Campus_12 (Math - Aptitude)

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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


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