A cask is full of wine but it has a leak in the bottom. When one-fourt

Question

A cask is full of wine but it has a leak in the bottom. When one-fourth of the cask empties out because of the leak, the cask is replenished with water. Next when half of the cask has leaked out, it is again filled with water. Finally when three-fourths of the cask leaks out, it is again filled with water. What is the percentage of wine in the cask now?

A. 9.375%
B. 8.33%
C. 7.2%
D. 7.5%
E. 6.66%

A. 9.375% 
B. 8.33% 
C. 7.2% 
D. 7.5% 
E. 6.66%

danbrown9445 · Accepted Answer

Let cask contain 1L of wine

1/4 cask empties due to leak
Wine is  \frac{3}{4}

1/2 cask empties due to leak
Wine is  \frac{3}{4}*\frac{1}{2} = \frac{3}{8}

3/4 cask empties due to leak
Wine is  \frac{3}{8}*\frac{1}{4}=\frac{3}{32}

Total Vol of cask is 1L,
So wine is  \frac{3}{32}*100  = Little less than 10% ( \frac{3}{30}*100 ) = 9.375%

tbhauer · Answer

% water = .25+(.75*.5)+((.75*.5)*.75)=.90625=90.625%
% wine=100-90.625=9.375%
Ans=A

arockia · Answer

Assume the initial volume of wine as 100L;
1/4th of the wine is leaked--> remaining amount of wine =75 L;
ci* vi=cf*vf;
1*75=100*cf;
cf1=75/100=0.75 -- concentration of wine after addition of water;
next, out of 100 litres, half is leaked--> remaining is 50 L;
50*0.75=cf2*100;
cf2=0.75/2;
now, 3/4th is leaked, remaining is 25 L;
25*0.75/2=cf3*100;
cf3=0.75/8;
amount of wine in final solution=0.75/8 *100=9.375 L which is 9.375 % of 100;
Hence , answer is A)

adityasuresh · Answer

A cask is full of wine but it has a leak in the bottom. When one-fourth of the cask empties out because of the leak, the cask is replenished with water. Next when half of the cask has leaked out, it is again filled with water. Finally when three-fourths of the cask leaks out, it is again filled with water. What is the percentage of wine in the cask now?

A. 9.375%
B. 8.33%
C. 7.2%
D. 7.5%
E. 6.66%

A. 9.375%
B. 8.33%
C. 7.2%
D. 7.5%
E. 6.66%

F/I  =  (1 - B/A)^N 
F = Final concentration of the element who's concentration is being reduced
I = Initial Concentration of the element who's concentration is being reduced
B = Amount of cask which empties out
A = amount in the cask after its replenished with water

Note : since B is different (one-fourth , one half , three fourth. the formula is used thrice)
Assuming Volume = 100
1.  F/100  =  1 -25/100   So Final concentration of wine = 75 , water = 25
2.  F/75  =  1 - 50/100  So Final concentration of wine = 37.5 Water = 63.5
3 .  F/37.5  = 1 -  75/100  So Final concentration of wine = 9.375

A

KarishmaB · Answer

Replacement in mixtures is discussed here:  https://anaprep.com/arithmetic-replacement-in-mixtures/ Pure wine means initial concentration of wine is 1.

After first replacement, concentration C2 = 1 * (3/4) (because volume after removal is 3 and after adding back is 4) 
After second replacement, concentration C3 = 1 * (3/4) *(1/2) (here, volume after removal is 1 and after adding back is 2)
After third replacement, concentration Cf = 1 * (3/4) *(1/2) * (1/4) (here, volume after removal is 1 and after adding back is 4)
Cf = 3/32

1/33.33 is 3% so 1/32 is a bit more than 3%. So 3/32 is a bit more than 9%.

Answer (A)

A cask is full of wine but it has a leak in the bottom. When one-fourt

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A cask is full of wine but it has a leak in the bottom. When one-fourt

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