BECE 1998 Mathematics Past Questions and Answers
The BECE 1998 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. Two sets which have no common members are known as ......
- A. equal sets
- B. equivalent sets
- C. empty sets
- D. disjoint sets
Answer: D. disjoint sets
Answer: D. Sets with no members in common are called disjoint sets.
Answer: D. Sets with no members in common are called disjoint sets.
2. P = {0, 2, 4, 6} and Q = {1, 2, 4, 5}. Find P ∩ Q.
- A. {0, 6}
- B. {2, 4}
- C. {0, 4}
- D. {0, 2, 6}
Answer: B. {2, 4}
Intersection (∩) are elements that can be found in both sets
P ∩ Q = {2, 4}
Intersection (∩) are elements that can be found in both sets
P ∩ Q = {2, 4}
3. Evaluate $\frac{1}{2}$[(4 – 1) – (5 – 6)]
- A. –4.0
- B. 1.0
- C. 2.0
- D. 3.0
Answer: C. 2.0
$\frac{1}{2}$[(4 – 1) – (5 – 6)] = $\frac{1}{2}$[(4 – 1) – (5 – 6)]
Simplify the brackets first
$\frac{1}{2}$[(4 – 1) – (5 – 6)] = $\frac{1}{2}$[(3) – (-1)]
Note: - (-) = - x - = +. -(-1) = + 1
$\frac{1}{2}$[(4 – 1) – (5 – 6)] = $\frac{1}{2}$[3 + 1]
$\frac{1}{2}$[(4 – 1) – (5 – 6)] = $\frac{1}{2}$[4]
$\frac{1}{2}$[(4 – 1) – (5 – 6)] = $\frac{1}{2}$ x 4
Note: 2 divides itself 1 time and 4, 2 times.
$\frac{1}{2}$[(4 – 1) – (5 – 6)] = 2
$\frac{1}{2}$[(4 – 1) – (5 – 6)] = $\frac{1}{2}$[(4 – 1) – (5 – 6)]
Simplify the brackets first
$\frac{1}{2}$[(4 – 1) – (5 – 6)] = $\frac{1}{2}$[(3) – (-1)]
Note: - (-) = - x - = +. -(-1) = + 1
$\frac{1}{2}$[(4 – 1) – (5 – 6)] = $\frac{1}{2}$[3 + 1]
$\frac{1}{2}$[(4 – 1) – (5 – 6)] = $\frac{1}{2}$[4]
$\frac{1}{2}$[(4 – 1) – (5 – 6)] = $\frac{1}{2}$ x 4
Note: 2 divides itself 1 time and 4, 2 times.
$\frac{1}{2}$[(4 – 1) – (5 – 6)] = 2
4. Find the highest(greatest) common factor of 35 and 70.
- A. 5
- B. 7
- C. 10
- D. 35
Answer: D. 35
The easiest way of finding the highest common factor (H.C.F) is using the prime factors method.
Prime factors of 35
$\begin{matrix} & 35 \\ 5 & 7 \\ 7 & 1\end{matrix}$
Prime factors of 35 = 5 x 7
Prime factors of 70
$\begin{matrix} & 70 \\ 2 & 35 \\ 5 & 7 \\ 7 & 1\end{matrix}$
Prime factors of 70 = 2 X 5 x 7
The highest (greatest) common factor is the product of the prime factors common in all in their lowest power.
The common prime factors are 5 and 7. Hence the H.C.F = 5 x 7 = 35
The easiest way of finding the highest common factor (H.C.F) is using the prime factors method.
Prime factors of 35
$\begin{matrix} & 35 \\ 5 & 7 \\ 7 & 1\end{matrix}$
Prime factors of 35 = 5 x 7
Prime factors of 70
$\begin{matrix} & 70 \\ 2 & 35 \\ 5 & 7 \\ 7 & 1\end{matrix}$
Prime factors of 70 = 2 X 5 x 7
The highest (greatest) common factor is the product of the prime factors common in all in their lowest power.
The common prime factors are 5 and 7. Hence the H.C.F = 5 x 7 = 35
5. Write 1101101$_{two}$ in base ten
- A. 31
- B. 43
- C. 108
- D. 109
Answer: D. 109
1101101$_{two}$ = 1 x 2$^{6}$ + 1 x 2$^{5}$ + 0 x 2$^{4}$ + 1 x 2$^{3}$ + 1 x 2$^{2}$ + 0 x 2$^{1}$ + 1 x 2$^{0}$
Any number raised to the power 0 is 1
Any number raised to the power 1 is the same number
2$^{2}$ = 2 x 2 = 4
2$^{3}$ = 2 x 2 x 2 = 8
2$^{4}$ = 2 x 2 x 2 x 2 = 16
2$^{5}$ = 2 x 2 x 2 x 2 x 2 = 32
2$^{6}$ = 2 x 2 x 2 x 2 x 2 x 2 = 64
1101101$_{two}$ = 1 x 64 + 1 x 32 + 0 x 16 + 1 x 8 + 1 x 4 + 0 x 2 + 1 x 1
1101101$_{two}$ = 64 + 32 + 0 + 8 + 4 + 0 + 1
1101101$_{two}$ = 109
1101101$_{two}$ = 1 x 2$^{6}$ + 1 x 2$^{5}$ + 0 x 2$^{4}$ + 1 x 2$^{3}$ + 1 x 2$^{2}$ + 0 x 2$^{1}$ + 1 x 2$^{0}$
Any number raised to the power 0 is 1
Any number raised to the power 1 is the same number
2$^{2}$ = 2 x 2 = 4
2$^{3}$ = 2 x 2 x 2 = 8
2$^{4}$ = 2 x 2 x 2 x 2 = 16
2$^{5}$ = 2 x 2 x 2 x 2 x 2 = 32
2$^{6}$ = 2 x 2 x 2 x 2 x 2 x 2 = 64
1101101$_{two}$ = 1 x 64 + 1 x 32 + 0 x 16 + 1 x 8 + 1 x 4 + 0 x 2 + 1 x 1
1101101$_{two}$ = 64 + 32 + 0 + 8 + 4 + 0 + 1
1101101$_{two}$ = 109
6. State the property used in the statement: p(q + r) = pq + pr
- A. Associative
- B. Commutative
- C. Distributive
- D. Identity
Answer and explanation: practise this paper on ePrep.
7. If n$^{2}$ + 1 = 50, find n
- A. 7
- B. 24.5
- C. 25
- D. 49
Answer and explanation: practise this paper on ePrep.
8. Evaluate $\frac{0.63 x 0.70}{9.00}$
- A. 0.0049
- B. 0.049
- C. 0.49
- D. 4.9
Answer and explanation: practise this paper on ePrep.
9. In the relation $\frac{1}{R}$ = $\frac{1}{{R}_{1}}$ + $\frac{1}{{R}_{2}}$, if R$_{1}$ = 1 and R$_{2}$ = 3,
find R
find R
- A. $\frac{1}{2}$
- B. $\frac{2}{3}$
- C. $\frac{3}{4}$
- D. $\frac{3}{2}$
Answer and explanation: practise this paper on ePrep.
10. Arrange the following from the highest to the lowest: $\frac{2}{3}$, -9, $\frac{3}{5}$ and 0.
- A. -9, $\frac{3}{5}$, $\frac{2}{3}$
- B. -9, 0, $\frac{3}{5}$, $\frac{2}{3}$
- C. $\frac{3}{5}$, $\frac{2}{3}$, 0, -9
- D. $\frac{3}{5}$, $\frac{2}{3}$, -9, 0
- E. $\frac{2}{3}$, $\frac{3}{5}$, 0, -9
Answer and explanation: practise this paper on ePrep.
Answer all 45 questions of the BECE 1998 Mathematics paper
Timed papers like the real exam, instant marking, worked explanations, topic tests and mock exams. Theory answers marked by real teachers. 3 weeks free.
Practise free on ePrep