WAEC Past Questions and Answers - Page 1018

5,086.

A particle moving with a velocity of 5m/s accelerates at 2m/s\(^2\). Find the distance it covers in 4 seconds.

A.

16m

B.

26m

C.

36m

D.

46m

Correct answer is C

from the equation of motion

u = 5m/s, a = 2m/s\(^2\), t = 4s 

s = ut + \(\frac{1}{2} at^2\)

s = 5*4 + \(\frac{1}{2} 2*4^2\)

s =  20 + 16

s = 36m

5,087.

Given that P = {x: x is a multiple of 5}, Q = {x: x is a multiple of 3} and R = {x: x is an odd number} are subsets of μ = {x: 20 ≤ x ≤ 35}, (P⋃Q)∩R.

A.

{20, 21, 25, 30, 33}

B.

{21, 25, 27, 33, 35}

C.

{20, 21, 25, 27, 33, 35}

D.

{21, 25, 27, 30, 33, 35}

Correct answer is B

P = { 20, 25, 30, 35}, Q = {21, 24, 27, 30, 33}, R = {21, 23, 25, 27, 29, 31, 33, 35}


(P⋃Q)∩R = {20, 21, 24, 25, 27, 30, 33, 35} ∩ {21, 23, 25, 27, 29, 31, 33, 35}


= {21, 25, 27, 33, 35}

5,088.

Find the coefficient of x\(^2\)in the binomial expansion of \((x + \frac{2}{x^2})^5\)

A.

10

B.

40

C.

32

D.

80

Correct answer is A

\((x + \frac{2}{x^2})^5\)

n = 5,  r = 4,  p = x  and q = \(\frac{2}{x^2}\)  

5C\(_4\)x\(^4\) (\(\frac{2}{x^2}\))1 = 5C\(_4\) \(\frac{2x^4}{x^2}\) 

5C\(_4\) 2x\(^2\) = \(\frac{5!}{[5-4]!4!}\) * 2x\(^2\)

\(\frac{5*4!}{4!} * 2x^2\) = 5 * 2x\(^2\) = 10x\(^2\)

The coefficient is 10.

5,089.

Evaluate \(4p_2 + 4C_2 - 4p_3\)

A.

18

B.

6

C.

-6

D.

-18

Correct answer is C

\(4p_2 + 4C_2 - 4p_3\)

\(np_r = \frac{n!}{[n-r]!} and  nC_r = \frac{n!}{[n-r]!r!} \)

= \(\frac{4!}{[4-2]!} + \frac{4!}{[4-2]!2!} - \frac{4!}{[4-3]!} = \frac{4!}{2!} + \frac{4!}{2!2!} - \frac{4!}{1!}\)

= \(\frac{4*3*2!}{2!} + \frac{4*3*2!}{2!2!} - \frac{4*3*2*1}{1!}\)

12 + 6 - 24 = -6

5,090.

A linear transformation T is defined by T: (x,y) → (3x - y, x + 4y). Find the image of (2, -1) under T.

A.

(7, -2)

B.

(5, -2)

C.

(-2, 7)

D.

(-7, 2)

Correct answer is A

Let (x1, y1) be the image of the point (x, y) under the given transformations.
x1 = 3x - y
y1 = x + 4y
\(\begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} \begin{vmatrix} x \\ y = \end{vmatrix}  \begin{vmatrix} x_1 \\ y_1 \end{vmatrix}\)

\(\begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} \begin{vmatrix} 2 \\ 1 = \end{vmatrix}  \begin{vmatrix} 7 \\ -2 \end{vmatrix}\)