A 30% solution of barium chloride is mixed with 10 grams of water to f

Question

A 30% solution of barium chloride is mixed with 10 grams of water to form a 20% solution. How many grams of original solution did we start with ?

A) 10
B) 15
C) 20
D) 25
E) 30

I am trained to complete above problem as per following technique. Is it possible to solve in following manner.

30% _________ 100%
 
 ______ 20%

ratio is 100-20 : 20-30, which is wrong. There should not be 20. Instead, value in the middle should be in between 30% to 100%.

Is there any way to solve above problem as per technique.?

Thank you

PareshGmat · Accepted Answer

Algebraic way:

Say x gms of barium chloride solution mixed with water

\frac{30}{100} x + 0 * 10  ............. (1)

Resultant is 20% of barium chloride

\frac{20}{100} (x + 10)  ................. (2)

Equating (1) & (2)

\frac{30}{100} x + 0 * 10 = \frac{20}{100} (x + 10)

x = 20

Answer = C

DesiGmat · Answer

just deal with unchanged quantities -> here unchanged quantity is Barium Chloride

Say original solution = X
New solution will be X + 10

Barium Chloride remains same in both the solution

0.3(X) = 0.2(X + 10)

0.1X = 2

X = 20

PareshGmat · Answer

10 grams of water means 0% of barium chloride solution

Take it in this way.....

30% _________ 0%
 
______ 20% ................

20 - 0= 20   ............. 30-20 = 10

Answer = 20 = C

thorinoakenshield · Answer

Let X gm be the starting solution
For 30% solution, the amount of BaCl in gm = 0.3X gm ----- (I)

Now, 10 g of H2O was added to the solution. Therefore the new quantity of solution =(X + 10) gm
This gives 20% solution. Hence the amount of BaCl in gm the new solution = 0.2 (X + 10) gm ----- (II)

Note that no new BaCl has been added. Hence the amount of BaCl should remain the same. Hence, we can equate I and II

0.3X = 0.2 (X + 10)
0.3X = 0.2X + 2
0.1X = 2
X = 20 = Answer

EMPOWERgmatRichC · Answer

Hi All,

These types of "weighted" average questions can be solved in a few different ways. You can use the Weighted Average formula, you can use Allegation (which is a ratio-based approach) and (at least in this question) you can TEST THE ANSWERS....

We're told that 10g of water are mixed with an unknown amount of 30% barium mixture to form a 20% barium final mixture. We're asked to figure out how much barium mixture we have....

Logically, if we had 10g of water mixed with 10g of barium mixture, we'd end up with a 15% overall mixture, so 10 is NOT the answer. We need MORE barium mixture to raise the final mixture up to 20%. Since the numbers in the prompt are all "round" numbers, it's highly likely that the correct answer will ALSO be a round number....

Let's TEST Answer C = 20g

(20g of barium mixture)(30% barium) = 6g barium
+ 10g water.....

(6g barium)/(30g total liquid) = .2 = 20% final barium mixture.

So, the initial barium mixture MUST be 20g

Final Answer:  C

GMAT assassins aren't born, they're made,
Rich

thorinoakenshield · Answer

Another quick way for these type of problems:

(old - new)/new = dilution
(30-20)/(20) = 1/2 = dilution = 10 gm
So, original solution = (1/2)*10 gm = 20 gm

Alex75PAris · Answer

Let's x be the original quantity

we have 20%(10+x) = 30%(x)
200 + 20x = 30x
x=20

Divyadisha · Answer

70% water (x gm) is mixed with 100% water (10gm) to give 80% water solution

x/10= 100-80/ 80-70= 20/10

x= 20
C is the answer

Ashishsteag · Answer

Simplest way is weighted average:Just remember that water has a 0% concentration.

and use the weighted average formula:
C1V1+C2V2/V1+V2=Final concentration.

0.3x+10*0/x+10=0.2

Solving,x=20.

law258 · Answer

0.70x+10=0.80(10+x)
0.70x+10=8+0.80x
2=0.10x
 x=20

bumpbot · Answer

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A 30% solution of barium chloride is mixed with 10 grams of water to f

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