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.

2,026.

Rationalize \(\frac{5\sqrt{7} - 7\sqrt{5}}{\sqrt{7} - \sqrt{5}}\)

A.

-2\(\sqrt{35}\)

B.

4\(\sqrt{7}\) - 6\(\sqrt{5}\)

C.

-\(\sqrt{35}\)

D.

4\(\sqrt{7}\) - 8\(\sqrt{5}\)

E.

\(\sqrt{35}\)

Correct answer is C

\(\frac{5\sqrt{7} - 7\sqrt{5}}{\sqrt{7} - \sqrt{5}}\) = \(\frac{5\sqrt{7} - 7\sqrt{5}}{\sqrt{7} - \sqrt{5}}\) x \(\frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} + \sqrt{5}}\)

= \(\frac{(5 \times 7) + (5 \sqrt{7} \times 5) - (7 \times \sqrt{5} \times 7) (-7 \times 5)}{(\sqrt{7})^2}\)

= \(\frac{5 \sqrt{35} - 7\sqrt{35}}{2}\)

= \(\frac{-2\sqrt{35}}{2}\)

= - \(\sqrt{35}\)

2,027.

A trader in country where their currency 'MONI'(M) is in base five bought 1035 oranges at M145 each. If he sold the oranges at M245 each, what will be his gain?

A.

1035

B.

10305

C.

1025

D.

20025

E.

30325

Correct answer is B

Total cost of 1035 oranges at N145 each

= 1035 x 145

= 20025

Total selling price at N245 each

= (103)5 x 245

= 30325

Hence his gain = 30325 - 20025

= 10305

2,028.

A variable point p(x, y) traces a graph in a two-dimensional plane. (0, 3) is one position of P. If x increases by 1 unit, y increases by 4 units. The equation of the graph is

A.

-3 = \(\frac{y + 4}{x + 1}\)

B.

4y = -3 + x

C.

\(\frac{y}{x}\) = \(\frac{-3}{4}\)

D.

4x = y + 3

Correct answer is D

P(x, y), P(0, 3) If x increases by 1 unit and y by 4 units, then ratio of x : y = 1 : 4

\(\frac{x}{1}\) = \(\frac{y}{4}\)

y = 4x

Hence the sign of the graph is y + 3 = 4x

2,029.

Find, without using logarithm tables, the value of \(\frac{log_3 27 - log_{\frac{1}{4}} 64}{log_3 \frac{1}{81}}\)

A.

\(\frac{-3}{9}\)

B.

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

C.

\(\frac{6}{11}\)

D.

\(\frac{43}{78}\)

Correct answer is B

\(\frac{\log_{3} 27 - \log_{\frac{1}{4}} 64}{\log_{3} (\frac{1}{81})}\)

\(\log_{3} 27 = \log_{3} 3^{3} = 3\log_{3} 3 = 3\)

\(\log_{\frac{1}{4}} 64 = \log_{\frac{1}{4}} (\frac{1}{4})^{-3} = -3\)

\(\log_{3} (\frac{1}{81}) = \log_{3} 3^{-4} = -4\)

\(\therefore \frac{\log_{3} 27 - \log_{\frac{1}{4}} 64}{\log_{3} (\frac{1}{81})} = \frac{3 - (-3)}{-4}\)

= \(\frac{6}{-4} = \frac{-3}{2}\)

2,030.

Bola choose at random a number between 1 and 300. What is the probability that the number is divisible by 4?

A.

\(\frac{1}{4}\)

B.

\(\frac{4}{4}\)

C.

\(\frac{6}{4}\)

D.

\(\frac{7}{4}\)

E.

\(\frac{37}{149}\)

Correct answer is E

Numbers divisible by 4 between 1 and 300 include 4, 8, 12, 16, 20 e.t.c.To get the number of figures divisible by 4, We solve by method of A.P

Let x represent numbers divisible by 4, nth term = a + (n - 1)d

a = 4, d = 4

Last term = 4 + (n - 1)4

296 = 4 + 4n - 4

= \(\frac{296}{4}\)

= 74
rn(Note: 296 is the last Number divisible by 4 between 1 and 300)

Prob. of x = \(\frac{74}{298}\)

= \(\frac{37}{149}\)