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A vessel ( capacity 10 ltrs) is filled entirely with petrol

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A vessel ( capacity 10 ltrs) is filled entirely with petrol  [#permalink]

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New post 28 Jul 2019, 04:18
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Question Stats:

65% (02:42) correct 35% (02:49) wrong based on 52 sessions

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A vessel ( capacity 10 ltrs) is filled entirely with petrol and kerosene in the ratio 3:2. Two ltrs 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%

(B)

nick1816
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Re: A vessel ( capacity 10 ltrs) is filled entirely with petrol  [#permalink]

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New post 28 Jul 2019, 04:38
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Initial volume of the petrol in vessel=(3/5)*10=6
Let volume of the petrol in vessel after 3 replacements=V

\(V=6(1-\frac{2}{10})^3\)
\(V=6*(\frac{4}{5})^3\)
V= 3.072L

% of petrol in the resulting solution= \(\frac{3.072}{10}*100=30.72\)%

shivansh9307 wrote:
A vessel ( capacity 10 ltrs) is filled entirely with petrol and kerosene in the ratio 3:2. Two ltrs 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%

(B)

nick1816
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Re: A vessel ( capacity 10 ltrs) is filled entirely with petrol  [#permalink]

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New post 28 Jul 2019, 08:32
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oohhh...nick1816 Thanks for the solution mate..
it took me a lot of time calculating Petrol amount after each iteration, though finally i made the correct option. But saw your post. This is good..
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Re: A vessel ( capacity 10 ltrs) is filled entirely with petrol  [#permalink]

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New post 11 Aug 2019, 07:41
nick1816 wrote:
Initial volume of the petrol in vessel=(3/5)*10=6
Let volume of the petrol in vessel after 3 replacements=V

\(V=6(1-\frac{2}{10})^3\)
\(V=6*(\frac{4}{5})^3\)
V= 3.072L

% of petrol in the resulting solution= \(\frac{3.072}{10}*100=30.72\)%

shivansh9307 wrote:
A vessel ( capacity 10 ltrs) is filled entirely with petrol and kerosene in the ratio 3:2. Two ltrs 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%

(B)

nick1816
Rahulihavenoclue


Hi, can you please explain , how did you come up with 6*(1-2/10)^3
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Re: A vessel ( capacity 10 ltrs) is filled entirely with petrol  [#permalink]

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New post 13 Aug 2019, 01:33
abhishek31 wrote:
nick1816 wrote:
Initial volume of the petrol in vessel=(3/5)*10=6
Let volume of the petrol in vessel after 3 replacements=V

\(V=6(1-\frac{2}{10})^3\)
\(V=6*(\frac{4}{5})^3\)
V= 3.072L

% of petrol in the resulting solution= \(\frac{3.072}{10}*100=30.72\)%

shivansh9307 wrote:
A vessel ( capacity 10 ltrs) is filled entirely with petrol and kerosene in the ratio 3:2. Two ltrs 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%

(B)

nick1816
Rahulihavenoclue


Hi, can you please explain , how did you come up with 6*(1-2/10)^3

I think it is:
current petrol in the vessel = 3/5 * 10 = 6l
so 6*(1-2/10)^3 is current petrol * (total - 2l being removed from total)^we do this for three times = petrol after the 3 replacements
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Re: A vessel ( capacity 10 ltrs) is filled entirely with petrol   [#permalink] 13 Aug 2019, 01:33
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