# 1994 - WAEC Mathematics Past Questions & Answers - page 1

1

If the 2nd and 5th terms of a G.P are 6 and 48 respectively, find the sum of the first for term

A
-45
B
-15
C
15
D
33
CORRECT OPTION: e

T$$_{2}$$ = ar = 6

T$$_{5}$$ = ar$$^{4}$$ = 48

$$\frac{T_5}{T_2}$$ = $$\frac{ar^{4}}{ar}$$ = $$\frac{48}{6}$$

= r$$^{3}$$ = 8

⇒ r = 2

S$$_{n}$$ = $$\frac{a((r^n) - 1)}{r - 1}$$

S$$_{4}$$ = $$\frac{a((r^4) - 1)}{r - 1}$$

S$$_{4}$$ = $$\frac{3((2^4) - 1)}{2 - 1}$$

= 3(16 -1)

= 45

2

If sin$$\theta$$ = K find tan$$\theta$$, 0° $$\leq$$ $$\theta$$ $$\leq$$ 90°.

A
1-K
B
$$\frac{k}{k - 1}$$
C
$$\frac{k}{\sqrt{1 - k^2}}$$
D
$$\frac{k}{1 - k}$$
CORRECT OPTION: c

$$\sin \theta = \frac{k}{1}$$

$$\implies 1^2 = k^2 + adj^2$$

$$adj = \sqrt{1 - k^2}$$

$$\therefore \tan \theta = \frac{k}{\sqrt{1 - k^2}}$$

3

Simplify: $$2\frac{1}{3} \div 2\frac{2}{3} \times 1\frac{1}{7}$$

A
0
B
1
C
2
D
3
CORRECT OPTION: b

$$2\frac{1}{3} \div 2\frac{2}{3} \times 1\frac{1}{7}$$

= $$\frac{7}{3} \div \frac{8}{3} \times \frac{8}{7}$$

= $$\frac{7}{3} \times \frac{3}{8} \times \frac{8}{7}$$

= 1

4

Which of the following is equal to $$\frac{72}{125}$$

A
$$\frac{2^3 \times 3^2}{5^3}$$
B
$$\frac{2^4 \times 3}{5^3}$$
C
$$\frac{2^3 \times 3}{5^3}$$
D
$$\frac{2^4 \times 3}{5^5}$$
CORRECT OPTION: a

$$\frac{72}{125} = \frac{8 \times 9}{5 \times 5 \times 5}$$

= $$\frac{2^3 \times 3^2}{5^3}$$

5

Express in $$\frac{8.75}{0.025}$$standard form

A
3.5 x 10-3
B
3.5 x 10-2
C
3.5 x 101
D
3.5 x 102
CORRECT OPTION: d

$$\frac{8.75}{0.025}$$

= $$\frac{8750}{25}$$

= $$350$$

= $$3.5 \times 10^2$$

6

Evaluate $$\frac{27^{\frac{1}{3}}}{16^{-\frac{1}{4}}}$$

A
6
B
5
C
4
D
3
CORRECT OPTION: a

$$\frac{27^{\frac{1}{3}}}{16^{-\frac{1}{4}}}$$

= $$\frac{(3^3)^{\frac{1}{3}}}{(2^4)^{-\frac{1}{4}}}$$

= $$\frac{3}{2^{-1}}$$

= 6

7

Simplify: 16$$^{\frac{5}{4}}$$ x 2$$^{-3}$$ x 3$$^0$$

A
0
B
2
C
4
D
10
CORRECT OPTION: c

16$$^{\frac{5}{4}}$$ x 2$$^{-3}$$ x 3$$^0$$

= $$(2^4)^{\frac{5}{4}} \times 2^{-3} \times 1$$

= $$2^5 \times 2^{-3} \times 1$$

= $$2^2$$

= 4.

8

simplify; 2log$$_{3}$$ 6 + log$$_{3}$$ 12 - log$$_{3}$$ 16

A
2
B
3
C
2 - 2log32
D
3 - log32
CORRECT OPTION: b

2log$$_{3}$$ 6 + log$$_{3}$$ 12 - log$$_{3}$$ 16

= $$\log_3 6^2 + \log_3 12 - \log_3 16$$

= $$\log_{3} (\frac{36 \times 12}{16})$$

= $$\log_{3} (27)$$

= $$\log_{3} 3^3$$

= $$3 \log_{3} 3$$

= 3

9

What is the number whose logarithm to base 10 is $$\bar{3}.4771$$?

A
3.0
B
0.3
C
0.03
D
0.003
CORRECT OPTION: d
10

A house bought for N100,000 was later auctioned for N80,000. Find the loss percent.

A
20%
B
30%
C
40%
D
50%
CORRECT OPTION: a

Loss = N(100,000 - 80,000)

= N20,000.

Loss percentage = $$\frac{20,000}{100,000} \times 100%$$

= 20%

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