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20 Apr 2019, 12:42
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
61% (02:18) correct
39%
(01:48)
wrong
based on 527
sessions
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From a cask of milk containing 30 litres, 6 litres are drawn out and is replaced with water. If the same process is repeated two more times, how many litres of milk will be left in the cask?
A. 10 litres
B. 12 litres
C. 14.38 litres
D. 15.36 litres
E. 20 litres
A. 10 litres
B. 12 litres
C. 14.38 litres
D. 15.36 litres
E. 20 litres
20 Apr 2019, 14:08
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30 liters = 24 liters of milk + 6 liters of water.
Next step: you draw 6 liters out from 30 liters. \(\frac{6}{30} = 0.2\). Hence, \(0.8(24 MILK + 6 WATER)\) liters are left. Milk left: 19.2
Since the process is repeated two times, the quantity of milk left is \(0.8 * 19.2 = 15.36\). Pick D.
Next step: you draw 6 liters out from 30 liters. \(\frac{6}{30} = 0.2\). Hence, \(0.8(24 MILK + 6 WATER)\) liters are left. Milk left: 19.2
Since the process is repeated two times, the quantity of milk left is \(0.8 * 19.2 = 15.36\). Pick D.
25 Jun 2019, 03:08
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This is a question on mixtures involving withdrawal and replacements. Naturally, you can use basic percentage change concepts to solve the question; but, specific to this kind of a question, there is a concept which is good to know, because you can directly apply it whenever you come across questions like these.
Let’s look at the approach using percentages, first.
The number 6, when compared to 30 is 20% or \(\frac{1}{5}\). So, every time, \(\frac{1}{5}\)th of the total volume is withdrawn and replaced.
To start off with, there are 30 litres of milk in the cask. If \(\frac{1}{5}\)th of this is removed (i.e. 6 litres is removed), the volume of milk left in the cask = (\(\frac{4}{5}\)) * 30.
6 litres of water is added to this to keep the total volume at 30 litres.
In the next step, \(\frac{1}{5}\)th of the total volume is removed.
Therefore, volume of milk left in the cask = (\(\frac{4}{5}\)) of (\(\frac{4}{5}\)) * 30.
Similarly, the volume of milk left in the cask, after the third operation = (\(\frac{4}{5}\)) of (\(\frac{4}{5}\)) of (\(\frac{4}{5}\)) * 30.
\(\frac{4^3}{5^3}\) = \(\frac{64}{125}\) is a little more than ½ (because \(\frac{64}{128}\) = \(\frac{1}{2}\)). Therefore, the above expression should give us a value that is a little more than 15. Clearly, that has to be option D.
From V units of pure liquid, if x units are removed and replaced with water, the volume of pure liquid remaining after ‘n’ such operations will be
V \((1 – \frac{x}{V})^n\)
If we take this approach, we have V = 30, x = 5 and n = 3. When substituted in the expression above, we have,
30 \((1 – \frac{6}{30}) ^3\), which simplifies to
\(30 (\frac{4}{5}) ^3\) which is the same as the expression we obtained in the other approach.
It’s good if you can remember the above expression because, then it means you just have to identify the relevant values and plug them in to get your answer. In case you prefer remembering minimal formulas, then the percentages approach works just fine.
Hope this helps!
Let’s look at the approach using percentages, first.
The number 6, when compared to 30 is 20% or \(\frac{1}{5}\). So, every time, \(\frac{1}{5}\)th of the total volume is withdrawn and replaced.
To start off with, there are 30 litres of milk in the cask. If \(\frac{1}{5}\)th of this is removed (i.e. 6 litres is removed), the volume of milk left in the cask = (\(\frac{4}{5}\)) * 30.
6 litres of water is added to this to keep the total volume at 30 litres.
In the next step, \(\frac{1}{5}\)th of the total volume is removed.
Therefore, volume of milk left in the cask = (\(\frac{4}{5}\)) of (\(\frac{4}{5}\)) * 30.
Similarly, the volume of milk left in the cask, after the third operation = (\(\frac{4}{5}\)) of (\(\frac{4}{5}\)) of (\(\frac{4}{5}\)) * 30.
\(\frac{4^3}{5^3}\) = \(\frac{64}{125}\) is a little more than ½ (because \(\frac{64}{128}\) = \(\frac{1}{2}\)). Therefore, the above expression should give us a value that is a little more than 15. Clearly, that has to be option D.
From V units of pure liquid, if x units are removed and replaced with water, the volume of pure liquid remaining after ‘n’ such operations will be
V \((1 – \frac{x}{V})^n\)
If we take this approach, we have V = 30, x = 5 and n = 3. When substituted in the expression above, we have,
30 \((1 – \frac{6}{30}) ^3\), which simplifies to
\(30 (\frac{4}{5}) ^3\) which is the same as the expression we obtained in the other approach.
It’s good if you can remember the above expression because, then it means you just have to identify the relevant values and plug them in to get your answer. In case you prefer remembering minimal formulas, then the percentages approach works just fine.
Hope this helps!
