2000 - WAEC Mathematics Past Questions and Answers - page 1

398753 \(\approxeq\) 399000 (to 3 s.f)

4321\(_7\) + 1234\(_7\) = 5555\(_7\)

Missing number = 12341\(_7\) - 5555\(_7\)

= 3453\(_7\)

\(2\frac{1}{6} + 2\frac{7}{12}\)

= \(\frac{13}{6} + \frac{31}{12}\)

= \(\frac{26 + 31}{12}\)

= \(\frac{57}{12} = \frac{19}{4}\)

\(\frac{19}{4} - 3\frac{1}{4}\)

= \(\frac{19}{4} - \frac{13}{4}\)

= \(\frac{6}{4}\)

= \(1\frac{1}{2}\)

\(104_x = 68\
1 \times x^2 + 0 \times x + 4 \times x^0 = 68\
x^2 = 68 - 4; x^2 = 64\
x = \sqrt{64}=8\)

Given that the ages are in the ratio 3: 4: 5.

Let the sum of their ages be t.

\(\therefore\) The youngest age = \(\frac{3}{12} t\)

The eldest age = \(\frac{5}{12} t\)

\(\implies \frac{5}{12} t - \frac{3}{12} t = 18\)

\(\frac{2t}{12} = 18 \implies t = \frac{18 \times 12}{2}\)

t = 108 years.

The sum of their ages = 108 years.

\(\log_4 x = -3\)

\(x = 4^{-3}\)

= \(\frac{1}{64}\)