A solution consists of only water and alcohol such that the ratio of

Question

A solution consists of only water and alcohol such that the ratio of alcohol to water in the solution is 7:3. How much amount of water should be added to the solution (in mililiters) so that the resulting solution contains 60% alcohol?

(1) Total quantity of the resulting solution is 350 mililiters.
(2) The original solution contains 10.5 mililiters of alcohol for every 4.5 mililiters of water.

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A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
D. EACH statement ALONE is sufficient to answer the question asked.
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

Thanks,
Saquib
Quant Expert
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Official Answer

A

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TheMechanic · Accepted Answer

A simple solution to this could be: C1 x V1 (initial) = C2 x V2 (final)

We are given, for concentration of alcohol: 
C1 = 70%
C2 = 60%.

Using Statement 1: We get V2= 350lits.
Putting this in above equation, 
we get V1 = 60 x 350/70 = 300 lits.

Hence the water that was added is 50 lits.

Stmt 1 is sufficient.

Stmt 2: It simply re-states the ratio of concentration of Alcohol and Water in actual quantities. It does not give any new relation. Hence it is not sufficient.

The answer is A.

I hope this helps.

EgmatQuantExpert · Answer

Solution

Step 1 & Step 2: Understanding the Question statement and Drawing Inferences

Given info:

⇒ Quantity of alcohol = \(\frac{7\mathrm x}{10}\)
⇒ Quantity of water = \(\frac{3\mathrm x}{10}\)

To find:

⇒ \(\frac{\frac{7\mathrm x}{10}}{(\mathrm x+\mathrm y)}\ast100=60\)
⇒ \(\frac{\frac{7\mathrm x}{10}}{(\mathrm x+\mathrm y)}=\frac35\)
Multiplying both sides of the equation by 5, we get
⇒ \(\frac{\left(\frac{7\mathrm x}{10}\right)\ast5}{(\mathrm x+\mathrm y)}=3\)
⇒ \(\frac{\left(\frac{7\mathrm x}2\right)}{(\mathrm x+\mathrm y)}=3\)
⇒\(3.5x = 3x+3y\)
⇒\(0.5x = 3y\) ...(Equation 1)

Step 3: Analyze statement 1 independently

⇒ \(x+y=350ml\) ...(Equation 2)

statement 1 is sufficient

Step 4: Analyze statement 2 independently

statement 2 is not sufficient

Step 5: Analyze the two statements together

Option A

Thanks,
Saquib
Quant Expert
e-GMAT

Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts

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rohit8865 · Answer

(in 350 ml --> 245 ml is alcohol + 105ml is water(given ratio 7:3)
let w be the quantity of water added to be 40% water(or 60% alcohol) in final solution
 thus 105 +w =40%(350+w)
thus w can be calculated
suff

(2) ony gives ratio 10.5/4.5= 7/3 which is given in question
No idea of volume...
insuff

Ans A

Madhavi1990 · Answer

Did it in 2:30 minutes (i guess way over time)
a/w = 7/3 (total parts 10)
we need a/w = 6/4

1) total ML is 350 --> easily we will know 350/10 = 35, divided each into 6/4 ratios. So exact amount of a/w can be achieved.

2) ratio of a/w= 10.5/ 4.5 --> great, but this would apply to each part of the mixture. How do we know the total milliliters? Hence insuff.

Ans A

fskilnik · Answer

In the solution below, the unit considered is always milliliters.

Let´s use the  k technique , one of the best tools of our method when dealing with ratios/proportions!

\left\{ \matrix{
 \,{m{water}}\,\,\left( w ight)\,\,\, = \,\,3k \hfill \cr 
 \,{m{alcohol}}\,\,\left( a ight) = 7k \hfill \cr} ight.\,\,\,\,\,\,\left( {k > 0} ight)\,\,\,\,\,\,\,\,\,\,\, 	o \,\,\,\,\,\,\,\,\,\left\{ \matrix{
 \,{m{water}}\,\,\left( w ight)\,\,\, = \,\,3k + x \hfill \cr 
 \,{m{alcohol}}\,\,\left( a ight) = 7k \hfill \cr} ight.\,\,\,\,\,\,\,{m{such}}\,\,{m{that}}\,\,\,\,\,\,{{7k} \over {10k + x}} = {3 \over 5}\,\,\,\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{{m{cross - multiply}}} \,\,\,\,\,\,5k = 3x\,\,\,\,\,\left( * ight)

? = x

\left( 1 ight)\,\,\,10k + x = 350\,\,\,\,\mathop \Rightarrow \limits^{\left( * ight)} \,\,\,\,\,6x + x = 350\,\,\,\,\, \Rightarrow \,\,\,\,\,{m{SUFF}}.\,\,\,\,\,\,\,

\left( 2 ight)\,\,{a \over w} = {{10.5} \over {4.5}}\,\,\left( { = {{105} \over {45}} = {7 \over 3}} ight)\,\,\,{m{already}}\,\,{m{known}}\,\,{m{pre - statements}}\,\,\,\,\, \Rightarrow \,\,\,\,\,{m{INSUFF}}.\,\,\,

This solution follows the notations and rationale taught in the GMATH method.

Regards, 
Fabio.

ScottTargetTestPrep · Answer

We can let the amount of water to be added be w and the original amount of alcohol and water be 7x and 3x, respectively. We can create the equation:

(3x + w)/(7x + 3x + w) = 6/10

(3x + w)/(10x + w) = 3/5

5(3x + w) = 3(10x + w)

15x + 5w = 30x + 3w

2w = 15x

We need to determine the value of w. We see that w = 15x/2 or x = 2w/15. Therefore, if we know one of the two variables, then we know the other.

Statement One Only:

Total quantity of the resulting solution is 350 milliliters.

This means 10x + w = 350. Since x = 2w/15, we have:

10(2w/15) + w = 350

4w/3 + w = 350

4w + 3w = 1050

7w = 1050

w = 150

Statement one alone is sufficient.

Statement Two Only:

The original solution contains 10.5 milliliters of alcohol for every 4.5 milliliters of water.

This means original amount of alcohol and water are 10.5y and 4.5y, respectively. Therefore, we have:
 
7x = 10.5y and 3x = 4.5y

Either way, we have x = 1.5y. However, since y can be any positive number (for example, if y = 2, then x = 3 and if y = 4, then x = 6), we can’t determine a unique value of x, and therefore, we can’t determine the value of w. Statement two alone is not sufficient.

Answer: A

ShreyaMittal1609 · Answer

We can use Alligation (0%,70% and avg is 60%). Ratio comes to be 1:6. Now St1 tells total quantity hence we can find water added (0%) to be 350/7

A solution consists of only water and alcohol such that the ratio of

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