2015 - WAEC Mathematics Past Questions and Answers - page 4

T = (\sqrt{\frac{TK - H}{K - H}})

Taking the square of both sides, give

T² = (\frac{TK - H}{K - H})

T²(K - H) = TK - H

T²K - T²H = TK - H

T²K - TK = T²H - H

K(T² - T) = H(T² - 1)

K = (\frac{H(T^2 - 1)}{T^2 - T})

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

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

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

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

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

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})

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

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%

(\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

mean = (\bar{x}) = (\frac{\sum x}{n})

= (\frac{5 + 3 + 0 + 7 + 2 + 1}{6})

(\frac{18}{6}) = 3

(\begin{array}{c|c}
x & x - \bar{x} & |x - \bar{x}|\ \hline 5 & -2 & 2 \ \hline 3 & 0 & 0 \ \hline 0 & -3 & 3 \ \hline 7 & 4 & 4 \ \hline 2 & -1 & 1 \ \hline 1 & -2 & 2 \ \hline & & & \sum|x - \bar{x}| \ \hline & & & 12
\end{array})

hence, MD = (\frac{\sum|x - \bar{x}}{n}|)

(\frac{12}{6})

= 2