# 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 = ${\mathrm{x\xb2}}^{}-5x-6$**B.**y = ${\mathrm{x\xb2}}^{}+5x-6$**C.**y = ${\mathrm{x\xb2}}^{}+x-6$**D.**y = ${\mathrm{x\xb2}}^{}-x-6$

**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.**$\frac{2}{}$/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