Mathematics Questions and Answers

First S.P = ₦230.00

% profit = 15%

% profit = $\frac{S.P - C.P}{C.P}$ x 100%

⇒ 15% = $\frac{230 - C.P}{C.P}$ x 100%

⇒ $\frac{15}{100}= \frac{230 - C.P}{C.P}$

⇒15C.P = 100 (230 - C.P)

⇒15C.P = 23000 - 100C.P

⇒15C.P + 100C.P = 23000

⇒115C.P = 23000

⇒ C.P = $\frac{23000}{115}$ = ₦200.00

Second S.P = ₦180.00

Since C.P is greater than S.P, therefore it's a loss

% loss = $\frac{C.P - S.P}{C.P}$ x 100%

⇒ $\frac{200 - 180}{200}$ x 100%

⇒ $\frac{20}{200}$ x 100%

⇒ $\frac{1}{10}$ x 100% = 10%

∴ It's a loss of 10%

$\frac {\sin A}{a} = \frac {\sin B}{b} = \frac {\sin C}{c}$

$\implies \frac {\sin 82^0}{43.2} = \frac {\sin 56^0}{AB}$

$\implies AB \times \sin 82^0 = 43.2 \times \sin 56^0$

$\therefore AB = \frac {43.2 \times \sin 56^0}{\sin 82^0}$ = 36.2cm (to 3 s.f)

The height of the second building H = h + 24

tan θ = $\frac {opp}{adj}$

tan 30$^o = \frac {h}{x}$

$\implies\frac{\sqrt 3}{3} = \frac {h}{x}$

$\implies x = \sqrt 3 = 3h$

$\implies x = \frac {3h}{\sqrt 3}$ ....(i)

tan 60$^o = \frac {24}{x}$

$\implies\sqrt 3 = \frac {24}{x}$

$\implies x\sqrt 3 = 24$

$\implies x = \frac {24}{\sqrt 3}$ ....(ii)

Equate equation (i) and (ii)

$\implies \frac {3h}{\sqrt 3} = \frac {24}{\sqrt 3}$

$\implies$ 3h = 24

$\implies h = \frac {24}{3}$ = 8m

∴The height of the second building = 8 + 24 = 32m

Let the third number = $x$

Then the first number = 100% $x + 35%x = 135$%$x = \frac {135x}{100} = 1.35x$ (Note: 100% $x = x$)

The second number = 180% $x = \frac {180x}{100} = 1.80x$

∴ The ratio of the first number to the second number = $1.35x : 1.80x = 3 : 4$

The locus of a point equidistant from two intersecting lines is pair of bisectors of the angles between the two lines.