2024 BECE Mathematics Sample Questions Released

 

The latest Ghana Education News update for final year students in Junior High School is that GES and NaCCA have released the 2024 BECE Mathematics Sample Questions to aid student’s preparation.

In this publication we share with candidates some of the questions  from both section A and B.

 

2024 BECE Mathematics Sample Questions – SECTION B

1. A group of 80 learners took a test in Mathematics or French or both. 15 of them passed in both Mathematics and French, but 11 more learners passed in French than Mathematics. Each learner passed in at least one of the two subjects.

a) Represent the above information on a Venn diagram.
b) Find the number that passed in
i) French.
ii) Only one subject

 

Suggested solution from Ghana Education News 

Let’s represent the information on a Venn diagram:

• Let the circle representing Mathematics be M, and the circle representing French be F.
• The overlapping region of the circles represents learners who passed both Mathematics and French.

Given:

• Total learners = 80
• Learners passing both Mathematics and French = 15
• Learners passing only French = 15 + 11 = 26 (since 11 more learners passed in French than Mathematics)
• Each learner passed at least one subject, so there are no learners who passed neither.

Now, let’s fill in the Venn diagram

a) Venn diagram representing the given information.

b) To find the number of learners that passed in French:

• Total learners passing French = Learners passing only French + Learners passing both
• Total learners passing French = 26 + 15 = 41

To find the number of learners that passed in only one subject:

• Learners passing only one subject = Learners passing only French + Learners passing only Mathematics
• Learners passing only one subject = 26 + 4 = 30

So, the number of learners that passed in: i) French = 41 ii) Only one subject = 30

2. a) Dufie spent 1/3 of her pocket money on breakfast and 1/6 of what still remained on sweets. If she still had GH¢ 55.00, how much originally was her pocket money?

Suggested Solution

Let’s denote Dufie’s original pocket money as  𝑥cedis.

1. She spent 1/3 of her pocket money on breakfast, leaving 2/3 of her pocket money.
2. She then spent 1/6​ of what still remained on sweets, leaving 5/6​ of 2/3 of her original pocket money.
3. Finally, she had GH¢ 55.00 left, which is 5/6×2/3​ of her original pocket money.

We can set up an equation to solve for $x$:

5/6×2/3×𝑥=55

First, simplify the left side of the equation:

5/6×2/3=10/18=5/9​

Now, the equation becomes:

5/9 𝑥=55

To solve for $x$, multiply both sides by 9/5

𝑥=55×9/5​

𝑥=99

So, Dufie’s original pocket money was GH¢ 99.00.

c) You are using a magnifying glass that shows the image of an object that is six times the object’s actual size.

Determine the length of the image of the spider seen through the magnifying glass.

3. a) The total surface area of a cube is 384 cm2. Find the:
i) Length of a side of the cube.
ii) Volume of the cube.
b. A pickup truck has a bucket that measures 4 ft by 8 ft by 17 in. What is the volume of the bucket?

4. a) Mr. Mensah is four times as old as his daughter, Kukua. In 14 years’ time, the sum of their ages will be 78.
Find their present ages.

Suggested Solution

Let’s denote the daughter’s age as $K$ and the father’s age as $M$.

Given:

1. Mr. Mensah is four times as old as his daughter: 𝑀=4𝐾
2. In 14 years’ time, the sum of their ages will be 78: (𝑀+14)+(𝐾+14)=78

Substitute the first equation into the second equation to find $K$:

(4𝐾+14)+(𝐾+14)=78

5𝐾+28=78

5𝐾=50

𝐾=10

So, Kukua’s present age is 10 years old. Now, substitute 𝐾=10 into the first equation to find Mr. Mensah’s age:

𝑀=4×10

𝑀=40

Therefore, Mr. Mensah is 40 years old.

2024 BECE Mathematics Sample Questions

b) Two soup tins are similar. Tin P can hold 500 grams of soup while tin Q can hold 750 grams of soup. The height of tin P is 11 cm.

i. Calculate the height of tin Q.
ii. Calculate the ratio of the heights of the tins

Suggested Solution

To find the height of tin Q, we can use the fact that the tins are similar, meaning their dimensions are proportional. Since the volumes of the tins are proportional to the cubes of their heights, we can set up a proportion:

Height of 𝑃/Height of 𝑄= (Volume of 𝑃Volume of 𝑄) superscript 1/3

Given:

• Volume of tin P = 500 grams
• Volume of tin Q = 750 grams
• Height of tin P = 11 cm

i. Calculate the height of tin Q:

11/Height of 𝑄=(500/750) superscript 1/3

11/Height of 𝑄=(2/3)superscript 1/3 

11/Height of 𝑄=2/3​

11=2/3×Height of 𝑄

Height of 𝑄=11 superscript 1/3 ​

Height of 𝑄=11×3/2​​

Height of 𝑄=16.5 cm

 

ii. Calculate the ratio of the heights of the tins:

Ratio=Height of 𝑃/Height of 𝑄​

Ratio=11/16.5​

Ratio=2/3​

So, the height of tin Q is 16.5 cm, and the ratio of the heights of the tins is 2:3.

Solve the remaining Questions Below

5. a) A box contains 60 mangoes of which some are not ripe. 

The probability of selecting at random a mango that is ripe is 3/4
How many mangoes are not ripe?   

b) Factorise completely : ( 𝑚 + 𝑛 )( 2𝑥 – 𝑦) – ( 𝑚 + 𝑛).

c) Given that 𝐶 = 𝑘 × 216 × 27, what is the smallest whole number k, that will make C a perfect square?
6. Copy and complete the magic square using any of the numbers 9,11,12, 14, 15, 16, 17, 18 and 19. 

Use each number once, such that the sum of the three numbers, horizontally, vertically or diagonally are equal.

b) Find the scale factor of the dilation. Then tell whether the dilation is a reduction or an enlargement.

Section A:

Read and write in number quantities up 10, 000,000. [Level 1]

Which of the following correctly reads the given number quantity 6,245,789? A. Six million, two hundred forty-five thousand, seven hundred eighty-nine. B. Six million, two hundred fifty-four thousand, seven hundred eighty-nine. C. Six million, two hundred forty-five thousand, seven hundred ninety. D. Six million, two hundred fifty-four thousand, seven hundred ninety