JAMB Mathematics Past Questions & Answers - Page 198

Let x represent the price of an orange and

y represent the number of oranges that can be bought

xy = 240k, y = $\frac{240}{x}$.....(i)

If the price of an oranges is raised by $\frac{1}{2}$k per orange, number that can be bought for N240 is reduced by 16

Hence, y - 16 = $\frac{240}{x + \frac{1}{2}$

= $\frac{480}{2x + 1}$

= $\frac{480}{2x + 1}$.....(ii)

subt. for y in eqn (ii) $\frac{240}{x}$ - 16

= $\frac{480}{2x + 1}$

= $\frac{240 - 16x}{x}$

= $\frac{480}{2x + 1}$

= (240 - 16x)(2x + 1)

= 480x

= 480x + 240 - 32x² - 16

480x = 224 - 32x²

x² = 7

x = $\sqrt{7}$

= 2.5

= 2$\frac{1}{2}$k

Let the selling price(SP from P to Q be represented by x

i.e. SP = x

When SP = x at 10% profit

CP = $\frac{100}{100}$ + 10 of x = $\frac{100}{110}$ of x

when Q sells to R, SP = N209 at loss of 5%

Q's cost price = Q's selling price

CP = $\frac{100}{95}$ x 209

= 220.00

x = 220

= $\frac{2200}{11}$

= 200

= N200.00

$\frac{0.0001432}{1940000}$ = k x 10ⁿ

where 1 $\leq$ k $\leq$ 10 and n is a whole number. Using four figure tables, the eqn. gives 7.38 x 10^-11

k = 7.381, n = -11

$\frac{2}{3} - \frac{1}{5}$ = $\frac{10 - 3}{15}$

= $\frac{7}{15}$

$\frac{1}{3}$ Of $\frac{2}{5}$ = $\frac{1}{3}$ x $\frac{2}{5}$

= $\frac{2}{15}$

($\frac{2}{3} - \frac{1}{5}$) - $\frac{1}{3}$ of $\frac{2}{5}$

= $\frac{7}{15} - \frac{2}{15}$ = $\frac{1}{3}$

3 - $\frac{1}{1 \frac{1}{2}}$ = 3 - $\frac{2}{3}$

= $\frac{7}{3}$

$\frac{\frac{2}{3} - \frac{1}{5} \text{of} \frac{2}{15}}{3 - \frac{1}{1 \frac{1}{2}}}$

= $\frac{\frac{1}{3}}{\frac{7}{3}}$

= $\frac{1}{3}$ x $\frac{3}{7}$

= $\frac{1}{7}$

a{$\frac{x + 1}{x - 2} - \frac{x - 1}{x + 2}$} = a{$\frac{(x + 1)(x + 2)- (x - 1)(x - 2)}{(x - 2)(x + 2)}$}

= 6

$\frac{6x}{x^2 - 4}$ = 6x

a = x² - 4