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.
\(\frac{8}{11}\)
\(\frac{3}{11}\)
\(\frac{4}{11}\)
\(\frac{5}{11}\)
Correct answer is B
Total numbers from 40 to 50 inclusive = 11 (40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50).
Prime numbers between 40 and 50 (inclusive) are 41, 43, and 47.
There are 3 prime numbers in this range.
therefore, Pr(prime numbers) = \(\frac{3}{11}\)
Arrange the following in ascending order of magnitude \(110_{two}, 31_{eight}, 42_{five}\)
\(110_{two}, 31_{five}, 42_{eight}\)
\(42_{five}, 110_{two}, 31_{eight}\)
\(42_{five}, 31_{eight}, 110_{two}\)
\(110_{two}, 42_{five}, 31_{eight}\)
Correct answer is D
Converting each number to base 10
\(110_{two} = 1 × 2^2 + 1 × 2^1 + 0 × 2^0\)
= 1 × 4 + 1 × 2 + 0 × 1
= 4 + 2 + 0
= \(6_{ten}\)
\(31_{eight} = 3 × 8^1 + 1 × 8^0\)
= 3 × 8 + 1 × 1
= 24 + 1
= \(25_{ten}\)
\(42_{five} = 4 × 5^1 + 2 × 5^0\)
= 4 × 5 + 2 × 1
= 20 + 2
= \(22_{ten}\)
Hence, \(31_{eight} > 42_{five} > 110_{two}\)
In ascending order, \(110_{two}, 42_{five}, 31_{eight}\)
91cm
7cm
13cm
57cm
Correct answer is C
given that diameter = 140cm ⇒radius = 70cm, volume of cylinder = 200litres = \(200,000cm^3\)
volume of cylinder = base area times height = \(\pi r^2h\)
\( 200,000cm^3 = \frac{22}{7} \times 70 \times 70 \times h\)
\(200,000 = \frac{107,800h}{7}\)
cross multiply
1,400,000 = 107,800h
\( h = \frac{1,400,000}{107,800}\)
= 13cm (to the nearest cm).
make x the subject of the relation \(y = \frac{ax^3 - b}{3z}\)
x = \(\sqrt[3] \frac{ax^3 - b}{3z}\)
x = \(\sqrt[3] \frac{3yz - b}{a}\)
x = \(\sqrt[3] \frac{3yz + b}{a}\)
x = \(\sqrt[3] \frac{3yzb}{a}\)
Correct answer is C
\(y = \frac{ax^3 - b}{3z}\)
cross multiply
\(ax^3 - b\) = 3yz
\(ax^3\) = 3yz + b
divide both sides by a
\(x^3 = \frac{3yz + b}{a}\)
take cube root of both sides
therefore, x = \(\sqrt[3] \frac{3yz + b}{a}\)
m:n = \(2\frac{1}{3} : 1\frac{1}{5}\) and n : q = \(1\frac{1}{2} : 1\frac{1}{3}\), find q : m.
35 : 18
16 : 35
18 : 35
35 : 16
Correct answer is B
m:n = \(2\frac{1}{3} : 1\frac{1}{5}\) = m : n = \(\frac{7}{3} : \frac{6}{5}\)
\(\frac{7}{3} : \frac{6}{5}\) = \(\frac{7}{3} \div \frac{6}{5}\)
\(\frac{m}{n}\) = \(\frac{7}{3} \times \frac{5}{6}\)
\(\frac{m}{n}\) = \(\frac{35}{18}\) = m = \(\frac{35n}{18}\)
n : q = \(1\frac{1}{2} : 1\frac{1}{3}\) = \(\frac{3}{2} : \frac{4}{3}\)
\(\frac{n}{q}\) = \(\frac{3}{2} \times\frac{3}{4}\)
\(\frac{n}{q}\) = \(\frac{9}{8}\) = q = \(\frac{8n}{9}\)
q : m = \(\frac{8n}{9}\) : \(\frac{35n}{18}\)
\(\frac{q}{m}\) = \(\frac{8n}{9} \div \frac{35n}{18}\)
\(\frac{q}{m}\) = \(\frac{8n}{9}\times\frac{18}{35n}\)
=\(\frac{q}{m} = \frac{16}{35}\) = q : m = 16 : 35
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