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.

2,821.

There are m boys and 12 girls in a class. What is the probability of selecting at random a girl from the class?

A.

\(\frac{m}{12}\)

B.

\(\frac{12}{m}\)

C.

\(\frac{12}{m+12}\)

D.

\(\frac{12}{m-12}\)

Correct answer is C

Prob. (a girl) \(=\frac{Number\hspace{1mm} of\hspace{1mm} girls}{Number \hspace{1mm} of \hspace{1mm} boys \hspace{1mm} and \hspace{1mm} girls\hspace{1mm} in \hspace{1mm} class}\\
= \frac{12}{m+12}\)

2,822.

In the diagram, \(R\hat{P}Q = Q\hat{R}Y, \hspace{1mm} P\hat{Q}R = R\hat{Y}Q, \\ \hspace{1mm}|QP| = 3cm \hspace{1mm}|QY| = 4cm \hspace{1mm}and \hspace{1mm}|RY | = 5cm. \hspace{1mm} Find \hspace{1mm}|QR|\)

A.

2.0cm

B.

2.5cm

C.

6.4cm

D.

10.0cm

Correct answer is C

\(\frac{PR}{QR} = \frac{QR}{QY} = \frac{QP}{YR}\) (SSS Congruence)

\(\frac{QR}{4} = \frac{8}{5}\)

\(QR = \frac{4 \times 8}{5}\)

= 6.4 cm

2,823.

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 pocket money be rep. by x. The amount spent on books = \(\frac{1}{4}of\hspace{1mm}x = \frac{x}{4}\)
The amount spent on dress \(=\frac{1}{3} of \hspace{1mm}x=\frac{x}{3}\)
∴The fraction that remains = \(\frac{x}{1}-\left(\frac{x}{4}+\frac{x}{3}\right)\\
\frac{3x+4x}{12}; = \frac{12x-7x}{12} = \frac{5x}{12} = \frac{5}{12}\)

2,824.

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

A.

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

B.

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

C.

5

D.

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

Correct answer is D

\(\frac{3}{4}t+\frac{1}{3}(21-t) = 11; \frac{3t}{4} + \frac{7}{1} - \frac{t}{3} = \frac{11}{1}\\
\frac{3 \times 3t + 7\times 12 – 4 \times t = 11 \times 12}{12}\\
9t + 84 – 4t = 132; 5t = 132 – 84\\
5t = 48; t = \frac{48}{5} = 9\frac{3}{5}\)

2,825.

Given that sin (5x - 28)o = cos (3x - 50)o,0 < x < 90o, find the value of x

A.

14o

B.

21o

C.

32o

D.

39o

Correct answer is B

Sin (5x – 28)o = cos (3x - 50)o
Since by the trigonometry relation
Sin(5x – 28)o = cos[90 – (5x – 28)]o
Hence cos(3x – 50)o = cos[90 – (5x – 28)]o
3x – 50 = 90 - (5x-28)
3x – 50 = 90 – 5x + 28
3x + 5x = 90 + 28 + 50
8x = 168
\(x = \frac{168}{8}=21^{\circ}\)