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jaswanth_ss

gmatophobia

pintukr
Rohan took pure milk and started to dilute it with water. He first replaced 16 liters from the beaker full of pure milk with 16 liters of water. He performed this process three more times. Finally, the ratio of milk left in the beaker to that of water became 81 : 175. Find out how much pure milk (in liters) was there in the beaker initially?

A. 48
B. 64
C. 72
D. 84
E. 96


from Unacademy Prep
Not a 'GMAT-like' question IMO :dontknow:

Ratio of milk left to the ratio of milk originally present = \((1-\frac{16}{x})^4\)

Let's assume that the beaker contained 81 + 175 = 256 units of liquid at the start of the mixing process. It's given that the beaker consisted of pure milk, hence all 256 units of liquid were pure milk.

Units of pure milk post the mixing process = 81 (given that ratio of milk left in the beaker to that of water became 81 : 175)

\(\frac{81}{256} = (1-\frac{16}{x})^4\)

Taking the fourth root on both sides of the equation

\(\frac{3}{4} = 1-\frac{16}{x}\)

\(\frac{16}{x} = 1 - \frac{3}{4}\)

\(x = 16 * 4 = 64\)

Option B

 
­gmatophobia Could you explain the reasoning behind \((1-\frac{16}{x})^4\)?
How should we approach this?
­Please Reply how (x-16/x)^4 is equal to post mixture ratio?
 
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pintukr
Rohan took pure milk and started to dilute it with water. He first replaced 16 liters from the beaker full of pure milk with 16 liters of water. He performed this process three more times. Finally, the ratio of milk left in the beaker to that of water became 81 : 175. Find out how much pure milk (in liters) was there in the beaker initially?

A. 48
B. 64
C. 72
D. 84
E. 96


from Unacademy Prep
­
The dilution formula is used to calculate the concentration or amount of a substance (like milk in our problem) after some of it is removed and replaced by another substance (like water). It applies in situations where a mixture is repeatedly diluted.
Attachments

dilution formula.png
dilution formula.png [ 15.1 KiB | Viewed 1460 times ]

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