Quantitative Aptitude

Multiplication of 2x5 results into 10. So number of zeroes depend on the number of pairs of 2 and 5. In any factorial, number of 5’s is lesser than the number of 2’s. So, we need to count the maximum power of 5 in 155! [155/5] + [155/5²] + [155/5³] =31 + 6 +1 =38 Hence, number of zeroes is 38.

Prime Factor of 45= 3²x5 We will count the number of 32 and 5 in 671!, and which one is lesser in number would be the answer. No of 3’s= 671/3 + 671/9 + 671/27 + 671/81 + 671/243 = 223 + 74 + 24 + 8 + 2 = 331 No of 3²= 331/2 = 165 No of 5= 671/5 + 671/25 + 671/125 + 671/625 = 134 + 26 + 5 + 1 = 166 165 will be the answer because 3² is lower in number than 5.

LCM = 15 * HCF We know that LCM + HCF = 480 16 * HCF = 480 HCF = 30 Then LCM = 450 LCM = 15 HCF 30 * x * y = 15 * 30 x * y = 15 Factors are {1 x 15} and { 3 x 5} Both numbers less than LCM so take {3 x 5} Hence numbers are 3 * 30 = 90 and 5 * 30 = 150

Find the LCM for 4, 6, 7, 9 LCM= 2² * 3² * 7 = 252 To become perfect square all factors should be in power of 2. So, multiply it by 7 LCM = 2² * 3² * 7² = 1764 And it is perfect square of 42.

Total work done is LCM(10, 12, 15)=60 unit A’s efficiency = 60/10= 6 B’s efficiency = 60/12= 5 C’s efficiency = 60/15= 4 First two days all work together So, the work completed in first two days= 15 x 2 = 30 unit Remaining work= 60 - 30 = 30 unit If B completes 3 day work also = 3 x 5 = 15 unit Total work remaining= 30 + 15 = 45 unit Number of days B and C works= 45/9=5 Total number of days to complete the work = 2 + 5 = 7 days.

Shortcut Let they earn 1 Rs/hr.

Woman     Child      Earns
  6    +    4     =   10
  |         |         |60%
 __        max 6   =  16

Workload increases by 60% from 10 to 16. Children can work maximum 6 hours Then women work per day 16 - 6 = 10 So, it increases by 4 hours/day extra.

First method 
Let A and B take x hours to complete a work together. 
A alone would take (x + 64) and B alone would take (x + 4)hours to complete the work. 
A( x + 64) = x (A + B) 
64A =x B …………(1) 
B(x + 4)= x(A + B) 
4B = x A……………(2) 
from (1)and (2) 
64A = x * x A/4 
x² = 256 
x = 16 
A alone = 16 + 64 = 80 hours 

Shortcut method - 
x² = more of A * more of B 
x²= 64 * 4 
x²= 256 
x= 16

Take LCM (10, 15) = 30 Let leak pipe is A and A’s efficiency = 30/10 = 3 Let inlet pipe B and B’s efficiency= 30/15 = 2 Pipe A is emptying at 3 units/hour and Pipe B is filling using then emptying rate down to 2 units/hour. So, filling pipe efficiency is 3 - 2 = 1 unit/ hour. Pipe B will fill tank in 30/1=30 hours Filling rate = 4 litre/minute It will fill 4 x 60 = 240 litre/hour. Total capacity= 240 x 30 = 7200 litres

Total unit water = LCM(20, 25) = 100 unit A’s efficiency = 100/20 = 5 unit/hour B’s efficiency =100/25 = 4 unit/hour After 5 hour the water filled by A and B together = 5 x 9 =45 unit Remaining unit = 100 - 45 = 55 unit Time taken by A alone = 55/5 = 11 hours

Time taken at 3 km/hr = Distance/speed = 10/3 Actual time is obtained by subtracting the late time So, Actual time = 10/3 - 1/3 = 9/3 = 3 hour Time taken at 4 km/hr = 10/4 hr Time difference = Actual time - time taken at 4 km/hr = 3 - 10/4 = 1/2 hour Hence, he will be early by 30 minutes.