Pumps A, B, and C operate at their respective constant rates. Pumps A

Generally, if we are told that:
A hours is needed for worker A (pump A etc.) to complete the job --> the rate of A=\(\frac{1}{A}\);
B hours is needed for worker B (pump B etc.) to complete the job --> the rate of B=\(\frac{1}{B}\);
C hours is needed for worker C (pump C etc.) to complete the job --> the rate of C=\(\frac{1}{C}\);

You can see that TIME to complete one job=Reciprocal of rate. eg 6 hours needed to complete one job (time) --> 1/6 of the job done in 1 hour (rate).

Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance.

Time*Rate=Distance
Time*Rate=Job

Also note that we can easily sum the rates:
If we are told that A is completing one job in 2 hours and B in 3 hours, thus A's rate is 1/2 job/hour and B's rate is 1/3 job/hour. The rate of A and B working simultaneously would be 1/2+1/3=5/6 job/hours, which means that the will complete 5/6 job in hour working together.

Time needed for A and B working simultaneously to complete the job=\(\frac{A*B}{A+B}\) hours, which is reciprocal of the sum of their respective rates. (General formula for calculating the time needed for two workers working simultaneously to complete one job).
Time needed for A and C working simultaneously to complete the job=\(\frac{A*C}{A+C}\) hours.

Time needed for B and C working simultaneously to complete the job=\(\frac{B*C}{B+C}\) hours.

General formula for calculating the time needed for THREE workers working simultaneously to complete one job is: \(\frac{A*B*C}{AB+AC+BC}\) hours. Which is reciprocal of the sum of their respective rates: \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\).

We have three equations and three unknowns:
1. \(\frac{1}{A}+\frac{1}{B}=\frac{5}{6}\)

2. \(\frac{1}{A}+\frac{1}{C}=\frac{2}{3}\)

3. \(\frac{1}{B}+\frac{1}{C}=\frac{1}{2}\)

Now the long way is just to calculate individually three unknowns A, B and C from three equations we have. But as we just need the reciprocal of the sum of relative rates of A, B and C, knowing the sum of \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{AB+AC+BC}{ABC}\) would be fine, we just take the reciprocal of it and bingo, it would be just the value we wanted.

If we sum the three equations we'll get:
\(2*\frac{1}{A}+2*\frac{1}{B}+2*\frac{1}{C}=\frac{5}{6}+\frac{2}{3}+\frac{1}{2}=2\)

\(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=1\), now we just need to take reciprocal of 1, which is 1.

So the time needed for A, B, and C working simultaneously to complete 1 job is 1 hour.

Hope it helps.
_________________

You are adding the times, when you should be adding the rates.

Rates of:
A+B = 5/6 (tank per hour)
A+C = 2/3 (tank per hour)
B+C = 1/2 (tank per hour)

Combining these rates is like seeing "how much work would all of these pairs of machines do in 1 hour?"

(A+B)+(A+C)+(B+C) = 5/6 + 2/3 + 1/2 = 5/6 + 4/6 + 3/6 = 12/6 (tank per hour)

Therefore, 2A's, 2B's and 2C's working together would fill 2 tanks in an hour.
A single A, B, and C working together would fill 1 tank in 1 hour.

A and B = 5/6 --> 1/A+1/B=5/6
A and C = 2/3 --> 1/A+1/C=2/3
B and C = 1/2 --> 1/B+1/C=1/2

Q 1/A+1/B+1/C=?

Add the equations: 1/A+1/B+1/A+1/C+1/B+1/C=5/6+2/3+1/2=2 --> 2*(1/A+1/B+1/A+1/C)=2 --> 1/A+1/B+1/A+1/C=1

Answer: D. (1)
_________________

Good technique..had another way but a few more unnecessary steps.

Ans e.

Let the rate of Pump A be a, Pump B b and Pump C c.

1/a + 1/b = 5/6

1/a + 1/c = 2/3

1/b + 1/c = 1/2

adding 2(1/a + 1/b + 1/c) = 5/6 + 2/3 + 1/2 = 5+4+3/6 = 2

=> 1/a + 1/b + 1/c = 1

or Pumps A,B and C take 1 hour to fill the tank.

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

a) 1/3
b) 1/2
c) 1/4
d) 1
e) 5/6


We knw that a+b are taking 6/5 hrs i.e. 1.2 hrs to fill a tank...
similarly B+C take 1.5 hrs and A+C take 2 hrs....

Adding all 3 equations we get

A+b + B+c + A+C = 1.2+1.5+2

2A+2B+2C = 4.7hrs
A+B+C = 2.35 hrs

can u please guide where am I going wrong???

Work done by A & B together in one hour = 5/6

Work done by B & C together in one hour = 2/3

Work done by C & A together in one hour = 1/2

=> 2(Work done by A , B & C in one hour) = 5/6 + 2/3 + 1/2

=> Work done by A , B & C in one hour = 1/2 [ 5/6 + 2/3 + 1/2] = 1

So together all three were able to finish the work in one hour.

Answer is E.

Nice Explanation.

Thanks for clearing the concepts.
_________________

Serioulsy.. Nice explanation..
It will clear the basics for solving Time and Rate questions..

We are given that pumps A and B, operating simultaneously, can fill a certain tank in 6/5 hours. Recall that rate = work/time. If we consider the work (filling the tank) as 1, then the combined rate of pumps A and B is 1/(6/5). We can express this in the following equation in which a = the time it takes pump A to fill the tank when working alone (thus 1/a is pump A’s rate) and b = the time it takes pump B to fill the tank when working alone (thus 1/b is pump B’s rate):

1/a + 1/b = 1/(6/5)

1/a + 1/b = 5/6

We are next given that pumps A and C, operating simultaneously, can fill the tank in 3/2 hours. We can create another equation in which c = the time it takes pump C to fill the tank alone.

1/a + 1/c = 1/(3/2)

1/a + 1/c = 2/3

Finally, we are given that pumps B and C, operating simultaneously, can fill the tank in 2 hours. We can create the following equation:

1/b + 1/c = ½

Next we can add all three equations together:

(1/a + 1/b = 5/6) + (1/a + 1/c = 2/3) + (1/b + 1/c = ½)

2/a + 2/b + 2/c = 5/6 + 2/3 + 1/2

2/a + 2/b + 2/c = 5/6 + 4/6 + 3/6

2/a + 2/b + 2/c = 12/6

2/a + 2/b + 2/c = 2

Since we need to determine the combined rate of all three machines when filling 1 tank, we can multiply the above equation by ½:

(2/a + 2/b + 2/c = 2) x ½

1/a + 1/b + 1/c = 1

Since the combined rate of all 3 pumps is 1 and time = work/rate, the time needed to fill the tank when all 3 pumps are operating simultaneously is 1/1 = 1 hour.

Answer: D
_________________

An alternative approach I prefer to use on most work/rate problems since I’m not a fan of fractions.

Assign the work a number of units. So 1 job in this case would equal 6 units work (based on LCM of the rates 5/6, 2/3 and 1/2)

Therefore multiplying the rates by these units of work gives us:
A+B = 5 units of work per hour (5/6*6)
B+C = 3 units of work per hour (1/2*6)
C+A = 4 units of work per hour (2/3*6)

Adding all of them would give us 12 which is equal to 2A + 2B + 2C = 12 therefore A+B+C=6 units of work per hour

Or one could substitute by finding one of the values (for example A = 5-B and B = 3-C therefore plugging in third equation
—> C + (5-B) = 4
—> C + 5 - (3-C) = 4
—> C + 5 - 3 + C = 4
—> 2C + 2 = 4
—> C = 1
Adding this to the rate of A+B gives us A+B+C = 6 units of work per hour.

Now since one task was 6 units of work and combined rate is 6 units/hour therefore it would take one hour to complete the task.

Posted from my mobile device

My strategy was:

1) 1/A + 1/B =5/6 so 1/A= 5/6-1/A
2) 1/B + 1/C = 1/2 so 1/C= 1/2 - 1/B
3)1/A + 1/C= 2/3 this last equation ill use to find B

1/B=1/3 and i will use this value to find 1/A and 1/C

then I will add 1/A + 1/B + 1/C and the result will be 1

Ra = Rate at which pump A works
Rb = Rate at which pump B works
Rc = Rate at which pump C works

Work done by pump A & B , W = (Ra + Rb)* 6/5
Work done by pump A & C, W = (Ra+Rc) * 3/2
Work done by pump B & C, W = (Rb + Rc) * 2

Ra +Rb = 5/6
Ra + Rc = 2/3
Rb + Rc = 1/2

2*Ra + 2*Rb + 2*Rc = 5/6 + 2/3 + 1/2
Ra + Rb + Rc = 1/2[ 5/6 + 2/3 + 1/2 ] = 1/2 [ 1/3(5/2+2)+1/2] = 1/2[ 3/2 + 1/2]
= 1/2[ 2] = 1

Thus, Ra + Rb + Rc = 1
Time taken by all the pump to fill the tank is obtain by

W = R*T
T = R / W = 1 hour

IMO(D)

let assume A, B, C be the rate of pipe if they fill the tank indvidually

A+B= \(\frac{5}{6}\)-----eqn 1

A+C= \(\frac{2}{3}\)------eqn 2

B+C= \(\frac{1}{2}\)-------eqn 3

now, adding all the 3 equations

A+B+A+C+B+C=\(\frac{5}{6}\)+\(\frac{2}{3}\)+\(\frac{1}{2}\)

2(A+B+C)= 2

A+B+C= 1

ANS D

Rate of each pairs: \(=\frac{5}{6}+\frac{2}{3}+\frac{1}{2}\\
\)
\(=\frac{12}{6}\)

\(=2\)

Time required for \(A+B+C=\frac{2}{2}=1\)

The answer is \(A\)
_________________

Also note that we can easily sum the rates:
If we are told that A is completing one job in 2 hours and B in 3 hours, thus A's rate is 1/2 job/hour and B's rate is 1/3 job/hour. The rate of A and B working simultaneously would be 1/2+1/3=5/6 job/hours, which means that the will complete 5/6 job in hour working together.

Posted from my mobile device

found this great article on Magoosh for work/rate problems -
https://magoosh.com/gmat/gmat-work-rate-problems/

You can't sum up time for work rate in this manner. The time decreases as the machine/resource increases

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 .


We can also use LCM method to solve this time and work related question.

Let's assume the total Volume = LCM ( 6,3,2) = 6 L

*Note: if the time taken to complete the work are in fraction form, assume total work/volume as LCM of the numerators of the fractions.

Per work of pipe A and B = A+ B= 6/(6/5) = 5 L / hour
Per work of pipe A and C = A+ C= 6/(3/2) = 4 L / hour.

Per work of pipe B and C = B + C = 6/2 = 3 L/hour

On adding all three equations,
2( A + B + C ) = 5 + 4 + 3 = 12 L
A + B + C = 12/2 = 6L / hour.

So the time taken by pumps A, B, and C, operating simultaneously, to fill the tank = Total Volume/ Per hour work of (A,B and C) = 6/6 = 1 hour

Option D is the answer.

Thanks,
Clifin J Francis