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]

31 Oct 2010, 05:20

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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 gallaons more than that of water in teh same solulution. Further, 50 gallons of Solution 2 contains 10 gallons more milk than water.

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Re: Mixture Problem [#permalink]

31 Oct 2010, 09:22

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udaymathapati wrote:

(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]

23 Apr 2011, 00:06

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]

23 Apr 2011, 09:43

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]

24 Apr 2011, 22:41

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: Mixture Problem [#permalink]

28 Apr 2011, 21:07

straight D, using the mixture formula.
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Re: Mixture Problem [#permalink] 28 Apr 2011, 21:07

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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

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

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