A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal tak

Question

A jar contains a mixture of 175 ml water and 700 ml alcohol. Randy takes out 10% of the mixture and substitutes it by water of the same amount. If the process is repeated once again, what will be the percentage of water in the mixture ?

A. 20.5
B. 25.4
C. 29.5
D. 30.3
E. 35.2

Are You Up For the Challenge: 700 Level Questions

A. 20.5
B. 25.4
C. 29.5
D. 30.3
E. 35.2

Are You Up For the Challenge: 700 Level Questions

Archit3110 · Accepted Answer

for every 10% of removal of mixture ; 90% of alcohol remains in the Jar
so at the end we have 700*.9*.9 ; 567 ltrs alcohol 
and water = 875 ( 700+175) - 567 = 308
% of water in jar = 308/875 ; 35.2 %
IMO E

preetamsaha · Answer

initial volume of Alcohol = 700, total volume = 700+ 175 = 875 , replaced quantity = 87.5 & n =2 
Using, final volume =initial volume ( 1-(replaced quantity)/(total volume))^n = 700(1-87.5/875)^2 = 700×81/100 = 567
So water in final mixture = 875 – 567 = 308
Required percentage = 308/875×100 = 35.2

correct answer E

preetamsaha · Answer

method II : 
As Randy takes out 10% of the mixture 
10% of alcohol is removed from 700 ml alcohol, 90% of alcohol is retained 
So alcohol remaining = 700 × 90% × 90% 
⟹ 700 × 0.9 × 0.9 = 567 
We totally have 875 ml overall mixture and of this 567 ml is alcohol. 
Remaining 875 – 567 = 308 is the amount of water. 
Approximately 308 is 30% of 1000 so by this we know that 308 is more than 30% 
Hence 35.2% is the percentage of water in the given mixture.

fireagablast · Answer

total ammount in mixture = 700+175 = 875

10% taken out, and replaced with water
175-175(0.1)+875(0.1) = 245

Process is repeated again
245-245(0.1)+875(0.1) = 308

New value of water is 308
Ratio of new water total is 308/875 = 35.2

ScottTargetTestPrep · Answer

Solution:

We see that the jar originally has 875 ml of the mixture. When Randy removes 10% (or 87.5 ml) of the mixture, he is removing 10% (or 17.5 ml) of water and 10% (or 70 ml) of alcohol. In other words, the jar has 175 - 17.5 = 157.5 ml of water and 700 - 70 = 630 ml of alcohol left. However, since he is replacing it with 87.5 ml water, the jar now has 157.5 + 87.5 = 245 ml of water and 630 ml of alcohol before the second replacement.

For the second replacement, he again removes 10% (or 87.5 ml) of the mixture, consisting of 10% (or 24.5 ml) water and 10% (or 63 ml) alcohol. In other words, the jar has 245 - 24.5 = 220.5 ml of water and 630 - 63 = 567 ml of alcohol left. However, since he is replacing it with 87.5 ml water, the jar now has 220.5 + 87.5 = 308 ml of water and 567 ml of alcohol after the second replacement. Therefore, the percentage of water in the mixture is 308/875 = 0.352 = 35.2%.

Alternate Solution:

Notice that the amount of alcohol decreases by 10% in each step. Thus, after the first replacement, the amount of alcohol in the mixture is 700 * 0.9 = 630 ml, and after the second replacement, the amount of alcohol decreases to 630 * 0.9 = 567 ml.

Notice also that the amount of water added back to the solution is the same as the amount of mixture removed at each step; thus, the total amount of mixture is always 700 + 125 = 825 ml.

After the second replacement, there are 875 - 567 = 308 ml of water in the mixture, and the percentage of water in the mixture is 308/875 = 0.352 = 35.2%.

Answer: E

Adnan123 · Answer

Total volume = 875
percentage of water = 175/875 = 20%
percentage of alcohol = 100 - 20 = 80%

CiVi = CfVf
Cf = 787.5/875 x 787.5/875 x 80
Concentration of alcohol = Cf = 64.8%
Concentration of water = 100 - 64.8 = 35.2%

A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal tak

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A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal tak

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