## 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 = √$\sqrt{169}$

= 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 $\cap $ N)

**A.**18**B.**20**C.**30**D.**38

**Correct Option: Answer is D**

Let n(M $\cup $ 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)