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.

2,421.

A binary operation \(\oplus\) is defines on the set of all positive integers by a \(\oplus\) b = ab for all positive integers a, b. Which of the following properties does NOT hold?

A.

closure

B.

identity

C.

positive

D.

inverse

Correct answer is D

a \(\oplus\) b = ab

The set of all national rules Q, is closed under the operations, additions, subtraction, multiplication and division. Since a \(\oplus\) b = ab; b \(\oplus\) a = ba = ab

The number 1 is the identity element under multiplication

2,422.

Find the nth term of the sequence 3, 6, 10, 15, 21.....

A.

\(\frac{n(n - 1)}{2}\)

B.

\(\frac{n(n + 1)}{2}\)

C.

\(\frac{(n + 1)(n + 2)}{2}\)

D.

n(2n + 1)

Correct answer is C

\(\frac{(n + 1)(n + 2)}{2}\)

If n = 1, the expression becomes 3

n = 2, the expression becomes 6

n = 4, the expression becomes 15

n = 5, the expression becomes 21

2,423.

Find the value of log10 r + log10 r2 + log10 r4 + log10 r8 + log10 r16 + log10 r32 = 63

A.

10-1

B.

10o

C.

10

D.

102

Correct answer is C

log10 r + log10 r2 + log10 r4 + log10 r8 + log10 r16 + log10 r32 = 63

log10r63 = 63

63 = 1063

∴ r = 10

2,424.

If the 6th term of an arithmetic progression is 11 and the first term is 1, find the common difference.

A.

\(\frac{12}{5}\)

B.

\(\frac{12}{5}\)

C.

-2

D.

2

Correct answer is D

In an AP, Tn = a + (n - 1)d

T6 = a + 5d = 11

The first term = a = 1

∴ T6 = 1 + 5d = 11

5d = 11 - 1

5d = 10

∴ d = 2

2,425.

Find the range of values of x for which \(\frac{1}{x}\) > 2 is true

A.

x < \(\frac{1}{2}\)

B.

x < 0 or x < \(\frac{1}{2}\)

C.

0 < x < \(\frac{1}{2}\)

D.

1 < x < 2

Correct answer is C

\(\frac{1}{x}\) > 2 = \(\frac{x}{x^2}\) > 2

x > 2x2

= 2x2 < x

= 2x2 - x < 0

= x(2x - 10 < 0

Case 1(+, -) = x > 0, 2x - 1 < 0

x > 0, x < \(\frac{1}{2}\) (solution)

Case 2(-, 4) = x < 0, 2x - 1 > 0

x < 0, x , \(\frac{1}{2}\) = 0