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A vessel (capacity 10 liters) is filled entirely with petrol and keros 28 Dec 2021, 05:45
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65% (hard)

Question Stats:

63% (02:42) correct 37% (02:46) wrong based on 164 sessions

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A vessel (capacity 10 liters) is filled entirely with petrol and kerosene in the ratio 3:2. Two liters of this solution is removed and replaced with 2 ltrs of kerosene. This procedure is repeated thrice. Find the % of petrol in the resulting solution.

(A) 20%
(B) 30.72%
(C) 40.23%
(D) 52.15%
(E) 63.2%


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A vessel (capacity 10 liters) is filled entirely with petrol and keros Updated on: 26 Mar 2023, 11:21
Originally posted by SaquibHGMATWhiz on 14 Jan 2023, 05:26.
Last edited by SaquibHGMATWhiz on 26 Mar 2023, 11:21, edited 1 time in total.
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Bunuel
A vessel (capacity 10 liters) is filled entirely with petrol and kerosene in the ratio 3:2. Two liters of this solution is removed and replaced with 2 ltrs of kerosene. This procedure is repeated thrice. Find the % of petrol in the resulting solution.

(A) 20%
(B) 30.72%
(C) 40.23%
(D) 52.15%
(E) 63.2%
Show more

Solution:

  • Capactity of vessel \(= 10\) liters
  • Inital amount of petrol \(=\frac{3}{2+3}\times 10=\frac{3}{5}\times 10=6\) liters
  • Inital amount of kerosene \(=\frac{2}{2+3}\times 10=\frac{2}{5}\times 10=4\) liters

  • Every time, \(\frac{2}{10}\) of the solution gets eliminated
  • So, after 3 removal volume of petrol will be \(=6(1-\frac{2}{10})^3\)
    \(=6(\frac{8}{10})^3\)
    \(=6(\frac{4}{5})^3\)
    \(=6\times \frac{4}{5}\times \frac{4}{5}\times \frac{4}{5}\)
    \(=3.072\)
  • Percentage in the final solution \(=\frac{3.072}{10}\times 100=30.72\%\)

Hence the right answer is Option B
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Himanshiog
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A vessel (capacity 10 liters) is filled entirely with petrol and keros 14 Jan 2023, 03:30
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Can some one please solve this question

Posted from my mobile device
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A vessel (capacity 10 liters) is filled entirely with petrol and keros 14 Jan 2023, 07:04
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SaquibHGMATWhiz
Bunuel
A vessel (capacity 10 liters) is filled entirely with petrol and kerosene in the ratio 3:2. Two liters of this solution is removed and replaced with 2 ltrs of kerosene. This procedure is repeated thrice. Find the % of petrol in the resulting solution.

(A) 20%
(B) 30.72%
(C) 40.23%
(D) 52.15%
(E) 63.2%

Solution:

  • Capactity of vessel \(= 10\) liters
  • Inital amount of petrol \(=\frac{3}{2+3}\times 10=\frac{3}{5}\times 10=6\) liters
  • Inital amount of kerosene \(=\frac{2}{2+3}\times 10=\frac{2}{5}\times 10=4\) liters

  • Every time, \(\frac{2}{10}\) of the petrol gets eliminated
  • So, after 3 removal volume of petrol will be \(=6(1-\frac{2}{10})^3\)
    \(=6(\frac{8}{10})^3\)
    \(=6(\frac{4}{5})^3\)
    \(=6\times \frac{4}{5}\times \frac{4}{5}\times \frac{4}{5}\)
    \(=3.072\)
  • Percentage in the final solution \(=\frac{3.072}{10}\times 100=30.72\%\)

Hence the right answer is Option B
Show more

Hi SaquibHGMATWhiz - I couldn't catch the below part quite well. Can you please let me know how did we conclude that

Quote:
Every time, \(\frac{2}{10}\) of the petrol gets eliminated
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Thanks in advance
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A vessel (capacity 10 liters) is filled entirely with petrol and keros 14 Jan 2023, 08:29
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SaquibHGMATWhiz
Bunuel
A vessel (capacity 10 liters) is filled entirely with petrol and kerosene in the ratio 3:2. Two liters of this solution is removed and replaced with 2 ltrs of kerosene. This procedure is repeated thrice. Find the % of petrol in the resulting solution.

(A) 20%
(B) 30.72%
(C) 40.23%
(D) 52.15%
(E) 63.2%

Solution:

  • Capactity of vessel \(= 10\) liters
  • Inital amount of petrol \(=\frac{3}{2+3}\times 10=\frac{3}{5}\times 10=6\) liters
  • Inital amount of kerosene \(=\frac{2}{2+3}\times 10=\frac{2}{5}\times 10=4\) liters

  • Every time, \(\frac{2}{10}\) of the petrol gets eliminated
  • So, after 3 removal volume of petrol will be \(=6(1-\frac{2}{10})^3\)
    \(=6(\frac{8}{10})^3\)
    \(=6(\frac{4}{5})^3\)
    \(=6\times \frac{4}{5}\times \frac{4}{5}\times \frac{4}{5}\)
    \(=3.072\)
  • Percentage in the final solution \(=\frac{3.072}{10}\times 100=30.72\%\)

Hence the right answer is Option B
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I thought the question stem said 2 liters of “solution” gets removed (not just “petrol”).

Posted from my mobile device
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A vessel (capacity 10 liters) is filled entirely with petrol and keros 15 Jan 2023, 02:14
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for this replacement questions use following
( 1- x/10)^n
x is qty of replacement and n is number of times replacement is being done
in this case x is 2 and n is 3
we get ( 1-2/10)^3
(8/10)^3
qty of petrol 6 and that of kerosene 4
% of petrol left will be
6 * ( .8)^3 /100
30.72%


Bunuel
A vessel (capacity 10 liters) is filled entirely with petrol and kerosene in the ratio 3:2. Two liters of this solution is removed and replaced with 2 ltrs of kerosene. This procedure is repeated thrice. Find the % of petrol in the resulting solution.

(A) 20%
(B) 30.72%
(C) 40.23%
(D) 52.15%
(E) 63.2%


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A vessel (capacity 10 liters) is filled entirely with petrol and keros 26 Mar 2023, 10:11
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ChanSu
SaquibHGMATWhiz
Bunuel
A vessel (capacity 10 liters) is filled entirely with petrol and kerosene in the ratio 3:2. Two liters of this solution is removed and replaced with 2 ltrs of kerosene. This procedure is repeated thrice. Find the % of petrol in the resulting solution.

(A) 20%
(B) 30.72%
(C) 40.23%
(D) 52.15%
(E) 63.2%

Solution:

  • Capactity of vessel \(= 10\) liters
  • Inital amount of petrol \(=\frac{3}{2+3}\times 10=\frac{3}{5}\times 10=6\) liters
  • Inital amount of kerosene \(=\frac{2}{2+3}\times 10=\frac{2}{5}\times 10=4\) liters

  • Every time, \(\frac{2}{10}\) of the petrol gets eliminated
  • So, after 3 removal volume of petrol will be \(=6(1-\frac{2}{10})^3\)
    \(=6(\frac{8}{10})^3\)
    \(=6(\frac{4}{5})^3\)
    \(=6\times \frac{4}{5}\times \frac{4}{5}\times \frac{4}{5}\)
    \(=3.072\)
  • Percentage in the final solution \(=\frac{3.072}{10}\times 100=30.72\%\)

Hence the right answer is Option B
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I thought the question stem said 2 liters of “solution” gets removed (not just “petrol”).

Same doubt.
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A vessel (capacity 10 liters) is filled entirely with petrol and keros 26 Mar 2023, 11:28
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ChanSu

I thought the question stem said 2 liters of “solution” gets removed (not just “petrol”).

Posted from my mobile device
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prachisaraf

I thought the question stem said 2 liters of “solution” gets removed (not just “petrol”).

Same doubt.
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Hi Prachi and ChanSu,

That was a typo. I have corrected it now. It is a solution only.

So what happens in such a question can be understood in the following manner.


1. First we have 10 liters of a solution with 6 liters of petrol and remove 2 from it so we have 8 liters of solution left.
    So both petrol and kerosene get reduced to \(\frac{8}{10}\) of the initial amount.
    This means we have 6 × \(\frac{8}{10}\) liter of petrol
    Next, we add kerosene. But it does not change the amount of petrol
2. Now we have 10 liters of solution with 6 × \(\frac{8}{10}\) liters of petrol.
    And we do a similar process as above so our net petrol becomes 6 × \(\frac{8}{10}\) × \(\frac{8}{10}\)

Once we repeat this process again we multiply it again by \(\frac{8}{10}\)

Hope this helps.
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A vessel (capacity 10 liters) is filled entirely with petrol and keros 26 Mar 2023, 23:26
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Bunuel
A vessel (capacity 10 liters) is filled entirely with petrol and kerosene in the ratio 3:2. Two liters of this solution is removed and replaced with 2 ltrs of kerosene. This procedure is repeated thrice. Find the % of petrol in the resulting solution.

(A) 20%
(B) 30.72%
(C) 40.23%
(D) 52.15%
(E) 63.2%


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Check out this post first: https://anaprep.com/arithmetic-replacement-in-mixtures/

While working with the ingredient that we are NOT adding back, i.e. petrol here,

\(Cf = Ci * \frac{Vi}{Vf}\)

Vi = Volume after removal
Vf = Volume after replacing back.

Since the process is repeated twice more, we get
\(Cf = Ci * (\frac{Vi}{Vf})^3\)

\(Cf = \frac{3}{5} * (\frac{8}{10})^3 \)

Now, \(\frac{8}{10} = \frac{4}{5}\) and \((\frac{4}{5})^3 = \frac{64}{125}\)

Note that \(\frac{64}{125}\) is slightly more than \(\frac{1}{2}\) because \(\frac{62.5}{125}\) will be exactly half.

So Cf = 60% * slightly more than \(\frac{1}{2}\)

So Cf = Slightly more than 30%

Answer (B)
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A vessel (capacity 10 liters) is filled entirely with petrol and keros 29 May 2024, 11:30
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