Last visit was: 21 Jul 2024, 01:23 It is currently 21 Jul 2024, 01:23
Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Math Expert
Joined: 02 Sep 2009
Posts: 94433
Own Kudos [?]: 642638 [18]
Given Kudos: 86715
Send PM
Most Helpful Reply
Math Expert
Joined: 02 Sep 2009
Posts: 94433
Own Kudos [?]: 642638 [6]
Given Kudos: 86715
Send PM
General Discussion
avatar
Intern
Intern
Joined: 09 Mar 2016
Posts: 2
Own Kudos [?]: 1 [0]
Given Kudos: 1
Send PM
Math Expert
Joined: 02 Sep 2009
Posts: 94433
Own Kudos [?]: 642638 [2]
Given Kudos: 86715
Send PM
Re: M13-16 [#permalink]
1
Kudos
1
Bookmarks
Expert Reply
Channarong22 wrote:
Bunuel : Why we minimize C=10???


We have that a + b = 16. Since a and b are distinct integers, then a = b = 8 is not possible, so the least value of b is 9 (7 + 9 = 16). Hence the least value of c is 10.
Math Expert
Joined: 02 Sep 2009
Posts: 94433
Own Kudos [?]: 642638 [1]
Given Kudos: 86715
Send PM
Re: M13-16 [#permalink]
1
Bookmarks
Expert Reply
I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
Intern
Intern
Joined: 03 May 2023
Posts: 22
Own Kudos [?]: 2 [0]
Given Kudos: 0
Location: Indonesia
Schools: Sloan '26
GRE 1: Q166 V163
GPA: 3.01
Send PM
Re: M13-16 [#permalink]
Hi Bunuel
Just a thought
I follow your logic and get that \(d=14\) and \(c = 10\), and I answered correctly. But I do notice some edge cases here

However, if we consider that \(a+b=2*8=16\), the distinct possibilities for the solutions for \(a + b\) are:
\([1, 15]\)
\([2, 14]\)
.
.
.
\([7, 9]\)

if we consider, for example, \(a = 1, b = 15\), which satisfy \(a + b = 16\), then \(c\) MUST BE greater than 15. I.E, \(c = 10\), the premise of If the average of the smaller two of these four integers is 8 breaks down, since now the two smallest integers are \(a = 1, c = 10\), and the average IS NOT 8.

Wouldn't it be better if the question stem explicitly say \(a < b < c < d\)? I know it not absolutely necessary, just to cross out said edge case

Thanks

Bunuel wrote:
Official Solution:


The average (arithmetic mean) of four distinct positive integers is 10. If the average of the smaller two of these four integers is 8, which of the following represents the maximum possible value of the largest integer?


A. 12
B. 14
C. 15
D. 16
E. 17


Assume the four distinct positive integers are \(a, b, c\), and \(d\), such that \(0 < a < b < c < d \).

Given that the average of these integers is 10, we know that \(a+b+c+d=4*10=40\);

Furthermore, we're told that the average of the smallest two of these integers is 8, which means that \(a+b=2*8=16\). So, \(16+c+d=40\), and thus \(c+d=24\).

To find the maximum possible value for \(d\), we need to minimize \(c\). The smallest value of \(c\) can be 10, for \(a=7\) and \(b=9\) (this is because \(a + b = 16\) and since \(a\) and \(b\) must be distinct integers, it's impossible for both to be 8. Thus, the smallest possible value for \(b\) is 9). So, with the smallest possible value for \(c\) as 10, we have \(10+d=24\) which gives us \(d=14\).


Answer: B
Math Expert
Joined: 02 Sep 2009
Posts: 94433
Own Kudos [?]: 642638 [1]
Given Kudos: 86715
Send PM
Re: M13-16 [#permalink]
1
Kudos
Expert Reply
r_putra_rp wrote:
Hi Bunuel
Just a thought
I follow your logic and get that \(d=14\) and \(c = 10\), and I answered correctly. But I do notice some edge cases here

However, if we consider that \(a+b=2*8=16\), the distinct possibilities for the solutions for \(a + b\) are:
\([1, 15]\)
\([2, 14]\)
.
.
.
\([7, 9]\)

if we consider, for example, \(a = 1, b = 15\), which satisfy \(a + b = 16\), then \(c\) MUST BE greater than 15. I.E, \(c = 10\), the premise of If the average of the smaller two of these four integers is 8 breaks down, since now the two smallest integers are \(a = 1, c = 10\), and the average IS NOT 8.

Wouldn't it be better if the question stem explicitly say \(a < b < c < d\)? I know it not absolutely necessary, just to cross out said edge case

Thanks

Bunuel wrote:
Official Solution:


The average (arithmetic mean) of four distinct positive integers is 10. If the average of the smaller two of these four integers is 8, which of the following represents the maximum possible value of the largest integer?


A. 12
B. 14
C. 15
D. 16
E. 17


Assume the four distinct positive integers are \(a, b, c\), and \(d\), such that \(0 < a < b < c < d \).

Given that the average of these integers is 10, we know that \(a+b+c+d=4*10=40\);

Furthermore, we're told that the average of the smallest two of these integers is 8, which means that \(a+b=2*8=16\). So, \(16+c+d=40\), and thus \(c+d=24\).

To find the maximum possible value for \(d\), we need to minimize \(c\). The smallest value of \(c\) can be 10, for \(a=7\) and \(b=9\) (this is because \(a + b = 16\) and since \(a\) and \(b\) must be distinct integers, it's impossible for both to be 8. Thus, the smallest possible value for \(b\) is 9). So, with the smallest possible value for \(c\) as 10, we have \(10+d=24\) which gives us \(d=14\).


Answer: B


All is correct both in the question and the solution. To maximize d, we need to to minimize c, for that we need to minimize b. The minimum value of b such that 0 < a < b and a + b = 16, is 9 (since all are integers).
Intern
Intern
Joined: 24 Dec 2022
Posts: 48
Own Kudos [?]: [0]
Given Kudos: 9
Send PM
Re M13-16 [#permalink]
I think this is a high-quality question and I agree with explanation.
GMAT Club Bot
Re M13-16 [#permalink]
Moderators:
Math Expert
94433 posts
Founder
37858 posts