BECE 2002 June A Mathematics Past Questions and Answers
The BECE 2002 June A 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. If P = {multiples of 4 less than 16}, find P.
- A. {4,8,10}
- B. {4,8,12}
- C. {1,4,8,12}
- D. {4,8,12,16}
Answer: B. {4,8,12}
Multiples of 4
4 x 1 = 4
4 x 2 = 4 + 4 = 8
4 x 3 = 8 + 4 = 12
Multiples of 4
4 x 1 = 4
4 x 2 = 4 + 4 = 8
4 x 3 = 8 + 4 = 12
2. The addition below was carried out in base x. Find x.
| | 2 | 4 | 3 | |
|---|---|---|---|---|
| | 2 | 2 | 1 | |
| 1 | 0 | 1 | 4 | $_{x}$ |
| | 2 | 4 | 3 | |
|---|---|---|---|---|
| | 2 | 2 | 1 | |
| 1 | 0 | 1 | 4 | $_{x}$ |
- A. Four
- B. Five
- C. Six
- D. Seven
Answer: B. Five
4 + 2 = 6. In base 5, its 11. The 1 is written and the other 1 is carried forward.
2 + 2 = 4 and the carried forward is added making 5. 5 in base 5 is 10
4 + 2 = 6. In base 5, its 11. The 1 is written and the other 1 is carried forward.
2 + 2 = 4 and the carried forward is added making 5. 5 in base 5 is 10
3. A farmer left home at 4:35 am and arrived on his farm at 6:18 am. How long did he take to get to his farm?
- A. 1 hour 23 minutes
- B. 1 hour 43 minutes
- C. 2 hours 43 minutes
- D. 10 hours 53 minute
Answer: B. 1 hour 43 minutes
At 5:35, the farmer would had spent 1 hour
Remaining minutes to 6:00 is 25 minutes.
At 6:18 am, total minutes = 25 + 18 = 43 minutes.
Total time taken is 1 hour 43 minutes.
At 5:35, the farmer would had spent 1 hour
Remaining minutes to 6:00 is 25 minutes.
At 6:18 am, total minutes = 25 + 18 = 43 minutes.
Total time taken is 1 hour 43 minutes.
4. Express 2474.5 in standard form
- A. 2.4745 x 10$^{2}$
- B. 2.4745 x 10$^{3}$
- C. 2.4745 x 10$^{–2}$
- D. 2.4745 x 10$^{-3}$
Answer: B. 2.4745 x 10$^{3}$
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.
2474.5 = 2.⌒4⌒7⌒4.5 = 2.4745 x 10$^{3}$
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.
2474.5 = 2.⌒4⌒7⌒4.5 = 2.4745 x 10$^{3}$
5. Simplify 1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$.
- A. $\frac{1}{8}$
- B. $\frac{3}{8}$
- C. $\frac{3}{16}$
- D. $\frac{5}{16}$
Answer: A. $\frac{1}{8}$
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$
Method I
Solving the whole numbers and fractions differently.
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = 1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = [1 + 2 - 3]($\frac{1}{2}$ + $\frac{1}{4}$ - $\frac{5}{8}$)
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = [3 - 3]($\frac{4 x 1 + 2 x 1 - 1 x 5}{8}$)
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = [0]($\frac{4 + 2 - 5}{8}$)
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = $\frac{6 - 5}{8}$
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = $\frac{1}{8}$
Method II
Changing the mixed fractions to improper fractions.
1$\frac{1}{2}$ = $\frac{2 x 1 + 1}{2}$ = $\frac{2 + 1}{2}$ = $\frac{3}{2}$
2$\frac{1}{4}$ = $\frac{4 x 2 + 1}{4}$ = $\frac{8 + 1}{4}$ = $\frac{9}{4}$
3$\frac{5}{8}$ = $\frac{8 x 3 + 5}{8}$ = $\frac{24 + 5}{8}$ = $\frac{29}{8}$
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = 1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = $\frac{3}{2}$ + $\frac{9}{4}$ - $\frac{29}{8}$
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = $\frac{4 x 3 + 2 x 9 - 1 x 29}{8}$
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = $\frac{12 + 18 - 29}{8}$
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = $\frac{30 - 29}{8}$
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = $\frac{1}{8}$
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$
Method I
Solving the whole numbers and fractions differently.
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = 1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = [1 + 2 - 3]($\frac{1}{2}$ + $\frac{1}{4}$ - $\frac{5}{8}$)
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = [3 - 3]($\frac{4 x 1 + 2 x 1 - 1 x 5}{8}$)
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = [0]($\frac{4 + 2 - 5}{8}$)
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = $\frac{6 - 5}{8}$
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = $\frac{1}{8}$
Method II
Changing the mixed fractions to improper fractions.
1$\frac{1}{2}$ = $\frac{2 x 1 + 1}{2}$ = $\frac{2 + 1}{2}$ = $\frac{3}{2}$
2$\frac{1}{4}$ = $\frac{4 x 2 + 1}{4}$ = $\frac{8 + 1}{4}$ = $\frac{9}{4}$
3$\frac{5}{8}$ = $\frac{8 x 3 + 5}{8}$ = $\frac{24 + 5}{8}$ = $\frac{29}{8}$
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = 1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = $\frac{3}{2}$ + $\frac{9}{4}$ - $\frac{29}{8}$
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = $\frac{4 x 3 + 2 x 9 - 1 x 29}{8}$
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = $\frac{12 + 18 - 29}{8}$
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = $\frac{30 - 29}{8}$
1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = $\frac{1}{8}$
6. Find the next two numbers in the sequence 2, 5, 9, 14, 20, _ , _ .
- A. 26, 34
- B. 26, 35
- C. 27, 34
- D. 27, 35
Answer and explanation: practise this paper on ePrep.
7. The sum of three numbers is 28,542. Two of the numbers are 10,250 and 9,750. Find the third number.
- A. 8,452
- B. 8,542
- C. 9,452
- D. 9,542
Answer and explanation: practise this paper on ePrep.
8. 135 pencils were to be packed into boxes. Each box could take 12 pencils. Find the number of boxes that were fully packed.
- A. 10 boxes
- B. 11 boxes
- C. 12 boxes
- D. 13 boxes
Answer and explanation: practise this paper on ePrep.
9. Which of the fractions $\frac{13}{20}$, $\frac{3}{5}$, $\frac{3}{4}$ and $\frac{7}{10}$ is greatest?
- A. $\frac{3}{5}$
- B. $\frac{3}{4}$
- C. $\frac{7}{10}$
- D. $\frac{13}{20}$
Answer and explanation: practise this paper on ePrep.
10. Out of ₵550,000.00 given to a school, an amount of ₵325,000.00 was used. What fraction of the total amount was used?
- A. $\frac{4}{13}$
- B. $\frac{9}{13}$
- C. $\frac{9}{22}$
- D. $\frac{13}{22}$
Answer and explanation: practise this paper on ePrep.
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