WASSCE 2021 Mathematics Past Questions and Answers
The WASSCE 2021 Mathematics paper has 50 objective and 13 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. Correct 0.00798516 to three significant figures.
- A. 0.0109
- B. 0.0800
- C. 0.00799
- D. 0.008
Answer: C. 0.00799
Significant figures, any of the digits of a number beginning with the digit farthest to the left that is not zero and ending with the last digit farthest to the right that is either not zero or that is a zero but is considered to be exact.
The significant figures in 0.00798516 are 7,9,8,5,1 and 6.
The third significant figure is 8.
If the significant figure next to the position of the significant figure you seek is 5 or more, you add 1 to the last significant figures you seek.
The next significant to the third (8) is 5, hence 1 is added to the 8, making 9.
Significant figures, any of the digits of a number beginning with the digit farthest to the left that is not zero and ending with the last digit farthest to the right that is either not zero or that is a zero but is considered to be exact.
The significant figures in 0.00798516 are 7,9,8,5,1 and 6.
The third significant figure is 8.
If the significant figure next to the position of the significant figure you seek is 5 or more, you add 1 to the last significant figures you seek.
The next significant to the third (8) is 5, hence 1 is added to the 8, making 9.
2. Simplify: (11$_{two}$)$^{2}$
- A. 1001$_{two}$
- B. 1101$_{two}$
- C. 101$_{two}$
- D. 10001$_{two}$
Answer: A. 1001$_{two}$
Method I
A number raised to the power 2 (square) means the number multiplied by itself.
| 1 | 1 | | |
|---|---|---|---|
| | | 1 | 1 |
| | x | 1 | 1 |
| | | 1 | 1 |
| + | 1 | 1 | |
| 1 | 0 | 0 | 1 |
Note: 2 in base two is 10.
Method II
Change the binary (base two) to base 10 and simplify and change the result back to base two.
11$_{two}$ = 1 x 2$^{1}$ + 1 x 2$^{0}$
11$_{two}$ = 1 x 2 + 1 x 1
Note: any number raised to the power 0 is 1.
11$_{two}$ = 2 + 1 = 3
(11$_{two}$)$^{2}$ = 3$^{2}$ = 3 x 3 = 9
Change the 9 base 10 to base 2.
9 base 10 to base 2
$\begin{matrix} & 9 \\ 2 & 4 r 1 \\ 2 & 2 r 0 \\ 2 & 1 r 0 \\ 2 & 0 r 1\end{matrix}$
You list the remainders from the bottom to the top.
9$_{ten}$ = 1001$_{two}$
Method I
A number raised to the power 2 (square) means the number multiplied by itself.
| 1 | 1 | | |
|---|---|---|---|
| | | 1 | 1 |
| | x | 1 | 1 |
| | | 1 | 1 |
| + | 1 | 1 | |
| 1 | 0 | 0 | 1 |
Note: 2 in base two is 10.
Method II
Change the binary (base two) to base 10 and simplify and change the result back to base two.
11$_{two}$ = 1 x 2$^{1}$ + 1 x 2$^{0}$
11$_{two}$ = 1 x 2 + 1 x 1
Note: any number raised to the power 0 is 1.
11$_{two}$ = 2 + 1 = 3
(11$_{two}$)$^{2}$ = 3$^{2}$ = 3 x 3 = 9
Change the 9 base 10 to base 2.
9 base 10 to base 2
$\begin{matrix} & 9 \\ 2 & 4 r 1 \\ 2 & 2 r 0 \\ 2 & 1 r 0 \\ 2 & 0 r 1\end{matrix}$
You list the remainders from the bottom to the top.
9$_{ten}$ = 1001$_{two}$
3. Solve: ${2}^{\sqrt{2x + 1}}$ = 32
- A. 13
- B. 24
- C. 12
- D. 11
Answer: C. 12
Before you can simplify a power in an equation, ensure that both the left and the right have the same base.
Express the 32 to base 2.
32 = 2 x 2 x 2 x 2 x 2 = 2$^{5}$
Solve: ${2}^{\sqrt{2x + 1}}$ = 2$^{5}$
Since the bases are the same, you can equate the powers.
$\sqrt{2x + 1}$ = 5
Get rid of the square root by squaring both sides of the equation.
2*x* + 1 = 5$^{2}$
2*x* + 1 = 25
2*x* = 25 - 1
2*x* = 24
Divide both sides by 2.
x = $\frac{24}{2}$ = 12
Before you can simplify a power in an equation, ensure that both the left and the right have the same base.
Express the 32 to base 2.
32 = 2 x 2 x 2 x 2 x 2 = 2$^{5}$
Solve: ${2}^{\sqrt{2x + 1}}$ = 2$^{5}$
Since the bases are the same, you can equate the powers.
$\sqrt{2x + 1}$ = 5
Get rid of the square root by squaring both sides of the equation.
2*x* + 1 = 5$^{2}$
2*x* + 1 = 25
2*x* = 25 - 1
2*x* = 24
Divide both sides by 2.
x = $\frac{24}{2}$ = 12
4. If log$_{10}$ 2 = m and log$_{10}$ 3 = n, find log$_{10}$ 24 in terms of m and n.
- A. 3*m* + n
- B. m + 3*n*
- C. 4*mn*
- D. 3*mn*
Answer: A. 3*m* + n
24 = 8 x 3
8 = 2 x 2 x 2 = 2$^{3}$
24 = 2$^{3}$ x 3
log$_{10}$ $^{24}$ = log$_{10}$ 2$^{3}$ x 3
From law of logarithm,
1. log a$^{b}$ = b*log *a
2. log a x b = log a + log b
log$_{10}$ 2$^{3}$ x 3 = 3log$_{10}$ 2 x 3
3log$_{10}$ 2 x 3 = 3log$_{10}$ 2 + log$_{10}$ 3
But log$_{10}$ 2 = m and log$_{10}$ 3 = n
3log$_{10}$ 2 + log$_{10}$ 3 = 3 x m + n
3log$_{10}$ 2 + log$_{10}$ 3 = 3*m* + n
24 = 8 x 3
8 = 2 x 2 x 2 = 2$^{3}$
24 = 2$^{3}$ x 3
log$_{10}$ $^{24}$ = log$_{10}$ 2$^{3}$ x 3
From law of logarithm,
1. log a$^{b}$ = b*log *a
2. log a x b = log a + log b
log$_{10}$ 2$^{3}$ x 3 = 3log$_{10}$ 2 x 3
3log$_{10}$ 2 x 3 = 3log$_{10}$ 2 + log$_{10}$ 3
But log$_{10}$ 2 = m and log$_{10}$ 3 = n
3log$_{10}$ 2 + log$_{10}$ 3 = 3 x m + n
3log$_{10}$ 2 + log$_{10}$ 3 = 3*m* + n
5. Find the 5$^{th}$ term of the sequence 2, 5, 10, 17....?
- A. 22
- B. 24
- C. 36
- D. 26
Answer: D. 26
The rule of the sequence is n$^{2}$ + 1
Where n is the number of term.
When n = 1 → 1$^{2}$ + 1 = 1 + 1 = 2
When n = 2 → 2$^{2}$ + 1 = 4 + 1 = 5
When n = 3 → 3$^{2}$ + 1 = 9 + 1 = 10
When n = 4 → 4$^{2}$ + 1 = 16 + 1 = 17
When n = 5 → 5$^{2}$ + 1 = 25 + 1 = 26
The rule of the sequence is n$^{2}$ + 1
Where n is the number of term.
When n = 1 → 1$^{2}$ + 1 = 1 + 1 = 2
When n = 2 → 2$^{2}$ + 1 = 4 + 1 = 5
When n = 3 → 3$^{2}$ + 1 = 9 + 1 = 10
When n = 4 → 4$^{2}$ + 1 = 16 + 1 = 17
When n = 5 → 5$^{2}$ + 1 = 25 + 1 = 26
6. If P = {-3 < x < 1} and Q = {-1 < x <3}, where x is a real number, find P ∩ Q.
- A. {-1 ≤ x ≤ 1}
- B. {-3 < x < 1}
- C. {-3 ≤ x ≤ 1}
- D. {-1 < x < 1}
Answer and explanation: practise this paper on ePrep.
7. Factorize 6*pq* - 3*rs* - 3*ps* + 6*qr*.
- A. 3(r - p)(s - 2*q*)
- B. 3(r - p)(2*q* + s)
- C. 3(p - r)(2*q* - s)
- D. 3(p + r)(2*q* - s)
Answer and explanation: practise this paper on ePrep.
8. What number should be subtracted from the sum of 2$\frac{1}{6}$ and 2$\frac{7}{12}$ to give 3$\frac{1}{4}$?
- A. 1$\frac{1}{2}$
- B. 1$\frac{1}{6}$
- C. $\frac{1}{2}$
- D. $\frac{1}{3}$
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9. Mensah is 5 years old and Joyce is thrice as old as Mensah. In how many years will Joyce be twice as old as Mensah?
- A. 3 years
- B. 10 years
- C. 5 years
- D. 15 years
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10. If 16 x 2$^{(x + 1)}$ = 4$^{x}$ x 8$^{(1 − x)}$, find the value of x.
- A. -4
- B. 4
- C. 1
- D. -1
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