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.
Find the volume of the cylinder above
[Take \(\pi= ^{22}/_7\)]
9,856 cm\(^3\)
14,784 cm\(^3\)
4,928 cm\(^3\)
19,712 cm\(^3\)
Correct answer is B
Volume of the cylinder = \(\frac{θ}{360} \times \pi r^2h\)
θ = 360\(^o\) - 90\(^o\) = 270\(^o\)
∴ Volume of the cylinder = \(\frac{270}{360} \times \frac{22}{7} \times \frac{14^2}{1} \times \frac{32}{1} = \frac{37,255,680}{2520} = 14,784cm^3\)
Find the compound interest (CI) on ₦15,700 for 2 years at 8% per annum compounded annually.
₦6,212.48
₦2,834.48
₦18,312.48
₦2,612.48
Correct answer is D
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
1162 cm\(^2\)
1163 cm\(^2\)
1160 cm\(^2\)
1161 cm\(^2\)
Correct answer is A
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\(^2)\)
243
108
54
135
Correct answer is C
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
1931.25 m\(^2\) ≤ A < 2021.25 m\(^2\)
1950 m\(^2\) ≤ A < 2002 m\(^2\)
1957 m\(^2\) ≤ A < 1995 m\(^2\)
1931.25 m\(^2\) ≥ A > 2021.25 m\(^2\)
Correct answer is A
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\(^2\)
32.4 cm
30.6 cm
28.8 cm
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)
10% gain
10% loss
12% loss
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%
Calculate, correct to three significant figures, the length AB in the diagram above.
36.4 cm
36.1 cm
36.2 cm
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)
24 m
32\(\sqrt3\) m
24\(\sqrt3\)
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
Two numbers are respectively 35% and 80% more than a third number. The ratio of the two numbers is
7 : 16
3 : 4
16 : 7
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 = \(1.35x : 1.80x = 3 : 4\)
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