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WAEC Mathematics Past Questions & Answers - Page 113

561.

The n^th term of a sequence is T_n = 5 + (n - 1)². Evaluate T₄ - T₆

-16

-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

1011

10011

100011

11001

View answer & explanation

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

50:11

15:17

13:55

12:11

View answer & explanation

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; $\frac{3\sqrt{5} \times 4\sqrt{6}}{2 \sqrt{3} \times 3\sqrt{2}}$

$\sqrt{2}$

$\sqrt{5}$

2 $\sqrt{2}$

2 $\sqrt{5}$

View answer & explanation

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 $\sqrt{5}$