WAEC Mathematics Past Questions & Answers - Page 78

386.

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 answer is D

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

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

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

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

387.

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 answer is 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

388.

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 answer is D

No explanation has been provided for this answer.

389.

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 answer is 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 = \(\frac{H(T^2 - 1)}{T^2 - T}\)

390.

In the given diagram, \(\bar{QT}\) and \(\bar{PR}\) are straight lines, < ROS = (3n - 20), < SOT = n, < POL = m and < QOL is a right angle. Find the value of n.

A.

35o

B.

40o

C.

55o

D.

60o

Correct answer is A

In the diagram, QOR + 2m(vertically opposite angles)

So, m + 90° + 2m = 180°

(angles on str. line)

3m = 180° - 90°

3m = 90°

m = \(\frac{90^o}{3}\)

= 30°

substituting 30° for m in

2m + 4n = 200° gives

2 x 30° + 4n = 200°

60° + 4n = 200°

4n = 200° - 60°

= 140°

n = \(\frac{140°}{4}\)

= 35°