WAEC Mathematics Past Questions & Answers - Page 113

561.

The nth term of a sequence is Tn = 5 + (n - 1)2. Evaluate T4 - T6

A.

30

B.

16

C.

-16

D.

-30

Correct answer is C

T4 = 5 + (4 - 1)2; where n = 4

= 5 + (3)2 = 5 + 9

= 14

T = 5 + (6 - 1)2

where n = 6

= 5 + (5)2

= 5 + 25 = 30

T4 + T6 = 14 - 30

= -16

562.

Convert 3510 to number in base 2

A.

1011

B.

10011

C.

100011

D.

11001

Correct answer is C

No explanation has been provided for this answer.

563.

In 1995, the enrollment of two schools X and Y were 1,050 and 1,190 respectively. Find the ration of the enrollments of X and Y

A.

50:11

B.

15:17

C.

13:55

D.

12:11

Correct answer is B

Divide cash enrollment by their common factors then the answer of each division gives the required ratio. i.e. x = \(\frac{1050}{70}\) = 15

and y = \(\frac{1190}{70}\) = 17

ratio of enrollment of x to y = 15:17

564.

Simplify; \(\frac{3\sqrt{5} \times 4\sqrt{6}}{2 \sqrt{3} \times 3\sqrt{2}}\)

A.

\(\sqrt{2}\)

B.

\(\sqrt{5}\)

C.

2\(\sqrt{2}\)

D.

2\(\sqrt{5}\)

Correct answer is D

\(\frac{3\sqrt{5} \times 4\sqrt{6}}{2 \sqrt{3} \times 3\sqrt{2}}\)

= \(\frac{\sqrt{5} \times 2\sqrt{6}}{\sqrt{2} \times \sqrt{3}}\)

= \(\frac{\sqrt{5} \times 2 \sqrt{6}}{\sqrt{6}}\)

= 2\(\sqrt{5}\)

565.

Express 302.10495 correct to five significant figures

A.

302.10

B.

302.11

C.

302.105

D.

302.1049

Correct answer is A

No explanation has been provided for this answer.