Latest TCS Aptitude Question SOLUTION: a beaker contains 180 litres of alcohol. On day 1, 60 litres of alcohol is replaced with water. On 2nd and 3rd days 60 litres of the mixture in the beaker is replaced with water. W

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(#M40026370) TCS question Keep an EYE

a beaker contains 180 litres of alcohol. On day 1, 60 litres of alcohol is replaced with water. On 2nd and 3rd days 60 litres of the mixture in the beaker is replaced with water. What will be the quantity of alcohol in the beaker after 3rd day?

Asked In TCS tejal dobhal (10 years ago)

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kumar  (10 years ago)

ans =(160/3) lit
1st time 180-60=120
2nd time 120-60*120/180=80lit
3rd time 80-60*80/180= 160/3 lit

Is this solution Helpfull? Yes (33) | No (5)  

Palash Bansal  (10 years ago)

====================Alcohol=============================Water
0day> 180 0
1day> 120(60 lit reduced) 60(60 lit added)
2day> 80(2x i.e. 40 l reduced as 120=60) 100(that 40 added)
3day> 160/3 200/3 (similarly taken out ratio of 80 and 100)
ans=160/3 lit

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ipsita saha  (10 years ago)

180(1-60/180)^3=160/3lit

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Harish Marrapu  (10 years ago)

Initially:: 180alc
1st day:: 120alc+60wat
2nd day:: (2/3 of 120)alc+(1/3 of 60)wat is removed and 60 wat is added.
As a result by 2nd day ending:: 80alc + 40wat + 60wat = 80alc + 100wat
3rd day:: (4/9 of alc and 5/9 of wat is rem) and 60wat is added.
As a result by 3rd day ending:: 53 lit(approx)alc is present.

Is this solution Helpfull? Yes (10) | No (5)  

SHIVANG RANA  (10 years ago)

1st day- 120 al 60 water(after replacement)
2nd day- 80 al 100 water(after replacement)
3rd day- 160/3 al 380/3 water(after replacement)

use the concept of ratio.....

Ans 160/3...

Is this solution Helpfull? Yes (7) | No (2)  

kuldeep kumar  (10 years ago)

ans is 53.33 letter
starting a=180 w=0
1st day a=120 w=60
2nd day a=80 w=100
3rd day a=53.33 and w=126.67

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naveen garg  (10 years ago)

soln is total quantity*(1-fraction of wine withdrawn)^n
180*(1-1/3)^3

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Rajat Arora  (10 years ago)

160/3 answer

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ravendra tiwaree  (10 years ago)

60 litr alcohal

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  (4 years ago)

Initially, 180L of alcohol

removing 60L from 180L = 180* X/100=60
solving, X=33.33%(amount of prcentage to be removed)

alcohol water total
1st day- =180-60=120 +60 180

2nd day- =120-(120*33.33/100) 60-(60*33/100)
=80(approx) =40(approx) 180
=40+60
(adding 60L water)

3rd day- =80-(80*33.33/100) =40-(40*33.33/100) 180
ANS="53.33" =26.68+100
(adding 60L water)

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Abhijit  (10 years ago)

it'll be 80 litres if on 2nd and 3rd day,total 60 litres of mixture is replaced with water.......

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Guru Pranav  (10 years ago)

amount of alcohol left = 160/3 Ans.

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chaitanyasailesh  (7 years ago)

0th day 180 lt of alcohol,,
1st day due to replacement 120 lt of alcohol, 60 lt of water
2nd day:mix of 60 lt is to be replaced then { 120-60*(2/3):60-60*(1/3)+60}=80:100
3rd day: now alcohol and water are in 4:5 ratio then {80-60*(4/9):100-60*(5/9)+60}=53.5:126.5
so 53.5 lt of alcohol is present

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sobiya  (6 years ago)

53.3 litre

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BALAGOPAL M  (3 years ago)

For solving this problem you have to use a simple equation:

Final Concentration=(Initial Concentration)*(1-Volume Replaced/Final Quantity)

1st Day: FC=1*(1-60/180)=2/3
2nd Day: FC=(2/3)*(1-60/180)=2/3*2/3
3rd Day: FC=(2/3*2/3)*(1-60/180)=2/3*2/3*2/3=0.296

NOTE:Here the final volume is always 180l because once one take 60l out of the mixture,it is getting replaced by same amount of water.

So 0.296 of the entire volume is alcohol;
ie; 0.296*180=53.33

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