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

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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

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

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

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
_________________

It is beyond a doubt that all our knowledge that begins with experience.

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: \(\frac{\frac{5}{12}x+\frac{7}{9}y}{\frac{7}{12}x+\frac{2}{9}y}=\frac{3}{4}\) --> \(\frac{x}{y}=\frac{88}{3}\).

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

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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

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 !!
_________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

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

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 !
_________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

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 !
_________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

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

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

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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

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

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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
_________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

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

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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
_________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

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.
_________________

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

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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?
_________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

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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