John purchased large bottles of water for

This topic has expert replies
Master | Next Rank: 500 Posts
Posts: 366
Joined: Fri Jun 05, 2015 3:35 am
Thanked: 3 times
Followed by:2 members

John purchased large bottles of water for

by NandishSS » Wed Nov 16, 2016 3:04 am
John purchased large bottles of water for $2 each and small bottles of water for $1.50 each. What percent of the bottles purchased were small bottles?

(1) John spent $33 on the bottles of water
(2) The average price of bottles purchased was $1.65

OA:B

How to quickly find the Several integer solutions possible to satisfy this 2x+1.5y=33(From Stat 1)?

User avatar
Legendary Member
Posts: 2131
Joined: Mon Feb 03, 2014 9:26 am
Location: https://martymurraycoaching.com/
Thanked: 955 times
Followed by:140 members
GMAT Score:800

by MartyMurray » Wed Nov 16, 2016 4:33 am
NandishSS wrote:How to quickly find the Several integer solutions possible to satisfy this 2x+1.5y=33(From Stat 1)?
Notice that 33 is a combination of 30 and 3. So, 15x + 2y = 15(2) + 2(1.5) = 33.

Then 30 is also divisible by 3, and 3 = 2 * 1.5. So the x coefficient could be 0.

Done.

Alternatively, you could see that 1.5 goes evenly into 3 and any multiple of 3. So many different odd multiples of 3 combined with multiples of 2 can add up to 33.

3 + 30, 9 + 24, 15 + 18, and so forth.

Notice, that one way to work with integer combinations is to start at one end and work your way to the other end.

In other words, start with one 3 and see whether that works with a bunch of 2's. Then use two 3's and see whether that works. You continue the process until you get to a number of 3's equal to or close to 33.

Similarly if you were to have 7's and 2's with which to work to generate a sum of 33, you would start with one 7 and work your way up, seeing which ones work.

7 + 26 works. 14 + an even number = an even number and does not work. 21 + 12 works. 28 + an even number = even number and does not work. 35 is too big. Done.
Marty Murray
Perfect Scoring Tutor With Over a Decade of Experience
MartyMurrayCoaching.com
Contact me at [email protected] for a free consultation.

User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Wed Nov 16, 2016 4:58 am
NandishSS wrote:John purchased large bottles of water for $2 each and small bottles of water for $1.50 each. What percent of the bottles purchased were small bottles?

(1) John spent $33 on the bottles of water
(2) The average price of bottles purchased was $1.65
Statement 2:
To determine the ratio of L to S, use ALLIGATION -- a very efficient way to handle mixture problems.

Step 1: Plot the 3 values on a number line, with the prices for L and S on the ends and the average price in the middle.
L 200---------165----------150 S

Step 2: Calculate the distances between the percentages.
L 200----35----165----15----150 S

Step 3: Determine the ratio in the mixture.
The required ratio of L to S is equal to the RECIPROCAL of the distances in red.
L:S = 15:35 = 3:7.

Implication:
Of every 10 bottles, 3 were large and 7 were small.
Thus, the percentage of small bottles = 7/10 = 70%.
SUFFICIENT.

Statement 1:
Since the two statements cannot contradict each other, the percentage yielded by Statement 2 must also be possible in Statement 1.
Thus, it must be possible in Statement 1 that 70% of the bottles were small.

Equation implied by Statement 1:
2L + 1.5S = 33
4L + 3S = 66.

In the resulting equation, it's possible that S=2 and L=15, since 4(15) + 2(3) = 66.
In this case, 2 of every 17 bottles are small.

Since the percentage of small bottles can be different values, INSUFFICIENT.

The correct answer is B.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 7222
Joined: Sat Apr 25, 2015 10:56 am
Location: Los Angeles, CA
Thanked: 43 times
Followed by:29 members

by Scott@TargetTestPrep » Thu Nov 17, 2016 1:48 pm
NandishSS wrote:John purchased large bottles of water for $2 each and small bottles of water for $1.50 each. What percent of the bottles purchased were small bottles?

(1) John spent $33 on the bottles of water
(2) The average price of bottles purchased was $1.65
We are given that John purchased large bottles of water for $2 each and small bottles of water for $1.50 each. We can let s = the number of small bottles purchased, and b = the number of larger bottles purchased. We need to determine what percentage of the bottles purchased were small bottles, i.e. the value of s/(s+b) x 100.

Statement One Alone:

John spent $33 on the bottles of water.

Using the information in statement one, we can create the following equation:

2b + 1.5s = 33

We can multiply the entire equation by 2 and we have:

4b + 3s = 66

4b = 66 - 3s

4b = 3(22 - s)

b = [3(22 - s)]/4

Since b must be an integer, 3(22 - s) must be a multiple of 4.

3(22 - s) is a multiple of 4 when s = 2, 6, 10, 14, or 18.

Since we have multiple values of s, we will also have multiple values of b, and thus we do not have enough information to answer the question.

Statement Two Alone:

The average price of bottles purchased was $1.65.

Using the information in statement two, we can create the following equation:

1.65 = (2b + 1.5s)/(b + s)

1.65(b + s) = 2b + 1.5s

165(b + s) = 200b + 150s

165b + 165s = 200b + 150s

15s = 35b

3s = 7b

(3/7)s = b

We can now determine the value of s/(s+b) x 100 by substituting (3/7)s for b:

s/[s+(3/7)s] x 100

s/[(10/7)s] x 100

1/(10/7) x 100

7/10 x 100 = 70

So the small bottles account for 70 percent of the bottles purchased. We have answered the question.

Answer: B

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage

GMAT/MBA Expert

User avatar
Elite Legendary Member
Posts: 10392
Joined: Sun Jun 23, 2013 6:38 pm
Location: Palo Alto, CA
Thanked: 2867 times
Followed by:511 members
GMAT Score:800

by [email protected] » Thu Nov 17, 2016 7:32 pm
Hi NandishSS,

This DS question is ultimately about Weighted Averages and ratios, and you'll have to do a bit of math to get to the correct answer.

We're told that large bottles of water cost $2 each and small bottles of water cost $1.50 each. We're asked what PERCENT of the total bottles were small bottles.

L = number of large bottles
S = number of small bottles

Using these variables, the question can then be rewritten as... What is the value of S/(S+T)

1) John spent $33 on the bottles of water

Since small bottles cost $1.50 each, it's helpful to recognize that 2 small bottles total $3

IF there are...
2 small bottles and 15 large bottles, then the answer to the question is 2/17
6 small bottles and 12 large bottles, then the answer to the question is 6/18
Fact 1 is INSUFFICIENT

2) The average price of bottles purchased was $1.65

With this information, we can create an equation based on the average formula:

[(1.50)(S) + (2)(L)] / (S+L) = 1.65

From here, we can simplify...

1.5S + 2L = 1.65S + 1.65L
.35L = .15S
35L = 15S
7L = 3S
7/3 = S/L

This tells us that for every 7 small bottles there are 3 large bottles. Thus, the answer to the question is 7/10.
Fact 2 is SUFFICIENT

Final Answer: B

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
Image

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3008
Joined: Mon Aug 22, 2016 6:19 am
Location: Grand Central / New York
Thanked: 470 times
Followed by:34 members

by Jay@ManhattanReview » Wed Jan 18, 2017 11:25 pm
NandishSS wrote:
How to quickly find the Several integer solutions possible to satisfy this 2x+1.5y=33(From Stat 1)?
Though there are many smart ways to handle 2x+1.5y=33. One such is demonstrated below.

We are given that 2x+1.5y=33

=> x = (33 - 1.5y)/2

Since x is an integer, (33 - 1.5y)/2 must be an integer, or (33 - 1.5y) must be divisible by 2 or it is an EVEN number.

The minimum value of y = 2.

=> x = (33 - 1.5*2)/2 = 30/2 = 15 => Large bottles = 15 and Small bottle = 2

The next possible value of y = 6. The value y = 4 is inadmissible as that would make 1.5*4 = 6 (even), and (33 - 6) an ODD.

=> @ y =6, x = (33 - 1.5*6)/2 = 24/2 = 12 => Large bottles = 12 and Small bottle = 6

The other possible values of [y = 6+4=10 => x = 9]; [y = 10+4=14 => x = 6]; [y = 14+4=18 => x = 3]; [y = 18+4=22 => x = 0 (invalid]

Hope this helps!

-Jay
_________________
Manhattan Review GMAT Prep

Locations: New York | Paris | Shanghai | Munich | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.