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by Ozlemg » Sat Aug 20, 2011 3:41 am
Working together at their constant rates , A and B can fill an empty tank to capacity in1/2 hr.what is the constant rate of pump B?
1) A's constant rate is 25LTS / min
2) the tanks capacity is 1200 lts.

Source: gmatclub.com
OA : C
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by GmatKiss » Sat Aug 20, 2011 3:52 am
what is the constant rate of pump B

For calculation rate, we need to know,
i) capacity of the tank ( Which helps us to calculate combined rate )
ii) Rate of any one of the pump ( Which will give us the other pumps rate)

2 unknowns are available from choices 1 and 2
Hence C

Hope it helps!
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by knight247 » Sat Aug 20, 2011 4:02 am
Let Rates for A and B be R1 and R2

Time Taken by A+Time Taken by B=0.5
1/R1+1/R2=30 Since, Time=Work/Rate.....In this case we consider work=1

The question asks us to find the value of R2

(1)A's constant rate is 25LTS/min meaning R1=25LTS/min
Substituting this value in the above eqn we get, 1/25+1/R2=30
Solving, we can find the value of R2. Hence Sufficient
(2)Rate of A and B combined=1200/30=40Lts/min
T1+T2=30 mins
R1+R2=40 lts/min
1/R1+1/R2=30. Can't really solve any further. Hence Insufficient.

Answer is A
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by Ozlemg » Sat Aug 20, 2011 4:24 am
knight247 wrote:Let Rates for A and B be R1 and R2

Time Taken by A+Time Taken by B=0.5
1/R1+1/R2=30 Since, Time=Work/Rate.....In this case we consider work=1

The question asks us to find the value of R2

(1)A's constant rate is 25LTS/min meaning R1=25LTS/min
Substituting this value in the above eqn we get, 1/25+1/R2=30
Solving, we can find the value of R2. Hence Sufficient
(2)Rate of A and B combined=1200/30=40Lts/min
T1+T2=30 mins
R1+R2=40 lts/min
1/R1+1/R2=30. Can't really solve any further. Hence Insufficient.

Answer is A
@knight

this where the trick lies.

Statement 1 gives us the Rate of A, but we do not still know the A&B's rate to calculate B's rate from that equation. Thus, in order to figure out the Rate of A&B we need to know what the tank's capacity.
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by knight247 » Sat Aug 20, 2011 4:51 am
We already have A and B's combined rate. If both of them working together fill up the tank in 1/2 hr i.e. 30 minutes. Their combined time=30mins. Also, Rate=1/Time=1/30
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by Frankenstein » Sat Aug 20, 2011 11:04 am
Hi knight,
Let V be the volume of the tank.
Time taken is actually V/R1 + V/R2.
So, V/R1 + V/R2 = 30.
IF you are making it 1/R1 + 1/R2, then the assumption is that the volume of the tank is 1 Lt.
That is the reason, we need both statements.
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by GMATGuruNY » Sun Aug 21, 2011 3:39 am
Ozlemg wrote:Working together at their constant rates , A and B can fill an empty tank to capacity in1/2 hr.what is the constant rate of pump B?
1) A's constant rate is 25LTS / min
2) the tanks capacity is 1200 lts.

Source: gmatclub.com
OA : C
Statement 1: A = 25 liters per minute.
Let tank = 1200 liters.
Combined rate for A and B to fill the tank in 30 minutes = 1200/30 = 40 liters per minute.
Rate for B alone = (Rate for A and B together) - (Rate for A alone) = 40-25 = 15 liters per minute.

Let tank = 1500 liters.
Combined rate for A and B to fill the tank in 30 minutes = 1500/30 = 50 liters per minute.
Rate for B alone = (Rate for A and B together) - (Rate for A alone) = 50-25 = 25 liters per minute.
Since the rate for B can be different values, insufficient.

Statement 2: tank = 1200 liters.
Combined rate for A and B together = 1200/30 = 40 liters per minute.
No way to determine the rate for B alone.
Insufficient.

Statements 1 and 2 together:
As shown above, when the rate for A = 25 liters per minute and the tank contains 1200 liters, the rate for B alone = 15 liters per minute.
Sufficient.

The correct answer is C.
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by bblast » Sun Aug 21, 2011 3:51 am
Hi Mitch/Frank,

whats the takeaway from this problem ?

Lame analogy :
If a and b complete 1/2 of a given work in 10 days then combined rate = 1/20.

If a's rate is 1/80 then we can calculate b's rate = 1/20 - 1/80 = 3/80. - {sufficient}


Why does above hypothesis not work at the problem in the thread ?
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by Frankenstein » Sun Aug 21, 2011 6:40 am
bblast wrote:Hi Mitch/Frank,

whats the takeaway from this problem ?

Lame analogy :
If a and b complete 1/2 of a given work in 10 days then combined rate = 1/20.

If a's rate is 1/80 then we can calculate b's rate = 1/20 - 1/80 = 3/80. - {sufficient}


Why does above hypothesis not work at the problem in the thread ?
Hi,
If W is the total work required to complete a given work, combined work rate of a and b is W/20.
If a alone completes the work in 80 days, that means a's rate is W/80, then b's rate will be 3W/80.
In general, the question asked will be how many days did B alone take to complete the work(generally work rate of B is not asked), we calculate by using Work/Work rate = W/(3W/80) = 80/3 days.
As the number of days taken can be calculated independent of value of W, answer will not change if you take W=1, 2,...any number.
But, if we are asked to find the work rate of B(similar to this question), the value of b's rate will be 3W/80, which will depend on W. So, We cannot take W=1.
Same logic here.
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