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A set of 51 different integers has a median of 30 and a range of 60.

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A set of 51 different integers has a median of 30 and a range of 60.  [#permalink]

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Updated on: 13 Aug 2018, 00:54
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55% (hard)

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58% (02:09) correct 42% (01:55) wrong based on 152 sessions

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Solve any Median and Range question in a minute- Exercise Question #3

A set of 51 different integers has a median of 30 and a range of 60. What is the value of the least possible integer in this set?

Options

a) -10
b) -6
c) -5
d) 5
e) 10

To solve question 4: Question 4

To read the article: Solve any Median and Range question in a minute

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Originally posted by EgmatQuantExpert on 11 Jul 2018, 08:02.
Last edited by EgmatQuantExpert on 13 Aug 2018, 00:54, edited 2 times in total.
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Re: A set of 51 different integers has a median of 30 and a range of 60.  [#permalink]

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11 Jul 2018, 08:13
6
To find the least possible integer in the set of 51 different integers

The median of the set is 30

=> If the numbers are arranged in ascending order, the 26th integer is 30

We know that all the integers to the right of the median are $$\geq$$ median.

Since the set consists of all different integers, the integers to the right of the median are $$>$$ median

Lets make the integers to the right of the median greater than median by least possible value i.e 1

=> value at 27th place will be 31(27+4)
=> value at 28th place will be 32(28+4)
...... => value at 51st place will be 55(51+4)

Given range = 60

=> max_possible_integer - lowest_possible_integer = 60

=> 55 - lowest_possible_integer = 60

=> lowest_possible_integer = 55 - 60 = -5

Hence option C
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Re: A set of 51 different integers has a median of 30 and a range of 60.  [#permalink]

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Updated on: 11 Jul 2018, 21:59
1
1
EgmatQuantExpert wrote:
Solve any Median and Range question in a minute- Exercise Question #3

A set of 51 different integers has a median of 30 and a range of 60. What is the value of the least possible integer in this set?

Options

a) -10
b) -6
c) -5
d) 5
e) 10

To read the article: Solve any Median and Range question in a minute

Given set, Say
X={$$x_{min}$$,.....(25 members including $$x_{min}$$)......,30,.......(25 members including $$x_{max}$$).....,$$x_{max}$$}
---------------------------------------------------(26th term Median) --------------------------------- $$(51^{th} term)$$
Given, $$x_{max} - x_{min}=60$$ ----(1)
Since number of members after median is 25, hence x(max)=30+25=55 (Note:- all terms of the set are different as given)
So, substituting in (1), we have
$$55-x_{min}=60$$
Or, $$x_{min}=-5$$

Therefore, the value of the least possible integer in this set is -5.

Ans. (C)
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Originally posted by PKN on 11 Jul 2018, 13:21.
Last edited by PKN on 11 Jul 2018, 21:59, edited 1 time in total.
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Re: A set of 51 different integers has a median of 30 and a range of 60.  [#permalink]

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11 Jul 2018, 18:38
PKN wrote:
Therefore, the value of the least possible integer in this set is -5.

Ans. (B)

PKN

-5 corresponds to option C and not option B
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Re: A set of 51 different integers has a median of 30 and a range of 60.  [#permalink]

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11 Jul 2018, 23:54
Median is 30...thus there are 25 digits above 30...the smallest number 25 digits above 30 is 55.....the range is 60....therefore the smallest number possible has to be -5...option c for me
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Re: A set of 51 different integers has a median of 30 and a range of 60.  [#permalink]

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12 Jul 2018, 00:14
median is 30 which is the 26th number in the series.
In order to minimize the lowest number, we need to minimize the maximum number.
Because,
(X)(smallest highest number) - (Y)(least negative number) = 60 (to minimize Y)

the smallest highest number possible for this series is 55. Because $$30+25 = 55$$. 25 different integers ahead of the median.

Therefore,

$$55 - x = 60$$
$$x=-5$$
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Re: A set of 51 different integers has a median of 30 and a range of 60.  [#permalink]

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15 Jul 2018, 07:05
1
1

Solution

Given:
• 51 different integers have a median 30 and range 60.

To find:
• The value of least possible integer.

Approach and Working:
We will get the value of least possible integer if then the highest element of the set is equal to the median.
However, all the elements of the set are different.
So, highest element must be as small as possible.
Now, we know, median will be 51+1/2=1= 26 element.
• Hence, minimum value of the highest element= 30+25= 55
• Thus, highest element – lowest element= 60
o 55- Lowest element= 60
o Lowest element= -5

Hence, the correct answer is option C.

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Re: A set of 51 different integers has a median of 30 and a range of 60.   [#permalink] 15 Jul 2018, 07:05
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