Worked Solutions to Selected Questions from O-Level Maths 1966
1.
(i) A man walking at a steady speed travels 3 12 miles in 44 minutes. Calculate in feet the
distance he walks in 25 seconds.
There are 5280 feet in 1 mile and 60 seconds in 1 minute, so the man travels:
÷10
÷4
÷6
÷11
x25
18480 feet in 2640 seconds
1848 feet in 264 seconds
462 feet in 66 seconds
77 feet in 11 seconds
7 feet in 1 second
175 feet in 25 seconds
(ii) Use tables to evaluate the acute angles A and B to the nearest degree, given that
(a) cos A = 0.4436,
(b) tan 2B = 2.3146.
From the trigonometric tables, we can see that:
CosA  0.4436
A  64 
2.
Tan2 B  2.3146
2 B  66  ...
B  33
(i) Calculate the angle subtended at the centre of a circle of radius 5 in. by a chord of length
3.4 in.
Using the Cosine Rule on the triangle:
3.4 2  5 2  5 2  2(5)(5)Cos
11.56  50  50Cos
50Cos  38.44
38.44
50
Cos  0.7688
Cos 
  40 
(ii) Solve the equation x2 = 3x2 – 15x + 18.
x 2  3x 2  15 x  18
0  2 x 2  15 x  18
0  2 x  3x  6
Therefore
3.
Express
2x  3  0
3
x
2
or
8x
4x  1
+
2x  1 2  4x
as a single fraction in its simplest form.
2
x6  0
x6
4x  1
8x

2x  1 2  4x
4.

4 x  12  4 x   8 x2 x  1
2 x  12  4 x 

8 x  16 x 2  2  4 x  16 x 2  8 x
2 x  12  4 x 

 2  4x
2 x  12  4 x 

 2  4 x 
2 x  12  4 x 

1
2x  1

1
1  2x
Solve the equations
x y
 = 1,
2 4
x = 2y + 12.
2 y  12 y
 1
2
4
4 y  24  y  4
5 y  24  4
5 y  20
y  4
x  2  4  12
Therefore
x  8  12
x4
3
5.
(i) A cone of base-radius 7 in. and height 2 ft is divided into two parts by a plane parallel to
the base and 1 ft from it. Calculate in cubic inches the volume of the larger part.
[Take the volume of a cone of height h and base-radius r as
22
21
r 2h.]
By similar triangles, r = 3.5 inches
We will use the fact that 1ft = 12inches, so that all of the calculations are in inches.
For the large cone:
22 2
 7  24
21
V  22  7  8
V 
V  154  8
V  1232inches 3
For the small cone:
22 49
  12
21 4
V  22  7
V 
V  154inches 3
Therefore, for the frustrum:
V  1232  154
V  1078inches 3
4
(ii) A regular polygon has each interior angle seven times as large as each exterior angle.
Calculate the number of sides.
8 x  180
x  22.5
Therefore, the number of sides is
360
22.5
360 720

22.5 45
144

9
 16
The polygon has 16 sides.
6.
Calculate the area enclosed between the line y = 5, the curve y = 12 – 4x + 12x2 and the
ordinates x = –2, x = 3.
5
Integration can be used to calculate the shaded area:
3
A   12  4 x  12 x 2 dx  25
2


3
 12 x  2 x 2  4 x 3  2  25
 36  18  108   24  8  32  25
 126  64  25
 165units 2
7.
A man spends his holiday touring by car, using petrol which costs 5s 4d per gallon. He finds
that on average he travels at a speed of 36 m.p.h. for 5 14 hours per day, using a gallon of
petrol for every 28 miles. In addition to his petrol costs his daily expenses amount to 39s.
Calculate
(a) the cost of petrol as a percentage of his total expenses (to the nearest percent),
(b) the greatest number of complete days he can spend on holiday if his total expenses must
not exceed £65,
(c) the number of miles he would travel in this case.
The total distance travelled per day is 36  5 14
36  5 14
21
4
 9  21
 36 
 189miles
The cost per mile for petrol is 5s 4d divided by 28. Given that there are 12d in £1, the cost
per mile is
64
28
16

7
 2 72 d
Therefore, the cost of petrol per day is
6
16
7
 29  16
 464d
189 
Therefore, as a percentage of his total expenses:
464
 100
464  39  12

464
 100
932

83
 100
117
 70.940170940170...%
This is 71% to the nearest percent.
Total expenses per day is 932d.
There are 120d in £1, so there are 26000d in £65.
Therefore, the total number of days is
We know that
And
26000
932
20 x 932 = 18640
7 x 932 = 6524
18640 + 6524 = 25164
Therefore, the total number of complete days is 27.
189 miles x 27 days is 5103 miles.
7
8.
A man spent 7s 6d in buying a number of articles costing x pence each. The price was then
reduced by two pence per article and his wife bought 50 at the reduced price.
The number of articles bought by the man was 20 less than the number of pence spent by his
wife. Form an equation to find x and hence determine the original price of one article.
We can convert the information in the question to a table as follows:
Number of
Articles
Cost/Article
(pence)
Man
50x  2  20
x
Wife
50
x2
Therefore, we can set up and solve an equation as follows:
50x  2  20x  7 s6d
50 x  120x  90d
5 x 2  12 x  9
5 x 2  12 x  9  0
5 x  3x  3  0
Therefore:
x
5
3
or
x3
Since we know that x  0 , we know that the original price of one article is 3d.
9.
If f =
3 M
g, express M in terms of f and g.
3 M
f 3  M   g 3  M 
3 f  Mf  3 g  Mg
Mf  Mg  3 g  3 f
M  f  g   3 g  f 
3 g  f 
M 
g f
8
10.
The base angles of an isosceles triangle are each 73. If the base of the triangle is of length
2x inches, show that the area of the triangle is x2 tan 73 sq in. Hence, or otherwise, given that
the area is 29.44 sq in., calculate the length of the base.
By considering one of the right-angled triangles in the diagram above, we can say that:
h
x
h  xTan73
Tan73 
Therefore, the area of the isosceles triangle is:
2 x  xTan73
2
2
 x Tan73
Area 
Given that the area is 29.55 inches2 we can solve the equation:
x 2Tan73  29.44
29.44
3.271
2
x 9
x2 
x  3inches
9
11.
A child’s clockwork car, when wound up and placed on the ground, runs forward in a straight
line for a time and then becomes stationary. During the time that it is in motion, its velocity
when it has been moving for t seconds is 32 t – 12 t2 feet per second. Calculate
(a) for how many seconds the car runs,
(b) the distance it travels,
(c) its greatest velocity,
(d) its greatest acceleration.
(a) When v = 0, the car is at the beginning and the end of its motion.
3
1
t  t2
2
2
2
0  3t t
0
Therefore
0  t 3  t 
t 0
or
t 3
So the car runs for 3 seconds
(b) Integrating the velocity function gives us displacement, s:
s
3 2 1 3
t  t c
4
6
When t=0, s=0, which means that c=0.
s
3 2 1 3
t  t
4
6
When t=3,
3
1
 9   27
4
6
3
1
s  9  9
4
2
27 9
s

4 2
9
s   2 14 feet
4
s
(c) To find the maximum velocity, we need to consider what happens when the derivative of
v is zero. We also need to ensure that the second derivative is negative to confirm that v
is a maximum.
10
dv 3
 t
dt 2
3
0  t
2
3
t
2
d 2v
 1
dt 2
 Maximum
Substituting into v:
v MAX 
3 3 1 9
  
2 2 2 4
9 9

4 8
9
  1 18 ft / sec
8

(d) The derivative of velocity with respect to time is also known as acceleration, so:
a
3
t
2
Given that 0  t  3 , the maximum acceleration occurs when t = 0. Therefore, the
maximum acceleration is 1.5 ft/s2
12.
Two unequal circles touch externally at B. The diameters through B are AB and BC
respectively. A common tangent touches the circle, whose diameter is AB, at X, and touches
the other circle at Y. The lines AX and CY meet, when produced, at D. Prove that the
quadrilateral BXDY is a rectangle.
A diagram is definitely needed to approach this problem:
11
To prove that BXDY is a rectangle, we need to show that all the angles in the quadrilateral
are right angles.
AXˆD  90 (angles in a semicircle)
 BXˆD  90 (angles on a straight line)
(angles in a semicircle)
BYˆC  90

 BYˆD  90 (angles on a straight line)
XYˆB  BCˆ Y
YXˆB  BAˆ X
(Alternate Segment Theorem)
(Alternate Segment Theorem)
 Triangle ADC is similar to triangle XBY => ADˆ C  XBˆ Y
We know that the angles in BXDY must sum to 360o . We also know that it contains two
right angles. Since the remaining two angles are equal, it follows that ADˆ C  XBˆ Y  90
Therefore, BXDY is a rectangle
12