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In what ratio should Solution 1 and Solution 2 be mixed to

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In what ratio should Solution 1 and Solution 2 be mixed to [#permalink]

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In what ratio should Solution 1 and Solution 2 be mixed to get a solution which contains water and milk in the ratio of 3:7?

(1) Solution 1 contains water and milk in the ratio 1:9 and Solution 2 contains water and milk in the ratio 2:3
(2) The amount of milk in 100 gallon of solution 1 is 80 gallons more than that of water in the same solution. Further, 50 gallons of Solution 2 contains 10 gallons more milk than water.
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udaymathapati wrote:
In what ratio should Solution 1 and Solution 2 be mixed to get a solution which contains water and milk in the ratio of 3:7?
1. Solution 1 contains water and milk in the ratio 1:9 and Solution 2 contains water and milk in the ratio 2:3
2. The amount of milk in 100 gallon of solution 1 is 80 gallaons more than that of water in teh same solulution. Further, 50 gallons of Solution 2 contains 10 gallons more milk than water.


(1) Solution 1 contains water and milk in the ratio 1:9 and Solution 2 contains water and milk in the ratio 2:3

Given: \(\frac{w_1}{m_1}=\frac{x}{9x}\) and \(\frac{w_2}{m_2}=\frac{2y}{3y}\), for some multiples \(x\) and \(y\).

We want \(\frac{x+2y}{9x+3y}=\frac{3}{7}\). Question: \(\frac{x+9x}{2y+3y}=\frac{2x}{y}=?\)

From first equation we can express \(x\) in terms of \(y\) (or vise versa) substitute it in the second and get desired ratio: \(\frac{x+2y}{9x+3y}=\frac{3}{7}\) --> \(y=4x\) --> \(\frac{2x}{y}=\frac{2x}{4x}=\frac{1}{2}\). Sufficient.

(2) The amount of milk in 100 gallon of solution 1 is 80 gallaons more than that of water in teh same solulution. Further, 50 gallons of Solution 2 contains 10 gallons more milk than water.

Given: \(w_1+m_1=100\) and \(w_1+80=m_1\) ---> \(w_1=10\) and \(m_1=90\) --> \(\frac{w_1}{m_1}=\frac{x}{9x}\);

\(w_2+m_2=50\) and \(w_2+10=m_2\) ---> \(w_2=20\) and \(m_1=30\)--> \(\frac{w_2}{m_2}=\frac{2y}{3y}\);

The same info as in (1). Sufficient.

Answer: D.
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Re: Mixture Problem [#permalink]

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New post 22 Apr 2011, 23:06
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answer is D


REMEMBER QUE IS ASKING RATIO
1. sufficent
equation will be like
1/10 x + 2/5 y = 3/10 (x+y)
we can find ratio of x and y
2. sufficent similarly we can fin ratio
( READ ST 2 CAREFULLY)

hence D
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Re: Mixture Problem [#permalink]

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New post 23 Apr 2011, 08:43
We need the ratio of both the liquids to get the answer
Statement 1 and Statement 2 independently give us this information
So answer D
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Re: Mixture Problem [#permalink]

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New post 24 Apr 2011, 21:41
W/M = 3/7

W1/M1 = 1/9 W2/M2 = 2/3


So Q1/Q2 = (2/3 - 3/7)/(3/7 - 1/9)


= (14 - 9)/21/(27 - 7)/63 = 5/21 * 63/20 = 3/4

(1) is sufficient


(2)

For Solution 1

M = W + 80

M + W = 100

For Solution 2


M = W + 10

M + W = 50

So we can find the ratios of M:W in solutions and using above alligation technique find the required ratio.

Answer - D
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Re: In what ratio should Solution 1 and Solution 2 be mixed to [#permalink]

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Re: In what ratio should Solution 1 and Solution 2 be mixed to [#permalink]

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udaymathapati wrote:
In what ratio should Solution 1 and Solution 2 be mixed to get a solution which contains water and milk in the ratio of 3:7?

(1) Solution 1 contains water and milk in the ratio 1:9 and Solution 2 contains water and milk in the ratio 2:3
(2) The amount of milk in 100 gallon of solution 1 is 80 gallons more than that of water in the same solution. Further, 50 gallons of Solution 2 contains 10 gallons more milk than water.


Using aligation:

"get a solution which contains water and milk in the ratio of 3:7" => water = 30%

(1) Solution 1 contains water and milk in the ratio 1:9 and Solution 2 contains water and milk in the ratio 2:3
Solution 2 (in terms of water):\(40%\)
Desired solution ( in % water): \(30%\)
Solution 1 (in terms of water):\(10%\)

\(\frac{Solution_2}{Solution_1}=\frac{2}{1}\)

(2) The amount of milk in 100 gallon of solution 1 is 80 gallons more than that of water in the same solution. Further, 50 gallons of Solution 2 contains 10 gallons more milk than water.
Solution 1: M=90 W=10 => 10% W
Solution 2: M=30 W=20 => 40% W, both statement give us the same info

Solution 2 (in terms of water):\(40%\)
Desired solution ( in % water): \(30%\)
Solution 1 (in terms of water):\(10%\)

\(\frac{Solution_2}{Solution_1}=\frac{2}{1}\)
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Re: Mixture Problem [#permalink]

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New post 19 Jun 2013, 06:58
subhashghosh wrote:
W/M = 3/7

W1/M1 = 1/9 W2/M2 = 2/3


So Q1/Q2 = (2/3 - 3/7)/(3/7 - 1/9)


= (14 - 9)/21/(27 - 7)/63 = 5/21 * 63/20 = 3/4

(1) is sufficient


(2)

For Solution 1

M = W + 80

M + W = 100

For Solution 2


M = W + 10

M + W = 50

So we can find the ratios of M:W in solutions and using above alligation technique find the required ratio.

Answer - D



Hi Karishma,
Why is the ratio of S1 to S2 not equal to 1/2, by using this method:

W1/M1 = 1/9 W2/M2 = 2/3


So Q1/Q2 = (2/3 - 3/7)/(3/7 - 1/9)


= (14 - 9)/21/(27 - 7)/63 = 5/21 * 63/20 = 3/4

Which one is correct?
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Re: Mixture Problem [#permalink]

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New post 19 Jun 2013, 20:33
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cumulonimbus wrote:
subhashghosh wrote:
W/M = 3/7

W1/M1 = 1/9 W2/M2 = 2/3


So Q1/Q2 = (2/3 - 3/7)/(3/7 - 1/9)


= (14 - 9)/21/(27 - 7)/63 = 5/21 * 63/20 = 3/4

(1) is sufficient


(2)

For Solution 1

M = W + 80

M + W = 100

For Solution 2


M = W + 10

M + W = 50

So we can find the ratios of M:W in solutions and using above alligation technique find the required ratio.

Answer - D



Hi Karishma,
Why is the ratio of S1 to S2 not equal to 1/2, by using this method:

W1/M1 = 1/9 W2/M2 = 2/3


So Q1/Q2 = (2/3 - 3/7)/(3/7 - 1/9)


= (14 - 9)/21/(27 - 7)/63 = 5/21 * 63/20 = 3/4

Which one is correct?


Because you don't average out the ratio; you average out the concentration of any one component where the weights used will be volume. Understand that when you find the average of a quantity, it should make physical sense.

Say you know that milk:water = 1:9 in a 100 ml solution.
When you do 1/9 * 100 ml, what do you get? What is 11.11 ml? Nothing
What you have to do is 1/10 * 100 ml = 10 ml (amount of milk in the solution). 1/10 is the concentration of milk in the solution and you multiply that by the volume of solution.

So here, you have to work with any one component. Say we work with water.
Avg concentration of water = 3/10
Concentration of water in solution 1 = 1/10
Concentration of water in solution 2 = 2/5 = 4/10

w1/w2 = (4/10 - 3/10)/(3/10 - 1/10) = 1/2

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Re: Mixture Problem [#permalink]

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New post 16 Jul 2013, 13:43
Bunuel wrote:
udaymathapati wrote:
In what ratio should Solution 1 and Solution 2 be mixed to get a solution which contains water and milk in the ratio of 3:7?
1. Solution 1 contains water and milk in the ratio 1:9 and Solution 2 contains water and milk in the ratio 2:3
2. The amount of milk in 100 gallon of solution 1 is 80 gallaons more than that of water in teh same solulution. Further, 50 gallons of Solution 2 contains 10 gallons more milk than water.


(1) Solution 1 contains water and milk in the ratio 1:9 and Solution 2 contains water and milk in the ratio 2:3

Given: \(\frac{w_1}{m_1}=\frac{x}{9x}\) and \(\frac{w_2}{m_2}=\frac{2y}{3y}\), for some multiples \(x\) and \(y\).

We want \(\frac{x+2y}{9x+3y}=\frac{3}{7}\). Question: \(\frac{x+9x}{2y+3y}=\frac{2x}{y}=?\)

From first equation we can express \(x\) in terms of \(y\) (or vise versa) substitute it in the second and get desired ratio: \(\frac{x+2y}{9x+3y}=\frac{3}{7}\) --> \(y=4x\) --> \(\frac{2x}{y}=\frac{2x}{4x}=\frac{1}{2}\). Sufficient.

(2) The amount of milk in 100 gallon of solution 1 is 80 gallaons more than that of water in teh same solulution. Further, 50 gallons of Solution 2 contains 10 gallons more milk than water.

Given: \(w_1+m_1=100\) and \(w_1+80=m_1\) ---> \(w_1=10\) and \(m_1=90\) --> \(\frac{w_1}{m_1}=\frac{x}{9x}\);

\(w_2+m_2=50\) and \(w_2+10=m_2\) ---> \(w_2=20\) and \(m_1=30\)--> \(\frac{w_2}{m_2}=\frac{2y}{3y}\);

The same info as in (1). Sufficient.

Answer: D.




Hi Bunuel,
I didnt understand the part:
Question: \(\frac{x+9x}{2y+3y}=\frac{2x}{y}=?\)?
how did we get this?
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Re: Mixture Problem [#permalink]

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New post 16 Jul 2013, 22:23
up4gmat wrote:
Bunuel wrote:
udaymathapati wrote:
In what ratio should Solution 1 and Solution 2 be mixed to get a solution which contains water and milk in the ratio of 3:7?
1. Solution 1 contains water and milk in the ratio 1:9 and Solution 2 contains water and milk in the ratio 2:3
2. The amount of milk in 100 gallon of solution 1 is 80 gallaons more than that of water in teh same solulution. Further, 50 gallons of Solution 2 contains 10 gallons more milk than water.


(1) Solution 1 contains water and milk in the ratio 1:9 and Solution 2 contains water and milk in the ratio 2:3

Given: \(\frac{w_1}{m_1}=\frac{x}{9x}\) and \(\frac{w_2}{m_2}=\frac{2y}{3y}\), for some multiples \(x\) and \(y\).

We want \(\frac{x+2y}{9x+3y}=\frac{3}{7}\). Question: \(\frac{x+9x}{2y+3y}=\frac{2x}{y}=?\)

From first equation we can express \(x\) in terms of \(y\) (or vise versa) substitute it in the second and get desired ratio: \(\frac{x+2y}{9x+3y}=\frac{3}{7}\) --> \(y=4x\) --> \(\frac{2x}{y}=\frac{2x}{4x}=\frac{1}{2}\). Sufficient.

(2) The amount of milk in 100 gallon of solution 1 is 80 gallaons more than that of water in teh same solulution. Further, 50 gallons of Solution 2 contains 10 gallons more milk than water.

Given: \(w_1+m_1=100\) and \(w_1+80=m_1\) ---> \(w_1=10\) and \(m_1=90\) --> \(\frac{w_1}{m_1}=\frac{x}{9x}\);

\(w_2+m_2=50\) and \(w_2+10=m_2\) ---> \(w_2=20\) and \(m_1=30\)--> \(\frac{w_2}{m_2}=\frac{2y}{3y}\);

The same info as in (1). Sufficient.

Answer: D.




Hi Bunuel,
I didnt understand the part:
Question: \(\frac{x+9x}{2y+3y}=\frac{2x}{y}=?\)?
how did we get this?


We need to find the ratio of Solution 1 to Solution 2 \(\frac{solution \ 1}{solution \ 2}=\frac{x+9x}{2y+3y}=\frac{10x}{5y}=\frac{2x}{y}=?\), while having that \(\frac{x+2y}{9x+3y}=\frac{3}{7}\).

Hope it's clear.
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Re: In what ratio should Solution 1 and Solution 2 be mixed to   [#permalink] 08 Nov 2015, 03:13
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