Ok, there are multiple steps involved in this one, as with any tough GMAT question:
First let's find out time devoted by each pump during 4 hour exercise. We know that ratio of pumps is
1 : 1.5
so time devoted by each pump would be
1.6 hr : 2.4 hr
From here we can make out that speed of fastest pump is
2.4/4 = .6
Now the simple part, how long will it take to fill tank all alone by faster pump:
Thus 4/.6 or 40/6 or 20/3 (Answer!)
Hope this clarifies!
gmat prep tough rate question
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albertrahul
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VerbalAttack
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Hi vinviper1,
You can also use the formula
1/ha + 1/hb = 1/hab
ha - No of hours it takes for a to complete a task
hb - No of hours it takes for b to complete a task
hab - No of hours it takes for a & b together to complete a task
Say it takes x hours for the fastest pump to fill the tank and 1.5x hours for the slower pump to fill the tank.
1/x + 1/1.5x = 1/4
Solve for x. x = 20/3hrs.
Cheers
You can also use the formula
1/ha + 1/hb = 1/hab
ha - No of hours it takes for a to complete a task
hb - No of hours it takes for b to complete a task
hab - No of hours it takes for a & b together to complete a task
Say it takes x hours for the fastest pump to fill the tank and 1.5x hours for the slower pump to fill the tank.
1/x + 1/1.5x = 1/4
Solve for x. x = 20/3hrs.
Cheers
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AleksandrM
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Use algebra.
Start with both pumps.
P1 and P2:
w = 4(r + 1.5r)
w = 10r
r = w/10
For the faster pump, time will be:
w = 1.5(w/10)t
w = (1.5w/10)t
10w = 1.5wt
10w/1.5w = t
t = 100/15 or 20/3 [the "w"s cancel out]
Start with both pumps.
P1 and P2:
w = 4(r + 1.5r)
w = 10r
r = w/10
For the faster pump, time will be:
w = 1.5(w/10)t
w = (1.5w/10)t
10w = 1.5wt
10w/1.5w = t
t = 100/15 or 20/3 [the "w"s cancel out]
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II
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I like Alex's approach the best here. My approach was very similar:
So we have 2 water pumps: 1 "water pump was working at 1.5 rate of the other".
So with all "working together" problems we know that we need to add the rates to get the combined rate.
Since we dont know the rates, lets assign variables:
Pump 1's rate = a
Pump 2's rate = 1.5a
So a + 1.5a = 2.5a
So 2.5a is the combined working rate of both pumps.
It took 4 hours to fill 1 pool at the rate of 2.5a.
RT = W
(2.5a)(4) = 1
10a = 1
a = 1/10
So now we know the rate of pump 1. Pump was working at the rate of 1/10.
We can now work out the rate of pump 2: (1/10) * 1.5 or (1/10)*(3/2) = 3/20.
So pump 2 was working at the rate of 3/20.
So if pump 2 was working alone, we can work out how long it will take to fill 1 pool.
RT = W
(3/20)t = 1
t = 1 / (3/20)
t = 1/1 * 20/3 = 20/3
So we have 2 water pumps: 1 "water pump was working at 1.5 rate of the other".
So with all "working together" problems we know that we need to add the rates to get the combined rate.
Since we dont know the rates, lets assign variables:
Pump 1's rate = a
Pump 2's rate = 1.5a
So a + 1.5a = 2.5a
So 2.5a is the combined working rate of both pumps.
It took 4 hours to fill 1 pool at the rate of 2.5a.
RT = W
(2.5a)(4) = 1
10a = 1
a = 1/10
So now we know the rate of pump 1. Pump was working at the rate of 1/10.
We can now work out the rate of pump 2: (1/10) * 1.5 or (1/10)*(3/2) = 3/20.
So pump 2 was working at the rate of 3/20.
So if pump 2 was working alone, we can work out how long it will take to fill 1 pool.
RT = W
(3/20)t = 1
t = 1 / (3/20)
t = 1/1 * 20/3 = 20/3
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Stuart@KaplanGMAT
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If there are exactly 2 workers, this formula simplifies to:VerbalAttack wrote:Hi vinviper1,
You can also use the formula
1/ha + 1/hb = 1/hab
ha - No of hours it takes for a to complete a task
hb - No of hours it takes for b to complete a task
hab - No of hours it takes for a & b together to complete a task
hab = (ha * hb)/(ha + hb)
using the simplified formula, and letting ha = a and hb = 1.5a, we get:
4 = (a * 1.5a)/(a + 1.5a)
4 = (1.5(a^2))/2.5a
4*2.5 = 1.5a
10 = 1.5a
10/1.5 = a
20/3 = a
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netigen
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There is another way, which may be simpler
The formula that every book preaches i.e. 1/T = 1/T1 + 1/T2 is actually
total Rate = Rate1 + Rate2 (rates can be added)
given that faster rate is 1.5 times the slower we get
total rate = 2.5 R1 where R1 = rate of slower pump
OR
total rate = 5/3 R2 where R2 = rate of faster pump
from the stem we know that total rate = 1/4, equate to get the answer
5/3 R2 = 1/4
R2 = 3/20
since, time = 1/R2 = 20/3
The formula that every book preaches i.e. 1/T = 1/T1 + 1/T2 is actually
total Rate = Rate1 + Rate2 (rates can be added)
given that faster rate is 1.5 times the slower we get
total rate = 2.5 R1 where R1 = rate of slower pump
OR
total rate = 5/3 R2 where R2 = rate of faster pump
from the stem we know that total rate = 1/4, equate to get the answer
5/3 R2 = 1/4
R2 = 3/20
since, time = 1/R2 = 20/3
I solved the problem using the standard rate formula:
1/P1 + 1/P2 = 1/t
P1 = Pump 1
P2 = Pump 2
t = time it takes to fill the pool
Since one pump work at 1.5 times the rate of the other:
P1 = x
P2 = 1.5 x
Substituting these numbers in the equation:
1/x + 1/1.5x = 1/4
x= 20/3
However, after if found this answer I felt like I needed to do more to solve this problem. Why is that and why did this end up being the correct answer?
1/P1 + 1/P2 = 1/t
P1 = Pump 1
P2 = Pump 2
t = time it takes to fill the pool
Since one pump work at 1.5 times the rate of the other:
P1 = x
P2 = 1.5 x
Substituting these numbers in the equation:
1/x + 1/1.5x = 1/4
x= 20/3
However, after if found this answer I felt like I needed to do more to solve this problem. Why is that and why did this end up being the correct answer?
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logitech
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Another approach - I know how to apply the rate equation and such but in GMAT this is how I go with this question.vinviper1 wrote:tough rate one...thanks!!
RATES = are like Velocity ( you can add them, subtract them )
WORK DONE = is like distance , which is Time x Velocity
Let me start solving the question
One rate is 1.5 times bigger than the other. I can use 1 and 1.5 but dont want the deal with decimals so 2 and 3 ( my rates )
It takes 4 hrs! Well I can add up the rates and multiply by time:
(2+3)x4 = 5x4=20, so this is the work, or the distance!
Question is how long will it take for the fastest one to do job ( or the fastest one to travel the same distance )
Time = Distance / Velocity = Work / Rate = 20 / 3
Once you master this , you solve this type of questions less than 20 seconds!
Hope this helps!
LGTCH
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mals24
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logitech wrote:
Once you master this , you solve this type of questions less than 20 seconds!
Hope this helps!
So when are you starting your own Ace the GMAT maths prep school. I want to register for them.
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karmayogi
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@Logitechmals24 wrote:logitech wrote:
Once you master this , you solve this type of questions less than 20 seconds!
Hope this helps!
So when are you starting your own Ace the GMAT maths prep school. I want to register for them.
That is the best approach, which I have ever seen, to solve the problem.
@Mals
Count me in
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logitech
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I am glad that you found it useful. Smile to a rate problem when you see it in the testkarmayogi wrote:@Logitechmals24 wrote:logitech wrote:
Once you master this , you solve this type of questions less than 20 seconds!
Hope this helps!
So when are you starting your own Ace the GMAT maths prep school. I want to register for them.
That is the best approach, which I have ever seen, to solve the problem.
@Mals
Count me in
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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"DON'T LET ANYONE STEAL YOUR DREAM!"
