FREE GRE Practice Test Math (Section 1 Explained) by Voltage Test Preparation LLC

GRE Diagnostic Math Section 1 EXPLAINED.pages

1. ab = 0 a

A. The quantity on the left is greater B. The quantity on the right is greater C. Both are equal D. The relationship cannot be determined without further information Correct Answer: D Explanation: Since ab = 0, either a or b must be 0. If b = 0, then ‘a’ could be any number, negative or positive (or even 0). So this is a D answer. 2. -1 < x < 0 x!

x⁵

A. The quantity on the left is greater B. The quantity on the right is greater C. Both are equal D. The relationship cannot be determined without further information Correct Answer: B Explanation: Since x is a negative fraction, its square will be positive, its cube will again be negative (but closer to zero), the fifth power will also be negative, but even closer to zero, and hence will be further to the right on the number line (i.e. greater).

GRE Diagnostic Math Section 1 EXPLAINED.pages

3. The average of four numbers is 30. The average of three of these numbers is 20. The value of the fourth number

A. The quantity on the left is greater B. The quantity on the right is greater C. Both are equal D. The relationship cannot be determined without further information

Correct Answer: C Explanation: If the average of four numbers is 30, their total must be 4 x 30 = 120 The average of three of the numbers is 20, and so their total is 3 x 20 = 60 So the fourth number must be 120 - 60 =60

GRE Diagnostic Math Section 1 EXPLAINED.pages

4. AC > CB > AB b

GRE Diagnostic Math Section 1 EXPLAINED.pages

5. The line containing point A is rotated 90 degrees anticlockwise about the origin, O. The y coordinate of point A before rotation.

The y coordinate of point A after rotation.

A. The quantity on the left is greater B. The quantity on the right is greater C. Both are equal D. The relationship cannot be determined without further information Correct Answer: C Explanation: Rotating anticlockwise through 90 degrees is the equivalent of making a mirror image (left-right reversed). The x-coordinates will switch from positive to negative. But the y-coordinate will remain unchanged.

GRE Diagnostic Math Section 1 EXPLAINED.pages

6. Points x and y lie on line l The number of points on line l that are twice as far from x as from y

A. The quantity on the left is greater B. The quantity on the right is greater C. Both are equal D. The relationship cannot be determined without further information

Correct Answer: C Explanation: It is easy to see that a point to the right of y could be placed so that it is twice as far from x as from y. But do not forget that there will be a point between x and y, lying one third the way along the segment from x to y, that will also fit the definition. So there are two points.

GRE Diagnostic Math Section 1 EXPLAINED.pages

7. O is the centre of the circle, and angle POQ is a right angle MN/PQ

2/1

A. The quantity on the left is greater B. The quantity on the right is greater C. Both are equal D. The relationship cannot be determined without further information Correct Answer: B Explanation: First notice that MO, ON, OP and QO are all radii. If we assume that the radius =1, then MN = 2 Now we need to find PQ. PQ is the hypotenuse of a right triangle (POQ) with the other sides both = 1. This is a â&#x20AC;&#x2DC;special triangleâ&#x20AC;&#x2122; with side in the ratio x : x : x"2 Since PQ is "2, the ratio MN/PQ = 2/"2 which is less than 2/1

GRE Diagnostic Math Section 1 EXPLAINED.pages

8. # Total savings on 40 gallons of fuel bought at $1.152 per gallon instead of $1.245

$3.70

A. The quantity on the left is greater B. The quantity on the right is greater C. Both are equal D. The relationship cannot be determined without further information Correct Answer: A Explanation: 40 gallons at 1.152 = 46.08 40 gallons at 1.245 = 49.8 The savings will be 49.8 - 46.08 = 3.72 (It is slightly quicker to take the difference per gallon [1.245 - 1.152] and multiply by 40)

GRE Diagnostic Math Section 1 EXPLAINED.pages

9. The cost of hiring props for the school play was to be shared by 8 students. 4 more students decide to share the cost. The new cost per person is $2 less. Total cost of hiring props

$50

A. The quantity on the left is greater B. The quantity on the right is greater C. Both are equal D. The relationship cannot be determined without further information Correct Answer: B Explanation: Let the total cost be x. The cost per student originally = x/8 The cost per students after 4 more students join in = x/12 The difference is 2, so x/8 - x/12 = 2. Add the fractions by putting on acommon denominator (24) (3x - 2x)/ 24 = 2; x/24 = 2; x = 48

GRE Diagnostic Math Section 1 EXPLAINED.pages

10. A bag contains only red, white and blue balls. One third of the balls are red, one fifth of the balls are white. One ball is to be selected at random. Probability of drawing a white ball

Probability of drawing a blue ball

A. The quantity on the left is greater B. The quantity on the right is greater C. Both are equal D. The relationship cannot be determined without further information Correct Answer: B Explanation: Since the balls are either red, white or blue, if 1/3 are red and 1/5 are white the rest (1 - 1/3 - 1/5) are blue. But no need to work out the fraction. If we pick a number that fits in with this distribution it will be easier to see the probability. So if there are 15 balls in all, 1/3 = 5 will be red, 1/5 = 3 will be white and the rest (7) will be blue. The probability of drawing white = 3/15 The probability of drawing blue = 7/15 (which is greater) 11. The sides of a rectangular piece of card are each 10 per cent too long for a particular project. By what percentage is the area too large? Correct Answer: 21 Explanation: Let each side = 110, so area = 110 x 110 = 12,100 If each side is ten per cent too long then each side should be reduced to 100. So the correct area is 100 x 100 = 10,000. The area is 2100 units too large. 2100 as a percentage of the correct area = (2100 /10,000) 100 = 21

GRE Diagnostic Math Section 1 EXPLAINED.pages

12. Andy, Mark and Sean all have their birthdays today, but Andy is more than twice as old as Mark and Mark is more than four years older than Sean. If Andy is less than 16 years old, what is one possible value for Mark's age in years ? Correct Answer: 6 OR 7 Explanation: A > 2M M > S + 4 (where A, M and S are Andy’s, Mark’s, and Sean’s ages today) We can select numbers less than 16 for Andy’s age and check the equations. If A = 15, then M could be 7 (S is not needed here, but cannot be less than 1, which means that M cannot be less than 6)

13. ABCD is a square. Also AP=PQ=QB=BR=RS=SC=CT=TU=UD=DV=VW=WA. The area of the octagon PQRSTUVW is what fraction of the square? Correct Answer: 7/9 OR .777 OR .778 Explanation: The octagon is not a ‘regular’ octagon. But the sides of the square are divided into three equal parts. If we let one of these parts = 1, the area of the square = 9 The octagon is the square minus four equal 45-45-90 triangles. The area of one such triangle = $ x 1 x 1 = $. Four triangles = 2 The octagon area = 9 - 2 = 7 The fraction = 7/9

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GRE Diagnostic Math Section 1 EXPLAINED.pages

14. Triangle ABC is a right angled triangle. Also AC = 5, CB = 3 and angle ADB is a right angle. What is the length of DB? Correct Answer: 2.4 Explanation: This is another â&#x20AC;&#x2DC;special triangleâ&#x20AC;&#x2122; problem. The hypotenuse = 5, one leg = 3, so the other leg (AB) = 4 Now we can calculate the area = $ x 4 x 3 = 6 The line segment DB is also the height of the triangle if we take AC as the base. Area = $ x AC x DB; 6 = $ x 5 x DB; DB = 6/2.5 = 2.4

15. A football team has won 10 games and lost 5 games. If the team wins the remaining games of the season, it will have won 80 percent of its games. How many games in total will have been played? Correct Answer: 25 Explanation: Let the total number of games to be played = n Number lost = 5; number won = n-5 n-5 = 80/100 x n; n - 5 = 0.8n; 0.2n = 5; n = 25

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GRE Diagnostic Math Section 1 EXPLAINED.pages

16-20 Use the two charts above 16. Which of the following can be correctly deduced from the data? (Select ALL that apply) A. The total number of Techno magazines printed over the given 6-month period was less than 200,000 B. In May approximately 28,000 copies of Techno magazine were sold C. The average (arithmetic mean) monthly number of copies of Techno printed in this period was less than 30,000

Correct Answer: AC Explanation: Choice B is incorrect because data are provided for copies printed not sold. (This is a trap for the unwary reader!) To check choice A, there is no need to actually add up the six values. Inspection of the graph shows that the three highest values (30,000, 40,000 and 30,000) total 100,000 and hence the three lowest values must be less than 100,000 and the total must be less than 200,000. Similarly, to check choice C you can take a short cut. Instead of adding all the values and dividing by six you can look at the horizontal line that represents 30,000 and see that the vertical distance from that line to the point above is less than the sum of the distances to the points below, showing that the average must lie somewhere below the 30,000 line.

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GRE Diagnostic Math Section 1 EXPLAINED.pages

17. Between which two of the months shown was there a twenty percent rise in online advertising revenues? A. January and February B. February and March C. March and April D. April and May E. May and June Correct Answer: C Explanation: This is an easy question. Just look for the pair in which the second bar is twenty percent longer. In March the online revenues were $10,000 and twenty percent of that is $2,000 making the April figure of $12,000 fit the requirements of the question. 18. If in February each magazine printed cost $1.50 to produce, which of the following is closest to the fraction of the production costs that were covered by the advertising revenues from the print version? A. 1/5 B. 2/5 C. 3/5 D. 4/5 E. 9/10 Correct Answer: B Explanation: The number of magazines printed in February was approx. 30,000. At $1.50 each, the total cost to print was $45,000. The printed version gave advertising revenues of approx. $17,000 for the same month. The question now requires you to calculate 17,000 as a fraction of 45,000. If you use your calculator to divide 17 by 45 you will get 0.3777. This value is close to 0.4 which is equivalent to 4/10 or 2/5. Another approach is to use the answer choices. If we take 1/5 of 45,000 we get 9,000 which is too small, as we are trying to get 17,000. Two fifths would be twice as much i.e. 18,000 which is a bit larger than we need. But three fifths would be much too large. Hence 2/5 is the closest.

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GRE Diagnostic Math Section 1 EXPLAINED.pages

19. The total advertising revenues (printed version + online version) were highest in which of the following months? A. February B. March C. April D. May E. June Correct Answer: E Explanation: You need to add the length of the print bar to the length of the online bar for each month and find the longest total. You can simplify the numbers by reading the axis as 2 to 18 rather than 2,000 to 18,000 and approximate the readings. So for February we get 17 +5 = 22, March 13 + 10 =23 and so on. April and may are less than 23. When we look closely at June it reads as 9 + just over 14, which will give us just over 23, making June the correct answer. 20. During the six-month period shown, the median number of copies printed per month was most nearly A. 30,000 B. 29,000 C. 28,000 D. 25,000 E. 22,000

Correct Answer: B Explanation: The median of a set of numbers is the middle value when the numbers are ranked in order. In the case of an even number of values, the median is the average of the two middle values. In this case, the values for the six months are approximately 22, 30, 40, 30, 28 and 18 thousands. In ranked order this becomes 40, 30, 30, 28, 22, 18 and the median is the average of 28 and 30 = 29.

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