Water equal to 10% of a solution of milk and water is added to the sol

Consider the initial solution as 100 L;
10 % solution of water is added =110L;
again removed 10 % solution =100L which is in the ratio of 3:1;
Before removal as well, the solution would be in the 3: 1 ratio;
so, in 110 litres of milk and water --> milk=82.5 L, water=27.5 L;
The volume of milk is never changed, so in 100 L , milk=82.5 L and water =17.5 L;
now initial ratio = 82.5/17.5 --> 33/7;
answer is c)

Please correct me, if i am wrong

Easier approach:

Say milk is m% in the original solution of milk and water. An amount of water (milk 0%) is added to this solution such that the amount is equal to 10% of the original solution. So if the original solution is 100 ml, 10 ml of water is added.
This is in effect just mixing 100 ml of m% milk solution with 10 ml of 0% milk solution to give 75% milk solution (because milk : water = 3:1). When we remove some of this solution, it doesn't change the concentration and it stays at 75% milk solution.

100/10 = (0 - 75)/(75 - m)
750 - 10m = - 75
m = 82.5%

So milk: water in original solution = 82.5 : 17.5 = 33 : 7

Answer (C)

Assume initial solution to be 40 L

Then 40L solution is added to 10% which is 4 L of water . important thing to note that since 10% of solution is removed after mixing , concentration will remain same after removing 10% of solution.

Just to be in the safe side for easy calculation, assume 10% of solution is not removed then the final solution becomes 44 L where milk and water will be in the ratio of 3:1. Since after removing 10% of this solution , ratio will still remain same , we can find out easily the initial ratio

So 3 parts of milk in final solution is before removing = (3/4)*44 = 33 L
and 1 part of water = 44-33 = 11L

Now water we have added was 4 L . So initial volume of water must be 11-4 = 7 L as volume is not changing before removal.

So the milk in initial solution will be 40-7= 33 L

Thus ratio should be initial solution irrespective of removal of solution at later stage = 33: 7. hence answer is C

The final solution has 75% milk, as water was added to the first solution, the first solution must contain milk more than 75%.

Only C is satisfying this criteria.

Couldn't solve using conventional method.

let's say the ratio of milk to water is a:b
10% of solution = \(\frac{1}{10}*(a+b)\)

adding 10% of solution will cause the ratio to be:
a : b + \(\frac{1}{10}*(a+b)\)

with some algebra, the ratio becomes:
a : \(\frac{11b+a}{10}\)

which is equivalent to 3:1
cross multiply 3:1 and the previous ratio to get:
a = \(3*\frac{11b+a}{10}\)
10a = 33b + 3a
7a = 33b
\(\frac{a}{b}\) = \(\frac{7}{33}\)

so ratio a:b = 33:7

w = 70/4
m = 100 - 70/4 = 330/4

m : w = 33:7

Your problem was the same in the previous question in which you tagged me too - you did not complete the solution to get to the answer.
Ensure you understand what each variable is and what exactly you are looking for.
If m = amount of milk in original solution and w = amount of water in original solution, you need to find the values of m and w to get the ratio (which is asked)

Let the quantity of the initial solution (IS) be 10 ml and the quantity of water therein be 'x' ml which means the quantity of milk is (10 -x)ml. After 1 ml (10% of the vol of IS) is added, the quantity of water becomes (x + 1) ml. The qty of milk, of course, remains (10 - x)ml since no milk was added or removed. Since the ratio of milk to water in the resultant solution is 3:1, we can set up the following equation:
(10 - x)/(x +1) = 3/1....> x = 7/4. So the quantities of water and milk were 7/4 ml and (10 - 7/4) = 33/4 in the IS.
Ratio of Milk to Water in IS was (33/4):(7/4) = 33:7. ANS: C

ANS EXPLANATION:

lets say initial milk quantity is 3 (as the quantity of milk would not be changed) and water quantity is n. The equation will be:
3/(n+0.1(n+3))= 3/1
So n will be 0.6363
The initial ratio will be 3/0.6363 = 4.71 which is equivalent to 33/7

Let m:w be initial ratio.
Total = m + w
10% of Total = 0.1(m+w)

Note : 10% solution removed doesn't matter to ratio. Remaining solution will still have the same ratio. Think about it.

no change in milk , so it will remain "m".
Change in water , so it will become "w + 10%(m+w)"

m/(w+0.1(m+w)) = 3:1

m = 3w + 0.3m + 0.3w
0.7m = 3.3w
m/w = 33/7

Given: Water equal to 10% of a solution of milk and water is added to the solution. Now 10% of the solution is removed.

Asked: If now the ratio of milk and water is 3 : 1, find the ratio of milk and water in the initial solution.

Let the initial ratio of milk and water be k.
Milk = k
Water = 1
Solution = k+1

Water equal to 10% of a solution of milk and water is added to the solution.
Water = .1(k+1) + 1= .1k + 1.1
Milk = k

Now 10% of the solution is removed.
Ratio of milk & water remains unchanged.

Now the ratio of milk and water is 3 : 1
k = 3(.1k + 1.1) = .3k + 3.3
.7k = 3.3
k = 33/7

IMO C

Bunuel could you provide a detailed solution, please?

KarishmaB

I solved the same way as described above
but after removing 10%

0.9m+0.9(w+10)=99

i equated with the equation with new ratio

\(\frac{0.9m}{ 0.9(w+10)} =\frac{ 3}{1}\)

so i got m = 3w+30

and plugged in here m+w = 100

and got w = 70/4

whats wrong with my solution?

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