WAEC Mathematics Past Questions & Answers - Page 61

301.

Find the inter-quartile range of 1, 3, 4, 5, 8, 9, 10, 11, 12, 14, 16

A.

6

B.

7

C.

8

D.

9

Correct answer is C

\(Q_1 = \frac{1}{4}\) (N + 1)th

\(\frac{1}{4} \times 12^{th}\) no.

= 3rd no (\(\cong\) 4)

\(Q_3 = \frac{3}{4}\)  (N + 1)th

= \(\frac{3}{4}\) x 12th no.

= 9th no. (\(\cong\) 12)

Hence, interquartile range

= \(Q_3 - Q_1\)

= 12 - 4

= 8

302.

If 3x\(^o\) 4(mod 5), find the least value of x

A.

1

B.

2

C.

3

D.

4

Correct answer is C

3x \(\equiv\) 4(mod 5)

In modulo 5, multiples of 5 that give solution to the given equation are 5, 20, 35, 50,... but 5 will yield the leaast value of x.

Thus; 3x = 4 + 5 = 9

x = \(\frac{9}{3}\)

x = 3

303.

Find  the \(n^{th}\) term of the sequence 2 x 3, 4 x 6, 8 x 9, 16 x 12...

A.

2\(^n\) x 3(n + 1)

B.

2\(^n\) x 3n

C.

2\(^n\) x 3\(^n\)

D.

2\(^n\) x 3\(^n - 1\)

Correct answer is B

2 x 3, 4 x 6, 8 x 9, 16 x 12,...

2\(^1\) x 3 x 1, 2\(^2\) x 3 x 2, 2\(^3\) x 3 x 3, 2\(^4\) x 3 x 4,.... 2\(^n\) x 3n

304.

A.

(3x + 2)(1 - x)

B.

(3x + 2)(2x + 1)

C.

3\((x + 2)^2\)

D.

3(x + 1)(1 - x)

Correct answer is D

\((x + 2)^2\) - \((2x + 1)^2\)

= \((x^2 + 4x + 4) - (4x^2 + 4x + 1)\)

= \(x^2 \) + 4x + 4 - 4 \(x^2 \) - 4x - 1

= -3 \(x^2 \) + 3

= 3 - 3 \(x^2 \)

= 3(1 - \(x^2 \))

= 3(1 + x)(1 - x)

305.

Solved the equation \(2x^2 - x - 6\) = 0

A.

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

B.

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

C.

x = -3 or 2

D.

x = 3 or -2

Correct answer is A

\(2x^2 - x - 6\) = 0

\(2x^2 - 4x + 3x - 6\) = 0

2x(x - 2) + 3(x - 2) = 0

(2x + 3) (x - 2) = 0

Either; 2x + 3 = 0 or x - 2 = 0

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