Mathematics Questions and Answers

\(\sqrt{160r^2 + \sqrt{71r^4 + \sqrt{100r^8}}}\)

Simplifying from the innermost radical and progressing outwards we have the given expression

\(\sqrt{160r^2 + \sqrt{71r^4 + \sqrt{100r^8}}}\) = \(\sqrt{160r^2 + \sqrt{71r^4 + 10r^4}}\)

= \(\sqrt{160r^2 + \sqrt{81r^4}}\)

\(\sqrt{160r^2 + 9r^2}\) = \(\sqrt{169r^2}\)

= 13r

use "T" to represent the total profit. The first receives \(\frac{1}{3}\) T

remaining, 1 - \(\frac{1}{3}\)

= \(\frac{2}{3}\)T

The seconds receives the remaining, which is \(\frac{2}{3}\) also

\(\frac{2}{3}\) x \(\frac{2}{3}\) = \(\frac{4}{9}\)

The third receives the left over, which is \(\frac{2}{3}\)T - \(\frac{4}{9}\)T = (\(\frac{6 - 4}{9}\))T

= \(\frac{2}{9}\)T

The third receives \(\frac{2}{9}\)T which is equivalent to N12000

If \(\frac{2}{9}\)T = N12, 000

T = \(\frac{12 000}{\frac{2}{9}}\)

Total share[T] = N54, 000

The first receives \(\frac{1}{3}\) of T → \(\frac{1}{3}\) * N54, 000 = N18,000

The second receives \(\frac{4}{9}\) of T → \(\frac{4}{9}\) * N54, 000 = N24,000

The third receives \(\frac{2}{9}\)T which is equivalent to N12000.

Adding the three shares give total profit of N54,000

\(\frac{0.00275 \times 0.0064}{0.025 \times 0.08}\)

Removing the decimals = \(\frac{275 \times 64}{2500 \times 800}\)

= \(\frac{88}{10^4}\)

\(88 x 10^{-4} = 88 x 10^{1} x 10^{-4} = 8.8 x 10^{-3}\)

You can use any whole numbers (eg. 1. 2. 3) to represent all the mangoes in the basket.

If the first child takes \(\frac{1}{4}\) it will remain 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\)

Next, the second child takes \(\frac{3}{4}\) of the remainder

which is \(\frac{3}{4}\) i.e. find \(\frac{3}{4}\) of \(\frac{3}{4}\)

= \(\frac{3}{4}\) x \(\frac{3}{4}\)

= \(\frac{9}{16}\)

the fraction remaining now = \(\frac{3}{4}\) - \(\frac{9}{16}\)

= \(\frac{12 - 9}{16}\)

= \(\frac{3}{16}\)

Interest I = \(\frac{PRT}{100}\)

∴ R = \(\frac{100 \times 1}{100 \times 5}\)

= \(\frac{100 \times 7.50}{500 \times 5}\)

= \(\frac{750}{500}\)

= \(\frac{3}{2}\)

= 1.5%