Latest Elitmus Aptitude Question SOLUTION: there is cask full of milk.E litres are drawn from the cask,it is then filled with water.this process is repeated.now the ratio of milk to water is 16:9.What is the capacity of the

5E

if x is capacity of cask, then
initially
x ltr milk .. 0 ltr water
after first draw and water filling by E ltrs,
x-E ltr milk and E ltr water

after 2nd draw and water filling by E ltrs,
x- [2E-(E^2/x)] and 2E -( E^2)/x ltr water

As per condition

x- [2E-(E^2/x)] /[ 2E -( E^2)/x] = 16/9

solving, we get
x=5E

let capacity of container be V
the
milk to total ratio will be 16/25 which equal to initaial ratio((V-E)/V)^2
after solving al this we wiil get V=5E.

Simple Logic
If it is 100 litres of Milk then suppose 20 liters water added and 20l iters milk removed
So, Milk=80 liters and Water=20 liters
Now, again remove 20% milk and add 20% water
So,20% of (80 Litres Milk + 20 Litres Water) i.e. 16 litres of milk removed and 4 litres of water removed
So, New quantity will be [ (80-16) Liters of milk + ( 20-4) liters of water ] + 20 liters of water
i.e. 64 liters milk + 36 litres of water
So ratio of milk to water is (64/36) when total capacity of cask is 100 liters
Similarly,this ratio with 16/9 that is total capacity is 16+9 = 25 i.e. Multiple of 5 = 5*E
So, Answer=5E

let the quantity of milk initially=x litres
quantity of milk left after 2 operations= [x{1-E/x}^2] litres
since the ratio of milk to water is 16:9
so total quantity=25
so, [x{1-E/x}^2]/x =16/25
on solving we get x=5E

Pragat pal Can you explain the answer?

quantity of the milk remain in the container is = v*(1-(x/v))^p
v= container vol
x drawn liquid
p how many times
use this equation

options are
A)E+9
B)9E
C)5E
D)cant be determined

Use simple formula
(1- E/X)^2 = 16/25
You'll get x=5E

for those who want to know how e/x came in first place . See this solution
->first milk is x and water 0
therefore total volume of cask =x
and let operation be defined as the process when E amount of liquid is removed and E amount of
water is added.
->then milk is x-E and water E
so the percentage of each of the components now is as follows-
Percentage of milk=(x-E)/x
Percentage of water=(E)/x [same E/x which many people are asking]
-> the process would be repeated many times till the required ratio reaches
so when E amount of this mixture is removed then
-> amount of milk removed=[(x-E)/x]*E [ as ((x-E)/x) is the percentage of milk in mixture obtained
after first operation ]
amount of milk remaining={ (x-E)- [(x-E)/x]*E }=[ { (x-E)^2/x } ] (answer can be calculated
from this only)

->Similarly amount of water removed=(E/x)*E=(E^2)/x (not needed for answer)
amount of water now when E water is not added= E - (E^2)/x
amount of water now(after operation) in the cask = E(added in new operation) + [ amount already in ]
= E(added in new operation) + E - (E^2)/x
= 2E - (E^2)/x
Percentage of milk now = (amount of milk / total volume)
= [ { (x-E)^2/x } ]/x
= [ (x-E)/x ]^2
this is equal to (16/(16+9)) the required percentage of milk.
therefore on solving the equality [ [ (x-E)/x ]^2 ]=( 16/25 )
you'll get x=5E.

1-9/50
41/50

1-9/50
41/50

@Pragat Pal Singh
how did you solve the equation?
I am not getting it.Please Explain.

A)E+9
B)9E
C)5E
D)cant be determined are the options.

For sure answer is not option (d).
So, go from options in reverse order (Put E=1 for easy calculations). 5E satisfies

how E^2/x came in first solution

it cannot be determined. and it not mentioned in the question that milk is taken out on percentage basis,so stop assuming this in your solutions guys,just take 'e' litres for every iteration and solve the question.

let z be the smallest natural number (>1) which give an integral value if the square root, cube root