JSCE Maths Free Practice Questions with Solutions

1. a. Express 132_four in Binary

Solution

First, convert 132_four to Base 10

Place Value	2	1	0
	1	3	2

= 1 × 4² + 3 × 4¹ + 2 × 4⁰

= 16 + 12 + 2

= 30₁₀

Now convert 30₁₀ to base 2

Answer = 11110₂

1. b. Solve the equation \( \frac{3x \: – \: 2}{8} = \frac{8 \: + \: 5x}{2} \)

Solution

Multiply both sides by 8

\( \normalsize \frac{3x \: – \: 2}{8} \scriptsize \: \times \: 8 = \normalsize \frac{8 \: + \: 5x}{2} \scriptsize \: \times \: 8\) \( \normalsize \frac{3x \: – \: 2}{\not{8}} \scriptsize \: \times \: \not{\!8}^1 = \normalsize \frac{8 \: + \: 5x}{\not{2}} \scriptsize \: \times \: \not{\!8}^4\)

(3x – 2) = 4(8 + 5x)

Open the brackets

3x – 2 = 32 + 20x

Collect like terms

3x – 20x = 32 + 2

-17x = 34

Divide both sides by -17

\( \frac{-17x}{-17} = \frac{34}{-17} \)

\( \scriptsize x = \normalsize \frac{34}{-17} \)

x = -2

2. a. Calculate the size of each angle of a regular pentagon

Solution

Regular pentagon = 5 sides

Each interior angle = \( \frac {n \: – \: 2 \: \times\: 180^o}{n} \)

= \( \frac {5 \: – \: 2 \: \times\: 180^o}{5} \)

= \( \frac {3 \: \times\: 180^o}{5} \)

\( \frac{540^o}{5} \)

= 108º

Solution

(i) Modal score = highest frequency = 18

(ii) Number of students = 3 + 5 + 8 + 7 + 12 = 35 students

(iii) Median score (middle score)

First list out the scores

13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18

Total frequency = 35

median = \( \frac{n \: + \: 1}{2} \)

= \( \frac{35 \: + \: 1}{2} \)

= \( \frac{36}{2} \)

= 18

The median score is the 18th frequency or score which is 16

13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18

3. Find the largest angle in the figure below

Solution

Sum of angles in a triangle = 180º

∴ 2xº + 3xº + 4xº = 180º

9xº = 180º

Divide both sides by 9

\( \frac{9x}{9} = \frac{180}{9} \)

x = 20º

From the triangle the biggest angle has to be 4x

Substitute x = 20º into 4x

4x = 4 × 20º

= 80º

4. The height of an equilateral triangle is 15 cm and its perimeter is 36 cm. Find the area of the triangle.

Solution

Permieter = 36 cm

∴ 36 = x + x + x

36 = 3x

3x = 36

Divide both sides by 3

\( \frac{3x}{3} = \frac{36}{3} \)

x = 12

Area of the triangle = \( \frac{1}{2} \scriptsize \: \times \: base \: \times \: height \)

= \( \frac{1}{2} \scriptsize \: \times \: 12 \: \times \: 15 \)

= 6 × 15

= 90 cm²

5. Calculate /DC/ using the similar shapes given below

Solution

The two triangles are similar

∴ \( \frac{|PQ|}{|QR|} = \frac{|DC|}{|BC|} \)

= \( \frac{12}{8} = \frac{|DC|}{6} \)

Cross multiply

8 × DC = 12 × 6

8DC = 72

Divide both sides by 8

\( \frac{8DC}{8} = \frac{72}{8} \)

DC = 9 cm

6. What is the average rainfall in mm² from the bar chart below?

Solution

Jan	40
Feb	80
Mar	120
Apr	60
May	40

Average Rainfall

= \( \frac{40+80+120+60+40}{5} \)

= \( \frac{340}{5} \)

= 68 mm²

7. Find the missing angles l, m and n in that order from the diagram below

Solution

The shape is a trapezoid.

Opposite angles of a trapezoid are supplementary. Supplementary angles of a trapezoid is defined as the sum of two opposite angles is 180

∴ n + 125 = 180

n = 180 – 125 = 55º

Also,

(25 + m) + (95 + l) = 180

But m = l (alternate angles)

so let’s say l and m = x

∴ (25 + x) + (95 + x) = 180

25 + x + 95 + x = 180

25 + 95 + x + x = 180

120 + 2x = 180

2x = 180 – 120

2x = 60

x = \( \frac{60}{2} \)

x = 30

∴ l and m = 30º

Answer: l, m and n = 30º, 30º and 55º

8. Simplify \( \frac{x^2 – 6x + 9}{x^2 – 9} \)

Solution

First step: Factorise the numerator

\( \scriptsize x^2 \: – \: 6x \: + \: 9 \)

=\( \scriptsize x^2 \: – \: 3x \: – \: 3x \: + \: 9 \)

= \( \scriptsize x(x \: – \: 3) \: – \: 3(x \: – \: 3) \)

= (x – 3)(x – 3)

Second step: Factorise the denominator

x² – 9

= x² – 3²

= (x – 3)(x + 3) (Difference of two squares)

∴ \( \frac{x^2 – 6x + 9}{x^2 – 9} = \frac{(x – 3)(x – 3)}{(x – 3)(x + 3)}\)

= \( \frac{(x – 3)}{(x + 3)}\)

9. In the diagram below, ABCD is a kite and ADEF is a rhombus. If BC = 4 cm and CD = 7 cm, find the perimeter of ABCDEF (Not drawn to scale)

Solution

We can redraw the diagram as shown below.

A kite has 2 pairs of adjacent equal sides

A rhombus has all sides equal

A perimeter of a shape is the length of its boundary or the total length around the shape

= 7 + 7 + 7 + 4 + 4 + 7

= 36 cm

Note: Perimeter is the total length around the shape so we ignore the side in the middle (7 cm)

10. An exercise book and two pens cost ₦90.00. If the difference between the cost of an exercise book and a pen is ₦45.00, determine the cost of a pen.

Solution

Let the cost of an exercise book = x

the cost of a pen = y

From the question

An exercise book and two pens cost ₦90.00

x + 2y = 90 ……………..(1)

The difference between the cost of an exercise book and a pen is ₦45.00

x – y = 45 …………………(2)

Subtract equation 2 from 1

x + 2y = 90

x – y = 45

2y – (-y) = 90 – 45

2y + y = 45

3y = 45

y = \( \frac{45}{3} \)

y = ₦15