Mathematics questions and answers

Mathematics Questions and Answers

How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.

86.

Calculate, correct to three significant figures, the length of the arc AB in the diagram above.
[Take \(\pi = ^{22}/_7\)]

A.

32.4 cm

B.

30.6 cm

C.

28.8 cm

D.

30.5 cm

Correct answer is B

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)

87.

An article when sold for ₦230.00 makes a 15% profit. Find the profit or loss % if it was sold for ₦180.00

A.

10% gain

B.

10% loss

C.

12% loss

D.

12% gain

Correct answer is B

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%

88.

Calculate, correct to three significant figures, the length AB in the diagram above.

A.

36.4 cm

B.

36.1 cm

C.

36.2 cm

D.

36.3 cm

Correct answer is C

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

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

\(\implies AB \times \sin 82^0 = 43.2 \times \sin 56^0\)

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

89.

The angle of elevation and depression of the top and bottom of another building, measured from the top of a 24 m tall building, is 30° and 60°, respectively. Determine the second building's height.

A.

24 m

B.

32\(\sqrt3\) m

C.

24\(\sqrt3\)

D.

32 m

Correct answer is D

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

90.

Two numbers are respectively 35% and 80% more than a third number. The ratio of the two numbers is

A.

7 : 16

B.

3 : 4

C.

16 : 7

D.

4 : 3

Correct answer is B

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