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# 15 lts are taken of from a container full of liquid A and replaced wit

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Manager
Joined: 21 Aug 2010
Posts: 161
Location: United States
GMAT 1: 700 Q49 V35
15 lts are taken of from a container full of liquid A and replaced wit  [#permalink]

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29 Sep 2011, 12:54
7
26
00:00

Difficulty:

65% (hard)

Question Stats:

64% (02:14) correct 36% (02:50) wrong based on 128 sessions

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15 lts are taken of from a container full of liquid A and replaced with Liquid B. Again 15 more lts of the mixture is taken and replaced with liquid B. After this process, if the container contains Liquid A and B in the ratio 9:16,What is the capacity of the container?

A:45
B:25
C:37.5
D:36
E:42

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Manager
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Joined: 07 Aug 2011
Posts: 182
Re: 15 lts are taken of from a container full of liquid A and replaced wit  [#permalink]

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30 Sep 2011, 04:58
7
1
9
i saw this formula on btg forum.
Let X be the capacity of the container.

% of Liq A left / % of A originally present = (1- (Vol of soln replaced/vol of container)) ^ n
[Note this formula is applied for repeated iterations such as these and 'n' = Number of iterations(in this case 2)]

9/25 = (1-15/x)^2 ...... [(9/9+16) / 1(Liq A present originally)]
3/5 = 1 - (15/X)
X=37.5
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Joined: 08 Jan 2009
Posts: 285
GMAT 1: 770 Q50 V46
Re: 15 lts are taken of from a container full of liquid A and replaced wit  [#permalink]

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29 Sep 2011, 18:21
3
2
Here is how I approached this question. It's actually pretty simple if we are methodical in our approach.

x = capacity

Initially:
x / x = 100%

Let's remove 15 liters:
(x - 15) / x = y
This should make sense - if we have 100 liters of A, take out 15, then we have 85% A. We don't know x though, so we have to express it in this form

Let's remove another 15 liters:
(xy - 15y) / x
This should make sense again - y is our proportion after removing 15 liters, x was our starting hence xy is how much of A we have. We remove 15 liters, but A is 15y, hence we deduct 15y. The total amount in the container is still x

Tie it all together (not as bad is it looks!):
(yx - 15y) / x = 9 / 25
(y (x - 15)) / x = 9 / 25
(((x - 15) / x) (x - 15)) / x = 9 / 25

(x-15)^2 / x^2 = 9 / 25

(x-15 / x) = 3 / 5
5x - 75 = 3x
2x = 75
x = 37.5
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Joined: 20 Dec 2015
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Concentration: Operations, Marketing
GMAT 1: 670 Q48 V36
GRE 1: Q157 V157
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Re: 15 lts are taken of from a container full of liquid A and replaced wit  [#permalink]

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07 Sep 2017, 02:12
3
1
Let the capacity of the container be X

Fist iteration removal of 15 liters of A and addition of 15 liters of B

Then the amount of liquid A in the mixture will be X-15 .
The capacity will quantity will remain same =X

Second iteration

Now first we are going to remove 15 liters of mixture that 15 liters will contain some amount of A and B
Final ratio of A/B =9/16
A to total capacity =9/25

Now X-15-15*(X-15)/X=9/25

(X-15)^2/x^2=9/25
Solving this we have 2X=75 X=37.5
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15 lts are taken of from a container full of liquid A and replaced wit  [#permalink]

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02 May 2016, 02:09
1
g106 wrote:
15 lts are taken of from a container full of liquid A and replaced with Liquid B. Again 15 more lts of the mixture is taken and replaced with liquid B. After this process, if the container contains Liquid A and B in the ratio 9:16,What is the capacity of the container?

A:45
B:25
C:37.5
D:36
E:42

working on formula (by karishma)
CiVi=CfVf
Let X be the volume of container then
here initial concentration is 100% and volume is (X-15)
final volume is again=x
so 1*(x-15)=Cf(x)
Cf=(x-15)/x
this process repeats again
here initial concentration is Cf= (x-15)/x and volume is (X-15)
final volume is again=x
{(x-15)/x}*(x-15)]=Cf1x

given Cf1=9/25
so [(x-15)/x]^2=9/25------>(x-15)/x=3/5
X=37.5
Ans C
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Joined: 02 Jun 2015
Posts: 12
Re: 15 lts are taken of from a container full of liquid A and replaced wit  [#permalink]

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25 Aug 2015, 01:14
2
You can also backsolve using the answer choices. Starting with choice C is efficient as its the median number.

If you have a 37.5 liter capacity, you start with 37.5 L of A and 0 L of B.

1st Replacement
After the first replacement you have 37.5-15=22.5 L of A and 15 L of B. The key is figuring out how many liters of A and B, respectively, are contained in the next 15 liters of mixture to be removed.

The current ratio of A to total mixture is 22.5/37.5; expressed as a fraction this becomes (45/2) / (75/2), or 45/2 * 2/75. Canceling the 2s and factoring out a 5 leaves the ratio as 9/15. Note, no need to reduce further as we're trying to figure out the amount of A and B in 15 L of solution. 9/15 of A means there must be 6/15 of B.

Multiply each respective ratio by 15 to get 9 L of A and 6 L of B in the next 15L removal.

Final Replacement
The next 15L removal means 9 liters of A and 6 liters of B are removed and replaced with 15 liters of B. 22.5-9=13.5 liters of A. 15 liters of B - 6 liters + 15 more liters = 24 liters of B.

Test to the see if the final ratio = 9/16; 13.5/24 = (27/2) * (1/24) = 9/16. Choice C is correct.

Note that if the first trial wasn't the correct answer choice, choosing the median number allows us to only make one more attempt to get to the correct choice. If the final ratio was less than 9/16, the current capacity being tested would be too small, as removing A to add more B creates a smaller numerator and a larger denominator. Vice versa if the ratio were larger.
Director
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Posts: 915
Location: United States
Re: 15 lts are taken of from a container full of liquid A and replaced wit  [#permalink]

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08 Sep 2019, 04:14
g106 wrote:
15 lts are taken of from a container full of liquid A and replaced with Liquid B. Again 15 more lts of the mixture is taken and replaced with liquid B. After this process, if the container contains Liquid A and B in the ratio 9:16,What is the capacity of the container?

A:45
B:25
C:37.5
D:36
E:42

final concentration X = initial concentration X (initial volume / final volume)^reps
initial volume = total - removed; final volume = total - removed + added pure Y
initial concentration A = 100% = 1; final concentration A = 9/(16+9) = 9/25
total = t; removed = 15; added of B = 15; reps = 2
$$9/25 = 1(t-15/t-15+15)^2…9/25 = 1(t-15/t)^2…3/5=t-15/t…2t=75…t=75/2=37.5$$

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Re: 15 lts are taken of from a container full of liquid A and replaced wit  [#permalink]

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08 Sep 2019, 04:55
g106 wrote:
15 lts are taken of from a container full of liquid A and replaced with Liquid B. Again 15 more lts of the mixture is taken and replaced with liquid B. After this process, if the container contains Liquid A and B in the ratio 9:16,What is the capacity of the container?

A:45
B:25
C:37.5
D:36
E:42

Given:
1. 15 lts are taken of from a container full of liquid A and replaced with Liquid B.
2. Again 15 more lts of the mixture is taken and replaced with liquid B.
3. After this process, if the container contains Liquid A and B in the ratio 9:16,

Asked: What is the capacity of the container?

Let the capacity of the container be V ltr

After first iteration,
Liquid A = V - 15 lts
Liquid B = 15 lts
Ratio A: B = (V-15)/15
Ratio A: A+B = (V-15)/V
Ratio B: A+B = 15/V = [1-(V-15)/V]

After second iteration
Liquid A = (V-15) - 15(V-15)/V = (V-15) (1 -15/V) = (V-15)^2/V
Liquid B = 15 - 15*15/V + 15 = 30 - 225/V = 30V -15^2/V
Ratio A: A+B = (V-15/V)^2 = 9/25
V-15/V = 3/5
V = 5/2*15 = 37.5 ltr

IMO C
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Re: 15 lts are taken of from a container full of liquid A and replaced wit   [#permalink] 08 Sep 2019, 04:55
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