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.

76.

The difference between an exterior angle of (n - 1) sided regular polygon and an exterior angle of (n + 2) sided regular polygon is 6\(^o\), then the value of "n" is

A.

11

B.

13

C.

12

D.

14

Correct answer is B

An exterior angle of a n-sided regular polygon = \(\frac{360}{n}\)

For (n - 1) sided regular polygon = \(\frac{360}{n - 1}\)

For (n + 2) sided regular polygon = \(\frac{360}{n + 1}\)

⇒ \(\frac{360}{n - 1} - \frac{360}{n + 2}\) = 6 9Given)

⇒ \(\frac{360(n + 2) - 360(n - 1)}{(n - 1)(n + 2)}\)

⇒ \(\frac{360n + 720 - 360n + 360}{(n - 1)(n + 2)}\)

⇒ \(\frac{1080}{(n - 1)(n + 2)} = \frac{6}{1}\)

⇒ 1080 = 6 (n - 1)(n + 2)

⇒ 180 = (n - 1)(n + 2)

⇒ 180 = n\(^2\)+ 2n - n - 2

⇒ 180 = n\(^2\) + n - 2

⇒ n\(^2\)b+ n - 2 - 180 = 0

⇒ n\(^2\) + n - 182 = 0

⇒ n\(^2\) + 14n - 13n - 182 = 0

⇒ n (n + 14) - 13 (n + 14) = 0

⇒ (n - 13) (n + 14) = 0

⇒ n - 13 = 0 or n + 14 = 0

⇒ n = 13 or n = -14

∴ n = 13 (We can't have a negative number of side)

77.

A student is using a graduated cylinder to measure the volume of water and reports a reading of 18 mL. The teacher reports the value as 18.4 mL. What is the student's percent error?

A.

2.17%

B.

1.73%

C.

2.23%

D.

1.96%

Correct answer is A

% error = \(\frac{Error}{Actual Value}\) x 100%

Student's Value = 18mL

Actual Value = 18.4mL

Error = 18.4 - 18 = 0.4

∴ % error = \(\frac{0.4}{18.4}\) x 100% = 2.17%

78.

The graph above depicts the performance ratings of two sports teams A and B in five different seasons

In the last five seasons, what was the difference in the average performance ratings between Team B and Team A?

A.

1.2

B.

6.4

C.

4.6

D.

1.8

Correct answer is A

Average performance rating of Team B = \(\frac{7+9+1+9+6}{5} = \frac{32}{5}\) = 6.4

Average performance rating of Team A = \(\frac{5+3+6+10+2}{5} = \frac{26}{5}\) = 5.2

∴ The difference in the average performance ratings between Team B and Team A = 6.4 - 5.2 = 1.2

79.

Express 16.54 x 10\(^{-5}\) - 6.76 x 10\(^{-8}\) + 0.23 x 10\(^{-6}\) in standard form

A.

1.66 x 10\(^{-4}\)

B.

1.66 x 10\(^{-5}\)

C.

1.65 x 10\(^{-5}\)

D.

1.65 x 10\(^{-4}\)

Correct answer is A

16.54 x 10\(^{-5}\) - 6.76 x 10\(^{-8}\) + 0.23 x 10\(^{-6}\)

⇒ 1.654 x 10\(^{-4}\) - 6.76 x 10\(^{-8}\) + 2.3 x 10\(^{-7}\)

⇒ 1.654 x 10\(^{-4}\) - 0.000676 x 10\(^{-4}\) + 0.0023 x 10\(^{-4}\)

⇒ (1.654 - 0.000676 + 0.0023) x 10\(^{-4}\)

∴ 1.655624 x 10\(^{-4}\) ≃ 1.66 x 10\(^{-4}\)

80.

The third term of an A.P is 6 and the fifth term is 12. Find the sum of its first twelve terms

A.

201

B.

144

C.

198

D.

72

Correct answer is C

T\(_3\) = 6

T\(_5\) = 12

S\(_{12}\) = ?

T\(_n\) = a + (n - 1)d

⇒ T\(_3\) = a + 2d = 6 ----- (i)

⇒ T\(_5\) = a + 4d = 12 ----- (ii)

Subtract equation (ii) from (i)

⇒ -2d = -6

⇒ d\(\frac{-6}{-2}\) = 3

Substitute 3 for d in equation (i)

⇒ a + 2(3) = 6

⇒ a + 6 = 6

⇒ a = 6 - 6 = 0

S\(_n\) = \(\frac{n(2a + (n - 1)d)}{2}\)

⇒ S\(_{12}\) = \(\frac{12(2 \times 0 + (12 - 1)3)}{2}\)

⇒ S\(_{12}\) = 6(0 + 11 x 3)

⇒ S\(_{12}\) = 6(33)

∴ S\(_{12}\) = 198