JAMB Mathematics Past Questions & Answers - Page 245

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

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%