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.

731.

A car dealer bought a second-hand car for 250,000 and spent N 70,000 refurbishing it. He then sold the car for N400,000. What is the percentage gain?

A.

60%

B.

32%

C.

25%

D.

20%

Correct answer is C

Total Cost Price = N(250,000 + 70,000)

= N 320,000

Selling Price = N 400,000(Given)

Gain = Selling Price - Cost Price

= 400,000 - 320,000

= 80,000

% gain = \(\frac{\text{Gain}}{\text{Cost Price}}\) × 100

= \(\frac{80,000}{320,000}\) × 100

Gain % = 25%

732.

Divide 4x3 - 3x + 1 by 2x - 1

A.

2x2 -x + 1

B.

2x2 - x -1

C.

2x2 + x + 1

D.

2x2 + x -1

Correct answer is D

by method of long division, we get the answer.

733.

In the figure, find x

A.

40o

B.

55o

C.

50o

D.

60o

Correct answer is A

Sum of angle at a point = 360o

2x + 3x + 4x = 360

9x = 360

x = \(\frac{360}{9}\)

x = 40o

734.

A man covered a distance of 50 miles on his first trip, on a later trip he traveled 300 miles while going 3 times as fast. His new time compared with the old distance was?

A.

three times as much

B.

the same

C.

twice as much

D.

half as much

Correct answer is C

Let the speed of the 1st trip be x miles/hr

and the speed of the 2nd trip be 3x miles/hr

Speed = distance/time

∴ Time taken to cover a distance of 50 miles on the 1st trip

= \(\frac{50}{xhr}\)

time taken to cover a distance of 300 miles on the next trip

= \(\frac{300}{3xhr}\)

= \(\frac{100}{xhr}\)

∴the new time compared with the old time is twice as much

735.

Simplify 4\(\sqrt{27}\) + 5\(\sqrt{12}\) − 3\(\sqrt{75}\)

A.

7

B.

− 7

C.

− 7\(\sqrt{3}\)

D.

7\(\sqrt{3}\)

Correct answer is D

4\(\sqrt{27}\) + 5\(\sqrt{12}\) − 3\(\sqrt{75}\)

= 4\(\sqrt{3}\) × 9 + 5\(\sqrt{3}\) × 4 − 3\(\sqrt{3}\) × 25

= 4 × 3\(\sqrt{3}\) + 5 × 2\(\sqrt{3}\) − 3 × 5\(\sqrt{3}\)

= 12\(\sqrt{3}\) + 10\(\sqrt{3}\) − 15\(\sqrt{3}\)

= (12 + 10 − 15)\(\sqrt{3}\)

= 7\(\sqrt{3}\)