WAEC Mathematics Past Questions & Answers - Page 77

Cost price CP of the 100 oranges = $\frac{100}{5}$ x N40.00

selling price SP of the 100 oranges = $\frac{100}{20}$ x N120

= N600.00

so, profit or loss per cent

= $\frac{SP - CP}{CP}$ x 100%

= $\frac{600 - 800}{800}$ x 100%

= $\frac{-200}{800}$ x 100%

Hence, loss per cent = 25%

R $\alpha$ D²

R = D²K

R = 4 Litres when D = 15cm

thus; 4 = 15²k

4 = 225k

k = $\frac{4}{225}$

This gives R = $\frac{4D^2}{225}$

Where R = 9litres

equation gives

9 = $\frac{4D^2}{225}$

9 x 225 = 4d²

D² = $\frac{9 \times 225}{4}$

D = $\sqrt{9 \times 225}{4}$

= $\frac{3 \times 15}{2}$

= 22$\frac{1}{2}$km

Let x represent the entire farmland

then, $\frac{2}{5}$x + $\frac{1}{3}$[x - $\frac{2}{3}x$] + M = x

Where M represents the part of the farmland used for growing maize, continuing

$\frac{2}{5}$x + $\frac{1}{3}$x [1 - $\frac{2}{3}x$] + M = x

$\frac{2}{5}x + \frac{1}{3}$x [$\frac{3}{5}$] + M = x

$\frac{2}{5}$x + $\frac{1x}{5}$ + M = x

$\frac{3x}{5} + M = x$

M = x - $\frac{2}{5}$x

= x[1 - $\frac{3}{5}$]

= x[$\frac{2}{5}$] = $\frac{2x}{5}$

Hence the part of the land used for growing maize is

$\frac{2}{5}$

$\frac{3x - y}{xy} - \frac{2x + 3y}{2xy} + \frac{1}{2}$

= $\frac{2(3x - y) - 1(2x + 3y) + xy}{2xy}$

= $\frac{6x - 2y - 2x - 3y + xy}{2xy}$

= $\frac{4x - 5y + xy}{2xy}$

Diagonal |AC| = (2x + 1)cm

In the diagram,area of $\Delta$ABC

is $\frac{110}{2}$ = $\frac{1}{2}$ x |AC| x |HB|

55 = $\frac{1}{2}$ x (2x + 1) x 10

55 = (2x + 1)5

55 = 10x + 5

55 - 5 = 10x

50 = 10x

x = $\frac{50}{10}$

= 5.0