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

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)

3

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

5

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

7

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

 h 165 170 175 180 185 190 F 8 26 53 75 85 90

The median is at F = 45, and is about 173.5 cm

9

10

6(x – 3) = x + 5

6x – 18 = x + 5

6xx = 5 + 18

5x = 23

x = 23/5 = 4.6

11

12

x =   0.121212121212 .....

100x = 12.1212121212 .......

99x = 12

x = 12/99 = 4/33

13

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

15

16

15 – 4x < 7

15 – 7 < +4x

4x > 8

x > 2

You need a circle at 2 and a line to the right

17

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

19

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°)