GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Oct 2019, 22:45 GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  M22-07

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 58443

Show Tags

1
7 00:00

Difficulty:   65% (hard)

Question Stats: 62% (01:18) correct 38% (01:32) wrong based on 171 sessions

HideShow timer Statistics

If two different solutions of alcohol with a respective proportion of water to alcohol of 3:1 and 2:3 were combined, what is the concentration of alcohol in the new solution if the original solutions were mixed in equal amounts?

A. 30.0%
B. 36.6%
C. 42.5%
D. 44.4%
E. 60.0%

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58443

Show Tags

1
2
Official Solution:

If two different solutions of alcohol with a respective proportion of water to alcohol of 3:1 and 2:3 were combined, what is the concentration of alcohol in the new solution if the original solutions were mixed in equal amounts?

A. 30.0%
B. 36.6%
C. 42.5%
D. 44.4%
E. 60.0%

The strength of the first solution $$= \frac{1}{3 + 1} = 25\%$$. The strength of the second solution $$= \frac{3}{2 + 3} = 60\%$$. Because the solutions were mixed in equal amounts, the strength of the new solution is $$\frac{60\% + 25\%}{2} = 42.5\%$$.

_________________
Intern  Joined: 14 Oct 2015
Posts: 29
GMAT 1: 640 Q45 V33 Show Tags

1
I think this is a high-quality question and I agree with explanation. I like it! Another way of viewing this. If you are going to use 5 L of the 2:3 solution, you must use 5L of the 3:1 solution. Thus you will be adding 3.75:1.25 L to 2:3 L which results in a 10 L solution that has 5.75:4.25 ratio. You want to know what 4.25L is as a % of 10L, so just times 4.25 / 10 by 10 -> 42.5 / 100
Intern  Joined: 19 Dec 2015
Posts: 28

Show Tags

2
Bunuel's solution is pretty awesome, I didn't think one could solve it that simply. I took a different approach and worked with actual numbers:

1) We know that the ratio of water:alcohol in the first solution is $$\frac{3}{1}$$ (which give 4 parts) and we know that the ratio of water:alcohol in the second solution is $$\frac{2}{3}$$ (which give 5 parts).

2) 20 is the LCM of 4 and 5, so let's assume that both solutions have a volume of 20 units. From the first solution, we have that water is $$\frac{3}{4}$$ of the solution (which we said had a volume of 20), which gives 15 units of water and thus 5 units of alcohol. From the second solution, we have that water is $$\frac{2}{5}$$ of the solution, which gives 8 units of water and 12 units of alcohol.

3) Since we are mixing both solutions with an equal amount, 20 units, our new solution has a total volume of 40 units. From 2), combining the water and alcohol in both solutions, we get that our new solution has 17 units of alcohol (5 units from the first solution and 12 units from the second solution) and 23 units of water (15 units from the first solution and 8 units from the second solution). Thus, the amount of alcohol in the mixed solution is $$\frac{17}{40}$$ = 42.5%.

PS: Time-wise I managed to do this slightly north of 2 minutes, but Bunuel's approach is of course much faster if one can identify the validity of that approach right off the bat.

/S
Intern  B
Joined: 05 Jun 2017
Posts: 7
Location: India
GPA: 3.67
WE: Science (Non-Profit and Government)

Show Tags

my method is simple alligation

25% and 60% mix leading to x but the ratio of both original solutions in mix is 1:1. therefore:

x-25=60-x means 2x=85. finally x=42.5%
Intern  B
Joined: 09 Aug 2017
Posts: 22

Show Tags

Bunuel wrote:
Official Solution:

If two different solutions of alcohol with a respective proportion of water to alcohol of 3:1 and 2:3 were combined, what is the concentration of alcohol in the new solution if the original solutions were mixed in equal amounts?

A. 30.0%
B. 36.6%
C. 42.5%
D. 44.4%
E. 60.0%

The strength of the first solution $$= \frac{1}{3 + 1} = 25\%$$. The strength of the second solution $$= \frac{3}{2 + 3} = 60\%$$. Because the solutions were mixed in equal amounts, the strength of the new solution is $$\frac{60\% + 25\%}{2} = 42.5\%$$.

\frac{60\% + 25\%}{2} = 42.5\ how did you got equation ? Why you have divided by 2 ? @bunuel

Sent from my Redmi 3S using GMAT Club Forum mobile app
Math Expert V
Joined: 02 Sep 2009
Posts: 58443

Show Tags

Spongebob02 wrote:
Bunuel wrote:
Official Solution:

If two different solutions of alcohol with a respective proportion of water to alcohol of 3:1 and 2:3 were combined, what is the concentration of alcohol in the new solution if the original solutions were mixed in equal amounts?

A. 30.0%
B. 36.6%
C. 42.5%
D. 44.4%
E. 60.0%

The strength of the first solution $$= \frac{1}{3 + 1} = 25\%$$. The strength of the second solution $$= \frac{3}{2 + 3} = 60\%$$. Because the solutions were mixed in equal amounts, the strength of the new solution is $$\frac{60\% + 25\%}{2} = 42.5\%$$.

\frac{60\% + 25\%}{2} = 42.5\ how did you got equation ? Why you have divided by 2 ? @bunuel

Sent from my Redmi 3S using GMAT Club Forum mobile app

Because we are told that the original solutions were mixed in EQUAL amounts. So, the percentage of alcohol in the resulting mixture would be the average of the percentages in two mixtures.

Check other discussion here: https://gmatclub.com/forum/if-two-diffe ... 99529.html

Hope it helps.
_________________
Intern  B
Joined: 17 Dec 2017
Posts: 17
Location: United States
GMAT 1: 720 Q49 V40 GPA: 3.31
WE: Consulting (Consulting)

Show Tags

Sallyzodiac wrote:
Bunuel's solution is pretty awesome, I didn't think one could solve it that simply. I took a different approach and worked with actual numbers:

1) We know that the ratio of water:alcohol in the first solution is $$\frac{3}{1}$$ (which give 4 parts) and we know that the ratio of water:alcohol in the second solution is $$\frac{2}{3}$$ (which give 5 parts).

2) 20 is the LCM of 4 and 5, so let's assume that both solutions have a volume of 20 units. From the first solution, we have that water is $$\frac{3}{4}$$ of the solution (which we said had a volume of 20), which gives 15 units of water and thus 5 units of alcohol. From the second solution, we have that water is $$\frac{2}{5}$$ of the solution, which gives 8 units of water and 12 units of alcohol.

3) Since we are mixing both solutions with an equal amount, 20 units, our new solution has a total volume of 40 units. From 2), combining the water and alcohol in both solutions, we get that our new solution has 17 units of alcohol (5 units from the first solution and 12 units from the second solution) and 23 units of water (15 units from the first solution and 8 units from the second solution). Thus, the amount of alcohol in the mixed solution is $$\frac{17}{40}$$ = 42.5%.

PS: Time-wise I managed to do this slightly north of 2 minutes, but Bunuel's approach is of course much faster if one can identify the validity of that approach right off the bat.

/S

Took the same approach. What if the question asked for the solutions to be mixed in different proportions instead of the same? How would you use this method with unequal proportions?
Senior Manager  S
Joined: 08 Jun 2015
Posts: 420
Location: India
GMAT 1: 640 Q48 V29 GMAT 2: 700 Q48 V38 GPA: 3.33

Show Tags

The answer is option C. The first soln has water and alcohol is the ration 75:25 and the second solution in the ratio 40:60. If equal amount of the two are taken, the ratio of water and alcohol is 115:85. The percentage of alcohol is (85/200)*100 or 42.5%
_________________
" The few , the fearless "
Manager  S
Joined: 26 Feb 2018
Posts: 51
Location: India
GMAT 1: 640 Q45 V34 GPA: 3.9
WE: Web Development (Computer Software)

Show Tags

I think this is a high-quality question and I agree with explanation.
Intern  B
Joined: 06 Dec 2018
Posts: 2

Show Tags

1
I just summed up the sums of the solutions which yielded 9 ( (1+3) + (3+2) = 9)
Afterwards I did the same with the alcohol solutions which yielded 4 (3 + 1)
THe percentage of alcohol would be 4/9 = 0.4444 , Can you tell me what is wrong? Re: M22-07   [#permalink] 03 Feb 2019, 05:38
Display posts from previous: Sort by

M22-07

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  