Mathematics Questions and Answers

Volume of the cylinder = \(\frac{θ}{360} \times \pi r^2h\)

θ = 360\(^o\) - 90\(^o\) = 270\(^o\)

∴ Volume of the cylinder = \(\frac{270}{360} \times \frac{22}{7} \times \frac{14^2}{1} \times \frac{32}{1} = \frac{37,255,680}{2520} = 14,784cm^3\)

Principal (P) = ₦15,700

Rate (R) = 8

Number of years (t) = 2

A = P \((1+\frac{R}{100})^t\)

⇒ A = 15700 \((1+\frac{8}{100})^2\)

⇒ A = 15700 (1 + 0.08)\(^2\)

⇒ A = 15700 (1.08)\(^2\)

⇒ A = 15700 x 1.1664

⇒ A = ₦18,312.48

Total amount, A = ₦18,312.48

A = P + CI

⇒ CI = A - P

⇒ CI = 18,312.48 - 15,700

∴ CI = ₦2,612.48

Let the length of the sides of triangle be 2x, 3x and 4x.

Perimeter of triangle = 180cm

⇒ \(2x +3x + 4x = 180\)

⇒ \(9x = 180\)

⇒ \(x = \frac{180}{9}\) = 20cm

Then the sides of the triangle are:

\(2x = 2\times20 = 40cm; 3x = 3\times20\) = 60cm and \(4x\) = 4\(\times20 \)= 80cm

Using Heron's formula

Area of triangle = \(\sqrt s(s-a)(s-b)(s-c)\)

Where s = \(\frac{a + b + c}{2}\)

Let a = 40cm, b = 60cm, c = 80cm and s = \(\frac{40 + 60 + 80}{2} = \frac{180}{2}\) = 90cm

⇒ A = \(\sqrt90 (90 - 40) (90 - 60) (90 - 80) = \sqrt90 \times 50 \times 30 \times 10 = \sqrt1350000\)

∴ A =1162cm\(^2\) (to the nearest cm\(^2)\)

Let the total number of items in the man's shop = \(y\)

Number of Brand A's items in the man's shop = \(\frac {1}{9} y\)

Remaining items = 1 - \(\frac {1}{9} y = \frac {8}{9} y\)

Number of Brand B's items in The man's shop = \(\frac{5}{8} of \frac{8}{9}y = \frac{5}{9}y\)

Total of Brand A and Brand B's items = \(\frac{1}{9}y + \frac{5}{9}y = \frac{2}{3}y\)

Number of Brand C's items in the man's shop = 1 - \(\frac{2}{3}y = \frac{1}{3}y\)

\(\implies\frac{1}{3}y\) = 81 (Given)

\(\implies y\) = 81 x 3 = 243

∴ The total number of items in the man's shop = 243

∴ Number of Brand B's items in the man's shop = \(\frac{5}{9}\) x 243 = 135

∴ The number of more Brand B items than Brand C = 135 - 81 =54

The sides have been given to the nearest meter, so

51.5 m ≤ length < 52.5

37.5 m ≤ width < 38.5

Minimum area = 37.5 x 51.5 = 1931.25 m\(^2\)

Maximum area = 38.5 x 52.5 = 2021.25 m\(^2\)

∴ The range of the area = 1931.25 m\(^2\) ≤ A < 2021.25 m\(^2\)

Consider ∆XOB and using Pythagoras theorem

13\(^2\) = 12\(^2\) + h\(^2\)

⇒ 169 = 144 + h\(^2\)

⇒ 169 - 144  =h\(^2\)

⇒ 25 = h\(^2\)

⇒ h = \(\sqrt25\) = 5cm

tan θ = \(\frac {opp}{adj}\)

⇒ tan θ = \(\frac{12}{5}\) = 2.4

⇒ θ = tan\(^{-1}\)(2.4)

⇒ θ = 67.38\(^0\)

∠AOB = 2θ = 2 x 67.38\(^o\) = 134.76\(^o\)

L = \(\frac{θ}{360^o} \times 2\pi r\)

⇒ L = \(\frac {134.76}{360} \times 2 \times \frac {22}{7} \times 13 = \frac {77082.72}{2520}\)

∴ L = 30.6cm (to 3 s.f)

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 = \(1.35x : 1.80x = 3 : 4\)