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Solution R contains water and fluoride ...

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Solution R contains water and fluoride ...

by Night reader » Mon Nov 29, 2010 6:41 am
Solution R contains water and fluoride in a 3 : 1 ratio, and Solution S contains water and fluoride in a 3 : 2 ratio. If 20 ounces of Solution R is to be combined with x ounces of Solution S to create a solution that contains water and fluoride in a 7 : 3 ratio, what is the value of x?

(GMAT drill: mixture word problems)
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by rishab1988 » Mon Nov 29, 2010 7:31 am
w/f =3/1 Solution R

w/f =3/2 solution S

4x=20 [solution R total added 20 ounces][the x here is a different x.Just to calculate how much is flourine and how much is water]
x=5

w=15
f=5

[ I hope you understood it.If not PM me]

5y=x [solution S]
y=x/5

water =3x/5
flourine =2x/5

(3x/5 +15)/(2x/5+5) =7/3
(3x+75)/(2x+25)=7/3
9x+225 =14x+175
50=5x
x=10

Answer is x=10
Last edited by rishab1988 on Mon Nov 29, 2010 7:45 am, edited 1 time in total.
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by jaymw » Mon Nov 29, 2010 7:38 am
Once you understand how to express that ratio algebraicly, this question becomes pretty easily solvable.

A 3:1 ratio of water and fluoride means that of this solution 3 quarters are water. 3 water / 1 floride --> 3 water / (3 water +1 flouride, which is the entire solution). The same goes for the second ratio which contains 60% or 3/5 water.

The rest is transforming the information into an equation.

3/4*20+3/5*x=7/10*(20+x)

You have 20 ounces of the 3/4 solution and x ounces of the 3/5 solution. What you want is a 7/10 solution which definitely contains the 20 of the first solution and the x ounces of the second solution that you are supposed to add to the first one.

Then solve for x:

15+0.6x=14+0.7x
1=0.1x
x=10

You have to add 10 ounces of the second solution.


Also, you could have used interpolation to solve this problem.


The first solution is 75% water.

The second solution is 60% water.

You want a solution with 70% water.

Obviously 70% is twice as far away from 60% as is 75%, therefore you need twice as much of the solution that is closer to the target percent. So 20 is twice as much as what you need of the 60% solution. Hence, you need 10 ounces of the 60% solution.

Clearly, the second way is a lot faster but demands a greater insight into the theoretical concept behind.

Hope this was helpful!
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by Night reader » Mon Nov 29, 2010 7:47 am
jaymw wrote:Once you understand how to express that ratio algebraicly, this question becomes pretty easily solvable.

A 3:1 ratio of water and fluoride means that of this solution 3 quarters are water. 3 water / 1 floride --> 3 water / (3 water +1 flouride, which is the entire solution). The same goes for the second ratio which contains 60% or 3/5 water.

The rest is transforming the information into an equation.

3/4*20+3/5*x=7/10*(20+x)

You have 20 ounces of the 3/4 solution and x ounces of the 3/5 solution. What you want is a 7/10 solution which definitely contains the 20 of the first solution and the x ounces of the second solution that you are supposed to add to the first one.

Then solve for x:

15+0.6x=14+0.7x
1=0.1x
x=10

You have to add 10 ounces of the second solution.


Also, you could have used interpolation to solve this problem.


The first solution is 75% water.

The second solution is 60% water.

You want a solution with 70% water.

Obviously 70% is twice as far away from 60% as is 75%, therefore you need twice as much of the solution that is closer to the target percent. So 20 is twice as much as what you need of the 60% solution. Hence, you need 10 ounces of the 60% solution.

Clearly, the second way is a lot faster but demands a greater insight into the theoretical concept behind.

Hope this was helpful!


this is an ideal equation; very focused approach!
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by GMATGuruNY » Mon Nov 29, 2010 1:30 pm
Night reader wrote:Solution R contains water and fluoride in a 3 : 1 ratio, and Solution S contains water and fluoride in a 3 : 2 ratio. If 20 ounces of Solution R is to be combined with x ounces of Solution S to create a solution that contains water and fluoride in a 7 : 3 ratio, what is the value of x?

(GMAT drill: mixture word problems)
This is a weighted average question. Solution R is 3/4 = 75% water. Solution S is 3/5 = 60% water. We want to combine the two solutions to get 7/10 = 70% water. A great method for solving is called alligation.

To combine a 75% entity with a 60% entity so that the combined entity is 70%:

The proportion needed of each starting percentage is the positive difference between the other 2 percentages.

Proportion needed of the 75% entity (Solution R) = 70-60 = 10.
Proportion needed of the 60% entity (Solution S) = 75-70 = 5.

Thus, Solution R: Solution S = 10:5 = 2:1.

Since the amount of Solution R must be double the amount of Solution S, and we are to use 20 ounces of Solution R, we'll need 10 ounces of Solution S. Thus, x=10.
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