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

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Math Expert
Joined: 02 Sep 2009
Posts: 52231
Carolyn bought 16 gumballs, Lew bought 12 gumballs, and Carey bought X  [#permalink]

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05 Jan 2015, 06:00
00:00

Difficulty:

15% (low)

Question Stats:

86% (01:25) correct 14% (01:36) wrong based on 148 sessions

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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

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Re: Carolyn bought 16 gumballs, Lew bought 12 gumballs, and Carey bought X  [#permalink]

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05 Jan 2015, 06:44
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.

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Re: Carolyn bought 16 gumballs, Lew bought 12 gumballs, and Carey bought X  [#permalink]

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05 Jan 2015, 16:36
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
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Re: Carolyn bought 16 gumballs, Lew bought 12 gumballs, and Carey bought X  [#permalink]

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05 Jan 2015, 19:17

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

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

Difference = 47 - 29 = 18
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Re: Carolyn bought 16 gumballs, Lew bought 12 gumballs, and Carey bought X  [#permalink]

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08 Jan 2015, 08:51
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).
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Re: Carolyn bought 16 gumballs, Lew bought 12 gumballs, and Carey bought X  [#permalink]

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22 Dec 2018, 07:55
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.
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Re: Carolyn bought 16 gumballs, Lew bought 12 gumballs, and Carey bought X &nbs [#permalink] 22 Dec 2018, 07:55
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