Two taps can fill a cistern in 20 minutes and 30 minutes. The first ta

Ratio A=1/20
Ratio B= 1/30
Combined ratio= 1/20+1/30=1/12

Now, since we know that tap A was opened before B, we can create the function:

1/20*x + 1/12*(15-x)=1

Substitute the values and you'll find that the correct answer is B.

Let, Total Work Unit = 60
Work done by First tap in 20 minutes = 60 units
Work done by First tap in 1 minutes = 60/20 units = 3 units
Work done by First tap in 15 minutes = 3*15 = 45 units

Work done by Second tap in 30 minutes = 60 units
Work done by Second tap in 1 minutes = 60/30 units = 2 units
Work done by Second tap in 15-x minutes = 2(15-x) = 30-2x units

Total Work done by both taps in 15 mins = 45 +30-2x = 60 Units
i.e. x = 7.5 Minutes

Answer: Option B
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hey GMATinsight
Why did you take 60 as total work done ? i guess its LCM of 20 and 30, but then let me ask you a question isnt in denominator given time 1/20 where 1 is work and 20 is time
thank you

bonjurno Italiandrummer97 , isnt tthere a mistake in your equation ?

why didnt you write equation like this 1/20*x + 1/30*(15-x)=1/12

from which source did you learn work and rate problems

Hey No, I don't think so: the reasoning behind the equation is this:

1)\(1/20x\) Is the ratio of A * time it is activated
2)\(1/12*(15-x)\) Is the ratio after you open tap B, which working together with tap A gives a combined ratio of 1/12, *time they are activated together(i.e. total time - x = 15-x).
3)Equate their sum to 1, which is the cistern we have to fill.

Then solve the equation and the result will be \(15/2\), hence the correct answer is B.

I learned to create models in high school and university, I just applied this knowledge to this work and rate problem I'm sure there are other ways, but that's the one that came up in my mind at the moment.

If you need any further clarification just quote me and I'll be happy to help
And please, give kudos if you found this solution helpful

dave13
This method requires us to assume the total work unit.

In order to keep work units done by workers in unit time (i.e. Rate) as Integer, we assume the total work unit a number which is common multiple of times i.ee. common multiple of 15 and 20 in this case. Hence one could have chosen 60 or 120 or 180 as the total work units.

I hope this helps!!!
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hey Italiandrummer97

thank you ! got it now! happy international drumming day!

\(\frac{x}{20} +\frac{15-x}{12} = 1\)

\((60)(\frac{x}{20} +\frac{15-x}{12}) = 1(60)\)

\(3x+5(15-x) =60\)

\(3x+75-5x=60\)

\(15=2x\)

\(x = 7.5\)

GMATinsight , thank you ! today i know TWO methods and yesterday i didnt know any of the methods

Solution

Given:

• Two taps can fill a cistern in 20 minutes and 30 minutes.
• First tap was opened initially for x minutes.
• After first x minutes, both the taps were open.

To find:

• The value of x

Approach and Working out:

Let us take the capacity of cistern= LCM(20, 30)=60 litres.

Hence, Tap A fills the 3 litres in 1 min

• And, Tap B fills the 2 litres in 1 min.

Thus, for first x min, the capacity of cistern filled = 3x
After, x min, both the taps were open for (15-x) min.

• Therefore, the amount filled by both the taps=5*(15-x)= 75-5x

Now, the cistern got filled.

• Hence, 3x+75-5x= 60
• 2x= 15
• x= 7.5 min

Answer: B
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Hi All,

This is my approach that I tend to find to be the quickest:

RateA x 20mins = 1
Ra = 1/20

RateB x 30mins = 1
Rb = 1/30

In 15 mins, Tap A fills the cistern to 3/4 capacity:
1/20 x 15 mins = 3/4
But the problem says that in 15 mins cistern will be completely full, not 3/4 full. Hence Tap B must have been turned on at some time.

1/30 x T = 1/4 (amount left to be filled)
T = 1/4 x 30/1
T = 30/4 = 7 and 1/2 = 7.5

Ans: B

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