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# When there are consecutive integers and if their range is equal to the

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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When there are consecutive integers and if their range is equal to the [#permalink]

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29 Feb 2016, 19:39
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When there are consecutive integers and if their range is equal to their median, what is the number of them?

1) The smallest number of them is 5
2) The largest number of them is 15

* A solution will be posted in two days.
[Reveal] Spoiler: OA

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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Re: When there are consecutive integers and if their range is equal to the [#permalink]

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02 Mar 2016, 21:51
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

When consecutive integers are there and if their range is equal to their median, what is the number of them?

1) The smallest number of them is 5
2) The largest number of them is 15

In the original condition, there are 2 variables(you need to figure out the starting number and the number of them) and 1 equation(range=median), which should match with the number of equations. So you need 1 variable. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
For 1), only 5,6,7,8,9,10,11,12,13,14,15 are possible and so as 2).
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Re: When there are consecutive integers and if their range is equal to the [#permalink]

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04 Aug 2016, 13:17
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MathRevolution wrote:
When there are consecutive integers and if their range is equal to their median, what is the number of them?

Let $$a$$ be the first element of the set, $$b$$ be the final element of the set, and $$n$$ be the range.

The median of a set of consecutive numbers is the same as the mean: $$\frac{\text{first term} + \text{last term}}{2}$$

1) The smallest number of them is 5

$$a = 5\\ b = 5 + n\\ \text{median} = \frac{a + b}{2} = \frac{10 + n}{2} = n\\ 2n = 10 + n\\ n = 10\\$$

Sufficient

2) The largest number of them is 15

$$b = 15\\ a = 15 - n\\ \text{median} = \frac{a + b}{2} =\frac{15 + 15 - n}{2} = n\\ 30 - n = 2n\\ 30 = 3n\\ n = 10$$

Sufficient

Knowing $$n$$ and a start point, there are 11 elements from 5 to 15.

[Reveal] Spoiler:
(D) each statement alone is sufficient

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Re: When there are consecutive integers and if their range is equal to the [#permalink]

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17 Sep 2017, 12:14
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Re: When there are consecutive integers and if their range is equal to the [#permalink]

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25 Sep 2017, 20:02
This seems to be easier if we apply the basic logic for consecutive integers.

We know that for a set of consecutive integers, mean=median.
Consider the first term as 'a' and last term as 'b'.

Range= b-a
Median=Mean= (a+b)/2.
Given: Range=Median or Mean
So, we can say that
b-a = (a+b)/2
b = 3a or a = b/3

Statement 1 : given smallest no is 5 or a = 5
So, b= 3a = 3*5 = 15.
Hence, we get a set {5,6,7,.....,15} which will satisfy all given conditions
Statement 1 is sufficient.

Statement 2: Similarly, given largest no is 15 or b=15
So, a=b/3 = 5
Hence, we get the same set again as {5,.......,14,15}.
Statement 2 is sufficient.

Ans :D
Re: When there are consecutive integers and if their range is equal to the   [#permalink] 25 Sep 2017, 20:02
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