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

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

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

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

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 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

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

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?

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)

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)

if the initial milk/water= x/1, solution is a liter
then we can know that... milk is [x/x+1]*a, water is [1/x+1] *a

After add some water, now water is [1/x+1] *a+0.1a

noted that pouring out the solution won't change the ratio of milk to water.
(They mix together and are pulled together.)

so before it, milk/water = 3= [x/x+1]*a / ( [1/x+1] *a+0.1a)
...[x/x+1] / ( [1/x+1] +0.1)= 3 ...x/(1+0.1(x+1))=3 ... x/(1.1+0.1x)=3...x=3.3+0.3x

x=3.3/0.7