A solution of 60 ounces of sugar and water is 20% sugar. How

Question

A solution of 60 ounces of sugar and water is 20% sugar. How much water must be added to make a solution that is 5% sugar?

A. 20 ounces
B. 80 ounces
C. 100 ounces
D. 120 ounces
E. 180 ounces

fluke · Accepted Answer

20% sugar in 60 oz solution;

Sugar = 60*20/100 = 12 oz
Water = 60-12 = 48 oz

Now; let's add "x" oz of water in the solution;

Total solution = 60+x oz
Sugar = 12 oz.(Sugar's amount will not change as we are not adding or subtracting sugar)

12 = (60+x) * 5/100
60+x = 240
x = 240 - 60 = 180 oz

Ans: "E"

gmat1220 · Answer

This is easy if you can visualize the balance.

0% sol (pure water) < 5% sol < 20% sol
So ratio 
pure water : 20% sol = (20 - 5) : (5 - 0) = 3:1

water to be added = 3 * 20% sol = 3*60 = 180

OldFritz · Answer

Straight E. Alligative method (as neatly performed in gmat1220's post above) precipitates (forgive the pun) the solution (another pun) to the question faster.

Quick arithmetic works too, but the important advice is to decide quickly and firmly what method you want to use. And stick with it.

Cheers,
Der alte Fritz.

Ousmane · Answer

sugar : 0.2*60 = 12
water: 0.8*60 = 48

If new solution is 5% of sugar then it is 95% water

(water + x)/ (60 + x) = 0.95
(48 + x) / (60 + x) = 0.95 
x = 180

Brother Karamazov

mbaiseasy · Answer

Fast Technique. Thanks to GMATClub!

dpvtank · Answer

I've done mixtures questions before, and maybe this is just me not getting it because my brain is all curdled up right now, so forgive me for my ignorance.

Shouldn't x = 48 oz (which is the amount of water in it)?

And therefore, 3x = 144 oz?

I too get the 1:3 ratio. But maybe I'm missing something here.

Please help.

OldFritz · Answer

No sir. Everything in the above diagram is set up in terms of Total weights of solutions (water + sugar) and not the weight of water alone. The individual weights of sugar or sugar are implied in the percentage concentrations (5%, 0%, and 20%). Therefore x= Total weight of 20%
solution = 60 oz. Not 48 oz., which represents the weight of water alone in the 20% solution

dpvtank · Answer

Of course! How silly of me. Thank you good sir.

shelrod007 · Answer

Amount of sugar remains the same throught so

(20/100)*60 = (x*5/100)0

Water to be added = x - 60
 = 240 - 60

law258 · Answer

Solve in terms of H20

0.80(60) + x = 0.95(60+x)
48 + x = 57 + 0.95x
0.05x = 9 --> x = 180

Geekie2411 · Answer

Since amount of sugar does not change:
12 = 0.05(60+x)
-> x= 180

gracie · Answer

A solution of 60 ounces of sugar and water is 20% sugar. How much water must be added to make a solution that is 5% sugar?

A. 20 ounces
B. 80 ounces
C. 100 ounces
D. 120 ounces
E. 180 ounces

let w=water to be added
.8*60+w=.95(60+w)
w=180 ounces
E.

JeffTargetTestPrep · Answer

We can let n = the amount of water added and create the equation:

(60 x 0.2)/(60 + n) = 1/20

20(12) = 60 + n

240 = 60 + n

180 = n

Alternate Solution:

We start with 60 ounces of solution that is 20% sugar. We add n ounces of water, which is 0% sugar. We end up with (60 + n) ounces of solution that is 5% sugar.

We can create an equation for the above scenario:

60 x 0.2 + n x 0 = (60 + n) x 0.05

12 + 0 = 3 + 0.05n

9 = 0.05 n

900 = 5n

180 = n

Answer: E

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A solution of 60 ounces of sugar and water is 20% sugar. How

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A solution of 60 ounces of sugar and water is 20% sugar. How

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