saurabhmahajan wrote:Please explain

in OG said that if A does x hour to finish a job and B does y hour to finish the same job so together they will finish the same job in 1/x+1/y=1/h (h is the number both A and B will do together to finish the job)
so in this problem, we can solve the same
1/a+1/B=1/6/5=5/6
1/b+1/c= 2/3
1/c+1/b=1/2
add 3 aquations above : (1/a+1/b+1/c)=1/2(5/6+4/6+3/6)=12/12=1
so answer is 1

Thanks diebeatsthegmat.
But is there any way where we can solve this problem by pluging some value ?

saurabhmahajan wrote:Thanks diebeatsthegmat.
But is there any way where we can solve this problem by pluging some value ?

take the LCM of 6/5 , 3/2 , 2

It will come as 6, ( LCM of fractions is LCM of Numerator and HCF of denominator )

So can A and B will will the tank 5 times
A and C will fill 4 times
B and C will it 3 times

in total tank is filled 12 times

6 ( A + B ) + 6 ( A + C ) + 6 ( B + C ) = 12
A + B + C = 12/12 = 1

Hope this helps.

saurabhmahajan wrote:Thanks diebeatsthegmat.
But is there any way where we can solve this problem by pluging some value ?

yes, honey, you can solve it by plugging some value, but i am not good at it. i just select a number to calculate in my head when the numbers are easy and even for me to do it.

thanks diebeatsthegmat and goyalsau,

i tried this way.please correct me if i am wrong. and if right then hope it helps you too.

let the tank size be 12.
then A+B = 12/(6/5) = 10 per hr
B+C = 12/(3/2) = 8 per hr
A+C = 12/(2/1) = 6 per hr
10+8+6= 24
Now 2(A+B+C) = 24... equals to 12 per hr
So ABC working together to fill 12 lit tank will take 1 hr (12/12)

Hope i did it correctly

Pumps A, B, and C operate at their respective constant rates. Pumps A and B, operating simultaneously, can fill a certain tank in 6/5 hours; pumps A and C, operating simultaneously, can fill the tank in 3/2 hours; and pumps B and C, operating simultaneously, can fill the tank in 2 hours. How many hours does it take pumps A, B, and C, operating simultaneously, to fill the tank?

A. 1/3
B. 1/2
C. 2/3
D. 5/6
E. 1

When the job is undefined, we can plug in a value for the job.

We should plug in a value that will make the math easy. To determine the rates for the various pumps, we'll be dividing the value of the job by the times given in the problem. Two of the times given are fractions. To divide by a fraction, we multiply by the reciprocal of the fraction. Thus, we should plug in a value that is divisible by the numerators of the fractions.

Plug in tank = 6.
Rate for A+B = w/t = 6/(6/5) = 5/hour
Rate for A+C = w/t = 6/(3/2) = 4/hour
Rate for B+C = w/t = 6/2 = 3/hour
Combining the rates, we get:
(A+B) + (A+C) + (B+C) = 5+4+3
2(A+B+C) = 12
A+B+C = 6/hour
Time for A+B+C = w/r = 6/6 = 1 hour.

The correct answer is E.

Perfect Answer...cheers !!

saurabhmahajan wrote:thanks diebeatsthegmat and goyalsau,

i tried this way.please correct me if i am wrong. and if right then hope it helps you too.

let the tank size be 12.
then A+B = 12/(6/5) = 10 per hr
B+C = 12/(3/2) = 8 per hr
A+C = 12/(2/1) = 6 per hr
10+8+6= 24
Now 2(A+B+C) = 24... equals to 12 per hr
So ABC working together to fill 12 lit tank will take 1 hr (12/12)

Hope i did it correctly