Mathematics 2018 Past Questions | WAEC
Study the following Mathematics past questions and answers for JAMB, WAEC NECO and Post-JAMB. Get prepared with official past questions and answers for upcoming examinations.
11. \(\frac{2 – 18m^2}{1 + 3m}\)
- A. \(2 (1 + 3m)\)
- B. \(2 (1 + 3m^2)\)
- C. \(2(1 – 3m)\)
- D. \(2(1 – 3m^2)\)
Correct Option: Answer is C
\(\frac{2 – 18m^2}{1 + 3m}\) = \(\frac{2(1 – 9)m^2}{1 + 3m}\)
= \(\frac{2(1 + 3m)(1 – 3m)}{1 + 3m}\)
= \(2(1 – 3m)\)
12. A curve is such that when y = 0, x = -2 or x = 3. Find the equation of the curve.
- A. y =
- B. y =
- C. y =
- D. y =
Correct Option: Answer is D
Since the curve cuts the x-axis at x = -2 and x = 3,
(x + 2)(x – 3) = 0
\(x^2 – 3x + 2x – 6\) = 0
\(x^2 – x – 6\) = 0
Hence, the equation of the curve is
y = \(x^2 – x – 6\)
13. The volume of a cylindrical tank, 10m high is 385 m2. Find the diameter of the tank. [Take π=22/7]
A. 14m
B. 10m
C. 7m
C. 5m
Correct Option: Answer is C
Volume of a cylinder = \( \pi r^2\)h
385 = \(\frac{22}{7}\) x \(r^2\) x 10
385 x 7 = 22 x \(r^2\) x 10
\(r^2\) = \(\frac{385 \times 7}{22 \times 10}\)
= 12.25
r = \(\sqrt{12.25}\)
= 3.5m
Hence, diameter of tank = 2r
= 2 x 3.5 = 7m
14. The surface area of a sphere is 792/7cm2. Find, correct to the nearest whole number, its volume. [Take π=22/7]
A. 113cm3
B. 131cm3
C. 311cm3
D. 414cm3
Correct Option: Answer is A
Surface area of a sphere = \(4 \pi r^2\)
\(4 \pi r^2\) = \(\frac{792}{7}cm^2\)
4 x \(\frac{22}{7}\) x \(r^2\) = \(\frac{792}{7}\)
\(r^2\) = \(\frac{792}{7}\) x \(\frac{7}{4 \times 22}\)
= 9
r = \(\sqrt{9}\)
= 3cm
Hence, volume of sphere
= \(\frac{4}{3} \pi r^3\)
= \(\frac{4}{3} \times \frac{22}{7} \times 3 \times 3 \times 3 \)
= \(\frac{4 \times 22 \times 9}{7}\)
\(\approx\) = 113.143
= 113\(cm^3\) (to the nearest whole number)
15. The angles of a polygon are x, 2x, 2x, (x + \(30^o\)), (x + \(20^o\)) and (x – \(10^o\)). Find the value of x
- A. \(45^o\)
- B. \(95^o\)
- C. \(84^o\)
- D. \(85^o\)
Correct Option: Answer is C
x + 2x + 2x + (x + \(30^o\)) + (x + \(20^o\)) + (x – \(10^o\)) = (2n – 4) x \(90^o\)
8x + 50 \(^o\) – 10\(^o\) = (2 x 6 -4) x 90\(^o\)
8x + 40\(^o\) = 8 x 90\(^o\) = 720\(^o\)
8x = 720\(^o\) – 40\(^o\) = 680\(^o\)
x = \(\frac{680^o}{8}\)
= 85\(^o\)
16. If M and N are the points (-3, 8) and (5, -7) respectively, find |MN|
- A. 8 units
- B. 11 units
- C. 15 units
- D. 17 units
Correct Option: Answer is D
|MN| = \(\sqrt{(x_1 – x_2)^2 + (y_1 – y_2)^2}\)
= \(\sqrt{(-3 -5)^2 + (8 – 7)^2}\)
= \(\sqrt{(-8)^2 + (8 + 7)^2}\)
= \(\sqrt{64 + (15)^2}\)
= \(\sqrt{64 + 225}\)
= \(\sqrt{289}\)
= 17 units
17. The equation of the line through the points (4,2) and (-8, -2) is 3y = px + q, where p and q are constants. Find the value of p.
- A. 1
- B. 2
- C. 3
- D. 9
Correct Option: Answer is A
Using the two – point from
\(\frac{y – y_1}{y_2 – y_1} = \frac{x – x_1}{x_2 – x_1}\)
\(\frac{y – 2}{-2 – 2} = \frac{x – 4}{-8 – 4}\)
\(\frac{y – 2}{-4} = \frac{x – 4}{-12}\)
\(\frac{-12(y -2)}{-4}\) = x – 4
3(y -2) = x -4
3y – 6 = x – 4
3y = x – 4 + 6
3y = x + 2…
By comparing the equations;
3y = px + , p = 1
18. The mean of 1, 3, 5, 7 and x is 4. Find the value of x
- A. 2
- B. 4
- C. 6
- D. 8
Correct Option: Answer is B
Mean = \(\frac{\sum x}{n}\)
4 = \(\frac{1 + 3 + 5 + 7 + x}{5}\)
4 x 5 = 16 + x
20 – 16 = x
4 = x
x = 4
19.
The table shows the distribution of goals scored by 25 teams in a football competition. Calculate the probability that a team selected at randon scored either 4 or 7 goals.
- A. 9/25
- B. 1/5
- C. 6/25
- D. /5
Correct Option: Answer is A
Probability (team scored 4 goals) = Probability (team scored 7 goals) = 3/25
Hence, probability that a team selected at random scored either 4 or 7 goals;
= 6/25+3/25
= 9/25
20.
The table shows the distribution of goals scored by 25 teams in a football competition. Calculate the probability that a team selected at random scored at most 3 goals.
- A. 3/25
- B. 1/5
- C. 6/25
- D. 2/5
Correct Option: Answer is D
Number of teams that scored at most 3 goals = 3 + 1 + 6 = 10
Hence, probability that a team selected at random scored at most 3 goals
= 10/25 = 2/5