Mathematics Questions and Answers

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