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

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Kudos [?]: 105 [2] , given: 22

Jug contains water and orange juice in the ratio 5:7 . anoth [#permalink]  01 Apr 2011, 05:22
2
KUDOS
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Difficulty:

35% (medium)

Question Stats:

72% (03:43) correct 27% (02:14) wrong based on 107 sessions
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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Kudos [?]: 261 [2] , given: 36

Re: Mixture [#permalink]  01 Apr 2011, 05:48
2
KUDOS
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

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Kudos [?]: 3618 [3] , given: 144

Re: Mixture [#permalink]  02 Apr 2011, 06:14
3
KUDOS
Expert's post
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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Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Director Status: GMAT Learner Joined: 14 Jul 2010 Posts: 655 Followers: 29 Kudos [?]: 152 [0], given: 32 Re: Mixture [#permalink] 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. _________________ I am student of everyone-baten Collections:- PSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html DS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html 100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html Director Joined: 01 Feb 2011 Posts: 774 Followers: 12 Kudos [?]: 75 [1] , given: 42 Re: Mixture [#permalink] 02 Apr 2011, 12:12 1 This post received KUDOS 5x/12 + 7y/9 = 3/7(x+y) solving the equation, we have x/y = 88/3. Director Joined: 03 Aug 2012 Posts: 607 Concentration: General Management, General Management GMAT 1: 630 Q47 V29 GPA: 3.7 Followers: 10 Kudos [?]: 89 [0], given: 209 Re: Mixture [#permalink] 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 Rgds, TGC !! _________________ _____________________________________________________________________________ I Assisted You => KUDOS Please Now I am a sky diver too _____________________________________________________________________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4032 Location: Pune, India Followers: 860 Kudos [?]: 3618 [0], given: 144 Re: Mixture [#permalink] 23 Jul 2013, 22:34 Expert's post 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Director
Joined: 03 Aug 2012
Posts: 607
Concentration: General Management, General Management
GMAT 1: 630 Q47 V29
GPA: 3.7
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Kudos [?]: 89 [0], given: 209

Re: Mixture [#permalink]  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

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,
TGC !
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Re: Mixture [#permalink]  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

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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Kudos [?]: 89 [0], given: 209

Re: Mixture [#permalink]  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

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

Rgds,
TGC !
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Kudos [?]: 985 [1] , given: 219

Re: Jug contains water and orange juice in the ratio 5:7 . anoth [#permalink]  24 Jul 2013, 01:05
1
KUDOS

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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Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

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Kudos [?]: 125 [0], given: 8

Re: Mixture [#permalink]  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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Kudos [?]: 5 [0], given: 10

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

Cheers
Re: Jug contains water and orange juice in the ratio 5:7 . anoth   [#permalink] 10 Oct 2013, 09:55
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