WAEC Mathematics Past Questions & Answers - Page 145

721.

Find the quadratic equation whose roots are c and -c

A.

x2 - c2 = 0

B.

x2 + 2cx = 0

C.

x2 + 2cx + c2 = 0

D.

x2 - 2cx + c2 = 0

Correct answer is A

Explanation
Roots; x and -c

sum of roots = c + (-c) = 0

product of roots = c x -c = -c2

Equation; x2 - (sum of roots) x = product of roots = 0

x2 - (0)x + (-c2) = 0

x2 - c2 = 0

722.

If p = \(\frac{1}{2}\) and \(\frac{1}{p - 1} = \frac{2}{p + x}\), find the value of x

A.

-2\(\frac{1}{2}\)

B.

-1\(\frac{1}{2}\)

C.

1\(\frac{1}{2}\)

D.

2\(\frac{1}{2}\)

Correct answer is B

p = \(\frac{1}{2}; \frac{1}{p - 1} = \frac{2}{p + x}\)

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

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

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

-2 = \(\frac{4}{1 + 2x} -2(1 + 2x) = 4\)

1 + 2x = \(\frac{4}{-2}\)

1 + 2x = -2

2x = -2 - 1

2x = -3

x = -\(\frac{3}{2}\)

x = -1\(\frac{1}{2}\)

723.

XY is a chord of circle centre O and radius 7cm. The chord XY which is 8cm long subtends an angle of 120o at the centre of the circle. Calculate the perimeter of the minor segment. [Take \(\pi = \frac{22}{7}\)]

A.

14.67cm

B.

22.67cm

C.

29.33cm

D.

37.33cm

Correct answer is B

perimeter of minor segment = Length of arc xy + chord xy

where lxy = \(\frac{120}{360} \times 2x \times \frac{22}{7} \times 7\)

= 14.67cm

perimeter of minor segment = 14.67 + 8 = 22.67cm

724.

What is the length of an edge of a cube whose total surface area is X cm2 and whose total surface area is \(\frac{X}{2}\)cm3?

A.

3

B.

6

C.

9

D.

12

Correct answer is A

Total surface area of cube = 6s2

6s2 = x

s2 = x = \(\frac{x}{6}\).....(1)

volume of a cube = s2 = \(\frac{x}{2}\)

s2 = \(\frac{x}{2}\)......(2)

put(1) into (2)

s(\(\frac{x}{6}\)) = \(\frac{x}{2}\)

s = \(\frac{x}{2} \times \frac{6}{x}\)

= 3cm

725.

An arc of a circle, radius 14cm, is 18.33cm long. Calculate to the nearest degree, the angle which the arc subtends at the centre of the circle. [T = \(\frac{22}{7}\)]

A.

11o

B.

20o

C.

22o

D.

75o

Correct answer is D

Length of an arc = \(\frac{\theta}{360} \times 2\pi r\)

18.33 = \(\frac{\theta}{360} \times 2 \times \frac{22}{7} \times 14\)

\(\theta = \frac{18.33 \times 360 \times 17}{2 \times 22 \times 14}\)

= 75o (approx.)