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.

2,621.

The seconds term of a geometric series is 4 while the fourth term is 16. Find the sum of the first five terms

A.

60

B.

62

C.

54

D.

64

Correct answer is B

T2 = 4, T4 = 16

Tx = arn-1

T2 = ar2-1 = 4 i.e. ar3 = 16, i.e. ar = 4

T4 = ar4-1

therefore, \(\frac{T_4}{T_r}\) = \(\frac{ar^3}{ar}\) = \(\frac{16}{4}\)

r2 = 4 and r = 2

but ar = 4

a = \(\frac{4}{r}\) = \(\frac{4}{2}\)

a = 2

Sn = \(\frac{a(r^n - 1)}{r - 1}\)

S5 = \(\frac{2(2^5 - 1)}{2 - 1}\)

= \(\frac{2(32 - 1)}{2 - 1}\)

= 2(31)

= 62

2,622.

Find the sum of the first 18 terms of the series 3, 6, 9,..., 36.

A.

505

B.

513

C.

433

D.

635

Correct answer is B

3, 6, 9,..., 36.

a = 3, d = 3, i = 36, n = 18

Sn = \(\frac{n}{2}\) [2a + (n - 1)d

S18 = \(\frac{18}{2}\) [2 x 3 + (18 - 1)3]

= 9[6 + (17 x 3)]

= 9 [6 + 51] = 9(57)

= 513

2,623.

Solve the inequality x2 + 2x > 15.

A.

x < -3 or x > 5

B.

-5 < x < 3

C.

x < 3 or x > 5

D.

x > 3 or x < -5

Correct answer is B

x2 + 2x > 15

x2 + 2x - 15 > 0

(x2 + 5x) - (3x - 15) > 0

x(x + 5) - 3(x + 5) >0

(x - 3)(x + 5) > 0

therefore, x = 3 or -5

i.e. x< 3 or x > -5

2,624.

Solve the inequality -6(x + 3) \(\leq\) 4(x - 2)

A.

x \(\leq\) 2

B.

x \(\geq\) -1

C.

x \(\geq\) -2

D.

x \(\leq\) -1

Correct answer is B

-6(x + 3) \(\leq\) 4(x - 2)

-6(x +3) \(\leq\) 4(x - 2)

-6x -18 \(\leq\) 4x - 8

-18 + 8 \(\leq\) 4x +6x

-10 \(\leq\) 10x

10x \(\geq\) -10

x \(\geq\) -1

2,625.

T varies inversely as the cube of R. When R = 3, T = \(\frac{2}{81}\), find T when R = 2

A.

\(\frac{1}{18}\)

B.

\(\frac{1}{12}\)

C.

\(\frac{1}{24}\)

D.

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

Correct answer is B

T \(\alpha \frac{1}{R^3}\)

T = \(\frac{k}{R^3}\)

k = TR3

= \(\frac{2}{81}\) x 33

= \(\frac{2}{81}\) x 27

dividing 81 by 27

k = \(\frac{2}{2}\)

therefore, T = \(\frac{2}{3}\) x \(\frac{1}{R^3}\)

When R = 2

T = \(\frac{2}{3}\) x \(\frac{1}{2^3}\) = \(\frac{2}{3}\) x \(\frac{1}{8}\)

= \(\frac{1}{12}\)