If after 200 grams of water were added to the 24% solution

Question

After 200 grams of water were added to the 24%-solution of alcohol, the strength of the solution decreased by one-third. What was the weight of the original 24% solution ?

A. 180 grams
B. 220 grams
C. 250 grams
D. 350 grams
E. 400 grams

Bunuel · Accepted Answer

Official Solution:

After 200 grams of water were added to the 24%-solution of alcohol, the strength of the solution decreased by one-third. What was the weight of the original 24% solution ?

A. 180 grams
B. 220 grams
C. 250 grams
D. 350 grams
E. 400 grams

Let the weight of the 24% solution be  x  grams, then the weight of alcohol in it would be  0.24x  grams. Next, since in the final solution the strength decreased by  \frac{1}{3} , then the strength of the final solution is  24*\frac{2}{3}=16 \% .
 
Since the amount of alcohol in the original and the final solutions is the same, we can equate them and set the equation:  0.24x=0.16(x+200) .
 
Solving gives:  x=400 .

Answer: E

jessetoronto · Answer

Some times I wonder if I should have Hermione Obliviate all Mixture memory and learning I have since beginning to study and start over!!! The easy ones are getting harder...ahhh

muralimba · Answer

For any mixture problem in GMAT, please use the alligation technique:
trigger points:
From any mixture problem, note down the below 3 pointers first.
1) What is solution 1
2) what is solution2
3) what is the resulting solution

For example, i have a solution with 40% milk (==> rest i.e. 60% is water) and i have one more solution which is with 70% milk ==> 30% water
qtn: if we mix these two solutions in a particulr ratio to get a resultant solution with 50% milk, how much % solution1 exists in the resultant solution?

40------------70
------50
20------------10 ==> use the solution 1 and solution2 in the ratio 2:1 to get a resultant solution with 50% milk

means if i use 200liters of sultion 1, then i have to use 100liters of solution 2 resulting 300liters of 50%milk..hope it is clear.

Back to the Original qtn:

solution1: water ==> 100% water
solution2: 24% alcohol ==> 76% water

resulting solution - alcohol concentarion is decreased by 1/3 ==> 24 - 24*1/3 = 16% alcohol==> 84% water

100------------76
-------84
8--------------16 ==> water:24%-alcohol solution = 1:2

OR

solution1: water ==> 0% alcohol
solution2: 24% alcohol

resulting solution - alcohol concentarion is decreased by 1/3 ==> 24 - 24*1/3 = 16% alcohol

0------------24
-------16
8------------16 ==> water:24%-alcohol solution = 1:2
(SAME AS ABOVE)

given that 200liters of water ==> 2*200 = 400 liters of 24%-alcohol solution

ANSWER "E".

Regards,
Murali.

Kudos?

KarishmaB · Answer

The key to solving mixture problems is to properly identify the solutions you want to mix. Make circles. They let you see what is it that you are mixing with what. e.g. Ques1.jpg Now simply use weighted averages formula: C_{avg} = \frac{C_1*V_1 + C_2 * V_2}{V_1 + V_2} 16 = \frac{0*200 + 24 * x}{200 + x} (You can ignore % because it will just get canceled since it is there on both sides of the equation) x = 400 gm I do prefer the scale method (alligation shown by muralimba above) using the number line since it is faster and much more intuitive but if someone is facing problems in mixtures, I strongly recommend to stick with the formula for as long as it takes to get comfortable.

Werewolf · Answer

Thanks Karishma. Looking at the solution, I think this 'alligation method' is quite handy and cool but I'm not sure I understand completely. Could you please explain? Thanks in advance.

KarishmaB · Answer

The alligation method, or the scale method as we call it, is based on the weighted averages formula itself: C_{avg} = \frac{C_1*V_1 + C_2 * V_2}{V_1 + V_2} If I re-arrange the formula, I get \frac{V_1}{V_2} = \frac{C_2 - C_{avg}}{C_{avg} - C_1} So I get that volume will be in the same ratio as difference between higher concentration and average concentration and difference between average concentration and lower concentration. How does this help? Knowing this, we can directly make a diagram and get the answer. e.g. 9 liters of 40% solution of sugar is mixed with a 90% solution of sugar to obtain 60% solution of sugar. How much 90% solution was used? Draw: Ques1.jpg On a scale, mark 40% as concentration1, 60% as average and 90% as concentration2. Now, distance between 40 and 60 is 20 and distance between 60 and 90 is 30, The ratio of Vol1/Vol2 will be 3:2 (Note, the numbers 20 and 30, ratio of 2:3 give ratio 3:2 for V1:V2 as seen by the formula) So if 9 liters of V1 was used, 6 liters of V2 was used. This method is especially useful what you have the average and need to find the volume. Now try the original question using this and get back if there is a doubt.

LalaB · Answer

using VeritasPrepKarishma's last method - 0%---16%(0.24*2/3)----24% v1/v2=(24-16)/(16-0)=1/2 1/3x=200 x=600 (total) 2/3*600=y y=400 (24%-solution) hope I got it right

KarishmaB · Answer

Yes you did. Good job! 
Please check my signature for the link to the relevant blog posts.

LalaB · Answer

we can do it this way too-
0%---16%(0.24*2/3)----24%

v1/v2=(24-16)/(16-0)=1/2

1/2=200/x

x=400

unceldolan · Answer

Weight of original solution: x
Alcohol of Original solution: 24/100x or 0.24 x

Alc. concentration in new solution decreases by 1/3 so its 2/3 of 24 % which is 16 %. 
Weight of new solution is x+200 (since 200 grams of water are added)

Lets say that Y is the amount of alc. in first solution: 24/100 * x = y
Since the amount of alcohol isn't changed, we can calculate the amount of alcohol in the second solution like this:
16/100 * (x+200) = y

Now we can simplify, equalize and calculate: 
simplify: 24/100x = 6/25x and 16/100*(x+200) = 4/25 * (x+200)
equalize: 6/25*x = 4/25 * (x+200)
calculate: 6/25x = 4x+800/25 => 6x = 4x + 800 => 2x = 800 => x=400
As we have defined X as the weight of the original soluition, we have our answer here, which is E!

Hope it helps!

vak3ee · Answer

Dear All,

Why doesn't this work?

24x ………. 16
---------- = ----
100x +200 100

I've been trying to figure this one out for an hour, just can't seem to understand why this method won't work! Thank you so much!
(PS Ignore the ……., that's just for formatting)

KarishmaB · Answer

When a solution is 24% alcohol solution, it means it has 24 parts alcohol and 76 parts water. In the denominator, you are adding 200 (pure water) to 100x (total solution). This doesn't make sense. You need to add 200 to only water.

So if you set up the equation like this, it will work:

\frac{24x}{(76x + 200)} = \frac{16}{84}

Again, remember that the final solution has 16% alcohol so that means 16 parts alcohol and 84 parts water.

You have assumed that the weight of the original solution is 100x. So once you solve this equation, you get x = 4 and hence weight of original solution is 100*4 = 400 gms.

email2vm · Answer

sol:

24 0
\ /
 16
 / \
16 8

why 16 in middle?
since the lost strength is 1/3 so remaining strength is 2/3 (which is resultant strength), there 2*24/3 = 16

why zero?
strength is amount of solute present in solvent. here we have 100% solvent and 0% solute. Hence strength of alcohol in pure water is 0.

ratio is 16: 8 or 2:1 = x: 200

there x = 400

20-sec approach!
Is this really a 700+ question?

SVaidyaraman · Answer

1. Let the quantity of the solution be x.
2. (Quantity of 24% solution)* (Strength of the solution )+ ( Quantity of water added)*(strength of water)/ (total quantity)=strength of the resultant solution
3. (x*0.24 ) + (200*0)/ (x+200) =16/100
x=400

oa7 · Answer

Is there a standard way to attack mixture problems? There are so many different ways of solving the problem its confusing.

SVaidyaraman · Answer

You are right. Most of the problems in a topic can be solved by using a common approach. I try to use such an approach. So even if for some problems there may be quicker solutions than a standard solution, trying to think of such quick solutions take time. In a standard approach, you automatically apply the same steps for seemingly different types of problems.

KarishmaB · Answer

Usually, weighted averages will help. Check Weighted Avg and Mixtures Basics here: https://anaprep.com/arithmetic-weighted-averages/ https://anaprep.com/arithmetic-mixtures/ and these videos: https://www.youtube.com/watch?v=_GOAU7moZ2Q https://www.youtube.com/watch?v=VdBl9Hw0HBg

ScottTargetTestPrep · Answer

Since 1/3 of 24 is 8, we know that after 200 grams of water were added to the 24%-solution of alcohol, the strength of the solution decreased to 16%. We can let the original amount of solution = x, and thus the original amount of alcohol = 0.24x. We can create the following equation and solve for x: 
 
0.24x/(x + 200) = 0.16
 
0.24x = 0.16(x + 200)
 
0.24x = 0.16x + 32
 
0.08x = 32
 
x = 32/0.08 = 3200/8 = 400
 
Answer: E

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