Mathematics Questions and Answers

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)