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Word Problem - A solution consists of only water and alcohol such that

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EgmatQuantExpert
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Word Problem - A solution consists of only water and alcohol such that [#permalink] New post   22 Jan 2017, 02:24
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Q.)
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.


    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.


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Re: Word Problem - A solution consists of only water and alcohol such that [#permalink] New post   29 Aug 2017, 13:06
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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%.
[V1= We need to find.]

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.
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Re: Word Problem - A solution consists of only water and alcohol such that [#permalink] New post   27 Mar 2017, 06:19
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Solution



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

Given info:

    • Ratio of alcohol to water in original solution = 7:3
    • Let the quantity of original solution be x ml.
⇒ Quantity of alcohol = \(\frac{7\mathrm x}{10}\)
⇒ Quantity of water = \(\frac{3\mathrm x}{10}\)

To find:

    • We have to find the quantity of water to be added to the solution, such that the resulting solution contains 60% of alcohol.
    • Let the quantity of water that is to be added to the solution be y millilitres.
    • So total quantity of final solution will be x+y millilitres
    • Now since only water is being added we can infer that the quantity of alcohol in the solution will not change.
    • So quantity of alcohol in the final solution will be
    • Percentage of alcohol in the final solution should be 60%
⇒ \(\frac{\frac{7\mathrm x}{10}}{(\mathrm x+\mathrm y)}\ast100=60\)
⇒ \(\frac{\frac{7\mathrm x}{10}}{(\mathrm x+\mathrm y)}=\frac35\)
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)
    • Since we do not know the value of x, we will not be able to determine the value of y.
    • So we need a relation in x and y, or value of x or y, to be able to determine the amount of water to be added to the solution.

Step 3: Analyze statement 1 independently

    • Total quantity of the resulting solution is 350ml.
    • Total quantity of the resulting solution = x+y ml (Already established above)
⇒ \(x+y=350ml\) ...(Equation 2)
    • Since we have another relation in x and y, we will be able to determine the value of y, and thus the amount of water to be added to the solution.
    • Hence statement 1 is sufficient to answer the question.

Step 4: Analyze statement 2 independently

    • The original solution contains 10.5ml of alcohol for every 4.5ml of water.
    • So, \(\frac{\mathrm{Alcohol}}{\mathrm{Water}}=\frac{10.5}{4.5}=\frac73\)
    • The ratio of alcohol to water from this statement comes out to be 7:3, which is the same as given to us in the question stem.
    • So we do not get any additional information from which we can determine the values of x or y.
    • Hence statement 2 is not sufficient to answer the question.

Step 5: Analyze the two statements together

    • Since from statement 1, we are able to arrive at a unique answer, combining and analysing statements together is not required.
    • Hence the correct answer is Option A

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Re: Word Problem - A solution consists of only water and alcohol such that [#permalink] New post   11 Nov 2018, 20:34
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EgmatQuantExpert wrote:
Q.)
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.


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
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Word Problem - A solution consists of only water and alcohol such that [#permalink] New post   Updated on: 27 Mar 2017, 06:20
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The official solution has been posted. Looking forward to a healthy discussion..:)

Originally posted by EgmatQuantExpert on 22 Jan 2017, 02:25.
Last edited by EgmatQuantExpert on 27 Mar 2017, 06:20, edited 1 time in total.
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Re: Word Problem - A solution consists of only water and alcohol such that [#permalink] New post   22 Jan 2017, 12:54
EgmatQuantExpert wrote:
Q.)
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.


    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
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e-GMAT

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(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
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Re: Word Problem - A solution consists of only water and alcohol such that [#permalink] New post   19 Nov 2017, 01:19
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
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Re: Word Problem - A solution consists of only water and alcohol such that [#permalink] New post   08 Nov 2018, 10:59
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EgmatQuantExpert wrote:
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.

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!

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

\(? = x\)


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


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


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

Regards,
Fabio.
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Re: Word Problem - A solution consists of only water and alcohol such that [#permalink] New post   11 Nov 2018, 03:47
A - sufficient
B - mistakanely considered it as sufficient - missed that its the same information as metnioned in the question - no new information
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Re: Word Problem - A solution consists of only water and alcohol such that [#permalink] New post   10 Apr 2023, 02:45
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