Question

Three consecutive multiples of $3$ have a sum of $36$ . What is the greatest number?

Answer 1

Hint: We have been given that three consecutive multiples of $3$ have a sum of $36$ .
So here, we will first assume the three consecutive multiples as $3x,3(x + 1)$ and $3(x + 2)$ .
After this we will add these three consecutive numbers equating to their sum i.e. $36$ and then we will find the value of $x$ , from which we can find the greatest number.

Complete step by step answer:
We have assumed the first multiple of $3$ be
$3x$ .
The second multiple is $3(x + 1)$ and the third multiple is
$3(x + 2)$ .
It is given that their sum is $36$. So we can write the equation as follow:
$3x + 3(x + 1) + 3(x + 2) = 36$
We will now add the values and simplify this:
$ \Rightarrow 3x + 3x + 3 + 3x + 6 = 36$
$ \Rightarrow 9x + 9 = 36$
We will divide the L.H.S and R.H.S by $9$:
$ \Rightarrow \dfrac{{9x}}{9} + \dfrac{9}{9} = \dfrac{{36}}{9}$
On simplifying the value, we have:
 $ \Rightarrow x + 1 = 4$
So it gives us
 $x = 4 - 1 = 3$
Now we will substitute the value $x = 3$ in the multiples. So the first multiple $3x$ can be written as
$3 \times 3 = 9$
The second multiple is $3(x + 1)$:
$3(3 + 1) = 3 \times 4$
So the value of second multiple is
 $12$
Similarly, in the third multiple $3(x + 2)$, we have:
$3(3 + 2)$
It gives us value:
 $3 \times 5 = 15$ .
Therefore we can see that out of all three multiples, the greatest number is $15$
Hence the required answer is $15$ .

Note:
We should also cross check our answer by adding the numbers i.e. $9 + 12 + 15$ . On adding it gives us value $36$ which is equal to the sum given in the question.
We should know that we represent the three consecutive numbers as $x,(x + 1),(x + 2)$ .
Here we have to find the multiples of $3$ , so we have multiplied the multiples by $3$.