Latest Elitmus Aptitude Question SOLUTION: Q. From a container having pure milk ,20% is replaced by water and the process is repeated thrice. At the end of the third operation , the milk is Option a) 40% pure b) 50% pure c)

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Q. From a container having pure milk ,20% is replaced by water and the process is repeated thrice. At the end of the third operation , the milk is
Option
a) 40% pure
b) 50% pure
c) 51.2% pure
d) none of these

Asked In Elitmus Ankit Boae (10 years ago)

Unsolved Read Solution (14)

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Read Solution (14) :

Ankit patel  (10 years ago)

Ans is C.As in 1st operation- milk is 80% and water is 20%.
2nd operation- milk is 64%(80-80*0.2) and water is 36%(20-20*0.2+20).
3rd operation- milk is 51.2% and water is 48.8%.

Is this solution Helpfull? Yes (28) | No (4)  

Simran  (10 years ago)

Let total quantity of original milk = 1000 gm
milk after first operation = 80% of 1000 = 800 gm
milk after second operation = 80% of 800 = 640 gm
milk after third operation = 80% of 640 = 512 gm
strength of final mixture = 51.2 %

Is this solution Helpfull? Yes (19) | No (1)  

AVIJIT  (10 years ago)

y=x(1-fraction atken out)^n
here y= remaining liquid
x=initial liquid
if initial liquid(x) is 100
fraction will be 20/100=1/5
y=100(1-1/5)^3
y=51.2%

Is this solution Helpfull? Yes (6) | No (1)  

manivikas  (10 years ago)

Milk : Water

step 0:-) 100 : 0
step 1:-) 80 : 20
step 2:-) 64 : 32
step 3:-) 51.2 : 48.8

Is this solution Helpfull? Yes (4) | No (2)  

rahul mishra  (10 years ago)

c: 51.2%
solve by using successive discount method

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basanagouda  (10 years ago)

right answer is = 40.96 %.
method 1) :q o pure milk left = 100(1-20/100)^4 = 100(0.8)^4 = 40.96 %.
method 2) : q of milk left after each iteration=80% then 64 % then 51.2 % and then finally 40.96 %.................................

Is this solution Helpfull? Yes (2) | No (3)  

Trisha  (10 years ago)

It will result into 4 operations...as in question it is given that , after one operation it is repeated thrice, so final answer =40% pure.

Is this solution Helpfull? Yes (1) | No (9)  

neha kumari  (9 years ago)

d none of these

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Arun Kumar  (9 years ago)

See every time 0.8 is milk left
=>0.8^2=0.512

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kirankumar  (8 years ago)

in first operation we will have 80% pure milk,
in second operation 80*20% =>80*0.2=> 16 now milk percentage is 80-16=64%,
in third operation 64*20%= > 64*0.2 => 12.8 now milk percentage is 64-12.8=51.2%

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jitendra idpache  (6 years ago)

ans : 51.2 %

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souharda  (9 years ago)

its 40.96% pure

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Ramesh Singh  (9 years ago)

it has a simple formula REMAINING MIXTUTRE=TOTAL QUANTITY*[(TOTAL-REPLACE QUANTITY)/TOTAL]^n
where n is no.of operation
so we will get 51.2%

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SANTOSH KUMAR  (8 years ago)

51.2%
let assume there is 100 lit pure milk in a container when 1st time 20% milk is replaced by water than there is 80 lit milk and 20 lit. water in container then in 2nd time there 80(100-20) % of 80 will be pure milk left in a container i.e 64 lit. and in 3rd time again 80% of 64 pure milk left in container i.e 51.2%.

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