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

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Jug contains water and orange juice in the ratio 5:7 . anoth [#permalink] New post 01 Apr 2011, 05:22
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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 22 Jul 2013, 23:20, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: Mixture [#permalink] New post 02 Apr 2011, 06:14
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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/
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Re: Mixture [#permalink] New post 01 Apr 2011, 05:48
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Let x ml of Jug 1 be mixed with y ml of jug 2

5x/12 + 7y/9 = 3/7*(x+y)

7x/12 + 2y/9 = 4/7 *(x+y)


=> (15x + 28y)/36 = 3/7 * (x+y)

=> 105x + 196y = 108x + 108y

=> 88y - 3x = 0

=> x/y = 88/3

Answer - C
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Re: Mixture [#permalink] New post 02 Apr 2011, 12:12
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5x/12 + 7y/9 = 3/7(x+y)

solving the equation, we have x/y = 88/3.
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Re: Jug contains water and orange juice in the ratio 5:7 . anoth [#permalink] New post 24 Jul 2013, 01:05
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Your solution is not complete.

You arrive here x=22y, is correct BUT is not the end.

The solutions must be mixed in the following ways \frac{12x}{9y} (because 5x+7x=12x and 7y+2y=9y)

A group/liter of the first solution must contain 12x, same reasoning for the second that must contain 9y.

\frac{x}{y}=22 but x must be "taken" in groups of 12 and y in groups of 9 so \frac{12x}{9y}=22*\frac{12}{9}=\frac{88}{3}.

x/y is not what we are looking for : \frac{12x}{9y}=?<===this is the question, and \frac{x}{y} does not answer it.

Hope I've explained myself well.
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Re: Jug contains water and orange juice in the ratio 5:7 . anoth [#permalink] New post 02 May 2014, 18:08
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gauravsoni wrote:
Hello ,

Can someone please show how to do this problem by allegation :?:


This solution: jug-contains-water-and-orange-juice-in-the-ratio-5-7-anoth-111775.html#p904568
and many others are based on alligation
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Re: Mixture [#permalink] New post 02 Apr 2011, 12:01
VeritasPrepKarishma wrote:
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/


Shortest way, what gmat need.
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Re: Mixture [#permalink] New post 22 Jul 2013, 23:17
Hi folks,

I solved this problem. However, I am not able to get where I messed up such that I am unable to reach to the solution.

J1: 5:7

J2: 7:2

J3: 3:4 (J1+J2)

Let us take x from J1 and y from J2

Water from X: 5x
Water from Y: 7y

Orange from X: 7x
Orange from Y: 2y

Given in the question is that in what ratio J1+J2 such that J3: 3:4 Implies we have to calculate x:y

=> (5x+7y)/(7x+2y) = 3/4

=> (5 (x/y) + 7)/( 7 (x/y) +2) = 3/4

=> ( 5A + 7 ) / (7A +2 ) = 3/4

=> A=22

=> (x/y) = 22

Please advice where I am committing mistake.

Thanks in advance

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Re: Mixture [#permalink] New post 23 Jul 2013, 22:34
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targetgmatchotu wrote:
Hi folks,

I solved this problem. However, I am not able to get where I messed up such that I am unable to reach to the solution.

J1: 5:7

J2: 7:2

J3: 3:4 (J1+J2)

Let us take x from J1 and y from J2

Water from X: 5x
Water from Y: 7y

Orange from X: 7x
Orange from Y: 2y

Given in the question is that in what ratio J1+J2 such that J3: 3:4 Implies we have to calculate x:y

=> (5x+7y)/(7x+2y) = 3/4

=> (5 (x/y) + 7)/( 7 (x/y) +2) = 3/4

=> ( 5A + 7 ) / (7A +2 ) = 3/4

=> A=22

=> (x/y) = 22

Please advice where I am committing mistake.

Thanks in advance

Rgds,
TGC !!


You need to understand what you are doing. What do you mean by 'take x from J1'? What is x? Volume in lts? Fraction of total mixture? Think about it.
Say it is volume in lts.

Then
Water from X: 5x/12 lts
Water from Y: 7y/9 lts

Orange from X: 7x/12 lts
Orange from Y: 2y/9 lts

Given that water:orange in J3 is 3:4, we get amount of water in J3 is 3(x + y)/7 lts
Amount of orange in J3 = 4(x+y)/7 lts

We get
(5x/12 + 7y/9)/(7x/12 + 2y/9) = 3/4
(45x + 84y)/(63x + 24y) = 3/4
180x + 336y = 189x + 72y
9x = 264y
x/y = 88/3

But this is an extremely convoluted way of solving this question. Check out the methods given above.
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Re: Mixture [#permalink] New post 23 Jul 2013, 23:21
VeritasPrepKarishma wrote:
targetgmatchotu wrote:
Hi folks,

I solved this problem. However, I am not able to get where I messed up such that I am unable to reach to the solution.

J1: 5:7

J2: 7:2

J3: 3:4 (J1+J2)

Let us take x from J1 and y from J2

Water from X: 5x
Water from Y: 7y

Orange from X: 7x
Orange from Y: 2y

Given in the question is that in what ratio J1+J2 such that J3: 3:4 Implies we have to calculate x:y

=> (5x+7y)/(7x+2y) = 3/4

=> (5 (x/y) + 7)/( 7 (x/y) +2) = 3/4

=> ( 5A + 7 ) / (7A +2 ) = 3/4

=> A=22

=> (x/y) = 22

Please advice where I am committing mistake.

Thanks in advance

Rgds,
TGC !!


You need to understand what you are doing. What do you mean by 'take x from J1'? What is x? Volume in lts? Fraction of total mixture? Think about it.
Say it is volume in lts.

Then
Water from X: 5x/12 lts
Water from Y: 7y/9 lts

Orange from X: 7x/12 lts
Orange from Y: 2y/9 lts

Given that water:orange in J3 is 3:4, we get amount of water in J3 is 3(x + y)/7 lts
Amount of orange in J3 = 4(x+y)/7 lts

We get
(5x/12 + 7y/9)/(7x/12 + 2y/9) = 3/4
(45x + 84y)/(63x + 24y) = 3/4
180x + 336y = 189x + 72y
9x = 264y
x/y = 88/3

But this is an extremely convoluted way of solving this question. Check out the methods given above.


Hi,

I think you didn't get my query. My query is what is the issue with my solutions , and I am not asking alternative solutions which by basics is my last plan.

Rgds,
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Re: Mixture [#permalink] New post 23 Jul 2013, 23:47
targetgmatchotu wrote:
Hi folks,

I solved this problem. However, I am not able to get where I messed up such that I am unable to reach to the solution.

J1: 5:7

J2: 7:2

J3: 3:4 (J1+J2)

Let us take x from J1 and y from J2

Water from X: 5x
Water from Y: 7y

Orange from X: 7x
Orange from Y: 2y

Given in the question is that in what ratio J1+J2 such that J3: 3:4 Implies we have to calculate x:y

=> (5x+7y)/(7x+2y) = 3/4

=> (5 (x/y) + 7)/( 7 (x/y) +2) = 3/4

=> ( 5A + 7 ) / (7A +2 ) = 3/4

=> A=22

=> (x/y) = 22

Please advice where I am committing mistake.

Thanks in advance

Rgds,
TGC !!


You arrive here x=22y<== this is not the end.

We know that 5x+7x=12x and that 7y+2y=9y. X can be seen as 12*22y=264y and now you have a fraction of this form \frac{264y}{9y}=\frac{88}{3}.

Or method #2

\frac{x}{y}=22 but x must be "taken" in groups of 12 and y in groups of 9 so \frac{12x}{9y}=22*\frac{12}{9}=\frac{88}{3}

Hope it's clear
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Re: Mixture [#permalink] New post 24 Jul 2013, 00:57
Zarrolou wrote:
targetgmatchotu wrote:
Hi folks,

I solved this problem. However, I am not able to get where I messed up such that I am unable to reach to the solution.

J1: 5:7

J2: 7:2

J3: 3:4 (J1+J2)

Let us take x from J1 and y from J2

Water from X: 5x
Water from Y: 7y

Orange from X: 7x
Orange from Y: 2y

Given in the question is that in what ratio J1+J2 such that J3: 3:4 Implies we have to calculate x:y

=> (5x+7y)/(7x+2y) = 3/4

=> (5 (x/y) + 7)/( 7 (x/y) +2) = 3/4

=> ( 5A + 7 ) / (7A +2 ) = 3/4

=> A=22

=> (x/y) = 22

Please advice where I am committing mistake.

Thanks in advance

Rgds,
TGC !!


You arrive here x=22y<== this is not the end.

We know that 5x+7x=12x and that 7y+2y=9y. X can be seen as 12*22y=264y and now you have a fraction of this form \frac{264y}{9y}=\frac{88}{3}.

Or method #2

\frac{x}{y}=22 but x must be "taken" in groups of 12 and y in groups of 9 so \frac{12x}{9y}=22*\frac{12}{9}=\frac{88}{3}

Hope it's clear


Still remains a query for me,here is the question:

In what proportion should these 2 liquids be mixed to give a water and orange juice in ratio 3 : 4 ????

X of J1 and Y of J2 to get J3 (3:4)

And X:Y = 22:1

Plz advise !

Rgds,
TGC !
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Re: Mixture [#permalink] New post 24 Jul 2013, 03:51
VeritasPrepKarishma wrote:
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/

This is a general concept and you can use it directly if applicable.
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Re: Jug contains water and orange juice in the ratio 5:7 . anoth [#permalink] New post 10 Oct 2013, 09:55
Hi There,
I want to show you my way: is it quite the same as the first one, but using one unknown nstead of two:
I know I have to mix the first quantity of water with the second quantity of water to obtain 3/7 for the final solution, but i don't know the percentage of the first and the second one (actually I want to get them!), so:
(5/12)X ---> first quantity of water for undefined percentage

(7/9) (1-X) ---> second quantity of water for the percentage remaining from the passage above

Now I can build an equation:

(5/12)X + (7/9) (1-X) = 3/7

solving this I'll obtain:

X = 88/91 and therefore (1-X) = 3/91 ----> Ratio between the two: 88 : 3
I know there's a little passage more in this way, but I like it more beacuse when I'm building the equations I don't trust them if they have more than one unknown and I don't have a system...I'm scared I could miss a part or could not see that I have to simplify something in order to reach the solution or to interpret what I have in front of me (in this case x : y = 88 : 3

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Re: Jug contains water and orange juice in the ratio 5:7 . anoth [#permalink] New post 18 Apr 2014, 21:25
Still on the same page , I got the answer 22 after a year following my instinct.

W:O
5:7 X
7:2 Y
3:4

{5X+7Y}/{7X+2Y} = 3/4

X/Y =22

Reasoning:

Take X from Sol1 and take Y from Sol2

The question is "In what proportion should these 2 liquids be mixed to give a water and orange juice in ratio 3 : 4"

And not "In what proportion should water and orange be mixed"

Please advise
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Re: Jug contains water and orange juice in the ratio 5:7 . anoth [#permalink] New post 20 Apr 2014, 02:46
Expert's post
TGC 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

Still on the same page , I got the answer 22 after a year following my instinct.

W:O
5:7 X
7:2 Y
3:4

{5X+7Y}/{7X+2Y} = 3/4

X/Y =22

Reasoning:

Take X from Sol1 and take Y from Sol2

The question is "In what proportion should these 2 liquids be mixed to give a water and orange juice in ratio 3 : 4"

And not "In what proportion should water and orange be mixed"

Please advise


I think I can see were you went wrong.

From your solution:
The volume of jug 1 = x liters. The ratio of water to orange there is 5:7;
The volume of jug 2 = y liters. The ratio of water to orange there is 7:2;

Then you want to make the ratio of water to orange in the mixture so that it will be 3 to 4. In your solution: (water)/(orange) = (5X+7Y)/(7X+2Y) = 3/4.

But, the amount of water in jug 1 is NOT 5x, it's 5/12*x and the amount of water in jug 2 is NOT 7y, it's 7/9*y.

Similarly the amount of orange in jug 1 is NOT 7x, it's 7/12*x and the amount of orange in jug 2 is NOT 2y, it's 2/9*y.

Hence the equation should be: \frac{\frac{5}{12}x+\frac{7}{9}y}{\frac{7}{12}x+\frac{2}{9}y}=\frac{3}{4} --> \frac{x}{y}=\frac{88}{3}.

Answer: C.

Does this make sense?
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Re: Jug contains water and orange juice in the ratio 5:7 . anoth [#permalink] New post 20 Apr 2014, 06:55
But, the amount of water in jug 1 is NOT 5x, it's 5/12*x and the amount of water in jug 2 is NOT 7y, it's 7/9*y.

I disagree with the above statement since if we want to know the actual quantity of the ingredient in the solution we multiply it by common factor, here taken as x.

So,

x is taken as common multiplying factor from first solution and so do y is taken for another solution.

So when we mix them resulting has a 3:4.

(5X+7Y)/(7X+2Y) = 3/4

Please suggest
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Re: Jug contains water and orange juice in the ratio 5:7 . anoth [#permalink] New post 20 Apr 2014, 10:44
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TGC wrote:
But, the amount of water in jug 1 is NOT 5x, it's 5/12*x and the amount of water in jug 2 is NOT 7y, it's 7/9*y.

I disagree with the above statement since if we want to know the actual quantity of the ingredient in the solution we multiply it by common factor, here taken as x.

So,

x is taken as common multiplying factor from first solution and so do y is taken for another solution.

So when we mix them resulting has a 3:4.

(5X+7Y)/(7X+2Y) = 3/4

Please suggest


What is x in your equation? If it is the amount of liquid in jag 1, then the amount of water in it is 5/12*x. For example, if the amount of liquid in jag 1 is 12 liters then the amount of water is 5/12*12=5 liters and amount of orange is 7/12*12=7 liters.
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Re: Jug contains water and orange juice in the ratio 5:7 . anoth [#permalink] New post 21 Apr 2014, 07:12
Consider below:

Water:oil = 3:4.......(1)

Water : oil = 5:3..........(2)

We have to make a mixture of water:oil = 4:5

I take x ltr from (1) and y ltr from (2)

So what is the quantity of water in new solution 3x+5y

What is quantity of oil in the new solution 4x+3y

What should be the new ratio for the given values of (x,y)? 4:5

(3x+5y)/(4x+3y) = 4/5

Please tell where I am going wrong?
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Re: Jug contains water and orange juice in the ratio 5:7 . anoth [#permalink] New post 21 Apr 2014, 07:24
Expert's post
TGC wrote:
Consider below:

Water:oil = 3:4.......(1)

Water : oil = 5:3..........(2)

We have to make a mixture of water:oil = 4:5

I take x ltr from (1) and y ltr from (2)

So what is the quantity of water in new solution 3x+5y

What is quantity of oil in the new solution 4x+3y

What should be the new ratio for the given values of (x,y)? 4:5

(3x+5y)/(4x+3y) = 4/5

Please tell where I am going wrong?


Water:oil = 3:4.

If you take x liters from this mixture, there will be 3/7*x liters of water there. For example, if you take 7 liters of that mixture there will be 3/7*7=3 liters of water and 4/7*7=4 liters of oil there

Sorry cannot explain any better...
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Re: Jug contains water and orange juice in the ratio 5:7 . anoth   [#permalink] 21 Apr 2014, 07:24
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