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Three positive integers a, b, and c are such that their average is 20 https://gmatclub.com/forum/threepositiveintegersabandcaresuchthattheiraverageis212496.html 
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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=(6011c)/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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