WAEC Further Mathematics Past Questions & Answers - Page 109

541.

Age in years 10 - 14 15 - 19 20 - 24 25 - 29 30 - 34
Frequency 6 8 14 10 12

 

In which group is the upper quartile?

A.

15 - 19

B.

20 - 24

C.

25 - 29

D.

30 - 34

Correct answer is C

The upper quartile occurs at the \(\frac{3N}{4}\) position.

= \(\frac{3 \times 50}{4} = 37.5\)

This occurs in the class 25 - 29.

542.

Age in years 10 - 14 15 - 19 20 - 24 25 - 29 30 - 34
Frequency 6 8 14 10 12

What is the class mark of the median class?

A.

17

B.

22

C.

27

D.

32

Correct answer is B

\(\text{Total frequency} = 50\)

Median occurs at \(\frac{50}{2} = \text{25th position}\)

= Class 20 - 24 with classmark \(\frac{20 + 24}{2} = 22\)

543.

The radius of a sphere is increasing at a rate \(3cm s^{-1}\). Find the rate of increase in the surface area, when the radius is 2cm. 

A.

\(8\pi cm^{2}s^{-1}\)

B.

\(16\pi cm^{2}s^{-1}\)

C.

\(24\pi cm^{2}s^{-1}\)

D.

\(48\pi cm^{2}s^{-1}\)

Correct answer is D

Surface area of sphere, \( A = 4\pi r^{2}\)

\(\frac{\mathrm d A}{\mathrm d r} = 8\pi r\)

The rate of change of radius with time \(\frac{\mathrm d r}{\mathrm d t} = 3cm s^{-1}\)

\(\frac{\mathrm d A}{\mathrm d t} = (\frac{\mathrm d A}{\mathrm d r})(\frac{\mathrm d r}{\mathrm d t})\)

= \(8\pi \times 2cm \times 3cm s^{-1} = 48\pi cm^{2}s^{-1}\)

544.

A function is defined by \(h : x \to 2 - \frac{1}{2x - 3}, x \neq \frac{3}{2}\). Find \(h^{-1}(\frac{1}{2})\).

A.

\(6\)

B.

\(\frac{11}{6}\)

C.

\(\frac{11}{4}\)

D.

\(\frac{5}{3}\)

Correct answer is B

\(h : x \to 2 - \frac{1}{2x - 3}\)

\(h(x) = \frac{2(2x - 3) - 1}{2x - 3} = \frac{4x - 7}{2x - 3}\)

Let x = h(y)

\(x = \frac{4y - 7}{2y - 3}\)

\(x(2y - 3) = 4y - 7  \implies 2xy - 4y = 3x - 7\)

\(y = \frac{3x - 7}{2x - 4}\)

\(h^{-1}(x) = \frac{3x - 7}{2x - 4}\)

\(\therefore  h^{-1}(\frac{1}{2}) = \frac{3(\frac{1}{2}) - 7}{2(\frac{1}{2}) - 4}\)

= \(\frac{\frac{-11}{2}}{-3} = \frac{11}{6}\)

545.

A function is defined by \(h : x \to 2 - \frac{1}{2x - 3}, x \neq \frac{3}{2}\). Find \(h^-1\), the inverse of h.

A.

\(\frac{3x - 4}{2x - 7}, x \neq \frac{7}{2}\)

B.

\(\frac{3x - 7}{2x - 4}, x \neq 2\)

C.

\(\frac{2x - 7}{4x - 3}, x \neq \frac{3}{4}\)

D.

\(\frac{4x - 7}{2x - 4}, x \neq 2\)

Correct answer is B

\(h : x \to 2 - \frac{1}{2x - 3}\)

\(h(x) = \frac{2(2x - 3) - 1}{2x - 3} = \frac{4x - 7}{2x - 3}\)

Let x = h(y)

\(x = \frac{4y - 7}{2y - 3}\)

\(x(2y - 3) = 4y - 7  \implies 2xy - 4y = 3x - 7\)

\(y = \frac{3x - 7}{2x - 4}\)

\(h^{-1}(x) = \frac{3x - 7}{2x - 4}\)