Basic Quantitative Example Question-03

Basic Quantitative Example Question-03:

Question:

If sum of 15 consecutive integers is 450, what is the greatest of those 15 consecutive integers?

Solution:

Always use the General form of consecutive integers and equate the sum of those 15 consecutive integers to the given sum (i.e. 450).

n   +   n+1   +   n+2   +   n+3   +   ……   +   (15^th integer)   = 450

Note that the 15^th integer will NOT be n+15, because:

1^st integer is n (i.e. n+0),
2^nd integer is n+1,
3^rd integer is n+2,
  .       .       .       .
  .       .       .       .
  .       .       .       .
  .       .       .       .

Thus, 15^th integer would be n+14, rather than n+15.

As it is given that the sum of the fifteen consecutive integers is 450, so, you have:

n   +   n+1   +   n+2   +   n+3   +   ……   +   n+14   = 450

Here, we recommend you to simply use the following procedure to add sum from n to n+14:

First, you need to remember that for any evenly spaced series/set (i.e. equally spaced series of numbers), the average can be find using the below equation:

Average = ^{(Smallest number + Largest number)}⁄₂

This means, if you have a series of evenly spaced numbers (i.e. series of numbers in which any two adjacent numbers have same difference or common difference), you may find the average simply by taking the average of the least and greatest of those numbers. For instance,
The average of integers from 1 to 8 inclusive is:

Traditionally, Average = ^Sum⁄_{Number of integers} = ^{(1+2+3+4+5+6+7+8)}⁄₈ = ³⁶⁄₈ = 4.5

For evenly spaced series of numbers:

Alternatively, Average = ^{(Smallest number + Largest number)}⁄₂ = Average = ^{(1 + 8)}⁄₂ = Average = ⁹⁄₂ = 4.5

Important Note: This short way to find average is only applicable for evenly spaced series. If the series of integers are not evenly spaced, then you cannot use this short way to find average. In that case you have to use the traditional way, i.e. sum divided by numbers. But when you see an evenly spaced series, simply use this short way and save couple of seconds.

So, in similar way we can quickly find the average of the 15 consecutive integers as below:

Average = ^{(Smallest number + Largest number)}⁄₂ = Average = ^{(n + n+14)}⁄₂ = n + 7

As we know that:

Average = ^Sum⁄_{Number of integers}

➩  Sum = Average × Numbers

➩   450 = (n + 7) × 15

➩ n + 7 = 30

➩       n = 23

At this stage, just put the value of n in the supposed 15 consecutive integers (i.e. General form of consecutive integers):

n,       n+1,       n+2,       n+3,       ……       n+14

As you know the largest of these 15 consecutive integers is n+14

By substituting value of n,

➩ Greatest integer = 23 + 14 = 37 Answer

Also, you may answer any question from this series. For instance, if you are asked to determine the sum of first three consecutive integers of those 15 consecutive integers, simply add first three integers of our general form of consecutive integers i.e.

n,       n+1,       n+2,       n+3,       ……       n+14

➩ Sum of first three integers = n   +   n+1   +   n+2

Buy substituting value of n

➩ Sum of first three integers = 23   +   23+1   +   23+2
➩ Sum of first three integers = 23   +   24   +   25
➩ Sum of first three integers = 72 Answer

similarly, you may answer any question just by using general form of consecutive integers:
n,       n+1,       n+2,       n+3,       ……       n+14

And then plugin value of n where required:

23,     23+1,     23+2,     23+3,     ……     23+14

In order to see another example, click on the “Example Question-04” button below: