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Re: A set of 25 different integers has a median of 50 and a [#permalink]
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Excellent Official Question.
Here is what i did in this one=>

Median => 13th element = 50


Now range =50.
Let a and b be the minimum value and maximum value in set.

As range is given=> To maximise b => We must maximise a.
As the integers involved are different and all the element to the left of the median must be less than or equal to it=>
50
49
48
47
46
45
44
43
42
41
40
39
38


Hence the maximum value of a will be 38

Now b-38=50(Range)
Hence b=50+38=88

Hence D


Note=>If the question didn't mention that all integers are different that a would have been 50 and b would have been 100
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Re: A set of 25 different integers has a median of 50 and a [#permalink]
enigma123 wrote:
I think Bunuel , you meant "We want to maximize \(x_{25}\), hence we need to minimize \(x_{1}\). I still don't understand how did you get the below:

Since all integers must be distinct then the maximum value of \(x_{1}\) will be \(median-12=50-12=38\)



How did we assue the least number to be 50-12? why it can't be 0 or 1 or 2 or 34 for that matter?
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Re: A set of 25 different integers has a median of 50 and a [#permalink]
I didn't read the word "different" and assumed that the first 13 numbers were 50. Hence, applying a range of 50, I reached the highest number as 100.

100 was present as an answer choice and hence, I immediately chose it. Amazing how the question paper setters lay these traps.
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Re: A set of 25 different integers has a median of 50 and a [#permalink]
I have done silly mistake while solving this question and marked option A... I have calculated the first element of the series right, but then messed up in adding the value 50 to the first number of the set...
X+12=50..
Number N should be X+50 not X+24

Regards
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Re: A set of 25 different integers has a median of 50 and a [#permalink]
enigma123 wrote:
A set of 25 different integers has a median of 50 and a range of 50. What is the greatest possible integer that could be in this set?

(A) 62
(B) 68
(C) 75
(D) 88
(E) 100

Any idea how to solve this question please?


Given: A set of 25 different integers has a median of 50 and a range of 50.

Asked: What is the greatest possible integer that could be in this set?

To maximise greatest possible integer, all other numbers need to be minimised.

Let the number be {38,39,40,41,42,43,44,45,46,47,48,49,50,,,,,,,,,,,,,,,38+50}

Greatest possible integer = 38+50 = 88

IMO D
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Re: A set of 25 different integers has a median of 50 and a [#permalink]
The phrase "different integers" changes the conditions. if it there was not we would say 100 is the answer but now option D is the answer.

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Re: A set of 25 different integers has a median of 50 and a [#permalink]
MY LOGIC

MEDIAN 50 RANGE 50 No of terms =25 MEDIAN >>13TH TERM

Z,X,C,V,B,B,,N,,F,D,S,A,(50),R,T,Y,U,I,O,P,Q,G, 12 TERMS ON EITHER SIDE SO

50-12=38 MUST BE THE FIRST TERM TO GET THE MAX..... SO RANGE=HT-LT>>>50=HT-38

*HT=88*

```D```

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Re: A set of 25 different integers has a median of 50 and a [#permalink]
50-12=38 , 38+50= 88
ans d
life is short dont waste your time
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Re: A set of 25 different integers has a median of 50 and a [#permalink]
enigma123 wrote:
A set of 25 different integers has a median of 50 and a range of 50. What is the greatest possible integer that could be in this set?

(A) 62
(B) 68
(C) 75
(D) 88
(E) 100

Any idea how to solve this question please?


value of median is 50 which would be 12th term from 25 terms so 1st term is 38
largest value = 50+38 ; 88
option D
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Re: A set of 25 different integers has a median of 50 and a [#permalink]
We have 25 DIFFERENT Integers.

50 is the median -> This means that there are 24 different integers lower than 50, and 24 integers higher than 50.

To maximize the largest possible value, we need to minimize the least value in the list by subtracting 1 24 times from 50.
-> 50-12 = 38

To get maximum value, add the range to the minimum value

Max value = 38 + 50

Answer is 88
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Re: A set of 25 different integers has a median of 50 and a [#permalink]
Hi Bunuel

https://gmatclub.com/forum/seven-pieces ... 44452.html

In the aforesaid question, you said said "to maximize g, we need to minimize all other terms"

But in the present question, you have stated We want to maximize \(x_{25}\), hence we need to maximize \(x_{1}\).

Can you please help me with the logic in such questions?

Bunuel wrote:
enigma123 wrote:
A set of 25 different integers has a median of 50 and a range of 50. What is the greatest possible integer that could be in this set?

(A) 62
(B) 68
(C) 75
(D) 88
(E) 100

Any idea how to solve this question please?



Consider 25 numbers in ascending order to be \(x_1\), \(x_2\), \(x_3\), ..., \(x_{25}\).

The median of a set with odd number of elements is the middle number (when arranged in ascending or descending order), so the median of given set is \(x_{13}=50\);

The range of a set is the difference between the largest and the smallest numbers of a set, so the range of given set is \(50=x_{25}-x_{1}\) --> \(x_{25}=50+x_{1}\);

We want to maximize \(x_{25}\), hence we need to maximize \(x_{1}\). Since all integers must be distinct then the maximum value of \(x_{1}\) will be \(median-12=50-12=38\) and thus the maximum value of \(x_{25}\) is \(x_{25}=38+50=88\).

The set could be {38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 88}

Answer: D.

Hope it's clear.
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Re: A set of 25 different integers has a median of 50 and a [#permalink]
The key point is that A Set is a Collection that cannot contain duplicate elements.
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Re: A set of 25 different integers has a median of 50 and a [#permalink]
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