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11 Sep 2025, 00:37
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
71% (02:21) correct
29%
(02:17)
wrong
based on 186
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A vessel is full of a mixture of milk and water, with 9% milk. Nine litres are withdrawn and then replaced with pure water. If the milk is now 6%, how much does the vessel hold?
(A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 45 litres
(A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 45 litres
11 Sep 2025, 01:09
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Bunuel
Total Volume of Mixture = V
Milk = 0.09V (9%)
9 litres are removed, so milk removed = 9% of 9 litres (as milk will be removed as per ratio it was)
so milk removed = 9% of 9 L = 0.81 L
Now using below we can calculate
(0.09V - 0.81)/V = 0.06
0.09V-0.81 = 0.06V
0.03V = 0.81
V = 27
So Volume is 27
ANS B
11 Sep 2025, 01:16
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Say the total litres initially is x litres,
out of which milk is 9% . ie, 9x/100
Given that 9 litres of mixture is removed which means new milk content is 9x/100 -(9*9)/100
And 9 litres of pure water is added ie., total litres remains same as x
New milk % = 6%
which means (9x/100-(9*9)/100)/ x = 6/100
Solve for x= 27 litres, Option B
out of which milk is 9% . ie, 9x/100
Given that 9 litres of mixture is removed which means new milk content is 9x/100 -(9*9)/100
And 9 litres of pure water is added ie., total litres remains same as x
New milk % = 6%
which means (9x/100-(9*9)/100)/ x = 6/100
Solve for x= 27 litres, Option B
Bunuel
