WAEC Mathematics Past Questions & Answers - Page 77

\(\frac{1}{4}\)p + 3p = 10...(1)

2p - \(\frac{1}{3}\)q = 7...(2)

Multiply equation (2) by 3 to clear fraction

3 x 2p - 3 x \(\frac{1}{3}\)q = 3 x 7

6p - q = 21

6p - 21 = q....(3)

substituting 6p - 21 for q in (1)

\(\frac{1}{4}\)p + 3(6p - 21) = 10...(4)

Multiply equation (4) by 4 to clear fraction

4 x \(\frac{1}{4}\)p + 4 x 3(6p - 21) = 4 x 10

p + 12(6p - 21) = 40

p + 72p - 252

73p = 292

p = \(\frac{292}{73}\)

= 4

No explanation has been provided for this answer.

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

\(\frac{2}{5}\)