Question
Find three consecutive whole numbers whose sum is more than 45 but less than 54.
✓
Answer
Let the three consecutive whole numbers be $(x-1)$, $x$ and $(x+1)$.
$\therefore$ Sum of the three numbers
$=(x-1)+x+(x+1)$
$=3 x$
Given that, the sum of the three numbers is greater than 45 and less than 54 .
When the sum of the three numbers is 45 ,
$
\begin{aligned}
& 3 \mathrm{x}=45 \\
& \therefore \mathrm{x}=\frac{45}{3} \\
& \therefore \mathrm{x}=15
\end{aligned}
$
When the sum of the three numbers is 54,
$
\begin{aligned}
& \therefore 3 \mathrm{x}=54 \\
& \therefore \mathrm{x}=\frac{54}{3} \\
& \therefore \mathrm{x}=18
\end{aligned}
$
$\therefore$ the value of $\mathrm{x}$ is greater than 15 and less than 18 .
$\therefore$ the value of $\mathrm{x}$ is either 16 or 17
$\therefore$ Sum of the three numbers
$=(x-1)+x+(x+1)$
$=3 x$
Given that, the sum of the three numbers is greater than 45 and less than 54 .
When the sum of the three numbers is 45 ,
$
\begin{aligned}
& 3 \mathrm{x}=45 \\
& \therefore \mathrm{x}=\frac{45}{3} \\
& \therefore \mathrm{x}=15
\end{aligned}
$
When the sum of the three numbers is 54,
$
\begin{aligned}
& \therefore 3 \mathrm{x}=54 \\
& \therefore \mathrm{x}=\frac{54}{3} \\
& \therefore \mathrm{x}=18
\end{aligned}
$
$\therefore$ the value of $\mathrm{x}$ is greater than 15 and less than 18 .
$\therefore$ the value of $\mathrm{x}$ is either 16 or 17
Case I:
If the value of $x$ is 16 , then the three consecutive whole numbers are $(16-1), 16,(16+1)$ i.e., $15,16,17$
Case II:
If the value of $x$ is 17 , then the three consecutive whole numbers are (17-1), $17,(17+1)$ i.e., $16,17,18$.
$\therefore$ The three consecutive whole numbers are $15,16,17$ or $16,17,18$.
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