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# Jug contains water and orange juice in the ratio 5:7 . anoth

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Director
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Re: Jug contains water and orange juice in the ratio 5:7 . anoth  [#permalink]

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21 Apr 2014, 10:06
Water: Oil = 3:4

When we deal with ratios, we consider a common multiplier to have absolute value of that quantity in a mixture.

I think I got my mistake.

Scenario 1:

When I want to get absolute value of the quantity in the mixture , I multiply by the factor X.

water is 3X and Oil is 4X in the mixture.

Scenario 2:

When I take X of this mixture then the water content contributed by this X ltrs from this mixture is

Water/Total * X = 3/7*X in this case

Thanks !

Well this was a too silly doubt to answer.

Rgds,
TGC!
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Re: Jug contains water and orange juice in the ratio 5:7 . anoth  [#permalink]

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21 Apr 2014, 21:21
Quantity of Orange Juice in

Jug 1 $$= \frac{7}{12}$$

Jug 2 $$= \frac{2}{9}$$

Requirement in the mixture$$= \frac{4}{7}$$

Using Alligation method

$$\frac{\frac{4}{7} - \frac{2}{9}}{\frac{7}{12} - \frac{4}{7}}$$

$$= \frac{88}{3}$$
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Re: Jug contains water and orange juice in the ratio 5:7 . anoth  [#permalink]

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02 May 2014, 03:29
Hello ,

Can someone please show how to do this problem by allegation
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Re: Jug contains water and orange juice in the ratio 5:7 . anoth  [#permalink]

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16 Sep 2016, 20:29
My 2 cents:
Mixture 1: w:o = 5:7 => w:t = 5:12
Mixture 2: w:o = 7:2 => w:t = 7:9
Final mixture: w:o = 3:4 => w:t = 3:7
x*(5/12) + (1-x)*(7/9) = 3/7
5x/12 + 7/9 - 7x/9 = 3/7
15x + 28 - 28x/(2*2*3*3) = 3/7
28 - 13x/(2*2*3*3) = 3/7
196 - 91x = 108
88 = 91x
x = 88/91 & (1-x) = 3/91
x:(1-x) = 88:3
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Jug contains water and orange juice in the ratio 5:7 . anoth  [#permalink]

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10 Jun 2017, 20:24
rxs0005 wrote:
Jug contains water and orange juice in the ratio 5:7 . another jug contains water and orange juice in ratio 7 : 2 . In what proportion should these 2 liquids be mixed to give a water and orange juice in ratio 3 : 4

A. 4 : 5
B. 85 : 3
C. 88 : 3
D. 2 : 3
E. 87 : 7

1. (Quantity of jug 1)* proportion of water + (Quantity of jug 2)* proportion of water / (Quantity of jug 1)* proportion of juice + (Quantity of jug 2)* proportion of juice = 3/4
2. Let x and y be the quantities in jug 1 and jug 2 resp to get the mixture
3. (x*5/12 + y*7/9) / (x*7/12 + y*2/9) = 3/4
x/y=88/3
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Re: Jug contains water and orange juice in the ratio 5:7 . anoth  [#permalink]

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10 Jun 2017, 22:55
Imo C
Setting up the equation and solving
5x/12+7y/9=3/7(x+y)
We get 88:3

Sent from my ONE E1003 using GMAT Club Forum mobile app
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Re: Jug contains water and orange juice in the ratio 5:7 . anoth  [#permalink]

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10 Jun 2017, 23:11
Attachment:

FullSizeRender (8).jpg [ 48.66 KiB | Viewed 967 times ]

Ans : C
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Re: Jug contains water and orange juice in the ratio 5:7 . anoth  [#permalink]

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01 Dec 2019, 07:57
rxs0005 wrote:
Jug contains water and orange juice in the ratio 5:7 . another jug contains water and orange juice in ratio 7 : 2 . In what proportion should these 2 liquids be mixed to give a water and orange juice in ratio 3 : 4

4 : 5

85 : 3

88 : 3

2 : 3

87 : 7

In such questions focus on one thing - either water or orange juice. Let's work with water
Jug1 - Water concentration is 5/12
Jug2 - Water concentration is 7/9
Mixture - Water concentration is 3/7

Now, $$\frac{w1}{w2} = \frac{\frac{7}{9} - \frac{3}{7}}{\frac{3}{7} - \frac{5}{12}} = \frac{88}{3}$$

A detailed explanation of the concept used above is given here:
http://www.veritasprep.com/blog/2011/03 ... -averages/

how do you ascertain when to use this method vs scale method?
Re: Jug contains water and orange juice in the ratio 5:7 . anoth   [#permalink] 01 Dec 2019, 07:57

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