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

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When 10% of a solution of milk and water is removed and replaced with  [#permalink]

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New post 20 Mar 2019, 11:13
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54% (01:56) correct 46% (02:15) wrong based on 35 sessions

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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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When 10% of a solution of milk and water is removed and replaced with  [#permalink]

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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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Re: When 10% of a solution of milk and water is removed and replaced with  [#permalink]

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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]

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New post 21 Mar 2019, 03:19
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]

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New post 21 Mar 2019, 09:47
1
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]

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New post 31 Mar 2019, 08:36
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]

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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] 31 Mar 2019, 10:24
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