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.

2,061.

Factorize abx2 + 8y - 4bx - 2axy

A.

(ax - 4)(bx - 2y)

B.

(ax + b)(x - 8y)

C.

(ax - 2y)(bx - 4)

D.

(bx - 4)(ax - 2y)

E.

(abx - 4)(x - 2y)

Correct answer is A

abx2 + 8y - 4bx - 2axy = (abx2 - 4bx) + (8y - 2axy)

= bx(ax - 4) 2y(ax - 4) 2y(ax - 4)

= (bx - 2y)(ax - 4)

2,062.

If 32y + 6(3y) = 27. Find y

A.

3

B.

-1

C.

2

D.

-3

E.

1

Correct answer is E

32y + 6(3y) = 27

This can be rewritten as (3y)2 + 6(3y) = 27

Let 3y = x

x2 + 6x - 27 = 0

(x + 9)(x - 3) = 0

when x - 3 = 0, x = 3

sub. for x in 3y = x

3y = 3

log33 = y

y = 1

2,063.

The factors of 9 - (x2 - 3x - 1)2 are

A.

-(x - 4)(x + 1) (x - 1)(x - 2)

B.

(x - 4)(x - 2) (x - 1)(x + 1)

C.

-(x - 2)(x + 1) (x - 2) (x - 1)

D.

(x - 2)(x + 2) (x - 1)(x + 1)

Correct answer is A

9 - (x2 - 3x - 1)2 = [3 - (x2 - 3x - 1)] [3 + (x2 - 3x - 1)]

= (3 - x2 + 3x + 1)(3 + x2 - 3x - 1)

= (4 + 3x - x2)(x2 - 3x + 2)

= (4 - x)(1 + x)(x - 1)(x - 2)

= -(x - 4)(x + 1) (x - 1)(x - 2)

2,064.

In a restaurant, the cost of providing a particular type of food is partly constant and partially inversely proportional to the number of people. If cost per head for 100 people is 30k and the cost for 40 people is 60k, Find the cost for 50 people?

A.

15k

B.

20k

C.

50k

D.

40k

E.

45k

Correct answer is C

C = a + k

\(\frac{1}{N}\) = c

= \(\frac{aN + k}{N}\)

CN = aN + K

30(100) = a(100) + k

3000 = 100a + k.......(i)

60(40) = a(40) + k

2400 = 40a + k.......(ii)

eqn (i) - eqn (ii)

600 = 60a

a = 10

subt. for a in eqn (i) 3000 = 100(10) + K

3000 - 1000 = k

k = 2000

CN = 10N + 2000. when N = 50,

50C = 10(50) + 2000

50C = 500 + 2000

C = \(\frac{2500}{50}\)

= 50k

2,065.

\((\sqrt[4]{3} + \sqrt[4]{2})(\sqrt[4]{3} - \sqrt[4]{2})(\sqrt{3} + \sqrt{2})\) is equal to

A.

1

B.

(\(\sqrt{2} + 4\sqrt{2}\))

C.

(6\(\sqrt{2}\)

D.

8

Correct answer is A

\((\sqrt[4]{3} + \sqrt[4]{2})(\sqrt[4]{3} - \sqrt[4]{2})(\sqrt{3} + \sqrt{2})\)

\((\sqrt[4]{3} + \sqrt[4]{2})(\sqrt[4]{3} - \sqrt[4]{2}) = \sqrt{3} - \sqrt[4]{6} + \sqrt[4]{6} - \sqrt{2}\)

= \(\sqrt{3} - \sqrt{2}\)

\((\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2}) = 3 + \sqrt{6} - \sqrt{6} - 2\)

= \(3 - 2 = 1\)