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Nilanjan Jana

· started a discussion

· 1 Months ago

Catching prob (Question: 96539.0 )

pls make a video

Question:
The diluted wine contains only 8 liters of wine and the rest is water. A new mixture whose concentration is 30%, is to be formed by replacing wine. How many liters of mixture shall be replaced with pure wine if there was initially 32 liters of water in the mixture?
Options:
A) 4
B) 5
C) 8
D) None of these
Solution:

Ans (b)                

Wine                     Water

8L                            32L

1              :               4

20%                        80% (original ratio)

30%                        70% (required ratio)

In this case, the percentage of water being reduced when the mixture is being replaced with wine.

So the ratio of left quantity to the initial quantity is 7 : 8.

Therefore \(\cfrac{7}{8}=\left [ 1-\cfrac{k}{40} \right ]\Rightarrow\cfrac{7}{8}=\left [ \cfrac{40-k}{40} \right ]\)

⇒ K = 5 liter

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