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.

2,141.

Make y the subject of the formula Z = x\(^2\) + \(\frac{1}{y^3}\)

A.

y = \(\frac{1}{(Z - x^2)^3}\)

B.

y = \(\frac{1}{(Z + x^2)^{\frac{1}{3}}}\)

C.

y = \(\frac{1}{(Z - x^2)^{\frac{1}{3}}}\)

D.

y = \(\frac{1}{\sqrt[3]{Z} - \sqrt[3]{x^2}}\)

Correct answer is C

Z = x\(^2\) + \(\frac{1}{y^3}\)

Z - x\(^2\) = \(\frac{1}{y^3}\)

y\(^3\) = \(\frac{1}{Z - x^2}\)

y = \(\sqrt[3]{\frac{1}{Z - x^2}}\)

∴ y = \(\frac{1}{\sqrt[3]{Z - x^2}}\)

y = \(\frac{1}{(Z - x^2)^{\frac{1}{3}}}\)

2,142.

If P varies inversely as V and V varies directly as R2, find the relationship between P and R given that R = 7 when P = 2

A.

P = 98R2

B.

PR2 = 98

C.

P = \(\frac{1}{98R2

D.

P = \(\frac{PR2}{98}\)

Correct answer is B

P = \(\frac{1}{v}\) and vR2 = P = \(\frac{k}{v}\)......(i)

and v KR2 .......(ii)

(where k is constant)

Subst. for v in equation (i) = p = \(\frac{1^2}{KR}\).....(ii)

when r = 7, p = 2

2 = \(\frac{k}{7^2}\)

k = 2 x 49

= 98

Subt. foe k in ....(iii)

P = \(\frac{98}{R^2}\)

PR2 = 98

2,143.

The formula Q = 1.5 + 0.5n gives the cost Q(in Naira)of feeding n people for a week. Find (in kobo) the extra cost of feeding one additional person

A.

350k

B.

200k

C.

150k

D.

50k

Correct answer is B

Q = 1.5 + 0.5n gives the cost 1(in Naira) of feeding n people for a week. Extra cost of feeding one additional person = n = 1 Subt. for n in the formula Q = 1.5 + 0.5(1) = 15 + 0.5 Q = N2 = 200k

2,144.

If a u2 - 3v2 and b = 2uv + v2 evaluate (2a - b)(a - b2), when u = 1 and v = -1

A.

9

B.

15

C.

27

D.

33

Correct answer is A

a = u2 - 3y2 = (1)2 - 3(-1) = -2

b = 2uu + v2 + v2

= 2(1)(-1) + (-1)2 = -1

∴ 2(2a - b)(a - b∴) = [2(-2) - 1](-2 - (-1)2)

= -3 - 3

= 9

2,145.

If P = \(\frac{2}{3}\) (\(\frac{1 - r^2}{n^2}\)), find n when r = \(\frac{1}{3}\) and p = 1

A.

\(\frac{3}{2}\)

B.

\(\frac{1}{3}\)

C.

3

D.

\(\frac{2}{3}\)

Correct answer is D

If P = \(\frac{2}{3}\) (\(\frac{1 - r^2}{n^2}\)), find n when r = \(\frac{1}{3}\) and p = 1

p = \(\frac{2(1 - r^2)}{3n^2}\) when r = \(\frac{1}{3}\) and p = 1

1 = \(\frac{2}{3}\) \(\frac{(1 - (\frac{1}{3})^2)}{n^2}\)

n2 = \(\frac{2(3 - 1)}{3 \times 3}\)

n2 = \(\frac{2 \times 2}{3 \times 3}\)

= \(\frac{4}{9}\)

n = \(\frac{4}{9}\)

= \(\frac{2}{3}\)