When 10% of a solution of milk and water is removed and replaced with

Question

When 10% of a solution of milk and water is removed and replaced with water, the ratio of milk and water becomes 3 : 1. Find the ratio of milk and water before the water is added to the solution.

1. 2 : 1
2. 4 : 1
3. 5 : 1
4. 5 : 2
5. 4 : 3

arimich · Accepted Answer

let, initially there was x unit milk and y unit water.
in the removed 10% mixture, milk is 0.1x unit and water is 0.1y unit. we also have to add equal amount of water, that is 0.1(x+y)
new milk amount= x-0.1x= 0.9x
new water amount= y-0.1y+ 0.1(x+y)= y+0.1x
as per the question, 0.9x/(y+0.1x) = 3/1
solution, x/y= 5/1
ans C

churuand · Answer

Hi, I can solve this question based on VeristasKarishma's weighted average formula. C2=C1*(V1/V2) in which C1 is the original ratio of the milk and total amount C2 is the later ratio which is 3/4 V1/V2=9/10 (because we take out 10% of the mixture) => 3/4=C1*9/10 => C1=5/6 => the original ratio of the milk and total 5:6 => the original ratio of the milk and water is 5:1 U can see the explanation behind this formula here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... -mixtures/

DakshaBafna · Answer

Why are we adding 0.1x when only water has to be added? Also, how did you get 5/1? Please help me understand. Thanks.

arimich · Answer

here, that '0.1x' is the amount not the material
solving the equation
0.9x= 3y +0.3x
=> 0.6x = 3y
=> x= 5y
=> x/y = 5/1

Archit3110 · Answer

VeritasKarishma  :
hello could you please advise on how to solve this question using weighted avg formula?

gracie · Answer

let m=volume of milk before removal
w=volume of water before removal
.9m/ =3/1➡
.9m/(w+.1m)=3➡
.2m=w➡
m/w=1/.2=5:1 ratio

Archit3110 · Answer

gracie 
hi could you please share the formula or logic behind use of the formula... i want to understand how did you deduce this formula...

coylahood · Answer

Cfinal = Cinitial x (v2/V1)^n
where cfinal = final concentrtaion
Cinitial = initial concentration
v2 = final volume
V1= initial volume
Assume an initial volume of 10liters
The replacement of 10 percent implies that the final volume is 9
The final ratio of 3:1 implies that the final concentration is 75%
Therefore 0.75=C1(9/10)
75=90c1
c1=75/90= 5/6
Therefore initial ratio = 5:1

effatara · Answer

Let's assume the total quantity of the solution is 40 ml (which,by the way, remains unchanged since the qty that is removed is replaced by water) and the initial qty of milk in the solution is 'x' ml. The quantity of milk in the final solution is thus 30 ml (since Milk:Water=3:1) and the quantity of solution replaced with water is 4 ml.
The remaining qty of milk after removal of 4 ml of solution = x - (x/40)*4 = 9x/10 = 30....> x=100/3
Thus, qty of water in the initial solution was 40 - 100/3 = 20/3
The ratio of Milk:Water in the initial solution was (100/3):(20/3) or 5:1. ANS: C

DrVanNostrand · Answer

The volume of the solution has been supposed to be 100 and the initial percentage of milk has been supposed to be x.

Then, the volume of milk after the replacement is x(1- 10/100).

In the final solution, Milk: Water ratio = 3:1. Therefore, Milk to total solution ratio = 3:4.

Now,

x(1-10/100) / 100 = 3/4

Therefore x = 250/3 and initial portion of water = 100 - (250/3) = 50/3

Thus, initial milk to water ratio = (250/3) : (50/3) = 5:1.

Hence, C is the answer.

GMATGuruNY · Answer

When the fractional amount of milk is reduced by 1/10 and replaced with pure water, the resulting fraction for milk is 3/4.

Thus, 9/10 of the original milk fraction is equal to 3/4:
 \frac{9}{10}m = \frac{3}{4}  
 m = \frac{5}{6}

Since milk constitutes 5/6 of the original solution, the original ratio of milk to water = 5:1

C

luisdicampo · Answer

Deconstructing the Question

Let the initial total solution be  100 .

Let initial milk =  M  and water =  W .

After removing 10%:

Milk becomes  0.9M 
Water becomes  0.9W

Then 10 units of water are added, so final water:

0.9W + 10

Final ratio is  3:1 :

\frac{0.9M}{0.9W + 10} = 3

Step-by-step

0.9M = 3(0.9W + 10)

0.9M = 2.7W + 30

Multiply by 10:

9M = 27W + 300

M = 3W + \frac{100}{3}

Using  M + W = 100 :

3W + \frac{100}{3} + W = 100

4W + \frac{100}{3} = 100

Multiply by 3:

12W + 100 = 300

12W = 200

W = \frac{50}{3}

M = 100 - \frac{50}{3} = \frac{250}{3}

Initial ratio:

\frac{250}{3} : \frac{50}{3} = 5:1

Answer: 3

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