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.

1,906.

A micrometer is defined as one millionth of a millimeter. A length of 12,000 micrometres may be represented as

A.

0.00012m

B.

0.0000012m

C.

0.000012m

D.

0.00000012m

E.

0.000000012m

Correct answer is C

1 UM = 10-6mm = 10-9m
1.2 x 104 x 10-9m = 1.2 x 10-5m (0.000012)

1,907.

An isosceles triangle of sides 13cm, 13cm, 10cm is inscribed in a circle. What is the radius of the circle?

A.

7cm

B.

12cm

C.

8cm

D.

36cm

E.

69cm

Correct answer is A

In \(\Delta DAC, \stackrel\frown{DAC} = \theta\)

\(\sin \theta = \frac{5}{13}\)

\(\theta = 22.6°\)

\(< DOC = 22.6° \times 2 = 45.2°\)

\(\sin 45.2 = \frac{5}{r} \implies r = \frac{5}{\sin 45.2}\)

\(r = 7.046cm\)

= \(7\frac{1}{24} cm\)

1,908.

By selling an article for N45.00 a man makes a profit of 8%. For how much should he have sold it in order to make a profit of 32%?

A.

N180.00

B.

N59.00

C.

N63.00

D.

N58.00

E.

N55.00

Correct answer is E

When C.P = N45.00 at 8% profit

C.P = \(\frac{100}{108}\) x 45

at profit of 32%, S.P = \(\frac{132}{100}\) x c.p

= \(\frac{132}{100}\) x \(\frac{100 \times 45}{108}\)

= \(\frac{5940}{108}\)

= N55.00

1,909.

A sum of money invested at 5% per annum simple interest amounts to $285.20 after 3 years. How long will it take the same sum to amount to $434.00 at 7\(\frac{1}{2}\)% per annum simple interest?

A.

7\(\frac{1}{2}\) years

B.

10 years

C.

5 years

D.

12 years

E.

14 years

Correct answer is B

\(A = P(1 + \frac{R}{100})^{T}\)

\(285.20 = P(1 + \frac{5}{100})^{3}\)

\(285.20 = P(1.05)^{3} \implies 1.16P = 285.20\)

\(P = \frac{285.20}{1.16} = $245.86\)

Given Amount = $434.00

Interest = Amount - Principal

Interest = $434.00 - $245.86 = $188.14

\(T = \frac{100I}{PR}\)

\(T = \frac{100 \times 188.14}{245.86 \times 7.5}\)

\(T = \frac{18814}{1843.95} = 10.2\)

Approximately 10 years.

1,910.

7 pupils of average age 12 years leave a class of 25 pupils of average age 14 years. If 6 new pupils of average age 11years join the class, what is the average age of the pupils now in the class?

A.

13years

B.

12 years 7\(\frac{1}{2}\) months

C.

13 years 5 months

D.

13 years 10 months

E.

11 years

Correct answer is D

Total age of the 7 pupils = 7 x 12 = 84

Total age of the 25 pupils = 25 x 14 = 350

Total age of the 6 pupils = 6 x 11 = 66

7 pupils leaving a class of 25 pupils given 18 pupils

Total age of the 18 pupils = 350 - 84 = 266

6 pupils joining 18 pupils = 24

Total age of 24 pupils = 266 + 66 = 332

Average age of 24 pupils now in class \(\frac{332}{24}\)

= 13yrs 10 months