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

Question

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

DAllison2016 · Accepted Answer

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{	ext{first term} + 	ext{last term}}{2}

1) The smallest number of them is 5

a = 5\
b = 5 + n\
	ext{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\
	ext{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.

(D) each statement alone is sufficient

MathRevolution · Answer

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).
Therefore, the answer is D.

monikaranchi26 · Answer

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

QuocHuynh · Answer

Hi guys, is it a common belief that when the question only mentions "consecutive integers" it means the step = 1?

because it doesn't mention explicitly so I asssumed there is also a scenario when step = 2 (5,7,9,11,13,15) => 6 numbers, answer E

Bunuel · Answer

When we see "consecutive integers" it ALWAYS means integers that follow each other in order with common difference of 1:  ... x-3, x-2, x-1, x, x+1, x+2, ...

-7, -6, -5 are consecutive integers.

2, 4, 6 ARE NOT consecutive integers, they are  consecutive even integers .

3, 5, 7 ARE NOT consecutive integers, they are  consecutive odd integers .

2, 5, 8, 11 ARE NOT consecutive integers, they are  terms of arithmetic progression with common difference of 3 .

Hope it's clear.

pappal · Answer

consecutive integers are of the form x,x+1,x+2--------x+(n-1) where n=no. of terms
given median=range also equals mean as property of consecutive no.'s
so mean=median=(F+L)/2
=>x+(n-1)/2=n-1=> x=(n-1)/2 and last no.=x+(n-1)=>3/2(n-1)
hence if any of the first or the last numbers are known then 'n' can be found.
hence D

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

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