Mathematics 2018 Past Questions | WAEC
Study the following Mathematics past questions and answers for JAMB, WAEC NECO and Post-JAMB. Get prepared with official past questions and answers for upcoming examinations.
41.
The diagram shows a trapezium inscribed in a semi-circle. If O is the mid-point of WZ and |WX| = |XY| = |YZ|, calculate the value of m
- A. 90º
- B. 60º
- C. 45º
- D. 30º
Correct Option: Answer is B
In the diagram, < WOZ = 180\(^o\) (angle on a straight line)
< WOX = < XOY = < YOZ
(|WX| = |XY| = |YZ|)
\(\frac{180^o}{3}\) = 60\(^o\)
= 60\(^o\)
M + m =2m (base angles of isosceles \(\bigtriangleup\), |OY| and |OZ| are radii)
< YOZ + 2m (base angles of a \(\bigtriangleup\))
60\(^o\) + 2m = 180\(^o\) (sum of a \(\bigtriangleup\))
60\(^o\) + 2m = 180\(^o\)
2m = 180\(^o\) – 60\(^o\)
2m = 120\(^o\)
m = \(\frac{120^o}{2}\)
= 60\(^o\)
42.
In the diagram, PQ//RS. Find x in terms of y and z
- A. x = 240\(^o\) – y – z
- B. x = 180\(^o\) – y – z
- C. x = 360\(^o\) + y -z
- D. x = 360\(^o\) – y – z
Correct Option: Answer is D
In the diagram,
a = z (alternate angles)
b = 180\(^o\) – a (angles on a straight line)
b = 180\(^o\) – z
c = 180\(^o\) – x (angles on a straight line)
y = b + c (sum of oposite interior angles)
y = 180\(^o\) – z + 180\(^o\) – x
y = 360\(^o\) – z – x
x = 360\(^o\) – y – z
43.
In the diagram, PQ is a straight line, (m + n) = 110\(^o\) and (n + r) = 130\(^o\) and (m + r) = 120\(^o\). Find the ratio of m : n : r
- A. 2 : 3 : 4
- B. 3 : 4 : 5
- C. 4 : 5 : 6
- D. 5 : 6 : 7
Correct Option: Answer is D
m + n = 110\(^o\), (n + r) = 130\(^o\)
(m + n) = 120\(^o\)
then, r = 130\(^o\) – n
and;
m + (130^o – n) = 120\(^o\)
m – n = -10\(^o\)
2m + (n + r) = 110 + 120 = 230
2m + 130 = 230
2m = 230 – 130
m = \(\frac{100}{2}\) = 50\(^o\)
n = 110\(^o\) – 50\(^o\)
= 60\(^o\)
r = 130\(^o\) – 60\(^o\) = 70\(^o\)
Hence, the ratio m : n : r
= 50 : 60 : 70
= 5 : 6 : 7
44.
Donations during the launching of a church project were sent in sealed envelopes. The table shows the distribution of the amount of money in the envelope. How much was the donation?
- A. N26,792.00
- B. N26,972.00
- C. N62.792.00
- D. N62,972.00
Correct Option: Answer is D
SOLUTION:
Total donation = 4 x 500 + 7 x 2000 + 20 x 1000 + 9 x 700 + 4 x 500 + 5 x 100 + 3 x 50 + 1 x 2 + 2 x 10
= 20000 + 14000 + 20000 + 6300 + 2000 + 500 + 150 + 2 + 20
= N62,972
45. A piece of thread of length 21.4cm is used to form a sector of a circle of radius 4.2cm on a piece of cloth. Calculate, correct to the nearest degree, the angle of the sector.
[Take \(\pi = \frac{22}{7}\)]
- A. 170\(^o\)
- B. 192\(^o\)
- C. 177\(^o\)
- D. 182\(^o\)
Correct Option: Answer is B
Length of arc, L = 21.4 – 2 x 4.2cm
= 21.4 – 8.4
= 13cm
But L = \(\frac{\theta}{360^o}\) x 2\(\pi r\)
i.e 13 = \(\frac{\theta}{360^o}\) x 2 x \(\frac{22}{7}\) x 4.2
= 13 x 360\(^o\) x 7
= \(\theta\) x 2 x 22 x 4.2
\(\theta\) = \(\frac{13 \times 360^o \times 7}{44 \times 4.2}\)
= \(\approx\) 177.27\(^o\)
\(\approx\) 177\(^o\) (to the nearest degree)
46. If tan x = \(\frac{4}{3}\), 0\(^o\) < x < 90\(^o\), find the value of sin x – cos x
- A. 1\10
- B. 1\5
- C. 5\12
- D. 1 2/5
Correct Option: Answer is B
From the diagram,
h\(^2\) = 4\(^2\) + 3\(^2\) (pythagoras’)
h\(^2\) = 16 + 9 = 25
h = \(\sqrt{25}\) = 5
Hence, sin x – cos x
= \(\frac{4}{5} – \frac{3}{5}\)
= \(\frac{2}{5}\)
47. Given that Y is 20cm on a bearing of 300o from x, how far south of y is x?
- A. 10cm
- B. 15cm
- C. 25cm
- D. 30cm
Correct Option: Answer is A
In △YSC, sin 30º = YS/20
|YS| = 20 sin 30º
= 20 x 0.5
10m
48. The diagonals of a rhombus WXYZ intersect at M. If |MW| = 5cm and |MX| = 12cm, calculate its perimeter
- A. 42cm
- B. 48cm
- C. 52cm
- D. 60cm
Correct Option: Answer is C
Let the length of a side of the rhombus be n
Then, n² = 5² + 12²
= 25 + 144 = 169
n = √
= 13cm
Hence, perimeter of rhombus = 4n = 4 x 13
= 52cm
49. M and N are two subsets of the universal set (U). If n(U) = 48, n(M) = 20, n(N) = 30 and n(MUN) = 40, find n(M N)
- A. 18
- B. 20
- C. 30
- D. 38
Correct Option: Answer is D
Let n(M N ) = x
Then 20 – x + x + 30
– x = n(M ∪ N)
50 – x = 40
50 – 40 = x
10 = x
x = 10
Hence, n(M υ(N)’ = 8 + (20 – 10) + (30 + 10)
= 8 + 10 + 20
= 38
50. The graph of y = x² and y = x intersect at which of these points?
- A. (0,0), (1,1)
- B. (0,0), (0,1)
- C. (1, 0), (0, 0)
- D. (0, 0) (0, 0)
Correct Option: Answer is A
y = x\(^2\) ….(1)
y = x ……(2)
y = y
x\(^2\) – x
x\(^2\) – x = 0
x(x – 1) = 0
x = 0 or x – 1 = 0
x = 0 or x = 1
when x = 0, y = 0\(^2\) = 0
when x = 1, y = 1\(^2\) = 1
Hence; the two graphs interest at (0, 0) and (1, 1)