Paper 1A
This paper is, I think, a
little on the easy side for a real GCSE.
I’ve just put the
answers to the even-numbered questions here. Later, I may add a
commentary in the right-hand column, and I may include the
odd-numbered questions.
1 |
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2 |
4(3
– 2t) = 12 – 8t (x – 4)(x + 2) = x2 – 4x + 2x – 8 = x2 – 2x – 8 8r – 20s = 4(2r – 5s) p2 – 3p = p(p – 3) |
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3 |
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4 |
p(1)
+ p(2) + p(3) + p(4) + p(5) = 1 0.08 + 0.13 + 0.35 + 0.28 + p(5) = 1 p(5) = 0.16 p(1 or 2) = p(1) + p(2) = 0.08 + 0.13 = 0.21 n(3) = ntotal ´ p(3) = 60 ´ 0.35 = 21 |
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5 |
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6 |
xR
= average of xP and xQ
= ½ (3 + 7) = 5 yR = average of yP and yQ = ½ (4 + 10) = 7 Co-ordinates of R are (5, 7) PQ2 = (7 – 3)2 + (10 – 4)2 = 16 + 36 = 52 PQ = Ö52 = 7.211 » 7.21 |
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7 |
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8 |
Number
of athletes taller than 180 cm = 10 + 5 = 15 As a percentage of 90, this is (15/90) ´ 100%
= 16.7% The table for plotting from is
The median is at F = 45, and is about 173.5 cm |
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9 |
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10 |
6(x
– 3) = x + 5 6x – 18 = x + 5 6x – x = 5 + 18 5x = 23 x = 23/5 = 4.6 |
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11 |
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12 |
x = 0.121212121212 ..... 100x = 12.1212121212 ....... 99x = 12 x = 12/99 = 4/33 |
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13 |
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14 |
Upper
bounds for the sides are 4.535 and 6.755 Upper bound for area = 4.535 ´ 6.755
= 30.634 » 30.6 Area = 4.53 ´ 6.75 = 30.58 » 30.6 |
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15 |
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16 |
15
– 4x < 7 15 – 7 < +4x 4x > 8 x > 2 You need a circle at 2 and a line to the right |
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17 |
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18 |
V
= 1/3 ´ base ´ height
= 1/3 ´ p r 2
h = 1/3 ´ ´ p ´ 4.52 ´ 10.8
= 229.02 » 229 cm3 A = p r 2 + 2 p r l, where l = Ö(r 2 + h 2) = (p ´ 4.52 ) + [2 p ´ 4.5 ´ Ö(4.52 + 10.82)] = 394.4 » 394 cm2 |
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19 |
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20 |
There
are several ways of doing this. Here are two: BDE = 180° – 100° = 80° (opp. angles of cyclic quadrilateral) BDC = 180° – 80° = 100° ( angles on a line add up to 180°) DBC = 35° ( alternate segment theorem) BCD = 180° – 100° – 35° = 45° (angles of triangle BDC = 180°) EBC = 100° (alternate segment theorem [on chord BE]) BCE = 180° – 100° – 35° = 45° (angles of triangle BEC = 180°) |