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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. 23B. 21C. 25D. 26E. 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=60a ≤ b ≤ cb=a+11c=60-(2a+11)b ≤ ca+11 ≤ 60-(2a+11)3a ≤ 38a ≤ 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. 23B. 21C. 25D. 26E. 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 geta+a+11+a+11=60 i.e a = 38/3 a non integerlet take c=b+1 from (2) we get c=a+12putting values in (1) we geta + (a + 11) + (a + 12) = 60 i.e a= 37/3 again a non integerlet take c= b+2putting values in (1) we geta + (a + 11) + (a + 13) = 60 i.e. a=36/3 = 12 b = 23c = 25so 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

 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 INTEGERS2) Their average is 20 3) A ≤ B ≤ C4) 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 = 23A+A+11 = 372A = 26A = 13The 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 = 24A+A+11 = 362A = 25A = 12.5This is NOT possible (the variables have to be INTEGERS)ELIMINATE 24.Let's TEST 25...IF....C = 25A+A+11 = 352A = 24A = 12The 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 = 20a = 9Then, form a grid and list down possible values.a b c9 20 3110 21 2911 22 2712 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. 23B. 21C. 25D. 26E. 24to minimize c, we have to maximize awe can suppose that we have: a+a+11+a+11=20*33a+22=603a=3838/3 is not an integer, thus, minimum value of c must be greater than a+11a+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. 23B. 21C. 25D. 26E. 24to minimize c, we have to maximize awe can suppose that we have: a+a+11+a+11=20*33a+22=603a=3838/3 is not an integer, thus, minimum value of c must be greater than a+11a+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 = 12.67Here 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. 23B. 21C. 25D. 26E. 24Responding 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 + 11What 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 = 23This is 8 below and 3 above the average, so we need 5 more above the average = 25.

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