WAEC Further Mathematics Past Questions & Answers

51.

Find correct to the nearest degree, the acute angle formed by the lines y = 2x + 5 and 2y = x - 6

76 $^∘$

53 $^∘$

37 $^∘$

14 $^∘$

View answer & explanation

Correct answer is C

tanθ

m1 - m2

1 + m1m2

y = 2x + 5
m1 = 2
2y = x - 6

tanθ

2 -

$\frac{1}{2}$

1+2(

$\frac{1}{2}$ )

tanθ = $\frac{3}{2}$ ÷ (1+1)

tanθ = $\frac{3}{2}$ ÷ 2

tanθ

θ = $tan^{-1} (\frac{3}{4})$

θ = 36.87º
θ = 37º

52.

The mean heights of three groups of students consisting of 20, 16 and 14 students each are 1.67m, 1.50m and 1.40m respectively. Find the mean height of all the students.

1.63m

1.54m

1.52m

1.42m

View answer & explanation

Correct answer is B

for 20 students, mean = 1.67

∑fx

∑fx = μf
∑fx = 20 × 1.67 = 33.4
for group 2
∑fx = 16 × 1.50 = 24
for group 3
∑fx = 14 × 1.40 = 19.6
33.4 + 24 + 19.6 = 77
∑f = 20+16+14 = 50

1.54m

53.

A body of mass 18kg moving with velocity 4ms-1 collides with another body of mass 6kg moving in the opposite direction with velocity 10ms-1. If they stick together after the collision, find their common velocity.

$\frac{1}{2}$ m/s

$\frac{1}{3}$ m/s

2m/s

3m/s

View answer & explanation

Correct answer is A

m1u1 + m2u2 = (m1 + m2)v

m1 = 18kg, m2 = 6kg, u1 = 4ms-1, u2 = -10m/s

18(4) + 6(-10) = (18+6)v

72 - 60 = 24v
12 = 24v
v = $\frac{1}{2}$ m/s

54.

The first, second and third terms of an exponential sequence (G.P) are (x - 4), (x + 2), and (3x + 1) respectively. Find the values of x.

$\frac{-1}{2}, 8$

$\frac{1}{2}, -8$

$\frac{-1}{2}, -8$

$\frac{1}{2}, 8$

View answer & explanation

Correct answer is A

U1 = x - 4
U2 = x + 2
U3 = 3x + 1

$\frac{u_2}{u_1} = \frac{u_3}{u_2}$

$\frac{x+2}{x-4} = \frac{3x+1}{x+2}$

(x+2)(x+2) = (x-4)(3x+1)
x $^2$ + 4x + 4 = 3x2 - 11x - 4
collecting like terms
2x $^2$ - 15x - 8 =0
2x $^2$ + x - 16x - 8 = 0
x(2x + 1) - 8(2x + 1) = 0
(x-8)(2x+1) = 0

x = ( $\frac{-1}{2}, 8$ )

55.

Find the coefficient of x $^3$ y $^2$ in the binomial expansion of (x-2y) $^5$

-80

View answer & explanation

Correct answer is C

x $^3$ y $^2$ in (x-2y) $^5$

n = 5, r = 3, p = x, q = -2y

5C $_3$ * x $^3$ -2y $^2$

5C $_3$ = $\frac{5!}{[5-3]!3!}$

$\frac{5*4*3!}{2! 3!}$ → $\frac{5*4}{2}$

5C $_3$ = 10

: 5C $_3$ * x $^3$ -2y $^2$ = 10 * x $^3$ 4y $^2$

40x $^3$ y $^2$
the coefficient is 40

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