M grams of water was added to 49 grams of a strong solution of acid. I

Question

GMATBusters’ Quant Quiz 2.0 Question -9 
 For questions from previous quizzes click   here

M grams of water was added to 49 grams of a strong solution of acid. If as a result, the concentration of acid in solution became 1/N of the initial concentration, what was the concentration of acid in the original solution?
1. M = 98
2. N = 3

fauji · Answer

Approach

- Strong Acid Solution doesn't mean 100% acid concentration, So as per given data we don't know the concentration of acid in the solution.
Assume Initial concentration of Acid and Water be A and W, and Final concentration be A2 and W2
- Given: Solution(A+W) = 49 gm, A2 = (1/N)A

S1 : Equation: A + W + M = A2+W2
 => 49+80 =  \frac{(A)}{N}  +(W+80) .. We cannot find Concentration of Acid from this statement -   Insufficient

S2 : Equation: A + W + M = A2+W2
 => 49+M =  \frac{(A)}{3}  + W2 ... Same as S1, We cannot find Concentration of Acid from this statement -   Insufficient

Combine S1 and S2:

=> 49+80 = (A1)/3 +(W+80) .. Can't get the concentration of Acid. -   Insufficient  
 
 IMO Option E!

numb007 · Answer

ANS: E
Sol1 : x gm acid + 49-x gm Water
Sol2: x gm acid + 49-x+M gm Water
Given: x/49+M=1/N*x/49

i) M=98 
 Hence, 49(N-1)=M
 N=3
 But we are not able to find the Value of X.Hence it does not provide a solution

ii) N=3
 We can find the value of M . But value of x is still not obtained.
Hence Hence it does not provide a solution.
On combining i and ii no new info appears. Hence it can't be solved.

rvarora · Answer

C1V1= C2V2
V1= 49
V2- 49+M
C2= C1/N

Therefore,
After solving C1 gets canceled
And we are left with 49N= 49+M

None of the statements are sufficient to answer the question

Hence E

pk123 · Answer

M grams of water was added to 49 grams of a strong solution of acid. If as a result, the concentration of acid in solution became 1/N of the initial concentration, what was the concentration of acid in the original solution?
1. M = 98
2. N = 3

Given M grams of Water
49 grams of strong solution of Acid......Strong solution can be 100% of acid or 80% of acid

Concentration of Acid in solution = 49/(49+M)
Given concentration becomes 1/N of initial concentration of acid

so 49/(49+M) = 1/N* Initial Concentration of Acid

Option A, if M= 98 ,
1/N* Initial Concentration of Acid = 1/3
Initial concentration still unknown

Option B:
N= 3
so 49/(49+M)= 1/3 * Initial concentration of Acid
Initial concentration still unknown

If we use both
1/3 * Initial concentration of Acid = 1/3
So Initial concentration of Acid = 1 = 100%

Option C is the answer

globaldesi · Answer

let original concentration be x
x
thus after m water addition
x/ 49+m = x/n

or 49+m = n
from he given equation it is not possible t find the value of X
thus both stmts are insufficient

QuantMadeEasy · Answer

solution = 49, acid amount = 0.49*x, water = 0.49*(100-x)
new solution = 49+M , Acid = 0.49*x; water = 0.49*(100-x) + M

New concentration of acid = 0.49*x / (49+M) = (1/N) (x/100)
x can not be calculated from this equation

IMO E

SUNNYRHODE002 · Answer

Answer : E

Solution attached.

Archit3110 · Answer

#1
M = 98
conc of acid in new solution not know insufficient 
#2
N = 3
weight and conc of solution not know ; insufficient
from 1 &2
both stmnts are not sufficient
IMO E

M grams of water was added to 49 grams of a strong solution of acid. If as a result, the concentration of acid in solution became 1/N of the initial concentration, what was the concentration of acid in the original solution?
1. M = 98
2. N = 3

vishumangal · Answer

Ans should be E as both statements are not sufficient to ans the Question stem

lacktutor · Answer

Let's say that x is the amount of acid in the solution 
-->  \frac{x}{(M+49)}= (\frac{1}{N})*(\frac{x}{49}) 
No matter whether you add water to the strong solution, the amount of acid will always be the same.

( Statement1 ): M = 98
-->  \frac{x}{(M+49)}= (\frac{1}{N})*(\frac{x}{49})  -->  \frac{1}{(98+49)} = (\frac{1}{N})*(\frac{1}{49})

N = \frac{(98+49)}{49}= 3 
x could be anything here. 
We cannot get the value of x.
Insufficient

( Statement2 ): N=3
-->  \frac{x}{(M+49)}= (\frac{1}{N})*(\frac{x}{49})  -->  \frac{1}{(M+49)}= (\frac{1}{3})*(\frac{1}{49}) 
M+49= 3*49 
M= 98
The same as Statement1
WE cannot get the value of x
Insufficient

Taken together 1&2,  
Clearly insufficient

The answer is E.

Aviral1995 · Answer

Consider the initial amount of acid in 49 gram of solution is a grams
amount of water in 49 gram of solution = (49-a) grams

Now M gram of water is added

Now concentration becomes a/((49-a)+M)
a/((49-a)+M) = 1/N * a/(49-a)
--> (49-a) + M = N * (49-a)

We need value of both M and N to find the value of a

Hence option C

Raxit85 · Answer

M grams of water was added to 49 grams of a strong solution of acid. If as a result, the concentration of acid in solution became 1/N of the initial concentration, what was the concentration of acid in the original solution?
1. M = 98
2. N = 3

It is given that the original solution is 49 g. If we add x gram of water, then new solution will be 49+M g
Let A be the acid amount in the solution.
The concentration before addition of water was A/49,
The new concentration after addition becomes A/(49+M).

A/49=(1/N)*(A/(49+M))
N=(49+M)/49
N=1+(M/49)

Here, both statements give the value of M or N, which in no way relate to acid or water quantity before addition.

Hence Ans is E

mohagar · Answer

E: as the concentration cannot e found

Kinshook · Answer

Given: M grams of water was added to 49 grams of a strong solution of acid. 
Asked: If as a result, the concentration of acid in solution became 1/N of the initial concentration, what was the concentration of acid in the original solution?

Let us assume the original concentration of acid to be x %
Weight of acid in 49 gm = 49x/100 gm

1. M = 98
Concentration of acid after addition of 98 gm of water = (49x/100)/147 = x/300 = x/3 % ; N=3
But value of x may be anything, and after addition of 98 gm of water, the concentration become x/3 %
NOT SUFFICIENT

2. N = 3 
If M gm of water is added to the solution contration become x/3 %
Water added = 49*3-49 = 98 gas; M = 98 gms
But value of x may be anything, and after addition of 98 gm of water, the concentration become x/3 %
NOT SUFFICIENT

(1) + (2) 
1. M = 98
Concentration of acid after addition of 98 gm of water = (49x/100)/147 = x/300 = x/3 % ; N=3
2. N = 3 
If M gm of water is added to the solution contration become x/3 %
Water added = 49*3-49 = 98 gas; M = 98 gms
Still value of x may be anything and can not be derived from information provided.
NOT SUFFICIENT

IMO E

hiranmay · Answer

M grams of water was added to 49 grams of a strong solution of acid. If as a result, the concentration of acid in solution became 1/N of the initial concentration, what was the concentration of acid in the original solution?
 Given: 
49 grams of acid solution, it has acid = a grams, so water has (49-a) grams => acid concentration = a/49
M grams of water added: so new acid concentration = a/(49+M) = (1/N)(a/49)
=> 49+M=49N

1. M = 98 -->  insufficient:49N = 49+98 => N= 3, but we can't find the value of a 
2. N = 3 -->  insufficient:49*3 = 49+M => M= 98, but we can't find the value of a

combining (1)+(2), also we can't find the value of a
So the  answer: E

AndreV · Answer

Initial solution=49 grams
Final solution=(49+M) grams
Initial concentration of acid=X
Final concentration of acid=X/N
The amount of acid did not change, then:
X(49) = (X/N)(49+M)
Then 49(N)=49+M
As can be seen, the value of X CANNOT be determined.
Answer: E

willacethis · Answer

Statement 1.
According to the statement, 98 gms of water added to 49 gms of acid solution.

Now, since we do not know the concentration of acid in the solution after mixing, we cannot find N and cannot reach to the concentration before mixing.
Hence, 1 is insufficient

Statement 2.
Now, we know that when M grams of water is added to 49 grams of initial solution, the concentration of acid becomes 1/3 of that in the initial concentration. Since we have no clue about the other 2 variables i.e. concentration of acid in end solution and the actual amount of water, we cannot reach the actual concentration after mixing through which we could have calculated the initial concentration.

Hence, statement 2 is insufficient.

Even when we combine both the statements,

it is to be noted that irrespective of what the after-mixing concentration is, the before-mixing concentration shall always be 1/3 of the end concentration when we add 98 gms of water to 49 gms of solution.

ex. if we assume initial concentration of acid= 21/49 , after adding water, it shall be 21/(49+98) = (1/3)*(21/49).

same would be the case for any other concentration.

Hence, both statements combined are insufficient.

Therefore, answer is option E.

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M grams of water was added to 49 grams of a strong solution of acid. I

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