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.

701.

A school girl spends \(\frac{1}{4}\) of her pocket money on books and \(\frac{1}{3}\) on dress. What fraction remains?

A.

\(\frac{5}{6}\)

B.

\(\frac{7}{12}\)

C.

\(\frac{5}{12}\)

D.

\(\frac{1}{6}\)

Correct answer is C

let the girls' initial pocket money be whole(1)

\(\frac{1}{4}\) of 1 = books, \(\frac{1}{3}\)  of 1 = dress 

fraction of her money left = 1 - \(\frac{1}{4}\) - \(\frac{1}{3}\)

                                   =   \(\frac{5}{12}\)

 

 

702.

Solve for t in the equation \(\frac{3}{4}\)t + \(\frac{1}{3}\)(21 - t) = 11

A.

\(\frac{9}{13}\)

B.

\(\frac{7}{13}\)

C.

5

D.

9\(\frac{3}{5}\)

Correct answer is D

\(\frac{3}{4}\) t + \(\frac{1}{3}\) (21 - t) = 11

  Multiply through by the LCM of 4 and 3 which is 12

  12 x(\(\frac{3}{4}\) t) + 12 x (\(\frac{1}{3}\) (21 - t)) = (11 x 12)

  9t + 4(21 - t) = 132

  9t + 84 - 4t = 132

  5t + 84 = 132

  5t = 132 - 84 = 48

  t = \(\frac{48}{5}\)

  t = 9 \(\frac{3}{5}\)

  Answer is D

703.

Find the value of x in the diagram

A.

10°

B.

28°

C.

36°

D.

40°

Correct answer is D

The diagram shows angles at a point, the total angle at a point is 360

  x - 10 + 4x - 50 + 2x + 3x + 20 = 360

  10x - 40 = 360

  10x = 360 + 40

  10x = 400

  x = \(\frac{400}{10}\)

  x = 40

704.

If y = 23\(_{five}\) + 101\(_{three}\) , find y, leaving your answer in base two

A.

1110

B.

10111

C.

11101

D.

111100

Correct answer is B

y = 23\(_{five}\) + 101\(_{three}\)

23\(_{five}\) = \(2 \times 5^1 + 3 \times 5^0\)

= 13\(_{ten}\)

101\(_{three}\) = \(1 \times 3^2 + 0 \times 3^1 + 1 \times 3^0\)

= 10\(_{ten}\)

y\(_{ten}\) = 13\(_{ten}\) + 10\(_{ten}\)

= 23\(_{ten}\)

= 10111\(_{two}\)

705.

If y = 23\(_{five}\) + 101\(_{three}\) , find y, leaving your answer in base two

A.

1110

B.

10111

C.

11101

D.

111100

Correct answer is B

y = 23\(_{five}\) + 101\(_{three}\)

23\(_{five}\) = \(2 \times 5^1 + 3 \times 5^0\)

= 13\(_{ten}\)

101\(_{three}\) = \(1 \times 3^2 + 0 \times 3^1 + 1 \times 3^0\)

= 10\(_{ten}\)

y\(_{ten}\) = 13\(_{ten}\) + 10\(_{ten}\)

= 23\(_{ten}\)

= 10111\(_{two}\)