WAEC Mathematics Past Questions & Answers - Page 219

1,091.

Evaluate \(\frac{1}{2}+\frac{3}{4}of\frac{2}{5}\div 1\frac{3}{5}\)

A.

\(\frac{15}{16}\)

B.

\(\frac{11}{16}\)

C.

\(\frac{49}{50}\)

D.

\(3\frac{1}{5}\)

Correct answer is B

\(\frac{1}{2} + (\frac{3}{4} \text{ of } \frac{2}{5}) \div 1\frac{3}{5}\)

= \(\frac{1}{2} + (\frac{3}{4} \times \frac{2}{5}) \div \frac{8}{5}\)

= \(\frac{1}{2} + \frac{3}{10} \div \frac{8}{5}\)

= \(\frac{1}{2} + (\frac{3}{10} \times \frac{5}{8})\)

= \(\frac{1}{2} + \frac{3}{16}\)

= \(\frac{11}{16}\)

1,092.

The nth term of a sequence is \(2^{2n-1}\). Which term of the sequence is \(2^9?\)

A.

3rd

B.

4th

C.

5th

D.

6th

Correct answer is C

\(T_{n} = 2^{2n - 1}\)

\(2^{2n - 1} = 2^9\)

\(2n - 1 = 9 \implies 2n = 9 + 1\)

\(2n = 10 \implies n = 5\)

The 5th term = 2\(^9\)

1,093.

Evaluate \(5\frac{2}{5}\times \left(\frac{2}{3}\right)^2\div\left(1\frac{1}{2}\right)^{-1}\)

A.

\(\frac{8}{25}\)

B.

\(\frac{12}{25}\)

C.

\(3\frac{3}{5}\)

D.

\(4\frac{1}{8}\)

Correct answer is C

\(5\frac{2}{5}\times \left(\frac{2}{3}\right)^2 ÷  \left(1\frac{1}{2}\right)^{-1}\\
\frac{27}{5}\times \frac{4}{9} \times \frac{3}{2}=3\frac{3}{5}
\)

1,094.

A boy estimated his transport fare for a journey as N190 instead of N200. Find the percentage error in his estimate

A.

95%

B.

47.5%

C.

5.26%

D.

5%

Correct answer is D

The percentage error is \(\frac{error}{actual}\times \frac{100}{1}\%\\
=\frac{200-190}{200}\times \frac{100}{1}\\
=\frac{10}{200}\times \frac{100}{1}\% = 5\%\)

1,095.

In a bag of oranges, the ratio of the good ones to the bad ones is 5:4. If the number of bad oranges in the bag is 36, how many oranges are there in the altogether?

A.

81

B.

72

C.

54

D.

45

Correct answer is A

Ratio of good ones to bad ones is 5:4; If 36 is bad;

∴ the good ones = \(\frac{5\times 36}{4}=45\) oranges. 

The total number of oranges is 36 + 45 = 81.