WAEC Past Questions and Answers - Page 1169

5,842.

A body is kept at rest by three forces \(F_{1} = (10N, 030°), F_{2} = (10N, 150°)\) and \(F_{3}\). Find \(F_{3}\).

A.

(12N, 090°)

B.

(10N, 270°)

C.

(10N, 180°)

D.

(10N, 120°)

Correct answer is B

No explanation has been provided for this answer.

5,843.

If \(\frac{^{n}C_{3}}{^{n}P_{2}} = 1\), find the value of n.

A.

8

B.

7

C.

6

D.

5

Correct answer is A

\(^{n}C_{3} = \frac{n!}{(n - 3)! 3!}\)

\(^{n}P_{2} = \frac{n!}{(n - 2)!}\)

\(\frac{^{n}C_{3}}{^{n}P_{2}} = \frac{n!}{(n - 3)! 3!} ÷ \frac{n!}{(n - 2)!}\)

\(\frac{n!}{(n - 3)! 3!} \times \frac{(n - 2)!}{n!} = \frac{(n - 2)!}{(n - 3)! 3!}\)

Note that \((n - 2)! = (n - 2) \times (n - 2 - 1)! = (n - 2)(n - 3)!\)

\(\frac{(n - 2)(n - 3)!}{(n - 3)! 3!} = 1\)

\(\frac{n - 2}{3!} = 1 \implies n - 2 = 6\)

\(n = 2 + 6 = 8\)

5,844.

Find the equation of the straight line that passes through (2, -3) and perpendicular to the line 3x - 2y + 4 = 0.

A.

2y - 3x = 0

B.

3y - 2x + 5 = 0

C.

3y + 2x + 5 = 0

D.

2y - 3x - 5 = 0

Correct answer is C

Given line: \(3x - 2y + 4 = 0 \implies 2y = 3x + 4\)

\(y = \frac{3}{2}x + 2\)

\(Gradient (\frac{\mathrm d y}{\mathrm d x}) = \frac{3}{2}\)

Gradient of perpendicular line = \(\frac{-1}{\frac{3}{2}} = \frac{-2}{3}\)

\(\implies \frac{y - (-3)}{x - 2} = \frac{-2}{3}\)

\(\frac{y + 3}{x - 2} = \frac{-2}{3} \)

\(3(y + 3) = -2(x - 2) \implies 3y + 2x + 9 - 4 = 0\)

= \(3y + 2x + 5 = 0\)