BECE 1995 Mathematics Past Questions and Answers
The BECE 1995 Mathematics paper has 40 objective and 5 theory questions. Here are sample objective questions with the correct answers and explanations. Sit the full paper with the real time limit on ePrep.
Sample questions with answers
1. P = {prime numbers less than 20} and Q = {odd numbers less than 10}.
Find P ∩ Q
Find P ∩ Q
- A. {2, 3}
- B. {1, 3, 5, 7, 11)
- C. {3, 5, 7, 9}
- D. {3, 5, 7}
Answer: D. {3, 5, 7}
P = {prime numbers less than 20}
Q = {odd numbers less than 10}
Prime numbers are numbers with only two factors (1 and itself). Thus only 1 and the number itself can divide the number without a remainder.
Note: 1 is not a prime number.
P = {2, 3, 5, 7, 11, 13, 17, 19}
Odd numbers are numbers not divisible by 2. Thus there is a remainder when divided by 2.
Q = {1, 3, 5, 7, 9}
Intersection (∩) are elements that can be found in both sets
P ∩ Q = {3, 5, 7}
P = {prime numbers less than 20}
Q = {odd numbers less than 10}
Prime numbers are numbers with only two factors (1 and itself). Thus only 1 and the number itself can divide the number without a remainder.
Note: 1 is not a prime number.
P = {2, 3, 5, 7, 11, 13, 17, 19}
Odd numbers are numbers not divisible by 2. Thus there is a remainder when divided by 2.
Q = {1, 3, 5, 7, 9}
Intersection (∩) are elements that can be found in both sets
P ∩ Q = {3, 5, 7}
2. Convert 104$_{ten}$ to a binary numeral.
- A. 1101000
- B. 1010100
- C. 1101100
- D. 1011010
Answer: A. 1101000
Binary means base 2
104$_{ten}$ to a binary numeral
$\begin{matrix} & 104 \\ 2 & 52 r 0 \\ 2 & 26 r 0 \\ 2 & 13 r 0 \\ 2 & 6 r 1 \\ 2 & 3 r 0 \\ 2 & 1 r 1 \\ 2 & 0 r 1\end{matrix}$
You list the remainders from the bottom to the top.
104$_{ten}$ = 1101000$_{two}$
Binary means base 2
104$_{ten}$ to a binary numeral
$\begin{matrix} & 104 \\ 2 & 52 r 0 \\ 2 & 26 r 0 \\ 2 & 13 r 0 \\ 2 & 6 r 1 \\ 2 & 3 r 0 \\ 2 & 1 r 1 \\ 2 & 0 r 1\end{matrix}$
You list the remainders from the bottom to the top.
104$_{ten}$ = 1101000$_{two}$
3. What is the H.C.F of 48, 30 and 18?
- A. 2
- B. 3
- C. 5
- D. 6
Answer: D. 6
The easiest way of finding the Highest Common Factor (H.C.F) is using the prime factors method.
Prime factors of 48
$\begin{matrix} & 48 \\ 2 & 24 \\ 2 & 12 \\ 2 & 6 \\ 2 & 3 \\ 3 & 1\end{matrix}$
Prime factors of 48 = 2 x 2 x 2 x 2 x 3
Prime factors of 48 = 2$^{4}$ x 3
Prime factors of 30
$\begin{matrix} & 30 \\ 2 & 15 \\ 3 & 5 \\ 5 & 1\end{matrix}$
Prime factors of 30 = 2 x 3 x 5
Prime factors of 18
$\begin{matrix} & 18 \\ 2 & 9 \\ 3 & 3 \\ 3 & 1\end{matrix}$
Prime factors of 18 = 2 x 3 x 3
Prime factors of 18 = 2 x 3$^{2}$
The highest common factor (H.C.F) is the product (multiplication) of the common prime factors in their lowest powers.
H.C.F of 48, 30 and 18 = 2 x 3 = 6
The easiest way of finding the Highest Common Factor (H.C.F) is using the prime factors method.
Prime factors of 48
$\begin{matrix} & 48 \\ 2 & 24 \\ 2 & 12 \\ 2 & 6 \\ 2 & 3 \\ 3 & 1\end{matrix}$
Prime factors of 48 = 2 x 2 x 2 x 2 x 3
Prime factors of 48 = 2$^{4}$ x 3
Prime factors of 30
$\begin{matrix} & 30 \\ 2 & 15 \\ 3 & 5 \\ 5 & 1\end{matrix}$
Prime factors of 30 = 2 x 3 x 5
Prime factors of 18
$\begin{matrix} & 18 \\ 2 & 9 \\ 3 & 3 \\ 3 & 1\end{matrix}$
Prime factors of 18 = 2 x 3 x 3
Prime factors of 18 = 2 x 3$^{2}$
The highest common factor (H.C.F) is the product (multiplication) of the common prime factors in their lowest powers.
H.C.F of 48, 30 and 18 = 2 x 3 = 6
4. Express 34m 5cm 6mm in millimetres
- A. 340506 mm
- B. 342506 mm
- C. 34056 mm
- D. 30456 mm
Answer: C. 34056 mm
1 metre = 1000 millimetres
1 centimetre = 10 millimetres
34 m = 34 x 1000 mm = 34000 mm
5 cm = 5 x 10 mm = 50 mm
34m 5cm 6mm = 34000 mm + 50 mm + 6 mm = 34056 mm
1 metre = 1000 millimetres
1 centimetre = 10 millimetres
34 m = 34 x 1000 mm = 34000 mm
5 cm = 5 x 10 mm = 50 mm
34m 5cm 6mm = 34000 mm + 50 mm + 6 mm = 34056 mm
5. Write 356.07 in standard form.
- A. 35.607 x 10
- B. 35.607 x 10$^{2}$
- C. 3.5607 x 10$^{2}$
- D. 3.5607 x 10$^{-2}$
Answer: C. 3.5607 x 10$^{2}$
A standard form should be in the form a.bcde x 10$^{n}$
The decimal point must be after the first non-zero digit.
The n is the number of times the decimal point has to be moved to be right after the first non-zero digit.
If it is moved to the left, the n is positive and if to the right, n is negative.
356.07 = 3.⌒5⌒6.07 = 3.5607 x 10$^{2}$
A standard form should be in the form a.bcde x 10$^{n}$
The decimal point must be after the first non-zero digit.
The n is the number of times the decimal point has to be moved to be right after the first non-zero digit.
If it is moved to the left, the n is positive and if to the right, n is negative.
356.07 = 3.⌒5⌒6.07 = 3.5607 x 10$^{2}$
6. Divide $(1\frac{1}{2}+\frac{1}{4})$ by $(1\frac{1}{2}-\frac{1}{4})$
- A. $\frac{5}{7}$
- B. 1
- C. 1$\frac{2}{5}$
- D. 1$\frac{3}{4}$
Answer and explanation: practise this paper on ePrep.
7. Which property is illustrated by the statement a × (b + c) = a × b + a × c?
- A. Inverse
- B. Identity
- C. Commutative
- D. Distributive
Answer and explanation: practise this paper on ePrep.
8. How many edges has a cube?
- A. 4
- B. 6
- C. 8
- D. 12
Answer and explanation: practise this paper on ePrep.
9. What is the mode of the following numbers: 4, 5, 3, 3, 4, 2, 7, 6, 5, 4, 4, 1?
- A. 3
- B. 4
- C. 5
- D. 6
Answer and explanation: practise this paper on ePrep.
10. What is the rule for the mapping below?
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| ↓ | ↓ | ↓ | ↓ | ↓ | ↓ |
| y | 2 | 4 | 8 | 16 | 32 |
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| ↓ | ↓ | ↓ | ↓ | ↓ | ↓ |
| y | 2 | 4 | 8 | 16 | 32 |
- A. y = 2*x* + 2
- B. y = 2$^{x}$
- C. y = x + 2
- D. y = x + 1
Answer and explanation: practise this paper on ePrep.
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