needed, all such information is stored in the memory. Briefly speaking, computer hardware is driven by the software.

For different kinds of work different kinds of software are needed.

A software which can draw graphs connot be used for calculation. There are various software or computer programs for carrying out calculations, like Calculator, Excel, Calc, ect .

You can use the program Calculator or Calc instead of your ordinary calculator.

136

Elementary Mathematics

The following examples are intended to help the students getting familiar in the use of calculator in solving porblems.

Example 1. Fill in the blanks : (a) 17

(b) 18

9

(a) Solution : 1 17

7 9

15 9

26

26.

The calculator is activated by pressing the ON button, then successively the buttons 1

7

9

and

are pressed.

Finally, The calculator is shut off by pressing the (b) Solution: 18

1

8

15

33

1

5

button 33

The calculator is activated by pressing the ON button, then successively the necessary buttons are pressed.

137

Elementary Mathematics

Exercise 14 Solve the following problems using calculator or computer : 1. Fill in the blanks :

2.

9

(a) 897

356

(b) 207 3.

(a) 26

(a) 9 (c) 555

5. Simplify :

40

44

65

what ?

200

5

(b) 587

204

what ?

(d) 793

546

what ?

what ?

34

what ? (d) 720

(a) 9 (c) 245

(c) 150

9

what ?

what ? (b) 128

4

46

what ?

28

13

(c) 400 4.

(b) 28

(a) 17

80

5 13

what ?

68 6

95

134 5

what ? (b) 77

83

56

7

9

10

6. Shahin bought from Newmarket fish for 340 taka, vegetables for 55 taka, onion for 34 taka and oil for 190 taka. He gave 650 taka to the shopkeeper . What amount of money will the shopkeeper refund him ?

7. Fatema bought from book-fair books for 328 taka, writing pads for 105 taka and eraser for 12 taka and gave 500 taka to the shopekeeper . What amount of money will Fatema get as refund ?

140

Elementary Mathematics Simplify : 64

Example 5.

Solution :

ON 64

6 8

ON 8

8

4

8

8

2

96

1 96

9

ON

50

8

8 12

12

6

5

0

46

96 50 46 The calculations are completed in three steps keeping the calculator on.

Example 6. Shabab went to Mohammadpur market with 600 taka and bought fish for 350 taka, vegetables for 76 taka and fruits for 82 taka. What amount of money remained with him ?

Solution : fish 350 taka vegetables 76 taka fruits 82 taka The calculator is activated by pressing the ON button, then successively the necessary buttons are pressed..

ON 350

3

5 76

ON

6

600

508

0 82

0

7

6

8

2

508

508 0

5

0

8

92

92

The calculation are completed in two steps keeping the calculator on.

The computer is an amazing invention of our time. The present age is often called the computer age. Computer technology has impacted and influenced our lives in manifold ways. One should therefore be knowledgable about computers from young age and help build 'Digital Bangladesh'.

139

Elementary Mathematics

Answers Exercise 1 1

4 Do yourself 5. 1371915

9. 2091200 taka

6. 961125 taka 7. 149000

8. 100375 taka

10. 894682 taka 11. 2835324 saplings 12. 5012252 kg

Exercise 2 1

3 Do yourself 4. 417 days

8. 87 ; 617

5. 1149 ; 36 6. 72 baskets 7. 14

9. 75 persons 10. 78 11. 124 ; 9

12. 261

Exercise 3 (A) 2. 20 days 3. 4 days

1 Do yourself

4. 12 days

5. 100 persons

6. 15 days

7. 36 persons 8. 4 persons 9. 25 days 10. 24days 11. 50 persons 12.15 persons 13. 336 kg.

Exercise 3 (B) 1. 887749

2. 863743

3. 1

4. 890001

5. 580851

6. 20334

7. 893 8. Proma 23, Rimi 18, Monisha 29 9. 122446 taka 11. 15 taka 12.8 taka 13. Father's age 72, son's age 24 years 16.103

17.98

18.7200 taka

21.8750 taka; 11200 taka

19. 725 taka

22.5520 taka

10. 266628 14.990 15. 102 20.388; 302

23.427 taka

Exercise 4 1. Do yourself 2. 46 taka 7. 107

8. 2169 persons

3. 79 4. 129 cm 5. 23 years, 25 years 6. 59 9. 65

10. 38 years 11. 26 mm

Exercise 5 1

3 Do yourself 4. 38 5. 30 persons, 2 mangoes and 5 lychees

19 buckets, 99 buckets respeceively 7. 15 11. 486 12. 174

8. 21 9.

13. 420 minutes 14. 122

141

6. 12 litre;

1 square metre 10. 360 15. 1433

Elementary Mathematics Exercise 6 3 Do yourself 4.(a) 10 (b) 15 (c) 48 (d) 5 (e) 4 5. (a) 10 (b) 1, 2, 3, 4, 5 (c) 10 years (d) 100 taka and 6 (c) 48 (d) 30 6. (a) 12 chocolates (b) 85 taka 1

Exercise 7 (A) 1

9 Do yourself 10. Green 11. Younger sister 12. Shafiq

13. Shetu

14. Shoumik Exercise 7 (B) 1

2 portion 5

5 Do yourself 6.

7.

1 portion 4

8.

1 portion 4

9.

1 bigha 2

4 portion 5 Exercise 7 (C)

10.

1. (a)

6 7

(b)

(g) 6

2. (a)

4.

(b) 2

19 20

17 24

(g) 1

1 20

3 taka 4

2 9

5 6 (c) 2

79 240 (h) 4

5.

10.

7 9

(d) 5

(i) 25

19 20

(i)

(b)

2 portion 3

9. 14

(c) 3

(h) 5

9 16

(h) 1

3. (a)

4 5

23 38

1 2

(c) 11 12

3 portion 4 1 portion 8

15 16

3 8

1 4

(j) 3

2 portion 3

7. 8

3 4

142

9 10

(k) 32

1 3

(f) 1

1 8

23 75 1 8 (g)

1 10

2 3

27 28

11. 11

(f) 1

(e) 1

(d) 7

(i) 3

6.

1 6

(k) 2

19 96

1 9

(e) 1 (j) 23

(d) 1

(j) 2

2 5

13 4 (f) 10 20 35 119 (k) 1 120

(e) 2

5 2 kilometre 8. portion 6 5

12. Do yourself

Elementary Mathematics Exercise 7 (D) 1. (a) 12 (b) 15 2. (a)

3

2

5. 12 years

1 4

1 8

(d)

8 3

3 4 7.

2

(e) 7

3 3

(d)

20

(c)

32 6. 6

3

(c)

15

(b)

3

(c)

4

(b) 8

20

3. (a)

3

3

5 12

9

(e)

8

8. 4

1 2

5

(f)

(e) 6

10

(d) 4

5

1

9

(f)

6 9.

1

1

(h) 4

2

49 9 10

4. 10 quintal

3 112

43

(h) 24

126

(g)

12 2

(f)

55

(g) 4

3 4

12. 3 metre 13. 5 bananas 16. 80,000 taka 17. 9600 taka Rony 450 marks, Panna 400 marks 19. 6000 taka 11. 40,000 taka 15. 5,000 taka

10. 7

14. 240 taka 18. Total marks 600

Exercise 8 (A) 1. (a)

(l) Do yourself

5. 650. 25 km

2. 472.44 inches 3. 21.59 cm

6. Urmi 10 and Moumi 8 lychees

8. Molly 16 eggs , price 128 taka.

4. 189.00 taka

7. 8.5 taka

Rumi 10 eggs, price 80 taka

9. 6000 taka

10. 20 metre Exercise 8 (B) 1.

(a)

(r) Do yourself 2. 4.30 taka

5. 5.5 hours 11. 6

6. 0.68 taka

7. 3 75

3. 38.75 taka 8. 18 25

9. 0.005

4. 39.37 inches 10. 63.75 taka

12. 33.05 and 37.55

Exercise 9 1 3 Do yourself 4. 810 persons 5. 2000 persons 6. 90% 7. 1% 8. 1000 persons 9. 3% 10. 12% 11. 10% 12. 300 taka 13. 1200 taka 14. 3400 taka 15. 32 taka 16. 32 taka 17. 8% 18. 3% 19. 343 taka 20. 192 taka 21 600 taka

22. 900 taka

23. 4 years

143

24. 135 taka

25. 4 years

Academic Year 2013, Math-5

Knowledge makes a man sound

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