Answers

Go to questions.

1.
(a) (i) 5.5 cm
(ii) 4.5 cm
(b) The scale is 1 : 20 000, so 1cm on the map = 20 000cm in real life. On the map, the greatest distance between the upper and lower bounds is 1cm (from part (a)). So in real life, the greatest distance is 20 000cm = 200m = 0.2km.

2.
Triangle is isosceles, therefore angle at C = 55°. Angle at A = 180° - 55° - 55° = 70°. Area of triangle = ½ ×AB×AC×sin70° = ½ × 12 × 12 × 0.93969 = 67.7 cm² (3s.f.)

3.
(i) 4y² – 81 = 0
\ 4y² = 81
\ y² = 81/4
\ y = 9/2 or –9/2
\ y = 4.5 or -4.5

(ii)    1     +   1  =  -1
     (x + 2)      3

multiply everything by 3(x + 2):
\ 3 + (x + 2) = -3(x + 2)
\ 3 + x + 2 = -3x –6
\ 11 + 4x = 0
\ x = -11/4

4.
(a) [The 'probability' column reading from top to bottom:] 5/14, 5/28, 5/28, 1/28, 5/28, 1/28, 1/28, 0
(b) 5/14
(c) 9/14

5. (a) 7x + 3 > 17 + 5x
\ 7x – 5x > 17 – 3
\ 2x > 14
\ x > 7
(b) (i) 12x⁵
(ii) 9y⁶
(c) 2x² - 7x + 3

6.
(a) 3n - 1
(b) n² + 1

7.

(a) mean = (sum of fx)/(sum of f)

= (3 + 16 + 65 + 126 + 184 + 56)/(1 + 2 + 5 + 7 + 8 + 2)
= 450/25 = 18
Hence the mean is 18 points

(b) 1, 3, 8, 15, 23, 25
(d) 19 points (approx.)

8.
(i) 12cm
(ii) 10cm

9.
Matthew £69.30, Nicola £25.20

10.
(a) a = 2, b = 3

11.
t = k/d² (where t is temperature and d is distance)
when d = 2, t = 50
\ 50 = k/4
\ k = 200
\ t = 200/d²
when d = 3.5, t = 200/12.25
=16.33 degrees Celsius (2d.p.)

12.
(a) 8.19 × 10¹³
(b) 2.7 × 10⁸
(c) 3125 days

13.
(b) (i) a + b
(ii) a - 1/2 b
(c) x = 5/3, y = 2/3

14.

Substitute the values given in the question into the equation to form two simultaneous equations:
58000 = a + 1000b (1)
64000 = a + 2000b (2)
(2) – (1): 6000 = 1000b
\ b = 6
sub in (1) a = 58000 - 6000
\ a = 52000