JSCE Maths Free Practice Questions with Solutions

jsce maths

1a. Express 132four in Binary

Solution

First, convert132four to Base 10

Place Value210
132

= 1 x 42 + 3 x 41 + 2 x 40

= 16 + 12 + 2

= 3010

Now convert3010 to base 2

Screen Shot 2021 08 29 at 11.34.13 AM

Answer = 111102

1b. 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

2a. 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

2b. The table shows the scores of students in a test.

Scores1314151618
No of Students358712

(i) What is the modal score?

(ii) How many students took part in the test?

(iii) Find the median score

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

Screen Shot 2021 08 29 at 2.15.16 PM

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 x 20º

= 80º

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

Solution

Screen Shot 2021 08 29 at 2.43.45 PM
The first step is to find x (equilateral triangle so x = x = x)

Perimeter of triangle = x + x + x

Permieter = 36cm

∴ 36 = x + x + x

36 = 3x

3x = 36

Divide both sides by 3

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

x = 12

Screen Shot 2021 08 29 at 2.54.56 PM

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

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

= 6 x 15

= 90cm2

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

Screen Shot 2021 08 29 at 3.17.56 PM Easy Resize.com

Solution

The two triangles are similar

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

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

Cross multiply

8 x DC = 12 x 6

8DC = 72

Divide both sides by 8

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

DC = 9cm

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

Screen Shot 2021 08 29 at 4.06.06 PM

Solution

Jan40
Feb80
Mar120
Apr60
May40

Average Rainfall

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

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

= 68mm2

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

Screen Shot 2021 08 29 at 4.39.33 PM

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 = 4cm and CD = 7cm, find the perimeter of ABCDEF (Not drawn to scale)

Screen Shot 2021 08 29 at 8.15.58 PM

Solution

We can redraw the diagram as shown below.

A kite has 2 pairs of adjacent equal sides

A rhombus has all sides equal

Screen Shot 2021 08 29 at 8.19.07 PM

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

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

= 36cm

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

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

Letthe 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

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