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.

726.

The pie chart shows the allocation of money to each sector in a farm. The total amount allocated to the farm is ₦ 80 000. Find the amount allocated to fertilizer

A.

₦ 35, 000

B.

₦ 40,000

C.

₦ 25,000

D.

₦ 20,000

Correct answer is D

Total angle at a point = 3600

∴ To get the angle occupied by fertilizer we have,

40 + 50 + 80 + 70 + 30 + fertilizer(x) = 360

270 + x = 360

x = 360 - 270

x = 90

Total amount allocated to the farm
= ₦ 80,000

∴Amount allocated to the fertilizer

= \(\frac{\text{fertilizer (angle) × Total amount}}{\text{total angle}}\)

= \(\frac{90}{360}\) × 80,000

= ₦20,000

727.

Find the principal which amounts to ₦ 5,500 at a simple interest in 5 years at 2% per annum

A.

₦ 4,900

B.

₦ 5,000

C.

₦ 4,700

D.

₦ 4,000

Correct answer is B

Principal = P, Simple Interest = I, Amount = A

Amount = Principal + Simple Interest

I = \(\frac{PRT}{100}\)

R = rate, T = time

I = \(\frac{P \times 5 \times 2}{100}\)

I = \(\frac{10P}{100}\)

I = \(\frac{P}{10}\)

Amount A = P + I

5500 = P + \(\frac{P}{10}\)

Multiply through by 100

5500 = 10P + P

5500 = 11P

p = \(\frac{5500}{11}\)

p = ₦5000

728.

If a rod 10cm in length was measured as 10.5cm, calculate the percentage error

A.

5%

B.

5%

C.

8%

D.

7%

Correct answer is A

Actual measurement = 10cm

approximated value of measurement = 10.5cm

% error = \(\frac{\text{Actual measurement − Approximated}}{\text{Actual measure}}\) × 100

= \(\frac{10 − 10.5}{10}\) × 100

= \(\frac{-0.5}{10}\) × 100

ignore -sign i.e take absolute value

= \(\frac{0.5}{10}\) × 100

= 5 %

729.

Evaluate \(\frac{0.00000231}{0.007}\) and leave the answer in standard form

A.

3.3 x 10-4

B.

3.3 x 10-3

C.

3.3 x 10-5

D.

3.3 x 10-8

Correct answer is A

\(\frac{0.00000231}{0.007}\) to standard form

= \(\frac{231 \times 10^{-8}}{7 \times 10^{-3}}\)

= 33 × 10\(^{-8 − (−3)}\)

= 33 × 10\(^{− 8 + 3}\)

= 33 × 10-5

= 3.3 x 10^-4

730.

Find the number of ways that the letters of the word EXCELLENCE be arranged

A.

\(\frac{10!}{2!2!2!}\)

B.

\(\frac{10!}{4!2!}\)

C.

\(\frac{10!}{4!2!2!}\)

D.

\(\frac{10!}{2!2!}\)

Correct answer is C

EXCELLENCE

It is a ten letter word = 10!

Since we have repeating letters, we have to divide to remove the duplicates accordingly. There are 4 Es, 2 Cs, 2 Ls

∴ there are
\(\frac{10!}{4!2!2!}\) ways to arrange