Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 18 Jul 2019, 23:15

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

Carolyn bought 16 gumballs, Lew bought 12 gumballs, and Carey bought X

Author Message
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 56260
Carolyn bought 16 gumballs, Lew bought 12 gumballs, and Carey bought X  [#permalink]

Show Tags

05 Jan 2015, 07:00
00:00

Difficulty:

15% (low)

Question Stats:

86% (01:55) correct 14% (02:10) wrong based on 165 sessions

HideShow timer Statistics

Tough and Tricky questions: Min/Max.

Carolyn bought 16 gumballs, Lew bought 12 gumballs, and Carey bought X gumballs. The average (arithmetic mean) number of gumballs the three bought is between 19 and 25, inclusive. What is the difference between the greatest number and the smallest number of gumballs Carey could have bought?

A. 18
B. 22
C. 24
D. 29
E. 33

Kudos for a correct solution.

_________________
Manager
Joined: 22 Oct 2014
Posts: 88
Concentration: General Management, Sustainability
GMAT 1: 770 Q50 V45
GPA: 3.8
WE: General Management (Consulting)
Re: Carolyn bought 16 gumballs, Lew bought 12 gumballs, and Carey bought X  [#permalink]

Show Tags

05 Jan 2015, 07:44
We are told that the average of the three is betwee 19 and 75 inclusive. Therefore, the total number of gumballs bought by all three together must be between 57 and 75 inclusive. Since 28 have already been bought by C & L, Carey might have bought between 29 and 47 inclusive. The difference is 18.

_________________
$$\sqrt{-1}$$ $$2^3$$ $$\Sigma$$ $$\pi$$ ... and it was delicious!

Please consider giving +1 Kudos if deserved!
Manager
Joined: 21 Aug 2010
Posts: 172
Location: United States
GMAT 1: 700 Q49 V35
Re: Carolyn bought 16 gumballs, Lew bought 12 gumballs, and Carey bought X  [#permalink]

Show Tags

05 Jan 2015, 17:36
2
Bunuel wrote:

Tough and Tricky questions: Min/Max.

Carolyn bought 16 gumballs, Lew bought 12 gumballs, and Carey bought X gumballs. The average (arithmetic mean) number of gumballs the three bought is between 19 and 25, inclusive. What is the difference between the greatest number and the smallest number of gumballs Carey could have bought?

A. 18
B. 22
C. 24
D. 29
E. 33

Kudos for a correct solution.

C = 16
L = 12
CR = x

Mean(min) = 19 = (C+L+CR) / 3 =>57 = 28+CR => CR = 29
Mean(Max) = 25 = (C+L+CR) / 3 => 75 = 28+CR => CR = 47

Diference = 18

Ans A
_________________
-------------------------------------
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1787
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: Carolyn bought 16 gumballs, Lew bought 12 gumballs, and Carey bought X  [#permalink]

Show Tags

05 Jan 2015, 20:17
1

Smallest gumballs = (19-16) + (19-12) + 19 = 29

Largest gumballs = (25-16) + (25-12) + 25 = 47

Difference = 47 - 29 = 18
_________________
Kindly press "+1 Kudos" to appreciate
Math Expert
Joined: 02 Sep 2009
Posts: 56260
Re: Carolyn bought 16 gumballs, Lew bought 12 gumballs, and Carey bought X  [#permalink]

Show Tags

08 Jan 2015, 09:51
Bunuel wrote:

Tough and Tricky questions: Min/Max.

Carolyn bought 16 gumballs, Lew bought 12 gumballs, and Carey bought X gumballs. The average (arithmetic mean) number of gumballs the three bought is between 19 and 25, inclusive. What is the difference between the greatest number and the smallest number of gumballs Carey could have bought?

A. 18
B. 22
C. 24
D. 29
E. 33

Kudos for a correct solution.

OFFICIAL SOLUTION:

(A) Here we could calculate the possible number of gumballs that Carey bought, but there is a shortcut. Since the average number bought by the three people has a range of 6, the sum of the number of gumballs bought by the three people must have a range of 18.

The correct answer is choice (A).

Alternative Strategy (Algebra):

Let’s find X if the average is 19:
(16 + 12 + X)/3 = 19
28 + X = 57
X = 29.

Now let’s find X if the average is 25:
(16 + 12 + X)/3 = 25
28 + X = 75
X = 47.

The difference between these two values is:
47 – 29 = 18.

Again, we see that the correct answer is choice (A).
_________________
Manager
Joined: 14 Feb 2016
Posts: 71
Re: Carolyn bought 16 gumballs, Lew bought 12 gumballs, and Carey bought X  [#permalink]

Show Tags

22 Dec 2018, 08:55
Here's how I did it. Set-up the inequality as follows and then solve for x.

- 19 < 16 + 12 + x / 3 < 25
- 57 < 28 + x < 75 (x3)
- 29 < x < 47 (-28)

Solve for x:

x = 18.

Can be easily solved in under 2 minutes, ofcourse you're mental calculus game needs to be on point.
_________________
Please provide kudos if you liked my post! I would really like to use the PDF file!!
Re: Carolyn bought 16 gumballs, Lew bought 12 gumballs, and Carey bought X   [#permalink] 22 Dec 2018, 08:55
Display posts from previous: Sort by