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JAMB Mathematics Past Questions & Answers - Page 78

386.

Simplify 4 $\sqrt{27}$ + 5 $\sqrt{12}$ − 3 $\sqrt{75}$

− 7

− 7 $\sqrt{3}$

7 $\sqrt{3}$

Correct answer is D

4 $\sqrt{27}$ + 5 $\sqrt{12}$ − 3 $\sqrt{75}$

= 4 $\sqrt{3}$ × 9 + 5 $\sqrt{3}$ × 4 − 3 $\sqrt{3}$ × 25

= 4 × 3 $\sqrt{3}$ + 5 × 2 $\sqrt{3}$ − 3 × 5 $\sqrt{3}$

= 12 $\sqrt{3}$ + 10 $\sqrt{3}$ − 15 $\sqrt{3}$

= (12 + 10 − 15) $\sqrt{3}$

= 7 $\sqrt{3}$

387.

A cylindrical tank has a capacity of 3080m3. What is the depth of the tank if the diameter of its base is 14m? Take pi = 22/7.

23m

25m

20m

22m

View answer & explanation

Correct answer is C

Capacity = Volume = 3080m³

base diameter = 14m

radius = $\frac{\text{diameter}}{2}$

= 7m

Volume of Cylidner = Capacity of cylinder

πr²h = 3080

$\frac{22}{7}$ × 7 × 7 × h = 3080

h = $\frac{3080}{22 \times 7}$

h = 20m

388.

In a regular polygon, each interior angle doubles its corresponding exterior angle. Find the number of sides of the polygon

View answer & explanation

Correct answer is B

2x + x = 180^o

3x = 180^o

x = 60^o (exterior angle of the polygon)

angle = $\frac{\text{total angle}}{\text{number of sides}}$

60 = $\frac{360}{n}$

n = $\frac{360}{60}$

n = 6 sides