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 = (3 + 7) = 5

yR = average of yP and yQ = (4 + 10) = 7

Co-ordinates of R are (5, 7)

 

PQ2 = (7 – 3)2 + (10 – 4)2 = 16 + 36 = 52

PQ = 52 = 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 = (r 2 + h 2)

    = (p 4.52 ) + [2 p 4.5 (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)