So the basic "rate" equation is:

1/A + 1/B = 1/T

You are given 3 scenarios:
1/A + 1/B = 1 / (6/5) = 5/6
1/B + 1/C = 1 / (3/2) = 2/3
1/A + 1/C = 1 / 2 = 1/2

Once you've set this up, simply use your algebra skills to determine that:
A = 2, B = 3, and C = 6

Therefore, 1/A + 1/B + 1/C = 1/T
or
1/2 + 1/3 + 1/6 = 1 / T
T = 1. E.

havok wrote:So the basic "rate" equation is:

1/A + 1/B = 1/T

You are given 3 scenarios:
1/A + 1/B = 1 / (6/5) = 5/6
1/B + 1/C = 1 / (3/2) = 2/3
1/A + 1/C = 1 / 2 = 1/2

Once you've set this up, simply use your algebra skills to determine that:
A = 2, B = 3, and C = 6

there is no need to determine the value of A,B and C, simply add three equations to get the desired result...!!!

there is no need to determine the value of A,B and C, simply add three equations to get the desired result...!!!

This is fine if you are comfortable with the concept that if A,B and C are the rate, then 1/A, a/b and 1/C are the time it takes them to fill one tank, and 2/A, 2/B and 2/C are "twice the time" - which is true, but not an easy point to conceptualize automatically. If you are fine with this concept, and are also comfortable working with fractions so that 5/6 + 2/3 + 1/2 is a breeze, then the following post is not meant for you - keep doing what you do.

For the rest of the world:
when the work is not defined (a pool, a job, a wall, a tank), plug in a good number for the work. A good number will be divisible numbers in the question: In this case, we have 6/5, 3/2 and 2/1, so we'll use 6*5*2 = 60 for the work. The actual number does not really matter - the goal is to use a large nubmber for the work, so that the rate is given in real integers, rather than the fraction 1/A. To aid in conceptualizing the problem, it helps to think of "units": 60 liters in the tank, for example.

Use the plugged in work to find the rate. So
A and B can do 60 liters in 6/5 hours, so their rate is 60 / 6/5 = 60*5/6 = 50 li/hr.
A+C = 60/3/2 = 60*2/3 = 20*2 = 40 li/hr.
B+C = 60/2=30 li/hr.

Add all three equations together:

A+B+A+C+B+C = 50+40+30 = 120 li/hr.
2A+2B+2C = 120 li/hr /:2
A+B+C = 60 li/hr.

so A, B, and C will take 1 hour to finish the 60 liter tank together.

Just to get an indication...what difficulty rating would you give to this problem?

Ryandmitri wrote:Just to get an indication...what difficulty rating would you give to this problem?

It ain't easy, that's for sure. I'd say a 47+.

Geva@MasterGMAT wrote:

there is no need to determine the value of A,B and C, simply add three equations to get the desired result...!!!

This is fine if you are comfortable with the concept that if A,B and C are the rate, then 1/A, a/b and 1/C are the time it takes them to fill one tank, and 2/A, 2/B and 2/C are "twice the time" - which is true, but not an easy point to conceptualize automatically. If you are fine with this concept, and are also comfortable working with fractions so that 5/6 + 2/3 + 1/2 is a breeze, then the following post is not meant for you - keep doing what you do.

For the rest of the world:
when the work is not defined (a pool, a job, a wall, a tank), plug in a good number for the work. A good number will be divisible numbers in the question: In this case, we have 6/5, 3/2 and 2/1, so we'll use 6*5*2 = 60 for the work. The actual number does not really matter - the goal is to use a large nubmber for the work, so that the rate is given in real integers, rather than the fraction 1/A. To aid in conceptualizing the problem, it helps to think of "units": 60 liters in the tank, for example.

Use the plugged in work to find the rate. So
A and B can do 60 liters in 6/5 hours, so their rate is 60 / 6/5 = 60*5/6 = 50 li/hr.
A+C = 60/3/2 = 60*2/3 = 20*2 = 40 li/hr.
B+C = 60/2=30 li/hr.

Add all three equations together:

A+B+A+C+B+C = 50+40+30 = 120 li/hr.
2A+2B+2C = 120 li/hr /:2
A+B+C = 60 li/hr.

so A, B, and C will take 1 hour to finish the 60 liter tank together.

hi geva, thanks a lot for sharing an alternative method, its always good to learn new things...!!!