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WAEC Mathematics Past Questions & Answers - Page 84

416.

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(1+r100)n

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

A=126 (1+4100)2Using LCM

=126 (100+4100)2=126(104100)2


=126 (1.042)

= 126 * 1.04 * 1.04

=136.28

A = 136.30 (approx.)

The Amount A, = N136.30k

417.

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

A.

y = 3x4

B.

y = 4x3

C.

y = 2x3

D.

y = x2

Correct answer is B

The slope of the line from (0, 0) passing through (-3, -4) = 4030

= 43

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=43x+b

0=43(0)+b

b=0

The equation of the line is y=43x

418.

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

419.

Find the total surface area of a cylinder of base radius 5cm and length 7cm ( π = 3.14)

A.

17.8cm2

B.

15.8cm2

C.

75.4cm2

D.

54.7cm2

E.

377.0cm^{2}

Correct answer is E

The total surface area of a cylinder = 2πrl + 2πr2

= 2πr(l + r)

= 2 × 3.14 x 5(7+5)

2 × 3.14 × 12 x 5

= 377.1cm (1DP)

420.

X and Y are two sets such that n(X) = 15, n(Y) = 12 and n{X ∩ Y} = 7. Find ∩{X ∪ Y}

A.

21

B.

225

C.

15

D.

20

Correct answer is D

n(X ∪ Y) = n(X) + n(Y) − n(X ∩ Y) = 15 + 12 − 7 ∴ n(X ∪ Y) = 20