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

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cfc198
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When 10% of a solution of milk and water is removed and replaced with [#permalink] New post   20 Mar 2019, 11:13
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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
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C
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arimich
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When 10% of a solution of milk and water is removed and replaced with [#permalink] New post   20 Mar 2019, 11:40
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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
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When 10% of a solution of milk and water is removed and replaced with [#permalink] New post   15 May 2019, 13:33
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Archit3110 wrote:
cfc198 wrote:
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


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


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/
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Re: When 10% of a solution of milk and water is removed and replaced with [#permalink] New post   20 Mar 2019, 23:50
arimich wrote:
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




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.
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Re: When 10% of a solution of milk and water is removed and replaced with [#permalink] New post   21 Mar 2019, 03:19
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DakshaBafna wrote:
arimich wrote:
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




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.


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
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Re: When 10% of a solution of milk and water is removed and replaced with [#permalink] New post   21 Mar 2019, 09:47
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cfc198 wrote:
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


VeritasKarishma :
hello could you please advise on how to solve this question using weighted avg formula?
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Re: When 10% of a solution of milk and water is removed and replaced with [#permalink] New post   31 Mar 2019, 08:36
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cfc198 wrote:
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


let m=volume of milk before removal
w=volume of water before removal
.9m/[.9w+.1(m+w)]=3/1➡
.9m/(w+.1m)=3➡
.2m=w➡
m/w=1/.2=5:1 ratio
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Re: When 10% of a solution of milk and water is removed and replaced with [#permalink] New post   31 Mar 2019, 10:24
gracie wrote:
cfc198 wrote:
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


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


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...
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Re: When 10% of a solution of milk and water is removed and replaced with [#permalink] New post   27 Oct 2019, 04:06
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
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Re: When 10% of a solution of milk and water is removed and replaced with [#permalink] New post   21 Nov 2019, 23:43
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
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When 10% of a solution of milk and water is removed and replaced with [#permalink] New post   10 Jun 2021, 12:56
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.
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When 10% of a solution of milk and water is removed and replaced with [#permalink] New post   10 Jun 2021, 14:12
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cfc198 wrote:
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.

A) 2 : 1
B) 4 : 1
C) 5 : 1
D) 5 : 2
E) 4 : 3


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

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Re: When 10% of a solution of milk and water is removed and replaced with [#permalink] New post   07 Nov 2023, 18:28
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Re: When 10% of a solution of milk and water is removed and replaced with [#permalink]
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