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.

2,261.

Solve the equation (x - 2) (x - 3) = 12

A.

2, 3

B.

3, 6

C.

-1, 6

D.

1, -6

Correct answer is C

(x - 2) (x - 3) = 12

x2 - 3x - 2x + 6 = 12

x2 - 5x - 6 = 0

(x +1)(x - 6) = 0

x = -1 or 6

2,262.

What is the nth term of the progression 27, 9, 3,......?

A.

27\(\frac{1}{3}\) n - 1

B.

3n + 2

C.

27 + 18(n - 1)

D.

27 + 6(n - 1)

Correct answer is A

Given 27, 9, 3,......this is a G.P

r = \(\frac{9}{27}\)

= \(\frac{1}{3}\)

T = arn - 1

= 27\(\frac{1}{3}\) n - 1

2,263.

Find the positive number n, such that thrice its square is equal to twelve times the number

A.

1

B.

2

C.

3

D.

4

Correct answer is D

3n2 = 12n

= 3n2 - 12n = 0

= 3n(n - 4) = 0

∴ n = 4

2,264.

Find the range of values of x which satisfy the inequality \(\frac{x}{2}\) + \(\frac{x}{3}\) + \(\frac{x}{4}\) < 1

A.

x < \(\frac{12}{13}\)

B.

x < 13

C.

x < 9

D.

\(\frac{13}{12}\)

Correct answer is A

\(\frac{x}{2}\) + \(\frac{x}{3}\) + \(\frac{x}{4}\)< 1

= \(\frac{6x + 4x + 3x < 12}{12}\)

i.e. 13 x < 12 = x < \(\frac{12}{13}\)

2,265.

Find the two values of y which satisfy the simultaneous equation 3x + y = 8, x\(^2\) + xy = 6.

A.

-1 and 5

B.

-5 and 1

C.

1 and 5

D.

1 and 1

Correct answer is A

\(3x + y = 8 ... (i)\)

\(x^{2} + xy = 6 ... (ii)\)

From (i), \(y = 8 - 3x\)

From (ii), \(xy = 6 - x^{2} \implies y = \frac{6 - x^{2}}{x}\)

Equating the two values of y, we have

\(8 - 3x = \frac{6 - x^{2}}{x} \implies x(8 - 3x) = 6 - x^{2}\)

\(8x - 3x^{2} = 6 - x^{2} \implies 6 - x^{2} - 8x + 3x^{2} = 0\)

\(2x^{2} - 8x + 6 = 0\)

\(x^{2} - 4x + 3 = 0\)

\(x^{2} - 3x - x + 3 = 0 \implies x(x - 3) - 1(x - 3) = 0\)

\((x - 1)(x - 3) = 0 \therefore \text{x = 1 or 3}\)

\(y = 8 - 3x \)

When x = 1, \(y = 8 - 3(1) = 5\)

When x = 3, \(y = 8 - 3(3) = -1\)

\(\therefore \text{y = -1 or 5}\)