10 positive integers are arranged in ascending order.

Question

10 positive integers are arranged in ascending order. The range of first seven is 25 and last 7 is 30. What will be the difference in minimum range of the set in following 2 cases
1) when 10 integers are distinct.
2) when the integers can be repeated in the set?

A) 3
B) 5
C) 7
D) 25
E) 30

Kurtosis · Answer

Let's assume the set starts from 1.
Case 1) when 10 integers are distinct.
1st position = 1
4th position = 4
7th position = 26
10th position = 34
Minimum Range of the set = 33

Case 2) when the integers can be repeated in the set.
1st position = 1
4th position = 1
7th position = 26
10th position = 31
Minimum Range of set = 30

Difference in Minimum Range = 33 - 30 = 3

Answer: A

nilboy · Answer

Please elaborate on the answer. How you defined the values in the set below?

Case 1) when 10 integers are distinct.
1st position = 1
4th position = 4
7th position = 26
10th position = 34
Minimum Range of the set = 33

Kurtosis · Answer

You can start with any number. Let's start with 1 so that calculation is easy.

We know that the numbers are distinct and the range must be minimum.

Range of first 7 integers = 25 = 7th position - 1st position --> 7th position = 25 + 1 = 26

Range of last 7 integers = 30 = 10th position - 4th position --> 10th position = 30 + 4th position

10th position's value must be less; so 4th position's value must also be minimum.

Since the integers are arranged in ascending order, 2nd position = 2; 3rd position = 3; 4th position = 4 (4 is the minimum value the 4th position can take to keep the range minimum).

Therefore 10th position = 30 + 4 = 34

Minimum Range = 34 - 1 = 33.

chetan2u · Answer

Explanation:- 
there are 10 positive integers..
so let these numbers be A,B,C,D,E,F,G,H,I,J....

we are given
J-D=30
G-A=25..

lets see the two cases given 
 1) when 10 integers are distinct. 
J-D is 30 and A, B and C are distinct..
For range to be minimum let these 3 (A, B and C) and D be consecutive numbers...
 so Range = 30 +3=33

2) when the integers can be repeated in the set 
J-D=30 and let G-D be 25,
so A,B, and C are same as D, to get the minimum range
 so Range= 30 + 0=30

Difference = 33-30= 3 
A

stonecold · Answer

Excellent Question  chetan2u 
Here is what i did 
let first integer be a
now in the First Case => all integers are distinct => seventh term will be a+25 and to keep the least value for the range => fourth term will be a+3
hence last term will be a+33 so the minimum range in statement 1 is => a+33-a = 33
now in statement 2 => integers can be repeated => fourth term will be a => last term will be a+30 hence the range will be a+30-a =30
so the difference is 3

hence A is correct.

Honestly ; This is such an awesome question => Looking forward to solving more of these

S.C.S.A

bumpbot · Answer

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10 positive integers are arranged in ascending order.

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10 positive integers are arranged in ascending order.

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