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

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

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

22 Feb 2008, 16:33
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)

### Show Tags

22 Feb 2008, 19:13
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

### Show Tags

22 Feb 2008, 18:22
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

### Show Tags

22 Feb 2008, 19:38
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

### Show Tags

22 Feb 2008, 20:15
good explanation gajare.
_________________
-Underline your question. It takes only a few seconds!
-Search before you post.
Manager
Joined: 06 Jan 2008
Posts: 211

### Show Tags

22 Feb 2008, 20:21
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)

### Show Tags

22 Feb 2008, 20:53
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

16 Feb 2014, 23:26
X gallons @20% Salt (S) & 80% Water (W)

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

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

18 Apr 2014, 14:01
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
Joined: 07 Dec 2014
Posts: 1223
A tank holds x gallons of a saltwater solution that is 20%  [#permalink]

### Show Tags

Updated on: 23 Sep 2017, 20:25
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
GMAT 1: 600 Q33 V40
GPA: 3.3
Re: A tank holds x gallons of a saltwater solution that is 20%  [#permalink]

### Show Tags

19 Nov 2015, 19:17
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
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

05 Dec 2015, 12:07
oh yeah...
0.2x+20 = 1/3*(0.8x+30)

solving this, we get x = 150.
Senior Manager
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

27 Jun 2016, 07:12
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?

_________________
Spread some love..Like = +1 Kudos
Senior Manager
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

06 Sep 2017, 01:15
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

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

_________________
Give Kudos for correct answer and/or if you like the solution.
Director
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

06 Sep 2017, 02:46
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
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

11 Sep 2017, 11:09
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
122 Reviews

5-star rated online GMAT quant
self study course

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

### Show Tags

17 Sep 2017, 06:32
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

04 Oct 2018, 17:31
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