WAEC Mathematics Past Questions & Answers - Page 84

416.

Find at which rate per annum simple interest N525 will amount to N588 in 3 years.

A.

3%

B.

2%

C.

5%

D.

4%

Correct answer is D

I = A − P

= N588 − N525

∴I = N63

I = PRT ÷ 100

R = [100I ÷PT]

R = [(100 × 63 ) ÷ (525 × 3)]

= (6300 ÷ 1575) = 4

∴ The rate = 4 %

417.

Make x the subject of the equation
s = 2 + \(\frac{t}{5} \)(x + ⅗y)

A.

x = 5[(s − 2) ÷ t] - 3/5y

B.

x = 25[(s − 2) ÷ t] − 3ty

C.

x = [1 ÷ (s − 2)3ty]

D.

x = [5(s − 2) 2 ÷ 3ty] × t

Correct answer is A

No explanation has been provided for this answer.

418.

The amount A to which a principal P amounts at r% compound interest for n years is given by the formula A = P(1 + (r ÷ 100)\(^n\). Find A, if P = 126, r = 4 and n = 2.

A.

N132.50K

B.

N136.30K

C.

N125.40K

D.

N257.42K

Correct answer is B

\( A = P \left(1 + \frac{r}{100}\right)^n \)

Where P = 126, r = 4,n = 2

A=126 \( \left(1 + \frac{4}{100}\right)^2 \text{Using LCM} \)

=126 \( \left(\frac{100+4}{100}\right)^2 = 126 \left(\frac{104}{100}\right)^2 \)


=126 \( \left(1.04^2 \right) \)

= 126 * 1.04 * 1.04

=136.28

A = 136.30 (approx.)

The Amount A, = N136.30k

419.

Find the equation of a line which is form origin and passes through the point (−3, −4)

A.

y = \( \frac{3x}{4} \)

B.

y = \( \frac{4x}{3} \)

C.

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

D.

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

Correct answer is B

The slope of the line from (0, 0) passing through (-3, -4) = \(\frac{-4 - 0}{-3 - 0}\)

= \(\frac{4}{3}\)

Equation of a line is given as \(y = mx + b\), where m = slope and b = intercept.

To get the value of b, we use a point on the line, say (0, 0).

\(y = \frac{4}{3} x + b\)

\(0 = \frac{4}{3}(0) + b\)

\(b = 0\)

The equation of the line is \(y = \frac{4}{3} x\)

420.

If x + y = 90 simplify \((sinx + siny)^2\)−2sinxsiny

A.

1

B.

0

C.

2

D.

-1

Correct answer is A

Given: \(x + y = 90° ... (1)\)

\((\sin x + \sin y)^{2} - 2\sin x \sin y = \sin^{2} x + \sin^{2} y + 2\sin x \sin y - 2\sin x \sin y\)

= \(\sin^{2} x + \sin^{2} y ... (2)\)

Recall: \(\sin x = \cos (90 - x) ... (a)\)

From (1), \(y = 90 - x ... (b)\)

Putting (a) and (b) in (2), we have

\(\sin^{2} x + \sin^{2} y \equiv \cos^{2} (90 - x) + \sin^{2} (90 - x)\)

= 1