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Three positive integers a, b, and c are such that their average is 20
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Author:  Bunuel [ 30 Jan 2016, 01:58 ]
Post subject:  Three positive integers a, b, and c are such that their average is 20

Three positive integers a, b, and c are such that their average is 20 and a ≤ b ≤ c. If the median is (a + 11), what is the least possible value of c?

A. 23
B. 21
C. 25
D. 26
E. 24

Author:  DmitryFarber [ 30 Jan 2016, 23:38 ]
Post subject:  Re: Three positive integers a, b, and c are such that their average is 20

We can solve algebraically without testing values:

a+b+c=60
a ≤ b ≤ c
b=a+11
c=60-(2a+11)

b ≤ c
a+11 ≤ 60-(2a+11)
3a ≤ 38
a ≤ 12 (since it must be an integer)

So the maximum values of a & b are 12 and 23, making the minimum value of c 25.

Author:  cutegirlsimran [ 30 Jan 2016, 23:02 ]
Post subject:  Re: Three positive integers a, b, and c are such that their average is 20

Three positive integers a, b, and c are such that their average is 20 and a ≤ b ≤ c. If the median is (a + 11), what is the least possible value of c?

A. 23
B. 21
C. 25
D. 26
E. 24

Solution: (a+b+c)/3 = 20 so a+b+c=60 --------------- (1)
a<=b<=c
median = a+11

b is the middle value so median is b
we have b = a + 11 ----------------------------(2)

the least value of c is when c=b so we get c= a+ 11 ------(3)

putting values in (1) we get
a+a+11+a+11=60 i.e a = 38/3 a non integer

let take c=b+1 from (2) we get c=a+12
putting values in (1) we get
a + (a + 11) + (a + 12) = 60 i.e a= 37/3 again a non integer

let take c= b+2
putting values in (1) we get
a + (a + 11) + (a + 13) = 60 i.e.

a=36/3 = 12
b = 23
c = 25

so C must be the correct answer

Author:  FightToSurvive [ 30 Jan 2016, 02:55 ]
Post subject:  Re: Three positive integers a, b, and c are such that their average is 20

a+b+c=60 & a<b<c

By hit and trial, c=25 gives b=23 and a = 12

Hence C=25.

Author:  EMPOWERgmatRichC [ 01 Feb 2016, 10:28 ]
Post subject:  Re: Three positive integers a, b, and c are such that their average is 20

Hi All,

As 'thick' as this question might appear, it can be solved with a bit of 'brute force' and a combination of basic arithmetic skills. We can use the answer choices and TEST THE ANSWERS, but we'll also need to 'play around' a bit with the information that we're given (so we can determine what answers are possible and what answers are not).

We're given a number of facts about the variables A, B and C:
1) They're POSITIVE INTEGERS
2) Their average is 20
3) A ≤ B ≤ C
4) The median is (A + 11)

We're asked for the LEAST possible value of C.

To start, since there are 3 variables, we know that B is the median. So we can think of the three variables as A, (A+11) and C. By extension, we know that A is NOT 'close' to C; this means that A is probably a lot less than the average and C is well ABOVE the average.

The average of the variables is 20, so their sum is 60.

Let's TEST 23...

IF....
C = 23
A+A+11 = 37
2A = 26
A = 13
The numbers would be 13, 24 and 23...
This is NOT possible (B ≤ C is not true in this case)
ELIMINATE 23.

Let's TEST 24...

IF....
C = 24
A+A+11 = 36
2A = 25
A = 12.5
This is NOT possible (the variables have to be INTEGERS)
ELIMINATE 24.

Let's TEST 25...

IF....
C = 25
A+A+11 = 35
2A = 24
A = 12
The numbers would be 12, 23 and 25...
This MATCHES everything we were told, so this MUST be the answer.

Final Answer:

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

Author:  Donnie84 [ 02 Feb 2016, 22:33 ]
Post subject:  Re: Three positive integers a, b, and c are such that their average is 20

According to the condition, a, b and c can be equal to each other so let them be. They will form an equidistant set whose average = median = b = 20.

a + 11 = 20
a = 9

Then, form a grid and list down possible values.

a b c
9 20 31
10 21 29
11 22 27
12 23 25 -> bingo!
13 24 23 -> rejected because it will violate main condition (a <= b <= c). We can stop here.

Answer (C).

Author:  mvictor [ 04 May 2016, 18:41 ]
Post subject:  Re: Three positive integers a, b, and c are such that their average is 20

Bunuel wrote:
Three positive integers a, b, and c are such that their average is 20 and a ≤ b ≤ c. If the median is (a + 11), what is the least possible value of c?

A. 23
B. 21
C. 25
D. 26
E. 24


to minimize c, we have to maximize a
we can suppose that we have: a+a+11+a+11=20*3
3a+22=60
3a=38
38/3 is not an integer, thus, minimum value of c must be greater than a+11
a+12 - doesn't work (37/3)
a+13 - works (36/3=12)
a=12, b=23, c=25.
minimum value of c is 25.

Author:  KarishmaB [ 04 May 2016, 21:28 ]
Post subject:  Re: Three positive integers a, b, and c are such that their average is 20

mvictor wrote:
Bunuel wrote:
Three positive integers a, b, and c are such that their average is 20 and a ≤ b ≤ c. If the median is (a + 11), what is the least possible value of c?

A. 23
B. 21
C. 25
D. 26
E. 24


to minimize c, we have to maximize a
we can suppose that we have: a+a+11+a+11=20*3
3a+22=60
3a=38
38/3 is not an integer, thus, minimum value of c must be greater than a+11
a+12 - doesn't work (37/3)
a+13 - works (36/3=12)
a=12, b=23, c=25.
minimum value of c is 25.


Good logic.

Assume the numbers are a, a+11 and a+11 (to minimize the greatest value c)
You get a = 38/3 = 12.67
Here itself, you see that a cannot be 13 since that will add up to a higher value than 60. So a must be 12 - the highest integer value it can take.
This gives c = 60 - 12 - 12 - 11 = 25

Author:  Shishou [ 23 Aug 2019, 08:56 ]
Post subject:  Re: Three positive integers a, b, and c are such that their average is 20

This means the median is b=a+11. Since the sum is 60,a=(60-11-c)/2. The only value of c that gives us an integer value for a and makes the median a+11 is 25. Answer is C

Author:  KarishmaB [ 06 Jan 2021, 00:57 ]
Post subject:  Re: Three positive integers a, b, and c are such that their average is 20

Bunuel wrote:
Three positive integers a, b, and c are such that their average is 20 and a ≤ b ≤ c. If the median is (a + 11), what is the least possible value of c?

A. 23
B. 21
C. 25
D. 26
E. 24


Responding to a pm:

Quote:
I was wondering where the problem was with my reasoning. In my head to minimise the median we can have b= a+11 and a and c equal a+10 and a+12. This would be the final order a+10, a+11 and a+12. This should minimise c more than just a, a+11 and a+13. If you set this equal to 60 you find that a=9. Therefore, you get that 19, 20 and 21. This satisfies all the conditions and the lowest value is 21. I don't understand how I can satisfy all the conditions and get an answer which according to the OA is wrong.


We need to minimise c.

Since median is 'a + 11', and since b is the middle value as per a ≤ b ≤ c, we know that 'b' is the median. So
b = a + 11
What this means is that 'b' is 11 more than 'a'. So in terms of 'a', we can write 'b = a + 11'.

The values will be a, a+11, c.
Note that 'a' cannot be 'a + 10'. What you are saying here is
a = a + 10
How is that possible? This gives 0 = 10 which is not true.

Now since we want to minimise c, we can give it the same value as b i.e. 'a + 11' because b ≤ c.
This doesn't give us an integer so we take the smallest value of c that does give us an integer.

Author:  Bambi2021 [ 25 Jun 2021, 18:28 ]
Post subject:  Re: Three positive integers a, b, and c are such that their average is 20

When a becomes 38/3, it means a must be less than 13.

If a = 12, then b = 23

This is 8 below and 3 above the average, so we need 5 more above the average = 25.

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