2000 - WAEC Mathematics Past Questions and Answers - page 1

(\frac{7}{19}) as a percentage will be
(\frac{7}{19}\times \frac{100}{1}% = 36.8%)

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

(\frac{10}{\sqrt{32}}=\frac{10}{\sqrt{16\times 2}} = \frac{10}{4\sqrt{2}}<br /> \frac{10\times \sqrt{2}}{4\sqrt{2}\times\sqrt{2}}=\frac{10\sqrt{2}}{4\times 2}= \frac{5}{4}\sqrt{2})

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

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

= 3453\(_7\)

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

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

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

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

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

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

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

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

(\left(\frac{16}{81}\right)^{-\frac{3}{4}}\times \sqrt{\frac{100}{81}}<br /> \frac{1}{\left(\sqrt[4]{\frac{16}{81}}\right)^3}\times \frac{10}{9}=\frac{1}{\left(\frac{2}{3}\right)^3}\times\frac{10}{9}<br /> =\frac{27}{8}\times \frac{10}{9}=\frac{15}{4})

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

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.

\(\log_4 x = -3\)

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

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