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Tap 1 takes....!

Expert replies
by chaitanya.bhansali » Thu Aug 07, 2014 10:13 am
Tap 1 takes 1.5 hours to fill a certain tank with water. Tap 2 fills the same tank with water at a rate X times as fast as tap 1. Tap 1 and tap 2 together take 15 minutes to fill the same tank with water. Find the value of X.

A) 1/9
B)1/5
C) 5
D) 9
E) 45
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Source: — Problem Solving |

by [email protected] » Thu Aug 07, 2014 10:33 am
Hi chaitanya.bhansali,

This is an example of a Work Formula question (2 entities sharing a job), which means that we'll need the Work Formula:

Work = (A)(B)/(A+B)

We're given….
Tap1 = A = 1.5 hours
Tap2 = B = B hours
Together = 15 minutes

Since the "units" of measure change, I'm going to convert everything into minutes:

A = 90 minutes
B = B minutes
Together = 15 minutes

(90)(B)/(90 + B) = 15

Now we do algebra…

90B = 1350 + 15B
75B = 1350
B = 18

So B takes 18 minutes to fill the tank on its own.

The question asks how much FASTER was B than A?

A = 90 minutes
B = 18 minutes

B does the job in 1/5 the time, so it's 5 times faster.

Final Answer: C

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by GMATGuruNY » Thu Aug 07, 2014 10:45 am
chaitanya.bhansali wrote:Tap 1 takes 1.5 hours to fill a certain tank with water. Tap 2 fills the same tank with water at a rate X times as fast as tap 1. Tap 1 and tap 2 together take 15 minutes to fill the same tank with water. Find the value of X.

A) 1/9
B)1/5
C) 5
D) 9
E) 45
Since Tap 2 fills the tank X times as fast as Tap 1, X = (Tap 2's rate)/(Tap 1's rate).

The times given are 1.5 hours -- the equivalent of 90 minutes -- and 15 minutes.
To make the math easy, let the tank = 90 gallons.

Since the 2 taps together take 15 minutes to fill the tank, the combined rate for Tap 1 and Tap 2 = 90/15 = 6 gallons per minute.
Since Tap 1 takes 90 minutes to fill the tank, Tap 1's rate = 90/90 = 1 gallon per minute.

Thus:
Tap 2's rate = (combined rate for both taps) - (Tap 1's rate) = 6-1 = 5 gallons per minute.
X = (Tap 2's rate)/(Tap 1's rate) = 5/1 = 5.

The correct answer is C.
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by shrivats » Thu Aug 07, 2014 9:32 pm
Tap 1 fills the tank in 1.5 hours = 90 mins. In one minute its fills 1/90 of the tank

Tap 2 fills X times as Tap 1, so fills X/90 of the tank in 1 minute.

together they take 15 minutes.

1/90 + X/90 = 1/15

=> X=5
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by GMATinsight » Fri Aug 08, 2014 6:03 am
chaitanya.bhansali wrote:Tap 1 takes 1.5 hours to fill a certain tank with water. Tap 2 fills the same tank with water at a rate X times as fast as tap 1. Tap 1 and tap 2 together take 15 minutes to fill the same tank with water. Find the value of X.

A) 1/9
B)1/5
C) 5
D) 9
E) 45
Tap 1's 1.5 Hour (90 Mins) work = 1 Tank
Tap 1's (1 Mins) work = 1/90 Tank

(Tap 1+Tap 2)'s (15 Mins) work = 1 Tank
(Tap 1+Tap 2)'s (1 Mins) work = 1/15 Tank

(Tap 2)'s (1 Mins) work = (Tap 1 + Tap 2)'s (1 Mins) work - (Tap 1)'s (1 Mins) work

Therefore, (Tap 2)'s (1 Mins) work = (1/15) - (1/90) = (5/90)

Tap 2 is X times faster than Tap 1, Therefore
i.e. (5/90) = X (1/90)
X = 5


Answer: Option C
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by Brent@GMATPrepNow » Fri Aug 08, 2014 3:12 pm
chaitanya.bhansali wrote:Tap 1 takes 1.5 hours to fill a certain tank with water. Tap 2 fills the same tank with water at a rate X times as fast as tap 1. Tap 1 and tap 2 together take 15 minutes to fill the same tank with water. Find the value of X.

A) 1/9
B)1/5
C) 5
D) 9
E) 45
Given that the answer choices are so spread apart, we can skip the formal approaches and apply some common sense.

Aside: The importance of ALWAYS checking the answer choices BEFORE choosing a specific approach is covered in our free video on General GMAT Math Strategies: https://www.gmatprepnow.com/module/gener ... es?id=1111

Okay, let's say that YOU are tap 1, and it takes you 90 minutes to complete some job (e.g., fill a tank). You're looking for someone to help you so that, TOGETHER, you can complete the job in 15 minutes.

Answer choice A) Joe offers to help, but he's 1/9 as fast as you are.
Yeesh, this guy is useless!
NOTE: if Joe worked at the SAME speed as you, then you'd be able to finish in half the time. In other words, the two of you would complete the job in 45 minutes. However, since Joe is A LOT slower than you, it will take LONGER than 45 minutes to complete the job. ELIMINATE A

Answer choice B) Al offers to help, but he's 1/5 as fast as you are.
Using the same logic that we used above, we can conclude that since Al is A LOT slower than you, it will take LONGER than 45 minutes to complete the job. ELIMINATE B

Answer choice E) Bob offers to help, and he's 45 times as fast as you.
NOTE: Bob works so fast that your contribution makes little difference. In fact, since Bob is 45 times as fast as you, he could work ALONE and finish the job in 1/45 of the time. In other words, working ALONE, Bob would complete the job in 2 minutes. Of course, with YOUR HELP, the two of you would take LESS than 2 minutes to complete the job. ELIMINATE E

Answer choice D) Gary offers to help, and he's 9 times as fast as you.
Since Gary is 9 times as fast as you, he could work ALONE and finish the job in 1/9 of the time. In other words, working ALONE, Gary would complete the job in 10 minutes. Of course, with YOUR HELP, the two of you would take LESS than 10 minutes to complete the job. ELIMINATE D

Answer choice C) Hal offers to help, and he's 5 times as fast as you.
Since Hal is 5 times as fast as you, he could work ALONE and finish the job in 1/5 of the time. In other words, working ALONE, Hal would complete the job in 18 minutes. Of course, with YOUR HELP, the two of you would take a little LESS than 18 minutes to complete the job. Perfect - 15 minutes is a little less than 18 minutes.

Answer: C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by manyaabroadtpr » Sat Aug 09, 2014 2:56 am
Let us assume that Tap-1 fills the tank at the rate of 2 tons/hour
Therefore capacity of the tank
= 2 X 1.5
= 3 tons

If (C) is the right answer,
then Tap-2 should fill the tank at the rate of 5 X 2 = 10 tons/hour

Time taken by both Tap 1 and tap 2 together to fill the same tank
= 3/(2+10)
= 3/12
= 0.25 hours
= 15 mins(this is the condition in the question)

Therefore answer is C.
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