BECE 1993 Mathematics Past Questions and Answers
The BECE 1993 Mathematics paper has 39 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. Expand (2*a* + b) (a + 2*b*)
- A. 2*a*$^{2}$ + 2*b*$^{2}$
- B. 2*a*$^{2}$ + b$^{2}$
- C. 5*a*$^{2}$ + 2*a*$^{2}$
- D. 2*a*$^{2}$ + 2*a* + 4*ab*$^{2}$
- E. 2*a*$^{2}$ + 5*ab* + 2*b*$^{2}$
Answer: E. 2*a*$^{2}$ + 5*ab* + 2*b*$^{2}$
(2*a* + b) (a + 2*b*) = (2*a* + b) (a + 2*b*)
(2*a* + b) (a + 2*b*) = 2*a*(a + 2*b*) + b(a + 2*b*)
(2*a* + b) (a + 2*b*) = 2*a*$^{2}$ + 4*ab* + ab + 2*b*$^{2}$
(2*a* + b) (a + 2*b*) = 2*a*$^{2}$ + 5*ab* + 2*b*$^{2}$
(2*a* + b) (a + 2*b*) = (2*a* + b) (a + 2*b*)
(2*a* + b) (a + 2*b*) = 2*a*(a + 2*b*) + b(a + 2*b*)
(2*a* + b) (a + 2*b*) = 2*a*$^{2}$ + 4*ab* + ab + 2*b*$^{2}$
(2*a* + b) (a + 2*b*) = 2*a*$^{2}$ + 5*ab* + 2*b*$^{2}$
2. Find the missing number in the following binary operation:
| | 1 | 1 | 0 | 0 | 1 | 1 | 0 |
|---|---|---|---|---|---|---|---|
| - | * | * | * | * | * | * | * |
| | | 1 | 1 | 1 | 0 | 1 | 1 |
| | 1 | 1 | 0 | 0 | 1 | 1 | 0 |
|---|---|---|---|---|---|---|---|
| - | * | * | * | * | * | * | * |
| | | 1 | 1 | 1 | 0 | 1 | 1 |
- A. 111011
- B. 101001
- C. 100011
- D. 101110
- E. 101011
Answer: E. 101011
Let x = the missing binary number.
1100110 - x = 111011
1100110 - 111011 = x
| | | 1 | 1 | 1 | | 1 | 1 |
|---|---|---|---|---|---|---|---|
| | 1 | 1 | 0 | 0 | 1 | 1 | 0 |
| | - | 1 | 1 | 1 | 0 | 1 | 1 |
| | | 1 | 0 | 1 | 0 | 1 | 1 |
Note: in binary, the value of a number borrowed is 2.
Let x = the missing binary number.
1100110 - x = 111011
1100110 - 111011 = x
| | | 1 | 1 | 1 | | 1 | 1 |
|---|---|---|---|---|---|---|---|
| | 1 | 1 | 0 | 0 | 1 | 1 | 0 |
| | - | 1 | 1 | 1 | 0 | 1 | 1 |
| | | 1 | 0 | 1 | 0 | 1 | 1 |
Note: in binary, the value of a number borrowed is 2.
3. If x = {1, 3, 5, 7, 9, 11, 13, 15}, find the truth set of x – 3 ≥ 10.
- A. {15}
- B. {13,15}
- C. {11,13,15}
- D. {9,11,13,15}
Answer: B. {13,15}
x – 3 ≥ 10
x ≥ 10 + 3
x ≥ 13
This is read as x is greater than or equal to 13. Or equal to means the number is also included.
List from the set {1, 3, 5, 7, 9, 11, 13, 15} the numbers greater than or equal to 13.
The truth set is {13, 15}
x – 3 ≥ 10
x ≥ 10 + 3
x ≥ 13
This is read as x is greater than or equal to 13. Or equal to means the number is also included.
List from the set {1, 3, 5, 7, 9, 11, 13, 15} the numbers greater than or equal to 13.
The truth set is {13, 15}
4. Which of these has the least number of lines of symmetry?
- A. An equilateral triangle
- B. A rectangle
- C. A square
- D. A circle
- E. An isosceles triangle
Answer: E. An isosceles triangle
p>A line of symmetry is the line that divides a shape or an object into two identical parts.
Note:
1. equilateral triangle has 3 lines of symmetry
2. circle has an infinite number of lines of symmetry
3. isosceles triangle has 1 line of symmetry
p>A line of symmetry is the line that divides a shape or an object into two identical parts.
Note:
1. equilateral triangle has 3 lines of symmetry
2. circle has an infinite number of lines of symmetry
3. isosceles triangle has 1 line of symmetry
5. Find 2$\frac{1}{2}$% of ₵2,000.00
- A. ₵40.00
- B. ₵50.00
- C. ₵100.00
- D. ₵800.00
Answer: B. ₵50.00
of means multiplication.
2$\frac{1}{2}$% of ₵2,000.00 = 2$\frac{1}{2}$% x ₵2,000.00
Change the mixed fraction to improper fraction.
2$\frac{1}{2}$ = $\frac{2 x 2 + 1}{2}$ = $\frac{4 + 1}{2}$ = $\frac{5}{2}$
2$\frac{1}{2}$% = $\frac{5}{2}$%
% means over 100
2$\frac{1}{2}$% = $\frac{5}{2}$ ÷ 100
Every whole number is being divided by 1
100 = $\frac{100}{1}$
2$\frac{1}{2}$% = $\frac{5}{2}$ ÷ $\frac{100}{1}$
Reciprocate the fraction (numerator becomes denominator and denominator becomes numerator) at the right and change the division (÷) to multiplication (x) sign.
2$\frac{1}{2}$% = $\frac{5}{2}$ x $\frac{1}{100}$
2$\frac{1}{2}$% = $\frac{5 x 1}{2 x 100}$ = $\frac{5}{200}$
2$\frac{1}{2}$% of ₵2,000.00 = $\frac{5}{200}$ x ₵2000 = 5 x ₵10 = ₵50
Note:
1. the zeros (0) at the end of 2000 and 200 cancel each other
2. 2 divides itself 1 time and 20, 10 times
of means multiplication.
2$\frac{1}{2}$% of ₵2,000.00 = 2$\frac{1}{2}$% x ₵2,000.00
Change the mixed fraction to improper fraction.
2$\frac{1}{2}$ = $\frac{2 x 2 + 1}{2}$ = $\frac{4 + 1}{2}$ = $\frac{5}{2}$
2$\frac{1}{2}$% = $\frac{5}{2}$%
% means over 100
2$\frac{1}{2}$% = $\frac{5}{2}$ ÷ 100
Every whole number is being divided by 1
100 = $\frac{100}{1}$
2$\frac{1}{2}$% = $\frac{5}{2}$ ÷ $\frac{100}{1}$
Reciprocate the fraction (numerator becomes denominator and denominator becomes numerator) at the right and change the division (÷) to multiplication (x) sign.
2$\frac{1}{2}$% = $\frac{5}{2}$ x $\frac{1}{100}$
2$\frac{1}{2}$% = $\frac{5 x 1}{2 x 100}$ = $\frac{5}{200}$
2$\frac{1}{2}$% of ₵2,000.00 = $\frac{5}{200}$ x ₵2000 = 5 x ₵10 = ₵50
Note:
1. the zeros (0) at the end of 2000 and 200 cancel each other
2. 2 divides itself 1 time and 20, 10 times
6. Find the highest common factor of 18, 36 and 120.
- A. 2$^{2}$ × 3$^{3}$ × 5
- B. 2 × 3 × 5
- C. 2 × 3
- D. 2$^{3}$ × 2$^{2}$
Answer and explanation: practise this paper on ePrep.
7. Arrange the fractions $\frac{3}{4}$, $\frac{2}{3}$, $\frac{3}{5}$ in ascending order.
- A. $\frac{2}{3}$, $\frac{3}{4}$, $\frac{3}{5}$
- B. $\frac{3}{4}$, $\frac{2}{3}$, $\frac{3}{5}$
- C. $\frac{3}{5}$, $\frac{2}{3}$, $\frac{3}{4}$
- D. $\frac{3}{5}$, $\frac{3}{4}$, $\frac{2}{3}$
Answer and explanation: practise this paper on ePrep.
8. Make b the subject of the relation $\frac{1}{a}$ = $\frac{1}{b}$ + c
- A. b = $\frac{ac - 1}{c}$
- B. b = $\frac{a}{a + ac}$
- C. b = $\frac{c}{1 - ac}$
- D. b = $\frac{a}{1 - ac}$
Answer and explanation: practise this paper on ePrep.
9. A man has three children whose ages are 9 years, 12 years and 24 years. Find the ratio of their ages.
- A. 1 : 2 : 3
- B. 1 : 2 : 4
- C. 2 : 3 : 6
- D. 3 : 4 : 8
Answer and explanation: practise this paper on ePrep.
10. Ten students in Kwamekrom JSS took 9 days to weed the school compound. How long would 15 students take to weed the compound if they worked at the same rate?
- A. 5 days
- B. 6 days
- C. 13$\frac{1}{2}$ days
- D. 14 days
Answer and explanation: practise this paper on ePrep.
Answer all 44 questions of the BECE 1993 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