Friend, try using allegation method in all such problems, it's way too easy and simplifies all the calculation you have to do to reach the answer!
In a nutshell: when you are mixing two solutions to make a mixture, the ratio of the difference in concentration of individual soln and the mixture is equal to The inverse of ratio of amount of solns being mixed.

Soln a: equal amount of water and acid => conc of acid is 50% ; however volume being mixed is not given! Say it is Va

Soln b: conc of acid is given as 80% and the volume being mixed is 40 quarts

Mixture: conc of acid in mixture is given as 70 %; have to find the amount of acid in the mixture!

Thus for amount of acid in the mixture we should know the amount of acid being mixed in soln 1, I.e. Va

Thus ratio of diffof conc=> (m-a)/(b-m) = (70-50)/(80-70)=2:1

This ratio is equal to the inverse of ratio of volumes being mixed= 40/ Va

40/Va = 2/1 => Va=20 quarts

Thus amount of acid in mixture = 50% of 20 + 80% of 40= 42
Hence E

Because of explanation the above method looks cumbersome, however it is the easiest approach to such problems, also when this represented in graphical form, it becomes much easier to visualise and understand.

jzw wrote:Can someone please assist with constructing the formula based on the chart method. Please tell me what I did wrong using THIS method. Thanks.

"Equal amounts of water and acid are added to 40 quarts of 80% acid solution. The new mixture has a concentration of 70%. How much acid is in the resulting solution?"

(a) 10 quarts
(b) 18 quarts
(c) 32 quarts
(d) 40 quarts
(e) 42 quarts

We can plug in the answers, which represent the total amount of acid in the mixture.
The amount of acid in the original solution = .8(40) = 32.
Since acid is being added, the total amount of acid in the mixture must be greater than 32.
Eliminate A, B and C.

Since the percentage of acid in the mixture is 70% -- a multiple of 10 -- the volume of the mixture is likely to be a multiple of 10.
Thus, the correct answer is almost certainly E, which implies that 10 quarts of acid are being added.

Answer choice E: 42 quarts
Since the amount of acid increases by 10 quarts, and the same amount of water is added:
Total volume = original solution + added acid + added water = 40+10+10 = 60 quarts.
Amount of acid in the mixture = .7(60) = 42 quarts.
Success!

The correct answer is E.

Note that we had to plug in only ONE answer choice -- a very efficient way to solve the problem.

As for your table, I would set it up this way:

---------------PERCENTAGE------- VOL --------AMOUNT OF ACID
ORIGINAL SO ------ .8 ------------- 40 -------------32
ADDED ACID ------ .5 ------------- X ------------- .5X
FINAL SLTN ------- .7 ----------- 40+X -------- 28 + .7X

32 + .5X = 28 + .7X
4 = .2X
X = 20.
Thus, the amount of acid in the final solution = 28 + .7(20) = 42.

Plugging in the answers seems easier and faster.

I did it in a different way gmatguru...

i converted the quart to a gallon and then got the answer... But the answer was a precise E...

took 3mins...

jzw wrote:Can someone please assist with constructing the formula based on the chart method. Please tell me what I did wrong using THIS method. Thanks.

"Equal amounts of water and acid are added to 40 quarts of 80% acid solution. The new mixture has a concentration of 70%. How much acid is in the resulting solution?"

(a) 10 quarts
(b) 18 quarts
(c) 32 quarts
(d) 40 quarts
(e) 42 quarts

---------------PERCENTAGE------- MIX --------AMOUNT
ORIGINALSO ------ .8 ----------- 40 -------- 32
ADDEDWATER ------ 1.0 ---------- X --------- X
ADDEDACIDD ------ 1.0 ---------- X --------- X
FINAL SLTN ------ .7 ----------- 40+2X ----- 28 + 1.4X

32 + 2X = 28 + 1.4X
320 + 20X = 28 + 14X

Problem is, in the answers I should have gotten: 320 + 10X = 280 + 14X and I'm not sure how that is. The only way would to not count the X twice when constructing the formula.

Feels like I'm missing some level of understanding of WHY we only are counting the X once when constructing the final formula, but why we counted it twice to get 40+2x.

Problem in your solution is : For added water you have taken percentage of acid as 1 which is wrong. percentage of acid in water is 0. So the corresponding "Amount" of acid for Added water should be "0" not "1". Then your total amount would be 32+X. which when equated to 28+1.4X gives the answer.
Hope it solves your query.

Let 'x' be the amount of the new solution added. Now acid constitutes 50% of this solution , therefore 0.5*x is the quantity of acid added.

Now,
0.5*x + 0.8*40 = 0.7(40+x), which implies x = 20 units.

Therefore the quantity of new solution added = 20 units.

Quantity of acid in final mixture = 0.5*20 + 0.8*40 = 42 units.