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Bunuel
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A vessel contains a mixture of milk and water in the ratio of 14 : 3. 08 Mar 2021, 09:15
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C
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65% (hard)

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67% (03:01) correct 33% (03:02) wrong based on 276 sessions

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A vessel contains a mixture of milk and water in the ratio of 14 : 3. Now, 25.5 litres of the mixture is taken out from the vessel and 2.5 litres of pure water and5 litres of pure milk is added to the mixture. If the resultant mixture contains 20% water, what was the initial quantity of mixture in the vessel before there placement ? (in litres)

(A) 54
(B) 60
(C) 68
(D) 81
(E) 98
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C
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A vessel contains a mixture of milk and water in the ratio of 14 : 3. 08 Mar 2021, 15:15
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Bunuel
A vessel contains a mixture of milk and water in the ratio of 14 : 3. Now, 25.5 litres of the mixture is taken out from the vessel and 2.5 litres of pure water and5 litres of pure milk is added to the mixture. If the resultant mixture contains 20% water, what was the initial quantity of mixture in the vessel before there placement ? (in litres)

(A) 54
(B) 60
(C) 68
(D) 81
(E) 98
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After 25.5 litres were removed, the ratio of milk and water would remain unchanged (removal doesn't change the ratio of the constituents)
So, milk to water remains 14:3 i.e. milk = 14x and water = 3x

Milk and water added = 5 lit and 2.5 lit

Final ratio of milk : water = 80:20 = 4:1

Thus: (14x+5)/(3x+2.5) = 4/1
x = 2.5

Thus, volume in the vessel after removing 25.5 litres = 14x + 3x = 17x
= 17 * 2.5 = 42.5 lit

This, initial volume = 42.5 + 25.5 = 68

Answer C

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A vessel contains a mixture of milk and water in the ratio of 14 : 3. 15 Apr 2024, 14:12
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Bunuel
A vessel contains a mixture of milk and water in the ratio of 14 : 3. Now, 25.5 litres of the mixture is taken out from the vessel and 2.5 litres of pure water and5 litres of pure milk is added to the mixture. If the resultant mixture contains 20% water, what was the initial quantity of mixture in the vessel before there placement ? (in litres)

(A) 54
(B) 60
(C) 68
(D) 81
(E) 98
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­Another way:

Initially let us have \(x\) litres: After removing \(25.5\) we have \(x-25.5\)

Water in remaining part \(=\frac{3}{17} (x-25.5)\) as it is homogenous mixture.

Water in resultant mixture after addition : \(\frac{20}{100}(x-25.5 +7.5 )\)

Equating litres of water, before and after:

\(\frac{3}{17}(x-25.5)+2.5 = \frac{20}{100} (x-18 )\)

Solving for \(x\), we get \(x=68\)

Hope it helped.
 ­
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A vessel contains a mixture of milk and water in the ratio of 14 : 3. 10 Mar 2022, 05:12
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\(\frac{milk}{water}\) = \(\frac{14}{3}\)

milk = \(\frac{14}{17}\)

water = \(\frac{3}{17}\)

25.5 litres of mixture is taken out = 25.5 = \(\frac{255}{10}\) = \(\frac{51}{2} \ litres\)

Remaining Milk = \(14x - \frac{14}{17}*\frac{51}{2}\) = \(14x-21 \ litres\)

Remaining Water = \(3x - \frac{3}{17}*\frac{51}{2}\) = \(3x-4.5 \ litres\)

Then, 2.5 litres of water and 5 litres of milk is added. Thus, final quantity of water and milk is:

Milk = \(14x-21+5\) = \(14x-16 \ litres\)

Water = \(3x-4.5+2.5\) = \(3x-2 \ litres\)

Resultant Mixture = \(14x-16+3x-2 \ litres\) = \(17x-18 \ litres\)

Given, resultant mixture contains 20% water:

\(0.2*(17x-18) = 3x-2\)
\(3.4x - 3.6 = 3x-2\)
\(0.4x = 1.6\)
\(x = 4\)

Thus, initial quantity of mixture was \(= 17x = 17*4 = 68 \ litres (C)\)
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A vessel contains a mixture of milk and water in the ratio of 14 : 3. 16 Apr 2024, 04:57
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Would it be wrong to straightforward the answer to the only multiple of 17 of the ACs?
Only 68 is divisible by 17, so the initial quantity must have been 68, otherwise that initial ratio would have been impossible.

Is my reasoning wrong (even though in this case the ans is correct)?
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A vessel contains a mixture of milk and water in the ratio of 14 : 3. 16 Apr 2024, 06:50
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Jsgambach
Would it be wrong to straightforward the answer to the only multiple of 17 of the ACs?
Only 68 is divisible by 17, so the initial quantity must have been 68, otherwise that initial ratio would have been impossible.

Is my reasoning wrong (even though in this case the ans is correct)?


 
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­Yes, I think you are correct. If there is only one option which is a multiple of 17 then that must be the answer, because the original quantity must be a multiple of 17.­
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A vessel contains a mixture of milk and water in the ratio of 14 : 3. 23 Dec 2024, 06:53
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Going by options, 14:3, 14+3 = 17, only 68 would be the possible value. Does GMAT give options so straightforward like this? Am I wrong in thinking that GMAT would me more careful while giving the options? Or does GMAT do have questions which when worked through options would be solved more faster?
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