Year : 
2020
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

21 - 30 of 50 Questions

# Question Ans
21.

A man is five times as old as his son. In four years' time, the product of their ages would be 340. If the son's age is y, express the product of their ages in terms of y.

A. 5y\(^2 - 16y - 380 = 0\)

B. 5y\(^2 - 24y - 380 = 0\)

C. 5y\(^2 - 16y - 330 = 0\)

D. 5y\(^2 - 24y - 324 = 0\)

Detailed Solution

Man = x, Son = y
x = 5y
(x + 4)(y + 4) = 350
(5y + 4)(y + 4) = 340
5y\(^2\) + 20y + 4y + 16 - 240 = 0
5y\(^2\) + 24y - 324 = 0
22.

Simplify; \(\frac{a}{b} - \frac{a}{a} - \frac{c}{b}\)

A. \(\frac{a - b + c}{ab}\)

B. \(\frac{ab - bc - ac}{ab}\)

C. \(\frac{a^2 - b^2 + ac}{ab}\)

D. \(\frac{a^2 - b^2 - ac}{ab}\)

Detailed Solution

\(\frac{a}{b} - \frac{b}{d} - \frac{c}{b}\)
\(\frac{a^2 - b^2- ac}{ab}\)
23.

Open Photo In the diagram, XYZ is an equilateral triangle of side 6cm and Y is the midpoint of \(\overline{XY}\). Find tan (< XZT)

A. \(\frac{1}{\sqrt{3}}\)

B. \(\frac{\sqrt{3}}{2}\)

C. \(\sqrt{3}\)

D. \(\frac{1}{2}\)

Detailed Solution

Open Photo
ZT = \(\sqrt{6^2 - 3^2}\)
ZT = \(\sqrt{27}\) = \(3\sqrt{3}\)
tan (< XZT) = \(\frac{3}{3\sqrt{3}}\)
= - \(\frac{1}{\sqrt{3}}\)
24.

A fence 2.4 m tall, is 10m away from a tree of height 16m. Calculate the angle of elevation of the top of the tree from the top of the fence.

A. 76.11\(^o\)

B. 53.67\(^o\)

C. 52.40\(^o\)

D. 51.32\(^o\)

Detailed Solution

Open Photo
Tan \(\theta\) = \(\frac{13.6}{10}\)
= tan\(^{-1}\)(1.36)
\(\theta\) = 53.67\(^o\)
25.

Fati buys milk at ₦x per tin sells each at a profit of ₦y. If she sells 10 tins of milk, how much does she receives from the sales?

A. ₦(xy + 10)

B. ₦(x + 10y)

C. ₦(10x + y)

D. ₦10(x + y)

Detailed Solution

Selling Price for each = ₦x + ₦y
= 10(₦x + ₦y)

26.

If tan y is positive and sin y is negative, in which quadrant would y lie?

A. First and third only

B. First and second only

C. Third only

D. Second only

C

27.

The dimension of a rectangular base of a right pyramid is 9 cm by 5cm. If the volume of the pyramid is 105 cm\(^3\), how high is the pyramid?

A. 10cm

B. 6cm

C. 8cm

D. 7cm

Detailed Solution

Volume = \(\frac{1}{3}\) x 9 x 5 x h
105 = \(\frac{1}{3}\) x 9 x 5 x h
\(\frac{105}{15} = \frac{15h}{15}\)
h = 7cm

28.

Each interior angle of a regular polygon is 168\(^o\). Find the number of sides of the polygon

A. 30

B. 36

C. 24

D. 18

Detailed Solution

Exterior angle = 180\(^o\) - 168\(^o\)
Number of sides = \(\frac{360^o}{12}\)
= 30\(^o\)

29.

In the diagram, \(\overline{MN}\)//\(\overline{PQ}\), < MNP = 2x, and < NPQ = (3x - 50\(\^o\)). Find the value of < NPQ

A. 200\(^o\)

B. 150\(^o\)

C. 120\(^o\)

D. 90\(^o\)

Detailed Solution

Open Photo
2x = 3x - 50 alternate angle
50\(^o\) = 3x - 2x
50\(^o\) = x
< NPQ = (3x - 50)\(^o\)
= 3(500)\(^o\) - 50
= 150 - 50 = 150\(^o\)
30.

The length of an arc of a circle of radius 3.5 cm is 1\(\frac{19}{36}\) cm. Calculate, correct to the nearest degree , the angle substended by the centre of the circle. [Take \(\pi = \frac{22}{7}\)]

A. 55\(^o\)

B. 36\(^o\)

C. 25\(^o\)

D. 22\(^o\)

Detailed Solution

L = \(\frac{\theta}{360^o}\) = 2\(\pi\)
\(\frac{55}{36} = \frac{\theta}{360^o} \times 2 \times \frac{22}{7} \times 3.5\)
\(\theta = \frac{360 \times 55 \times 44 \times 0.5}{3 \times 44 \times 0.5}\)
= \(\theta = 25^o\)
21.

A man is five times as old as his son. In four years' time, the product of their ages would be 340. If the son's age is y, express the product of their ages in terms of y.

A. 5y\(^2 - 16y - 380 = 0\)

B. 5y\(^2 - 24y - 380 = 0\)

C. 5y\(^2 - 16y - 330 = 0\)

D. 5y\(^2 - 24y - 324 = 0\)

Detailed Solution

Man = x, Son = y
x = 5y
(x + 4)(y + 4) = 350
(5y + 4)(y + 4) = 340
5y\(^2\) + 20y + 4y + 16 - 240 = 0
5y\(^2\) + 24y - 324 = 0
22.

Simplify; \(\frac{a}{b} - \frac{a}{a} - \frac{c}{b}\)

A. \(\frac{a - b + c}{ab}\)

B. \(\frac{ab - bc - ac}{ab}\)

C. \(\frac{a^2 - b^2 + ac}{ab}\)

D. \(\frac{a^2 - b^2 - ac}{ab}\)

Detailed Solution

\(\frac{a}{b} - \frac{b}{d} - \frac{c}{b}\)
\(\frac{a^2 - b^2- ac}{ab}\)
23.

Open Photo In the diagram, XYZ is an equilateral triangle of side 6cm and Y is the midpoint of \(\overline{XY}\). Find tan (< XZT)

A. \(\frac{1}{\sqrt{3}}\)

B. \(\frac{\sqrt{3}}{2}\)

C. \(\sqrt{3}\)

D. \(\frac{1}{2}\)

Detailed Solution

Open Photo
ZT = \(\sqrt{6^2 - 3^2}\)
ZT = \(\sqrt{27}\) = \(3\sqrt{3}\)
tan (< XZT) = \(\frac{3}{3\sqrt{3}}\)
= - \(\frac{1}{\sqrt{3}}\)
24.

A fence 2.4 m tall, is 10m away from a tree of height 16m. Calculate the angle of elevation of the top of the tree from the top of the fence.

A. 76.11\(^o\)

B. 53.67\(^o\)

C. 52.40\(^o\)

D. 51.32\(^o\)

Detailed Solution

Open Photo
Tan \(\theta\) = \(\frac{13.6}{10}\)
= tan\(^{-1}\)(1.36)
\(\theta\) = 53.67\(^o\)
25.

Fati buys milk at ₦x per tin sells each at a profit of ₦y. If she sells 10 tins of milk, how much does she receives from the sales?

A. ₦(xy + 10)

B. ₦(x + 10y)

C. ₦(10x + y)

D. ₦10(x + y)

Detailed Solution

Selling Price for each = ₦x + ₦y
= 10(₦x + ₦y)

26.

If tan y is positive and sin y is negative, in which quadrant would y lie?

A. First and third only

B. First and second only

C. Third only

D. Second only

C

27.

The dimension of a rectangular base of a right pyramid is 9 cm by 5cm. If the volume of the pyramid is 105 cm\(^3\), how high is the pyramid?

A. 10cm

B. 6cm

C. 8cm

D. 7cm

Detailed Solution

Volume = \(\frac{1}{3}\) x 9 x 5 x h
105 = \(\frac{1}{3}\) x 9 x 5 x h
\(\frac{105}{15} = \frac{15h}{15}\)
h = 7cm

28.

Each interior angle of a regular polygon is 168\(^o\). Find the number of sides of the polygon

A. 30

B. 36

C. 24

D. 18

Detailed Solution

Exterior angle = 180\(^o\) - 168\(^o\)
Number of sides = \(\frac{360^o}{12}\)
= 30\(^o\)

29.

In the diagram, \(\overline{MN}\)//\(\overline{PQ}\), < MNP = 2x, and < NPQ = (3x - 50\(\^o\)). Find the value of < NPQ

A. 200\(^o\)

B. 150\(^o\)

C. 120\(^o\)

D. 90\(^o\)

Detailed Solution

Open Photo
2x = 3x - 50 alternate angle
50\(^o\) = 3x - 2x
50\(^o\) = x
< NPQ = (3x - 50)\(^o\)
= 3(500)\(^o\) - 50
= 150 - 50 = 150\(^o\)
30.

The length of an arc of a circle of radius 3.5 cm is 1\(\frac{19}{36}\) cm. Calculate, correct to the nearest degree , the angle substended by the centre of the circle. [Take \(\pi = \frac{22}{7}\)]

A. 55\(^o\)

B. 36\(^o\)

C. 25\(^o\)

D. 22\(^o\)

Detailed Solution

L = \(\frac{\theta}{360^o}\) = 2\(\pi\)
\(\frac{55}{36} = \frac{\theta}{360^o} \times 2 \times \frac{22}{7} \times 3.5\)
\(\theta = \frac{360 \times 55 \times 44 \times 0.5}{3 \times 44 \times 0.5}\)
= \(\theta = 25^o\)