Problem Solving

@Zerks: Median is not the middle number, by number, in the set. Median is, after arranging the numbers of the set in descending order, the middle number, by value. I hope you understand this.

When you say 45 is the highest number, the lowest number must be 15 (range is 30). Agreed? All this means is after arranging 13 different integers in an order, and assuming highest and lowest ones to be 45 and 15 respectively, with 30 as median, there must be 6 integers whose values are lower than 30 but higher than 15. This condition can certainly be true with the highest value as 45.

Let's do the same with 54. Lowest number now must be 24. Median still remains 30. Again 6 mumbers must be below 30 and above 23. So the numbers can be 24, 25, 26, 27, 28, 29 (6 total). 30 would be a center or 7th number and then you can put any value to the remaining 6 that are above 30 but below 54.

As we saw, even 54 yields a correct answer. You pick the HIGHEST POSSIBLE VALUE, therefore 54 and not 45.

I hope you understood my exaplanation.

i had no idea how to solve these type of questions but after having look at the solutions provided by all of you i learnt a new method and they way to solve this question. thanks everyone

answer can be C only.
highest possible value can be 54 only.
since max value = min value + Range.
Range is constant, so only thing we can maximise is the min value which can be only 24.

7th term = 30
all terms are different
hence the terms must start from 30-6 = 24 making 30 the seventh term
range = 30=> max term = 24+30 = 54

median is 7th term. range = 30
Min number = 30-6 = 24
Max = 24+30 = 54

C !

fell for the trap did not notice "Different"

For me the answer must be AMO C.

=> We have 13 DIFFERENT intergers
=> The median is 30
=> Since 13 is odd, the median must be the 7th number of the set this means that the 6 smallest numbers are less than 30 (being different) and the 6 highest number are greater than 30.

=> The greatest smallest number of the set is 24 (The 7 smallest numbers of the set being {24;25;26;27;28;29;30;...}

=> The range of the set being 30, the greatest number of the set is then 24+30 = 54

AMO C

The OA please

(E) 60

Median = 30 (7th Value).
Range = 30 (Last - First).
24, 25, 26, 27, 28, 29, 30, _, _, _,_, _, Last

Remember that we are arranging the number in ascending order and not necessarily equally spaced.
Therefore if the highest first number is 24, therefore the Highest Last number will surely be 54.

24, 25, 26, 27, 28, 29, 30, _, _, _,_, _, 54

This is the way I would solve. If I am going wrong somewhere please correct.

Thanks
Ritesh

mabutls wrote:This my solution:
Call 13 different integers in order is a1<a2<a3<...a13 so we have:

a13 - a1= 30 and a1+a2+...a13=30*13=390

Note that all 13 integers in the set are different so a2>=a1+1; a3>=a2+1>=a1+2;...;a12>=a1+11
That's why we can induce a1+...+a12+a13>=a1+(a1+1)+(a1+2)+...+(a1+11)+(a1+30) = 13a1+96
SO a1=<(390-96)/13= 23
at the end a13=<53 ----> answer B

The Bold line is the mistake. The calculation is valid in case of mean and not median.

IMO C

A set of 13 different integers has a median of 30 and a range of 30. What is the greatest possible integer that could be in this set?

(A) 36
(B) 43
(C) 54
(D) 57
(E) 60
Median means 50% above and 50% below the median value.
(A) 36
(B) 43
(C) 54 Possible bcz the smallest no. should be 24 and the other difft. nos may be 25,26,27,28,29.
(D) 57 Not possible bcz the smallest no. must be 27, then other nos. before the median value should be difft and less that 30.
(E) 60 Not possible bcz if 60 is the greatest number then smallest no. must be 30.

So,the final Candidate is C

arora007 wrote:can't this be a 60 ?? E ??


twelve 30s in a row...and then 30+30 which gives a range of 30.


correct me if i am wrong!!

since it is mentioned in question as 13 different integers. we can't have 12 30's