2018 - WAEC Mathematics Past Questions and Answers - page 2

11

The volume of a cylindrical tank, 10m high is 385 m\(^2\). Find the diameter of the tank. [Take \(\pi = \frac{22}{7}\)]

A
14m
B
10m
C
7m
D
C. 5m
correct option: 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

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12

The surface area of a sphere is \(\frac{792}{7} cm^2\). Find, correct to the nearest whole number, its volume. [Take \(\pi = \frac{22}{7}\)]

A
113\(cm^3\)
B
131\(cm^3\)
C
311\(cm^3\)
D
414\(cm^3\)
correct option: 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)

 

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13

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
A. \(95^o\)
C
B. \(84^o\)
D
C. \(85^o\)
correct option: 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\(^o\)

 

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14

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: 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

 

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15

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: 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

 

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16

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: 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

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17

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
\(\frac{9}{25}\)
B
\(\frac{1}{5}\)
C
\(\frac{6}{25}\)
D
\(\frac{2}{5}\)
correct option: a

Prob. (team scored 4 goals) = Prob. (team scored 7 goals) = \(\frac{3}{25}\)

Hence, probability that a team selected at random scored either 4 or 7 goals;

= \(\frac{6}{25} + \frac{3}{25}\)

= \(\frac{9}{25}\)

 

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18

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
\(\frac{3}{25}\)
B
\(\frac{1}{5}\)
C
\(\frac{6}{25}\)
D
\(\frac{2}{5}\)
correct option: d

No. 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

= \(\frac{10}{25}\) = \(\frac{2}{5}\)

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19

The total surface area of a hemispher is 75\(\pi cm^2\). Find the radius.

A
5.0 cm
B
7.0 cm
C
8.5 cm
D
12.0 cm
correct option: a

Total surface area of hemisphere is

3\(\pi r^2\) = 75\(\pi cm^2\)

\(r^2\) = \(\frac{75 \pi}{3 \pi}\)

\(r^2\) = 25

r = \(\sqrt{25}\)

r = 5cm

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20

Find the value of x for which \(\frac{x - 5}{x(x - 1)}\) is defined

A
0 or 5
B
-5 or 5
C
-11 or 5
D
0 or 1
correct option: a

The expression \(\frac{x - 5}{x(x - 1)}\) is defined whten

x(x - 5) = 0.38

either x = 0 or x - 5 = 0

Hence, x = 0 or x = 5

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