Year : 
2015
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

31 - 40 of 45 Questions

# Question Ans
31.

Make K the subject of the relation T = \(\sqrt{\frac{TK - H}{K - H}}\)

A. K = \(\frac{H(T^2 - 1)}{T^2 - T}\)

B. K = \(\frac{HT}{(T - 1)^2}\)

C. K = \(\frac{H(T^2 + 1)}{T}\)

D. K = \(\frac{H(T - 1)}{T}\)

Detailed Solution

T = \(\sqrt{\frac{TK - H}{K - H}}\)

Taking the square of both sides, give

T2 = \(\frac{TK - H}{K - H}\)

T2(K - H) = TK - H

T2K - T2H = TK - H

T2K - TK = T2H - H

K(T2 - T) = H(T2 - 1)

K = \(\frac{H(T^2 - 1)}{T^2 - T}\)
32.

Which of the following is used to determine the mode of a grouped data?

A. Bar chart

B. Frequency polygon

C. Ogive

D. Histogram

D
33.

The area of a rhombus is 110 cm\(^2\). If the diagonals are 20 cm and (2x + 1) cm long, find the value of x.

A. 5.0

B. 4.0

C. 3.0

D. 2.5

Detailed Solution

Diagonal |AC| = (2x + 1)cm

In the diagram,area of \(\Delta\)ABC

is \(\frac{110}{2}\) = \(\frac{1}{2}\) x |AC| x |HB|

55 = \(\frac{1}{2}\) x (2x + 1) x 10

55 = (2x + 1)5

55 = 10x + 5

55 - 5 = 10x

50 = 10x

x = \(\frac{50}{10}\)

= 5.0
34.

Simplify: \(\frac{3x - y}{xy} - \frac{2x + 3y}{2xy} + \frac{1}{2}\)

A. \(\frac{4x + 5y - xy}{2xy}\)

B. \(\frac{5y - 4x + xy}{2xy}\)

C. \(\frac{5x + 4y - xy}{2xy}\)

D. \(\frac{4x - 5y + xy}{2xy}\)

Detailed Solution

\(\frac{3x - y}{xy} - \frac{2x + 3y}{2xy} + \frac{1}{2}\)

= \(\frac{2(3x - y) - 1(2x + 3y) + xy}{2xy}\)

= \(\frac{6x - 2y - 2x - 3y + xy}{2xy}\)

= \(\frac{4x - 5y + xy}{2xy}\)
35.

A farmer uses \(\frac{2}{5}\) of his land to grow cassava, \(\frac{1}{3}\) of the remaining for yam and the rest for maize. Find the part of the land used for maize

A. \(\frac{2}{15}\)

B. \(\frac{2}{5}\)

C. \(\frac{2}{3}\)

D. \(\frac{4}{5}\)

Detailed Solution

Let x represent the entire farmland

then, \(\frac{2}{5}\)x + \(\frac{1}{3}\)[x - \(\frac{2}{3}x\)] + M = x

Where M represents the part of the farmland used for growing maize, continuing

\(\frac{2}{5}\)x + \(\frac{1}{3}\)x [1 - \(\frac{2}{3}x\)] + M = x

\(\frac{2}{5}x + \frac{1}{3}\)x [\(\frac{3}{5}\)] + M = x

\(\frac{2}{5}\)x + \(\frac{1x}{5}\) + M = x

\(\frac{3x}{5} + M = x\)

M = x - \(\frac{2}{5}\)x

= x[1 - \(\frac{3}{5}\)]

= x[\(\frac{2}{5}\)] = \(\frac{2x}{5}\)

Hence the part of the land used for growing maize is
&
36.

The rate of consumption of petrol by a vehicle varies directly as the square of the distance covered. If 4 litres of petrol is consumed on a distance of 15km. how far would the vehicle go on 9 litres of petrol?

A. 22\(\frac{1}{2}\)km

B. 30km

C. 33\(\frac{1}{2}\)km

D. 45km

Detailed Solution

R \(\alpha\) D2

R = D2K

R = 4 Litres when D = 15cm

thus; 4 = 152k

4 = 225k

k = \(\frac{4}{225}\)

This gives R = \(\frac{4D^2}{225}\)

Where R = 9litres

equation gives

9 = \(\frac{4D^2}{225}\)

9 x 225 = 4d2

D2 = \(\frac{9 \times 225}{4}\)

D = \(\sqrt{9 \times 225}{4}\)

= \(\frac{3 \times 15}{2}\)

= 22\(\frac{1}{2}\)km
37.

A trader bought 100 oranges at 5 for N40.00 and 20 for N120.00. Find the profit or loss percent

A. 20% profit

B. 20% loss

C. 25% profit

D. 25% loss

Detailed Solution

Cost price CP of the 100 oranges = \(\frac{100}{5}\) x N40.00

selling price SP of the 100 oranges = \(\frac{100}{20}\) x N120

= N600.00

so, profit or loss per cent

= \(\frac{SP - CP}{CP}\) x 100%

= \(\frac{600 - 800}{800}\) x 100%

= \(\frac{-200}{800}\) x 100%

Hence, loss per cent = 25%
38.

Describe the shaded portion in the diagram

A. P' \(\cap\) Q \(\cap\) R'

B. (P' \(\cap\) R)' \(\cap\) R

C. P' \(\cap\) Q \(\cap\) R

D. (P \(\cap\) Q)' \(\cap\) R

A
39.

Find the value of p if \(\frac{1}{4}\)p + 3q = 10 and 2p - q = 7

A. 4

B. 3

C. -3

D. -4

Detailed Solution

\(\frac{1}{4}\)p + 3p = 10...(1)

2p - \(\frac{1}{3}\)q = 7...(2)

Multiply equation (2) by 3 to clear fraction

3 x 2p - 3 x \(\frac{1}{3}\)q = 3 x 7

6p - q = 21

6p - 21 = q....(3)

substituting 6p - 21 for q in (1)

\(\frac{1}{4}\)p + 3(6p - 21) = 10...(4)

Multiply equation (4) by 4 to clear fraction

4 x \(\frac{1}{4}\)p + 4 x 3(6p - 21) = 4 x 10

p + 12(6p - 21) = 40

p + 72p - 252

73p = 292

p = \(\frac{292}{73}\)

= 4
40.

Calculate the mean deviation of 5, 3, 0, 7, 2, 1

A. 0.0

B. 2.0

C. 2.5

D. 3.0

Detailed Solution

mean = \(\bar{x}\) = \(\frac{\sum x}{n}\)

= \(\frac{5 + 3 + 0 + 7 + 2 + 1}{6}\)

\(\frac{18}{6}\) = 3

\(\begin{array}{c|c}
x & x - \bar{x} & |x - \bar{x}|\\ \hline 5 & -2 & 2 \\ \hline 3 & 0 & 0 \\ \hline 0 & -3 & 3 \\ \hline 7 & 4 & 4 \\ \hline 2 & -1 & 1 \\ \hline 1 & -2 & 2 \\ \hline & & & \sum|x - \bar{x}| \\ \hline & & & 12
\end{array}\)

hence, MD = \(\frac{\sum|x - \bar{x}}{n}|\)

\(\frac{12}{6}\)

= 2
31.

Make K the subject of the relation T = \(\sqrt{\frac{TK - H}{K - H}}\)

A. K = \(\frac{H(T^2 - 1)}{T^2 - T}\)

B. K = \(\frac{HT}{(T - 1)^2}\)

C. K = \(\frac{H(T^2 + 1)}{T}\)

D. K = \(\frac{H(T - 1)}{T}\)

Detailed Solution

T = \(\sqrt{\frac{TK - H}{K - H}}\)

Taking the square of both sides, give

T2 = \(\frac{TK - H}{K - H}\)

T2(K - H) = TK - H

T2K - T2H = TK - H

T2K - TK = T2H - H

K(T2 - T) = H(T2 - 1)

K = \(\frac{H(T^2 - 1)}{T^2 - T}\)
32.

Which of the following is used to determine the mode of a grouped data?

A. Bar chart

B. Frequency polygon

C. Ogive

D. Histogram

D
33.

The area of a rhombus is 110 cm\(^2\). If the diagonals are 20 cm and (2x + 1) cm long, find the value of x.

A. 5.0

B. 4.0

C. 3.0

D. 2.5

Detailed Solution

Diagonal |AC| = (2x + 1)cm

In the diagram,area of \(\Delta\)ABC

is \(\frac{110}{2}\) = \(\frac{1}{2}\) x |AC| x |HB|

55 = \(\frac{1}{2}\) x (2x + 1) x 10

55 = (2x + 1)5

55 = 10x + 5

55 - 5 = 10x

50 = 10x

x = \(\frac{50}{10}\)

= 5.0
34.

Simplify: \(\frac{3x - y}{xy} - \frac{2x + 3y}{2xy} + \frac{1}{2}\)

A. \(\frac{4x + 5y - xy}{2xy}\)

B. \(\frac{5y - 4x + xy}{2xy}\)

C. \(\frac{5x + 4y - xy}{2xy}\)

D. \(\frac{4x - 5y + xy}{2xy}\)

Detailed Solution

\(\frac{3x - y}{xy} - \frac{2x + 3y}{2xy} + \frac{1}{2}\)

= \(\frac{2(3x - y) - 1(2x + 3y) + xy}{2xy}\)

= \(\frac{6x - 2y - 2x - 3y + xy}{2xy}\)

= \(\frac{4x - 5y + xy}{2xy}\)
35.

A farmer uses \(\frac{2}{5}\) of his land to grow cassava, \(\frac{1}{3}\) of the remaining for yam and the rest for maize. Find the part of the land used for maize

A. \(\frac{2}{15}\)

B. \(\frac{2}{5}\)

C. \(\frac{2}{3}\)

D. \(\frac{4}{5}\)

Detailed Solution

Let x represent the entire farmland

then, \(\frac{2}{5}\)x + \(\frac{1}{3}\)[x - \(\frac{2}{3}x\)] + M = x

Where M represents the part of the farmland used for growing maize, continuing

\(\frac{2}{5}\)x + \(\frac{1}{3}\)x [1 - \(\frac{2}{3}x\)] + M = x

\(\frac{2}{5}x + \frac{1}{3}\)x [\(\frac{3}{5}\)] + M = x

\(\frac{2}{5}\)x + \(\frac{1x}{5}\) + M = x

\(\frac{3x}{5} + M = x\)

M = x - \(\frac{2}{5}\)x

= x[1 - \(\frac{3}{5}\)]

= x[\(\frac{2}{5}\)] = \(\frac{2x}{5}\)

Hence the part of the land used for growing maize is
&
36.

The rate of consumption of petrol by a vehicle varies directly as the square of the distance covered. If 4 litres of petrol is consumed on a distance of 15km. how far would the vehicle go on 9 litres of petrol?

A. 22\(\frac{1}{2}\)km

B. 30km

C. 33\(\frac{1}{2}\)km

D. 45km

Detailed Solution

R \(\alpha\) D2

R = D2K

R = 4 Litres when D = 15cm

thus; 4 = 152k

4 = 225k

k = \(\frac{4}{225}\)

This gives R = \(\frac{4D^2}{225}\)

Where R = 9litres

equation gives

9 = \(\frac{4D^2}{225}\)

9 x 225 = 4d2

D2 = \(\frac{9 \times 225}{4}\)

D = \(\sqrt{9 \times 225}{4}\)

= \(\frac{3 \times 15}{2}\)

= 22\(\frac{1}{2}\)km
37.

A trader bought 100 oranges at 5 for N40.00 and 20 for N120.00. Find the profit or loss percent

A. 20% profit

B. 20% loss

C. 25% profit

D. 25% loss

Detailed Solution

Cost price CP of the 100 oranges = \(\frac{100}{5}\) x N40.00

selling price SP of the 100 oranges = \(\frac{100}{20}\) x N120

= N600.00

so, profit or loss per cent

= \(\frac{SP - CP}{CP}\) x 100%

= \(\frac{600 - 800}{800}\) x 100%

= \(\frac{-200}{800}\) x 100%

Hence, loss per cent = 25%
38.

Describe the shaded portion in the diagram

A. P' \(\cap\) Q \(\cap\) R'

B. (P' \(\cap\) R)' \(\cap\) R

C. P' \(\cap\) Q \(\cap\) R

D. (P \(\cap\) Q)' \(\cap\) R

A
39.

Find the value of p if \(\frac{1}{4}\)p + 3q = 10 and 2p - q = 7

A. 4

B. 3

C. -3

D. -4

Detailed Solution

\(\frac{1}{4}\)p + 3p = 10...(1)

2p - \(\frac{1}{3}\)q = 7...(2)

Multiply equation (2) by 3 to clear fraction

3 x 2p - 3 x \(\frac{1}{3}\)q = 3 x 7

6p - q = 21

6p - 21 = q....(3)

substituting 6p - 21 for q in (1)

\(\frac{1}{4}\)p + 3(6p - 21) = 10...(4)

Multiply equation (4) by 4 to clear fraction

4 x \(\frac{1}{4}\)p + 4 x 3(6p - 21) = 4 x 10

p + 12(6p - 21) = 40

p + 72p - 252

73p = 292

p = \(\frac{292}{73}\)

= 4
40.

Calculate the mean deviation of 5, 3, 0, 7, 2, 1

A. 0.0

B. 2.0

C. 2.5

D. 3.0

Detailed Solution

mean = \(\bar{x}\) = \(\frac{\sum x}{n}\)

= \(\frac{5 + 3 + 0 + 7 + 2 + 1}{6}\)

\(\frac{18}{6}\) = 3

\(\begin{array}{c|c}
x & x - \bar{x} & |x - \bar{x}|\\ \hline 5 & -2 & 2 \\ \hline 3 & 0 & 0 \\ \hline 0 & -3 & 3 \\ \hline 7 & 4 & 4 \\ \hline 2 & -1 & 1 \\ \hline 1 & -2 & 2 \\ \hline & & & \sum|x - \bar{x}| \\ \hline & & & 12
\end{array}\)

hence, MD = \(\frac{\sum|x - \bar{x}}{n}|\)

\(\frac{12}{6}\)

= 2