Question

Find three consecutive whole numbers whose sum is more than \[45\] but less than \[54\].

Answer 1

Hint: Whole numbers start from zero and end up to infinity. The numbers that follow each other continuously in order from smallest to largest are called consecutive numbers for examples \[1,2,3,4..............\] etc.
The properties of whole numbers are based on arithmetic operations such as addition, subtraction, division, and multiplication. Two whole numbers if added or multiplied will give a whole number itself. Subtraction of two whole numbers may not result in whole numbers, i.e. it can be an integer too. Also, the division of two whole numbers results in getting a fraction in some cases.

Complete step by step answer:

Let three consecutive whole number are x, x+1 and \[x + 2\]
It is given that sum is more than \[45\] and less than \[54\]
So,
\[45 < x + x + 1 + x + 2 < 54\]
\[45 < 3x + 3 < 54\]
Subtracting 3 to each sides
\[45 - 3 < 3x + 3 - 3 < 54 - 3\]
\[42 < 3x < 51\]
Dividing by 3 each side
\[\dfrac{{42}}{3} < \dfrac{{3x}}{3} < \dfrac{{51}}{3}\]
\[14 < x < 17\]
It means that the value of \[x\] is more than \[14\] but less than. So, the value of \[x\] is either \[15\,or16\]. If \[x = 15\], then three consecutive numbers are \[15,16,17\].
If \[x = 16\], then three consecutive numbers are \[16,17,18\]

Note: This question can also be solved as taking three number as \[x - 1,x.x + 1\]
So, \[3x > 45\] and \[3x < 54\] ……………….(1)
By satisfying equation 1, we get \[3x = 48\,or\,3x = 51\]
\[ \Rightarrow x = 16\] numbers are \[15,16,17\]
Or \[x = 17\] numbers are \[16,17,18\]