WASSCE 2019 Mathematics Past Questions and Answers
The WASSCE 2019 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. Express, correct to three significant figures, 0.003597.
- A. 0.00359
- B. 0.00360
- C. 0.004
- D. 0.359
Answer: B. 0.00360
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 are 3597 and the third significant figure is 9. The number next to the third significant is more than 4, hence 1 is added to the third significant figure resulting in 0.00360.
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 are 3597 and the third significant figure is 9. The number next to the third significant is more than 4, hence 1 is added to the third significant figure resulting in 0.00360.
2. Evaluate ${(0.064)}^{-\frac{1}{3}}$.
- A. -$\frac{5}{2}$
- B. -$\frac{2}{5}$
- C. $\frac{2}{5}$
- D. $\frac{5}{2}$
Answer: D. $\frac{5}{2}$
Express the decimal in the form a x 10$^{b}$ where a is a whole number
0.064 = 0.⌒0⌒6⌒4 = 64 x 10$^{-3}$
Note: the power is negative since the decimal is moved to the right.
${(0.064)}^{-\frac{1}{3}}$ = ${(64 x {10}^{-3})}^{-\frac{1}{3}}$
64 = 4$^{3}$
${(0.064)}^{-\frac{1}{3}}$ = ${({4}^{3}x{10}^{-3})}^{-\frac{1}{3}}$
The power outside affects each of the powers in the bracket.
${(0.064)}^{-\frac{1}{3}}$ = ${4}^{3 x -\frac{1}{3}}x{10}^{-3 x -\frac{1}{3}}$
Notes:
1. The 3 cancel each other.
2. -1 x -1 = 1
${(0.064)}^{-\frac{1}{3}}$ = 4$^{-1}$ x 10$^{1}$
Note:
a$^{-b}$ = $\frac{1}{{a}^{b}}$
4$^{-1}$ = $\frac{1}{{4}^{1}}$ = $\frac{1}{4}$
${(0.064)}^{-\frac{1}{3}}$ = $\frac{1}{4}$ x 10 = $\frac{5}{2}$
Express the decimal in the form a x 10$^{b}$ where a is a whole number
0.064 = 0.⌒0⌒6⌒4 = 64 x 10$^{-3}$
Note: the power is negative since the decimal is moved to the right.
${(0.064)}^{-\frac{1}{3}}$ = ${(64 x {10}^{-3})}^{-\frac{1}{3}}$
64 = 4$^{3}$
${(0.064)}^{-\frac{1}{3}}$ = ${({4}^{3}x{10}^{-3})}^{-\frac{1}{3}}$
The power outside affects each of the powers in the bracket.
${(0.064)}^{-\frac{1}{3}}$ = ${4}^{3 x -\frac{1}{3}}x{10}^{-3 x -\frac{1}{3}}$
Notes:
1. The 3 cancel each other.
2. -1 x -1 = 1
${(0.064)}^{-\frac{1}{3}}$ = 4$^{-1}$ x 10$^{1}$
Note:
a$^{-b}$ = $\frac{1}{{a}^{b}}$
4$^{-1}$ = $\frac{1}{{4}^{1}}$ = $\frac{1}{4}$
${(0.064)}^{-\frac{1}{3}}$ = $\frac{1}{4}$ x 10 = $\frac{5}{2}$
3. Solve: $\frac{y + 1}{2}$ - $\frac{2y - 1}{3}$ = 4
- A. y = 29
- B. y = -29
- C. y = -19
- D. y = 19
Answer: C. y = -19
$\frac{y + 1}{2}$ - $\frac{2y - 1}{3}$ = 4
Get rid of the fractions by multiplying both sides by the L.C.M of the denominators 2 and 3. Th L.C.M is 6
6 x $\frac{y + 1}{2}$ - $\frac{2y - 1}{3}$ x 6 = 4 x 6
3(y + 1) - 2(2*y* - 1) = 24
3*y* + 3 - 4*y* + 2 = 24
3*y* - 4*y* + 5 = 24
-y = 24 - 5
-y = 19
Divide both sids by -1
y = $\frac{19}{-1}$ = -19
$\frac{y + 1}{2}$ - $\frac{2y - 1}{3}$ = 4
Get rid of the fractions by multiplying both sides by the L.C.M of the denominators 2 and 3. Th L.C.M is 6
6 x $\frac{y + 1}{2}$ - $\frac{2y - 1}{3}$ x 6 = 4 x 6
3(y + 1) - 2(2*y* - 1) = 24
3*y* + 3 - 4*y* + 2 = 24
3*y* - 4*y* + 5 = 24
-y = 24 - 5
-y = 19
Divide both sids by -1
y = $\frac{19}{-1}$ = -19
4. Simplify, correct to three significant figures, (27.63)$^{2}$ - (12.37)$^{2}$.
- A. 610
- B. 611
- C. 612
- D. 614
Answer: A. 610
a$^{2}$ - b$^{2}$ = (a + b)(a - b) known as difference of two squares.
(27.63)$^{2}$ - (12.37)$^{2}$ = (27.63 + 12.37)(27.63 - 12.37)
(27.63)$^{2}$ - (12.37)$^{2}$ = (40)(15.26)
(27.63)$^{2}$ - (12.37)$^{2}$ = 610.4 ≈ 610 (three significant figures)
Note: 0 in the middle or end of a number is significant.
a$^{2}$ - b$^{2}$ = (a + b)(a - b) known as difference of two squares.
(27.63)$^{2}$ - (12.37)$^{2}$ = (27.63 + 12.37)(27.63 - 12.37)
(27.63)$^{2}$ - (12.37)$^{2}$ = (40)(15.26)
(27.63)$^{2}$ - (12.37)$^{2}$ = 610.4 ≈ 610 (three significant figures)
Note: 0 in the middle or end of a number is significant.
5. If 7 + y = 4(mod 8), find the least value of y. 10 ≤ y ≤ 30.
- A. 21
- B. 19
- C. 13
- D. 11
Answer: C. 13
4 (mod 8) = 8 x 1 + 4 = 12
Since y is greater than or equal to 10 (10 ≤ y, thus 10 is less than or equal to y), the minimum value should be 17 (7 + 10)
4 (mod 8) = 8 x 2 + 4 = 20
20 is more than the minimum value (17)
7 + y = 20
y = 20 - 7
y = 13
4 (mod 8) = 8 x 1 + 4 = 12
Since y is greater than or equal to 10 (10 ≤ y, thus 10 is less than or equal to y), the minimum value should be 17 (7 + 10)
4 (mod 8) = 8 x 2 + 4 = 20
20 is more than the minimum value (17)
7 + y = 20
y = 20 - 7
y = 13
6. If T = {prime numbers} and M = {odd numbers} are subset of µ = {x : 0 < x ≤ 10, and x is an integer}, find ( T$^{'}$ ∩ M$^{'}$ )
- A. {1,2,3,5,7,8,9}
- B. {1,2,4,6,8,10}
- C. {1,4,6,8,10}
- D. {4,6,8,10}
Answer and explanation: practise this paper on ePrep.
7. Evaluate:
| log$_{3}$ 9 - log$_{2}$ 8 |
|---|
| log$_{3}$ 9 |
| log$_{3}$ 9 - log$_{2}$ 8 |
|---|
| log$_{3}$ 9 |
- A. -$\frac{1}{2}$
- B. $\frac{1}{3}$
- C. $\frac{1}{2}$
- D. -$\frac{1}{3}$
Answer and explanation: practise this paper on ePrep.
8. If 23$_{y}$ = 1111$_{two}$, find the value of y.
- A. 7
- B. 6
- C. 5
- D. 4
Answer and explanation: practise this paper on ePrep.
9. If 6, p and 14 are consecutive terms in Arithmetic Progression (A.P), find the value of p.
- A. 8
- B. 6
- C. 10
- D. 9
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10. Evaluate : 2$\sqrt{28}$ - 3$\sqrt{50}$ + $\sqrt{72}$
- A. 4$\sqrt{7}$ + $\sqrt{2}$
- B. 4$\sqrt{7}$ - 9$\sqrt{2}$
- C. 4$\sqrt{7}$ - 11$\sqrt{2}$
- D. 4$\sqrt{7}$ - 21$\sqrt{2}$
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