2015 - WAEC Mathematics Past Questions and Answers - page 4
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 = \(\frac{H(T - 1)}{T}\)
correct option: a
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 = \(\frac{H(T^2 - 1)}{T^2 - T}\)
Users' Answers & CommentsTaking 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
correct option: d
Users' Answers & Comments33
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
correct option: a
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
\(\frac{4x - 5y + xy}{2xy}\)
correct option: d
\(\frac{3x - y}{xy} - \frac{2x + 3y}{2xy} + \frac{1}{2}\)
= \(\frac{2(3x - y) - 1(2x + 3y) + xy}{2xy}\)
= \(\frac{6x - 2y - 2x - 3y + xy}{2xy}\)
= \(\frac{4x - 5y + xy}{2xy}\)
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
\(\frac{4}{5}\)
correct option: b
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
\(\frac{2}{5}\)
Users' Answers & Commentsthen, \(\frac{2}{5}\)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
\(\frac{2}{5}\)
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
correct option: a
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
Users' Answers & CommentsR = 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
correct option: d
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%
Users' Answers & Commentsselling 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
correct option: a
Users' Answers & Comments39
Find the value of p if \(\frac{1}{4}\)p + 3q = 10 and 2p - q = 7
A
4
B
3
C
-3
D
-4
correct option: a
\(\frac{1}{4}\)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
Users' Answers & Comments2p - \(\frac{1}{3}\)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
correct option: b
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
Users' Answers & Comments= \(\frac{5 + 3 + 0 + 7 + 2 + 1}{6}\)
\(\frac{18}{6}\) = 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