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JSCE: MATHEMATICS

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  1. Free JSCE/BECE Mathematics Past Questions & Answers
    1 Quiz
  2. JSCE/BECE Mathematics Theory Practice Past Questions
    2 Topics

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Question 1

(a) Evaluate and express your answer in standard form: 0.0000585 ÷ 0.09

(b) Construct triangle ABC such that:
\( \scriptsize \bar{AB} = 7cm, \bar{AC} = 5cm \) and angle BAC = 60º. Measure BC

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Question 2

(a) There are 420 students in a school. \( \frac{1}{3} \) of the population is made up of girls
(i) How many boys are in the school?
(ii) Express the number of boys as a fraction of all the students.

(b) Find the product of \( \frac{1}{7} \) and the sum of \( \frac{3}{5}\) and \( \scriptsize 1\frac{1}{2} \)

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Question 3

Kola tossed a fair die 30 times and the number that showed up in each toss was given as follows:

Screenshot 2023 04 14 at 11.23.08

(a) Prepare a frequency table for the data
(b) Find the mode of the data
(c) Determine the median
(d) Calculate the mean

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Question 4

A boy sighted the school’s flag at a point which is 20 m from the foot of the pole, and due East of the pole. If the angle of elevation of the top of the flag pole from the boy is 21º, find the height of the flag pole.
[Sin 21º = 0.3544, Cos 21º = 0.936, Tan 21º = 0.3339]

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Question 1

(a) Evaluate and express your answer in standard form: 0.0000585 ÷ 0.09

convert both numbers to standard form

⇒ \( \frac{5.85 \: \times \: 10^{-5}}{9\:\times\: 10^{-2}} \)

⇒ \( \scriptsize 0.65 \: \times \: 10^{[-5 \:-\:(-2)]}\)

⇒ \( \scriptsize 0.65 \: \times \: 10^{(-5 + 2)}\)

⇒ \( \scriptsize 0.65 \: \times \: 10^{-3}\)

convert 0.65 to standard form

⇒ \( \scriptsize 6.5 \: \times \: 10^{-1}\: \times \: 10^{-3}\)

⇒ \( \scriptsize 6.5 \: \times \: 10^{[-1\: +\: (-3)]}\)

⇒ \( \scriptsize 6.5 \: \times \: 10^{-4}\)

Question 2

(a) There are 420 students in a school. \( \frac{1}{3} \) of the population is made up of girls

(i) How many boys are in the school?

Solution

If \( \frac{1}{3} \) of the population is made up of girls, Then the fraction of boys is;

⇒ \(\scriptsize 1 \: – \: \normalsize \frac{1}{3} \\ = \frac{3}{3} \: -\: \frac{1}{3}\\ = \frac{2}{3}\)

We can use this fraction to calculate the number of boys.

Given: Total number of students = 420

Number of boys = \( \frac{2}{3} \scriptsize \: \times \: 420 \\ = \scriptsize 2 \: \times \: 140 \\ = \scriptsize 280 \: boys \)

(ii) Express the number of boys as a fraction of all the students.

We have already calculated this.

Solution:

The fraction of boys is;

⇒ \(\scriptsize 1 \: – \: \normalsize \frac{1}{3} \\ = \frac{3}{3} \: -\: \frac{1}{3}\\ = \frac{2}{3}\)

 

(b) Find the product of \( \frac{1}{7} \) and the sum of \( \frac{3}{5}\) and \( \scriptsize 1\frac{1}{2} \)

Solution:

1st step:

Let’s find the sum of \( \frac{3}{5}\) and \( \scriptsize 1\frac{1}{2} \)

⇒ \( \frac{3}{5} \scriptsize \: + \: 1\frac{1}{2} \)

⇒ \( \frac{3}{5}  \: + \: \frac{3}{2} \)

L.C.M of the denominators = 10

⇒ \( \frac{6 \: + \: 15}{10}\)

⇒ \( \frac{21}{10}\)

2nd step:

⇒ \( \frac{1}{7} \: \times \: \frac{21}{10} \)

⇒ \( \frac{21}{70} \)

Simplify further by dividing both the numerator and denominator by 7

⇒ \( \frac{3}{10} \)

Question 3

Kola tossed a fair die 30 times and the number that showed up in each toss was given as follows:
 
 
(a) Prepare a frequency table for the data
 
Solution:

Score (x) Tally Frequency (f) fx
1 ││ 2 2
2 │││ 3 6
3 ││││ ││ 7 21
4 ││││ │││ 8 32
5 ││││ ││ 7 35
6 │││ 3 18
    30 114
 
(b) Find the mode of the data
 
Answer:
 
Mode ⇒ score with the most frequency = 4  
 
 
(c) Determine the median
 
Solution Total frequency = 30
 
Median is the average of the 15th and 16th score (Since the sequence is even) From the table, the 15th and 16th number falls on number 4.
 
= \(\frac{4\:+\:4}{2} \\ =\scriptsize 4 \)
 
Long method: We can find the median by listing all the numbers in ascending order.
 
1 1 2 2 2 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 6 6 6
 
Median = \(\frac{4\:+\:4}{2} \\ =\scriptsize 4 \)  
 
 
(d) Calculate the mean
 
Solution:
 
Mean = \(\frac{\sum fx}{f} \\ = \frac{114}{30} \\ = \scriptsize 3.8\)

Question 4

A boy sighted the school’s flag at a point which is 20 m from the foot of the pole, and due East of the pole. If the angle of elevation of the top of the flag pole from the boy is 21º, find the height of the flag pole.
[Sin 21º = 0.3544, Cos 21º = 0.936, Tan 21º = 0.3339]

Solution:

1st let’s sketch the diagram

 

From the diagram, we know the opposite and adjacent.

Using SOHCAHTOA

We will use the tan formula to find the ‘opposite’ which is the height of the flagpole

tan θ = \( \frac{opposite}{adjacent} \)

where θ = 21º
opposite = height of the flagpole = x
adjacent = 20 m

∴ tan 21º = \( \frac{x}{20} \)

x = tan 21º × 20

But Tan 21º = 0.3339

∴ x = 0.3339 × 20

x = 6.678

x ≅ 6.7 m

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