math-aptitude for placement / aptitude / interview/assessment/multiple choice/career/campus

math-aptitude for placement / aptitude / interview/assessment/multiple choice/career/campus



1. A trader has 100 Kg of wheat, part of which he sells at 5% profit and the rest at 20% profit. He gains 15% on the whole. Find how much is sold at 5% profit?
(A) 60
(B)50
(C)66.66
(D) 33.3

2. In 1978, a kg of paper was sold at Rs25/-. If the paper rate increases at 1.5% more than the inflation rate which is 6.5% a year, then what wil be the cost of a kg of paper after 2 years?

3. A driver drives at a speed of 72kmph in a highway. Another driver drives at the speed of 25 m/sec . find the difference in their speeds in m/sec. ?

4. In a mixture R is 2 parts and S is 1 part. In order to make
S to 25% of the mixture how r is needed ?






1. I am three times as old as my son. Five years later I shall be two and a half times as old as my son. What is my age?

Answer: Let my age be x years and age of my son be y years.
(i) x = 3y
Five years later
(ii) x+5 = 5/2(y+5)
From (i) and (ii)
y=15 and x=45

2. The L.C.M. (Lowest (or Least) Common Multiple) of two numbers is 45 times to their H.C.F (Highest Common Factor). If one of the numbers is 125 and sum of L.C.M. and H.C.F. is
1150, the other number is:
a) 215
b) 220
c) 225
d) 235

Answer: Let L.C.M. be l and H.C.F. be h. Then l = 45h
l + h =1150
This gives h =25 and l = 1125
Thus the second number = (25 * 1125)/125 = 225

3. Six bells commence tolling together and toll at intervals 2,4,6,8,10 and 12 seconds respectively. In 30 minutes how many times they toll together.
a) 4
b) 10
c) 15
d) 16

Answer: L.C.M. of 2,4,6,8,10,12 = 120
So they toll together after every 120 seconds i.e. 2 minutes. So in 30 minutes they toll together 30/2 + 1 = 16 times

4. The value of (0.625 * 0.0729 * 28.9)/(0.0017 * 0.025 * 8.1) is
a) 3825
b) 3.825
c) 38.25
d) 382.5

Answer: The sum of decimal places in numerator and denominator being the same, decimal point can be removed
(625*729*289) / (17*25*81) = 3825


5. The number of coins 1.5 cm in diameter and 0.2cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm is:
a) 380
b) 450
c) 472
d) 540

Answer: Volume of 1 coin = 22/7 *(1.5)/2*(1.5)/2 * 0.2 = 99/280 cm3
Volume of large cylinder = 22/7 *(4.5)/2*(4.5)/2 * 10 = (99*45)/28cm3
Number of coins = (99*45)/28 * 280/99 = 450