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

 It is currently 19 Jun 2018, 09:18

# LIVE NOW:

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Solution Y is 40 percent sugar by volume, and solution X is 20 percent

Author Message
TAGS:

### Hide Tags

Manager
Joined: 17 Oct 2012
Posts: 68
Location: India
Concentration: Strategy, Finance
WE: Information Technology (Computer Software)
Solution Y is 40 percent sugar by volume, and solution X is 20 percent [#permalink]

### Show Tags

28 Sep 2014, 05:01
5
00:00

Difficulty:

35% (medium)

Question Stats:

78% (01:23) correct 22% (01:59) wrong based on 366 sessions

### HideShow timer Statistics

Solution Y is 40 percent sugar by volume, and solution X is 20 percent sugar by volume. How many gallons of solution X must be added to 150 gallons of solution Y to create a solution that is 25 percent sugar by volume?

(A) 37.5
(B) 75
(C) 150
(D) 240
(E) 450
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1837
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent [#permalink]

### Show Tags

28 Sep 2014, 23:58
2
2
chetan86 wrote:
Solution Y is 40 percent sugar by volume, and solution X is 20 percent sugar by volume. How many gallons of solution X must be added to 150 gallons of solution Y to create a solution that is 25 percent sugar by volume?

(A) 37.5
(B) 75
(C) 150
(D) 240
(E) 450

Solution Y

Sugar ................ Water .............. Total

40 ............................ 60 .................. 100

60 .......................... 90 .................... 150

Say x litres of Solution x added; New equation would be

$$60+\frac{20x}{100}$$ .............. $$90+\frac{80x}{100}$$ ............ 150+x

New sugar concentration is 25%; setting up the equation

$$\frac{25}{100} (150+x) = 60 + \frac{20x}{100}$$

x = 450
_________________

Kindly press "+1 Kudos" to appreciate

Director
Joined: 23 Jan 2013
Posts: 597
Schools: Cambridge'16
Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent [#permalink]

### Show Tags

23 Oct 2014, 07:19
2
Differential method:

40----------25----20

15x=5y
x/y=5/15=1/3, so 1 part of 40% sol should be added 3 part of 20%.
So 150*3=450

E
Manager
Joined: 21 Jul 2014
Posts: 126
Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent [#permalink]

### Show Tags

23 Oct 2014, 10:09
1
4
In order to solve this problem you must be able to translate English into Algebra equations. Once you have written one or more algebraic equations it becomes a simple math problem to solve. You should look out for some keywords such as added, percent and so on while creating the equations. In this case, lets go step by step:

1. The problem states that there is 150 gallons of Y solution and X will be added to create a solution - this shows that the total number of gallons in the new solution will be $$150+x$$.
2. The new solution is 25 percent sugar by volume. So, that becomes $$\frac{25}{100}* (150+x)$$.
3. Solution Y is 40 percent sugar and we know that there are 150 gallons of Y solution. Solving for that gives $$\frac{40}{100} * 150$$. This gives us 60 gallons of sugar by volume in Solution Y out of the 150 total gallons
4. Solution X is 20 percent sugar and this will be added to 60 gallons of Y. So the translation here is $$60 + \frac{20}{100}(x)$$

Now we can form the equation -> $$60 + \frac{20}{100}(x) = \frac{25}{100}(150+x)$$

This is solving equation with one variable.

$$60 + 0.2x = 37.5 + .25x$$

$$.05x = 22.5$$

$$x = 450$$

Manager
Status: I am not a product of my circumstances. I am a product of my decisions
Joined: 20 Jan 2013
Posts: 127
Location: India
Concentration: Operations, General Management
GPA: 3.92
WE: Operations (Energy and Utilities)
Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent [#permalink]

### Show Tags

23 Oct 2014, 22:25
1
chetan86 wrote:
Solution Y is 40 percent sugar by volume, and solution X is 20 percent sugar by volume. How many gallons of solution X must be added to 150 gallons of solution Y to create a solution that is 25 percent sugar by volume?

(A) 37.5
(B) 75
(C) 150
(D) 240
(E) 450

Sol Y = 40% sugar---------------------Sol Mix = 25 % sugar-----------------Sol X = 20% sugar

Sol Y = +0.15 than Avg---------------Avg = 0.25------------------------------Sol X = -0.05 than avg

0.15 Y - 0.05 X = 0
0.15 Y = 0.05 X

X/Y = 0.15/0.05 = 3:1

X:Y:Total = 3:1:4

If Y = 150 Gallons then by ratio X:Y = 3:1

X/150= 3/1
X = 150 * 3 = 450 Gallons

Ans E
VP
Joined: 07 Dec 2014
Posts: 1018
Solution Y is 40 percent sugar by volume, and solution X is 20 percent [#permalink]

### Show Tags

07 Nov 2015, 13:04
1
(.4)(150)+.2x=.25(150+x)
x=450 gallons
Manager
Joined: 05 Jul 2015
Posts: 105
GMAT 1: 600 Q33 V40
GPA: 3.3
Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent [#permalink]

### Show Tags

19 Nov 2015, 16:47
1
I did a slightly different differential method.

40----25----20
--15------5---

So Y has to be 5/20 of the new solution. Given Y = 150 gallons

150 = 5/20(total)
Total = 600
X = 600 - 150

Ans E) 450
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 11801
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent [#permalink]

### Show Tags

10 Mar 2018, 12:05
1
Hi All,

Mixture questions can also be answered with the Part/Whole formula:

In this case, we're dealing with Sugar/Total

Solution X = 20% sugar
Solution Y = 40% sugar

We're told to mix a certain amount of Solution X with 150 gallons of Solution Y to get a 25% sugar solution:

X = number of gallons of Solution X

[(.2)(X) + (.4)(150)] / (X + 150) = .25

Now, it's just algebra steps:

.2X + 60 = .25X + 37.5

22.5 = .05X

450 = X

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

Study Buddy Forum Moderator
Joined: 04 Sep 2016
Posts: 1016
Location: India
WE: Engineering (Other)
Solution Y is 40 percent sugar by volume, and solution X is 20 percent [#permalink]

### Show Tags

24 Apr 2018, 18:57
Attachment:

2504 sol.jpeg [ 24.41 KiB | Viewed 727 times ]
Bunuel niks18 chetan2u gmatbusters VeritasPrepKarishma

Quote:
Solution Y is 40 percent sugar by volume, and solution X is 20 percent sugar by volume. How many gallons of solution X must be added to 150 gallons of solution Y to create a solution that is 25 percent sugar by volume?

I am trying to master concept of weighted average by scaling since I find it
very time efficient and applicable in multiple scenarios.

_________________

It's the journey that brings us happiness not the destination.

DS Forum Moderator
Joined: 27 Oct 2017
Posts: 564
Location: India
GPA: 3.64
WE: Business Development (Energy and Utilities)
Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent [#permalink]

### Show Tags

24 Apr 2018, 19:45
1
Your method is correct, and infact I find weighted average/mixture alligation method to be the best approach to solve such questions.

Attachment:
2504 sol.jpeg
Bunuel niks18 chetan2u gmatbusters VeritasPrepKarishma

Quote:
Solution Y is 40 percent sugar by volume, and solution X is 20 percent sugar by volume. How many gallons of solution X must be added to 150 gallons of solution Y to create a solution that is 25 percent sugar by volume?

I am trying to master concept of weighted average by scaling since I find it
very time efficient and applicable in multiple scenarios.

Posted from my mobile device

Posted from my mobile device
_________________
VP
Status: It's near - I can see.
Joined: 13 Apr 2013
Posts: 1043
Location: India
GMAT 1: 480 Q38 V22
GPA: 3.01
WE: Engineering (Consulting)
Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent [#permalink]

### Show Tags

28 Apr 2018, 04:30
chetan86 wrote:
Solution Y is 40 percent sugar by volume, and solution X is 20 percent sugar by volume. How many gallons of solution X must be added to 150 gallons of solution Y to create a solution that is 25 percent sugar by volume?

(A) 37.5
(B) 75
(C) 150
(D) 240
(E) 450

Weighted Average Method :

Let's add G gallons of solution X

So we have, $$\frac{G}{150}$$ = $$\frac{40 - 25}{25 - 20}$$

$$G = 150 * \frac{15}{5}$$

$$G = 450$$

Algebraic Method:

$$0.20X + 0.40Y = 0.25(X +Y)$$

$$0.20X + 0.40Y = 0.25X + 0.25Y$$

$$0.5X = 0.15Y$$

$$X = \frac{0.15*150}{0.5}$$

$$X = 450$$
_________________

"Success is not as glamorous as people tell you. It's a lot of hours spent in the darkness."

Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent   [#permalink] 28 Apr 2018, 04:30
Display posts from previous: Sort by