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

 It is currently 18 Oct 2019, 15:25 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.  A tank holds x gallons of a saltwater solution that is 20%

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

Hide Tags

Manager  Joined: 06 Jan 2008
Posts: 211
A tank holds x gallons of a saltwater solution that is 20%  [#permalink]

Show Tags

5
26 00:00

Difficulty:   75% (hard)

Question Stats: 64% (02:52) correct 36% (03:21) wrong based on 381 sessions

HideShow timer Statistics

A tank holds x gallons of a saltwater solution that is 20% salt by volume. One Fourth of the water is evaporated, leaving all of the salt. When 10 Gallons of water and 20 gallons of salt are added, the resulting mixture is 33 1/3 % salt by volume. What is the value of x?

A. 37.5
B. 75
C. 100
D. 150
E. 175

Can you show me how to solve this without backsolving, ie the real algabraic way?
Senior Manager  Joined: 02 Aug 2007
Posts: 332
Location: Greater New York City area
Schools: Tuck, Ross (R1), Duke, Tepper, ISB (R2), Kenan Flagler (R2)
Re: Salt Storage Tank  [#permalink]

Show Tags

10
5
Manbehindthecurtain wrote:
neelesh wrote:
Is it e. 175 ?

Nope, 150. I can only get it by following PR's backsolving explanation. I hate that.

Original mixture has 20% salt and 80% water.
Total = x
Out of which Salt = 0.2x and water = 0.8x
Now, 1/4 water evaporates and all salt remains.
So what remains is 0.2x salt and 0.6x water.
Now 20 gallons salt is added and 10 gallons of water is added.
So salt now becomes -> (0.2x + 20) and water --> (0.6x+10)
Amount of salt is 33.33% of total. So amount of water is 66.66%. So salt is half of the volume of water.
So (0.2x+20) = (0.6x+10)/2,
Solving, x = 150.
General Discussion
Manager  Joined: 06 Jan 2008
Posts: 211
Re: Salt Storage Tank  [#permalink]

Show Tags

neelesh wrote:
Is it e. 175 ?

Nope, 150. I can only get it by following PR's backsolving explanation. I hate that.
Manager  Joined: 20 Dec 2004
Posts: 207
Re: Salt Storage Tank  [#permalink]

Show Tags

rgajare14 wrote:
So (0.2x+20) = (0.6x+10)/2, Solving, x = 150.

This is what I could not establish and started back solving from the answers and still got it wrong...

Good one. _________________
Stay Hungry, Stay Foolish
Senior Manager  Joined: 19 Nov 2007
Posts: 342
Re: Salt Storage Tank  [#permalink]

Show Tags

good explanation gajare.
_________________
-Underline your question. It takes only a few seconds!
-Search before you post.
Manager  Joined: 06 Jan 2008
Posts: 211
Re: Salt Storage Tank  [#permalink]

Show Tags

rgajare14 wrote:
Manbehindthecurtain wrote:
neelesh wrote:
Is it e. 175 ?

Nope, 150. I can only get it by following PR's backsolving explanation. I hate that.

Original mixture has 20% salt and 80% water.
Total = x
Out of which Salt = 0.2x and water = 0.8x
Now, 1/4 water evaporates and all salt remains.
So what remains is 0.2x salt and 0.6x water.
Now 20 gallons salt is added and 10 gallons of water is added.
So salt now becomes -> (0.2x + 20) and water --> (0.6x+10)
Amount of salt is 33.33% of total. So amount of water is 66.66%. So salt is half of the volume of water.
So (0.2x+20) = (0.6x+10)/2,
Solving, x = 150.

Outstanding response. thank you. Can you describe how you started unpeeling this? I found the question wording really confusing.

Also, what level difficulty would you say this is?
Senior Manager  Joined: 02 Aug 2007
Posts: 332
Location: Greater New York City area
Schools: Tuck, Ross (R1), Duke, Tepper, ISB (R2), Kenan Flagler (R2)
Re: Salt Storage Tank  [#permalink]

Show Tags

Quote:
Nope, 150. I can only get it by following PR's backsolving explanation. I hate that.

Original mixture has 20% salt and 80% water.
Total = x
Out of which Salt = 0.2x and water = 0.8x
Now, 1/4 water evaporates and all salt remains.
So what remains is 0.2x salt and 0.6x water.
Now 20 gallons salt is added and 10 gallons of water is added.
So salt now becomes -> (0.2x + 20) and water --> (0.6x+10)
Amount of salt is 33.33% of total. So amount of water is 66.66%. So salt is half of the volume of water.
So (0.2x+20) = (0.6x+10)/2,
Solving, x = 150.

Quote:
Outstanding response. thank you. Can you describe how you started unpeeling this? I found the question wording really confusing.

Also, what level difficulty would you say this is?

Thanks. Actually , I did not solve it backwards. I solved it as per the flow given in the problem question. Honestly, I dont think this was more than a 650 level question.
Retired Moderator Joined: 20 Dec 2013
Posts: 166
Location: United States (NY)
GMAT 1: 640 Q44 V34 GMAT 2: 710 Q48 V40 GMAT 3: 720 Q49 V40 GPA: 3.16
WE: Consulting (Venture Capital)
Re: A tank holds x gallons of a saltwater solution that is 20%  [#permalink]

Show Tags

X gallons @20% Salt (S) & 80% Water (W)

By Gallon:
x/5 = S
4x/5 = W

Mixture composition adjusted for 1/4 W evaporation and additional S/W gallons:
x/5 + 20= S
3x/5 + 10= W

Using Salt @33.3% to solve for x:
(4x/5 + 30)/3 = x/5 + 20
x=150
_________________
Director  Joined: 03 Aug 2012
Posts: 660
Concentration: General Management, General Management
GMAT 1: 630 Q47 V29 GMAT 2: 680 Q50 V32 GPA: 3.7
WE: Information Technology (Investment Banking)
Re: A tank holds x gallons of a saltwater solution that is 20%  [#permalink]

Show Tags

4
2
Understand the logic of the flow of question.

Consider X initial solution.

0.2X = Salt
0.8X = Water

1/4(0.8X)= Evaporated Leaving 0.6X=water

0.6X+10=Water
0.2X+20=Salt

Salt concentration now becomes= 1/3 of the Solution

(0.2X+20)/(0.8X+30) = 1/3

X=150
VP  P
Joined: 07 Dec 2014
Posts: 1223
A tank holds x gallons of a saltwater solution that is 20%  [#permalink]

Show Tags

1
A tank holds x gallons of a saltwater solution that is 20% salt by volume. One Fourth of the water is evaporated, leaving all of the salt. When 10 Gallons of water and 20 gallons of salt are added, the resulting mixture is 33 1/3 % salt by volume. What is the value of x?

A. 37.5
B. 75
C. 100
D. 150
E. 175

after evaporation, x-(.25)(.8)x=.8x gallons
salt now=25% of .8x gallons
water now=75% of .8x gallons
(.25)(.8x)+20=(.8x+30)/3
x=150 gallons
D

Originally posted by gracie on 07 Nov 2015, 09:57.
Last edited by gracie on 23 Sep 2017, 20:25, edited 1 time in total.
Manager  Joined: 05 Jul 2015
Posts: 93
Concentration: Real Estate, International Business
GMAT 1: 600 Q33 V40 GPA: 3.3
Re: A tank holds x gallons of a saltwater solution that is 20%  [#permalink]

Show Tags

I did it a different way but still got the right answer:
Salt + Water = Xgallons
(1/5) + (4/5) = (5/5)

water left after evaporation: (4/5)*(3/4) = (3/5) then (3/5)(5/5) = (3/4) of total is water
Amount of salt in new mix = (1/4)

New mix:
10 gals of Water plus 20 Gals of salt added = 33% Salt to 66% Water = 1:2

Amount of water plus 10 gals / amount of salt plus 20 gals = 1/2
(3/4)X + 10 gals
------------- = (1/2)
(1/4)X+20 Gals

Cross multiply:
(6/4)X +20 = (1/4)X+20
X=4/5

So X (the old mix) is (4/5) of the combined old plus new mix so...

30gals = 1/5 of X
150gals = X
Board of Directors P
Joined: 17 Jul 2014
Posts: 2509
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30 GPA: 3.92
WE: General Management (Transportation)
Re: A tank holds x gallons of a saltwater solution that is 20%  [#permalink]

Show Tags

oh yeah...
0.2x+20 = 1/3*(0.8x+30)

solving this, we get x = 150.
Senior Manager  G
Joined: 03 Apr 2013
Posts: 264
Location: India
Concentration: Marketing, Finance
GMAT 1: 740 Q50 V41 GPA: 3
Re: A tank holds x gallons of a saltwater solution that is 20%  [#permalink]

Show Tags

Manbehindthecurtain wrote:
A tank holds x gallons of a saltwater solution that is 20% salt by volume. One Fourth of the water is evaporated, leaving all of the salt. When 10 Gallons of water and 20 gallons of salt are added, the resulting mixture is 33 1/3 % salt by volume. What is the value of x?

A. 37.5
B. 75
C. 100
D. 150
E. 175

Can you show me how to solve this without backsolving, ie the real algabraic way?

Got the right answer but the wording is super bad!
_________________
Spread some love..Like = +1 Kudos Senior Manager  P
Joined: 29 Jun 2017
Posts: 424
GPA: 4
WE: Engineering (Transportation)
Re: A tank holds x gallons of a saltwater solution that is 20%  [#permalink]

Show Tags

1
x gallon was solution = 0.2x salt and 0.8x water, 0.25% evaporated = 0.6x water remains
new solution is 0.2x salt and 0.6x water
20 gallon salt is added, new salt = 20+0.2x
10 gallon water added , new water = 10+0.6x
total solution = 30+0.8x

new solution is 33.33% salt by vol = 1/3

$$\frac{(20+0.2x)}{(30+0.8x)}$$ = $$\frac{1}{3}$$

D is the answer
_________________
Give Kudos for correct answer and/or if you like the solution.
Director  D
Joined: 13 Mar 2017
Posts: 728
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: A tank holds x gallons of a saltwater solution that is 20%  [#permalink]

Show Tags

Manbehindthecurtain wrote:
A tank holds x gallons of a saltwater solution that is 20% salt by volume. One Fourth of the water is evaporated, leaving all of the salt. When 10 Gallons of water and 20 gallons of salt are added, the resulting mixture is 33 1/3 % salt by volume. What is the value of x?

A. 37.5
B. 75
C. 100
D. 150
E. 175

Can you show me how to solve this without backsolving, ie the real algabraic way?

It can be easily solved by front solving.

Salt Water Total
0.2x 0.8x x
After Evaporation
0.2x 0.6x 0.8x
After adding 10 gallons of water and 20 gallons of salt
20+0.2x 10+0.6x 30+0.8x

20+0.2x )/ (30+0.8x = 1/3
x = 150

Very Simple . Will take around 1 minute 40 seconds
_________________
CAT 2017 (98.95) & 2018 (98.91) : 99th percentiler
UPSC Aspirants : Get my app UPSC Important News Reader from Play store.

MBA Social Network : WebMaggu

Appreciate by Clicking +1 Kudos ( Lets be more generous friends.)

What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish".
Target Test Prep Representative D
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8104
Location: United States (CA)
Re: A tank holds x gallons of a saltwater solution that is 20%  [#permalink]

Show Tags

Manbehindthecurtain wrote:
A tank holds x gallons of a saltwater solution that is 20% salt by volume. One Fourth of the water is evaporated, leaving all of the salt. When 10 Gallons of water and 20 gallons of salt are added, the resulting mixture is 33 1/3 % salt by volume. What is the value of x?

A. 37.5
B. 75
C. 100
D. 150
E. 175

We are given that a tank holds x gallons of a saltwater solution that is 20% salt by volume. Thus, the solution contains 0.2x salt and 0.8x water. After 1/4 of the water is evaporated, there is 0.6x water left.

We are also given that when 10 gallons of water and 20 gallons of salt are added, the resulting mixture is 33 1/3% (or 1/3) salt by volume. Thus:

(0.2x + 20)/(0.2x + 0.6x + 10 + 20) = 1/3

(0.2x + 20)/(0.8x + 30) = 1/3

Multiplying 10/10 on the left side of the equation, we have:

(2x + 200)/(8x + 300) = 1/3

8x + 300 = 6x + 600

2x = 300

x = 150

_________________

Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Manager  B
Joined: 11 Jun 2017
Posts: 62
Re: A tank holds x gallons of a saltwater solution that is 20%  [#permalink]

Show Tags

A tank holds x gallons of a saltwater solution that is 20% salt by volume. One Fourth of the water is evaporated, leaving all of the salt. When 10 Gallons of water and 20 gallons of salt are added, the resulting mixture is 33 1/3 % salt by volume. What is the value of x?

Initially the solution contains 0.8x of water and 0.2x of salt. 1/4th of the water is evaporated which is (1/4*0.8x) = x/5. Hence, 1-x/5= 4x/5 is the remaining content of solution.
20 gallons of salt is added , hence, new salt content in solution= 0.2x+20
Total content = 4x/5+30

0.2x+20/4x/5+30 = 1/3
Solving for x, we get x=150. (option D)
Non-Human User Joined: 09 Sep 2013
Posts: 13262
Re: A tank holds x gallons of a saltwater solution that is 20%  [#permalink]

Show Tags

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________ Re: A tank holds x gallons of a saltwater solution that is 20%   [#permalink] 04 Oct 2018, 17:31
Display posts from previous: Sort by

A tank holds x gallons of a saltwater solution that is 20%

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

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