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2/5 th of mixture of milk and water is replaced
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05 Feb 2021, 20:52
05 Feb 2021, 20:52
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2/5 th of mixture of milk and water is replaced by orange juice and the ratio of Milk, Water and Orange juice in the resulting mixture becomes 4:5:6 respectively. Initial quantity of milk in the mixture is 20 litres less than that of water. Then What % of resulting mixture should be taken out, due to which total quantity of milk and orange juice in the final mixture becomes 66 litres ?
A)35%
B)55%
C)45%
D)72%
E)64%
Posted from my mobile device
Carcass and GreenlightTestPrep please give a solution to this problem.
A)35%
B)55%
C)45%
D)72%
E)64%
Posted from my mobile device
Carcass and GreenlightTestPrep please give a solution to this problem.
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Re: 2/5 th of mixture of milk and water is replaced
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11 Feb 2021, 06:26
11 Feb 2021, 06:26
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Let
Milk = M lts
Water = (M + 20) lts
Total (initial) = (2M + 20) lts
Now, \(\frac{2}{5}^{th}\) of total = \(\frac{4}{5}\)M + 8 = Orange
Therefore, In resulting mixture;
Milk = \(\frac{3}{5}\)M
Water = \(\frac{3}{5}\)M + 12
Orange = \(\frac{4}{5}\)M + 8
Also, Milk : Water : Orange = 4 : 5 : 6
i.e. Milk = 4x
Water = 5x
Orange = 6x
We can relate both the expression of Milk or Water or Orange, I am gonna take Milk here,
Milk = 4x = \(\frac{3}{5}\)M ....... (1)
Water = 5x = \(\frac{3}{5}\)M + 12 ....... (2)
Put the value of (1) in (2);
Water = 5x = 4x + 12
Therefore, x = 12
Now,
Milk = 48 lts
Water = 60 lts
Orange = 72 lts
Total = 180 lts
Currently, Milk and Orange = 48 + 72 = 120 lts
But we require it to be 66 lts
We must take out 54 lts to do so
So, % = \(\frac{54}{120}\) x 100 = 45%
Hence, option C
Milk = M lts
Water = (M + 20) lts
Total (initial) = (2M + 20) lts
Now, \(\frac{2}{5}^{th}\) of total = \(\frac{4}{5}\)M + 8 = Orange
Therefore, In resulting mixture;
Milk = \(\frac{3}{5}\)M
Water = \(\frac{3}{5}\)M + 12
Orange = \(\frac{4}{5}\)M + 8
Also, Milk : Water : Orange = 4 : 5 : 6
i.e. Milk = 4x
Water = 5x
Orange = 6x
We can relate both the expression of Milk or Water or Orange, I am gonna take Milk here,
Milk = 4x = \(\frac{3}{5}\)M ....... (1)
Water = 5x = \(\frac{3}{5}\)M + 12 ....... (2)
Put the value of (1) in (2);
Water = 5x = 4x + 12
Therefore, x = 12
Now,
Milk = 48 lts
Water = 60 lts
Orange = 72 lts
Total = 180 lts
Currently, Milk and Orange = 48 + 72 = 120 lts
But we require it to be 66 lts
We must take out 54 lts to do so
So, % = \(\frac{54}{120}\) x 100 = 45%
Hence, option C
General Discussion
Re: 2/5 th of mixture of milk and water is replaced
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06 Feb 2021, 03:49
06 Feb 2021, 03:49
Expert Reply
I do not understand, BUT perhaps that is me, why you crossposting between gmatclub and greprep club when chetan2u gave you already a satisfactory explanation
https://gmatclub.com/forum/2-5-th-of-mi ... l#p2719626
Regards
https://gmatclub.com/forum/2-5-th-of-mi ... l#p2719626
Regards
