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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