GCSE MATHEMATICS PROVISIONAL MARK SCHEME – NOVEMBER 2002 – PAPER 5 No
Working … ≈ 40 ÷ 0.2
1
Answer ≈ 200
2
Mark 2
Notes B2 for integer answer 185 ≤ ans.≤ 200 400 390 (B1 for 40 or 0.2 seen or for oe or oe) 2 2
3 B1 for two relevant semicircles radii 3cm (±2mm, within guidelines) B2cao for pair of parallel lines equal in length to AB and 3cm from AB (±2mm, within guidelines) (B1 if 2 parallel lines 3cm (±2mm) from AB but not equal in length to AB ( ≥ 3.5 cm ) OR 1 parallel line equal in length to AB and 3cm from AB, within guidelines)
3
4
a b
Draw frequency polygon through the points (3,1) (8,3) (13,9) (18,8) and (23,11)
a
Volume = 130 × 20 = 2 600 cm3 24 Scale factor oe 16 x = 10 × 1.5 = 15
b
5
16 - 20
a
1 ÷2 2 1 1 3 = 3 + = 3 miles 2 4 4 7
2 2
2 600 cm3
3
15
2
3 4
2
3
M1 for sight of 32÷2 or 33÷2 oe (possibly implied) A1cao for 16 - 20 B2 for plotting correct points ± ½ sq and joining with line segments (B1 for joining points either plotted at correct midpoints of intervals, but incorrect heights or plotted at correct heights and positioned consistently within intervals or four correct points joined by line segments or five correct points not joined or joined with a curve) Ignore end closures M1 for 130 × 20 A1cao for 2600 B1(indep) for cm3. 16 24 M1 for or 1.5 oe OR for oe if consistent 16 24 16 10 = OR for oe 24 x A1cao for 15. 1 M1 for 7 ÷ 2 oe 2 3 A1 for 3 oe (accept decimal) 4
GCSE MATHEMATICS PROVISIONAL MARK SCHEME – NOVEMBER 2002 – PAPER 5 No b
c
6
7
a
8
i ii
Working 1 1 7 ÷ 5 × 3 OR 7 ÷ 5 × 2( = 3) 2 2 1 1 = 1 ×3 OR 7 − 3 2 2 1 = 4 miles 2 1 4 5 8 7 −3 = 7 −3 2 5 10 10 7 = 3 10 21 x 21 + x = 6x OR =x− 6 6 21 = 6x − x 21 x= 5 (i) (ii)
11 < x + 3 + x + 3 < 20 ⇒ 5< 2x < 14 5 14 <x< 2 2 x is integer so x = 3, 4, 5, 6
Answer 1 4 2
3
7 oe 10
Mark 3
2
Notes 1 1 ÷ 5 × 3 oe or 7 ÷ 5 × 2 oe 2 2
M2 for either 7
1 (M1 for 7 ÷5 oe) 2 1 A1 for 4 oe (accept decimal) 2 M1 for valid use of a common denominator that is a multiple of 10 or for 7.5 − 3.8 7 oe (accept decimal) 10 M1 for a correct first step A1 for 3
4.2
3
6 × 109 1.5 × 10−2
2
printed ans.
5
3, 4, 5, 6
M1ft (indep) for a valid second step method 21 A1cao for oe 5 B1cao B1cao.
M1 for forming 11< p <20 oe or for two correct separate inequalities in terms of x A1cao [Convincingly shown] 5 14 M1 for < x < (Can be implied by at least A1 2 2 below; condone extras within the interval) A2 cao (A1 for either 3, 4, 5, 6, 7 or for any three of 3, 4, 5, 6 only)
GCSE MATHEMATICS PROVISIONAL MARK SCHEME – NOVEMBER 2002 – PAPER 5 No 9 1 0
a
b
Working Blank; A ; Blank; V; A; V y=
x + 60 1 ; Gradient = 2 2
Sub coords of given pt in y = "0.5"x+c
1 1
1 2
a b(i) (ii)
c
{Reading top to bottom} LHS {0.7}; 0.3 oe RHS 0.6; 0.4; 0.6; 0.4 oe P(F & G) = 0.7 × “ 0.6” = 0.42 P({F & G} or {S & H}) = (0.7 × 0.6) + (0.3 × 0.4) = 0.42 + 0.12 = 0.54 P( not [F&G] ) = 1 − "0.42" (= 0.58) Estimate for not [F&G] = "0.58" × 200 = 116 OR Estimate for [F&G] =”0.42” × 200 (=84)
Answer
Mark 3
0.5 oe
2
y = 0.5x + 7
2
2x ( 4x + 5y)
2
2 0.42 oe
2
0.54 oe
3
116
3
Notes B3 for all 6 correct (mark each box) (B2 for 5 or 4 correct else B1 for 3 correct) B2 cao for 0.5 oe x + 60 (B1 for y = oe seen) 2 B2ft for y="0.5"x + 7 oe (B1 for an equation of the form either y ="0.5"x +c where c > 0 or 2y = x + k where 7 ≤ k < 14 or y=px+7 where p > 0) B2cao [B1 for x (8x +10y) only ]
B1 for LHS oe B1 for RHS oe M1 ft diagram for 0.7 × “ 0.6” oe A1cao M1 ft diagram for (" 0.7 × 0.6") and ("0.3" ×"0.4") oe M1(dep) for adding A1cao M1ft answer in (b)(i) for 1 − "0.42" or 0.58 seen M1(dep) ft for "0.58" × 200 A1cao OR M1ft answer in (b)(i) for ”0.42” × 200
GCSE MATHEMATICS PROVISIONAL MARK SCHEME – NOVEMBER 2002 – PAPER 5 No
Working Estimate for not [F&G] = 200 − “84” = 116
1 3 1 4
Answer
Enlargement centre (1,1) scale factor − 1.5 a b
1 5
"200" × [40 ÷ Σstudents] oe = 8000 ÷ 1160 oe
1 6
4 x − 8 = 3 y − 5x y
mean = 0.3 st. dev = 0.2 60x + y 40x 7
x=
3y +8 4 +5y
Mark
2 2 2 3
Notes M1(dep) ft for 200 − “84” A1cao B1cao for ‘Enlargement’ as the only transformation B1cao (dep on previous B1) B1cao for mean = 0.3 oe B1cao for st.dev. = 0.2 oe B1cao for mean = 60x + y oe B1cao for standard deviation = 40x oe M1 for "200" × [40 ÷ Σstudents] oe A2 cao for 7 200 (A1 for 8000 ÷ 1160 OR oe dec to 2sf) 29
M1 expanding or splitting into four correct terms 4
4x + 5x y = 3 y + 8
M1(indep) rearranging correctly to isolate x-terms from non x terms M1(indep) factorising x with other factor a function of y
x (4+5 y) = 3 y + 8
A1cao for x = 1 7
(a)
PB = PO + OB PB = b −
(b)
OC AC
oe
b−
1 a 2
2
1 a 2
=2b = AO +
OC
M1 for PB = PO + OB oe A1 for b −
Printed result
= − a + 2b
3
3y +8 oe 4 +5y
1 a oe 2
B1 (possibly implied by correct AC if clear) M1 valid method to find AC in terms of OA and OB or in terms of a and b
GCSE MATHEMATICS PROVISIONAL MARK SCHEME – NOVEMBER 2002 – PAPER 5 No
Working = 2 (b −
(c) 1 8
1 a ) ⇒ AC is parallel to PB 2
(a) (b)
5
1 9
(a)
Mark
16
1
B1cao for 16
x3
1
B1cao
2
M1 for raising all to power 6 A1cao correct order [SC if M0 award B1 for either a list with four in correct order or a correct list of five but reversed.] 4 M1 for either 24 × (√3)4 or ( 12 ) or 16 81 A1cao for 144
Raise all to power 6 gives 83 , 44 , 25 , 72 , 36,
26 (c)
Answer
(2
3
)
4
, 36
= 24 × (√3)4 .
1 (x + 3 + x + 5) (2x + 1) 2 = (x + 4) (2x + 1) = 2x2 + 8x + x + 4
Area =
(b)
Area of square = (2x+5)2 = 4x2 +20x +25 Shaded area = "4x2+20x+25"− ("2x2+9x+4") = 2x2 + 11x + 21
(c)
2x2 + 11x + 21 − 42 = 0 (2x − 3)(x +7) = 0
1 6
1 3
2
1
7 , 43 , 82 144
2
2x2 +9x + 4
3
Printed result
3
x = 1.5
3
(a)
M1 for any correct unsimplified form for the area A2 for 2x2 +9x + 4 [else B1 for either (2x+1)(x+3) = 2x2 + 7x + 3 or (2x+8)(2x+1) = 4x2 + 18x + 8 or 3 terms correct in expn (x+4) (2x+1) =2x2+8x +x+4 ] B1 for 4x2 +20x +25 M1 must both be in form ax2+bx+c A1cao (Printed result obtained convincingly) M1 (putting sh. area = 42 and rearranging appropriately) A1 for either (2x − 3)(x +7) or
Length cannot be negative so 2 0
Notes A1cao (Printed result convincingly shown)
(i) Draws horizontal line y = 0.2
13 and 167
2
(ii) EITHER completes the graph, as a curve, correctly OR uses 180+p (or 360−p) where sinp =0.6
218 and 322
2
−11 ± 289 4
A1cao for x = 1.5 [Must have formed and solved a quadratic algebraically] The M1…may be implied by one ‘correct’ solution A1cao for both 13 ± 3 and 167± 3 M1 for completed curve, scale correct, through (270, −1) and beyond y = −0.6 OR uses 180+p (or 360−p) where
GCSE MATHEMATICS PROVISIONAL MARK SCHEME – NOVEMBER 2002 – PAPER 5 No
(b)
Working
Answer
Use of x = [a (i) − 90] or x = [ 90 − a (i) ] oe
eg x = 77 , eg x = 283
Mark
2
ALT Draw y = cos x graph
2 1
(a)
(a)
1 1 x+y + = y x xy
=
(b) (c)
sinp =0.6 (M1…may be implied by one ‘correct’ solution) A1cao for both 218 ± 3 and 322 ± 3 M1 for use of x = [a (i) − 90] or x = [ 90 − a (i) ] oe (oe eg x = [a (i) − 90 + 360n] or x = [ 90 −a (i) + 360n] where n is an integer) A1 ft for two correct ft answers M1 for y = cos x graph drawn as a curve through (0 , 1) (90 , 0) (180 , −1) (270 , 0) and beyond y = 0.2 and also intersecting the y = sin x graph within ± 1 sq of horizontal 45 A1 for both 77 ± 3 and 283 ± 3 oe alternatives
(i)
( 4 , − 25)
1
B1cao for ( 4 , − 25)
(ii) (iii)
( 2 , − 3) ( 1 , − 25) a=1 b=9
1 1 2
B1cao for ( 2 , − 3) B1cao for ( 1 , − 25) B1cao B1cao [SC if 0/2 award M1 if either (i) indicates curve crossing +'ve x-axis at 4.5 or (ii) substitutes to get − 25 = (4+a)(4−b) oe or (iii) substitutes to get 0 = (9+a)(9−b) oe]
Printed result
2
(b)
2 2
Notes
rational which is always rational rational
Explanation eg (x − y) = (x + y) − 2y = rational − irrational = irrational eg x = 3 + √2 ; y = 3 − √2
B1 for
x+y xy
B1cao (Printed result convincingly explained) 2
M1 (accept equiv) A1cao (convincing valid argument)
2
B2 cao [B1 if both x and y are different irrational numbers AND satisfy two of the three conditions ie
GCSE MATHEMATICS PROVISIONAL MARK SCHEME – NOVEMBER 2002 – PAPER 5 No
Working
Answer
Mark (I) (II) (III)
Notes x and y are both positive xy is rational x + y is rational ]