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# A two digit number is ${\text{3}}$more than ${\text{4 }}$ times the sum of its digits. If ${\text{18}}$is added to the number, the digits are reversed. Find the number.

Last updated date: 20th Jul 2024
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Hint: Assume the number at one’s place, ten’s place. Then use the given conditions to write equations in two variables and solve them to find the number.

Let the digit in one’s place be $x$ and ten’s place be ${\text{ }}y$.
Original number is $10y + x$.
Number formed by reversing the digits is ${\text{ }}10x + y$
Given that the number is ${\text{3}}$ more than ${\text{4 }}$ times the sum of its digits
$\Rightarrow 10y + x = 4(x + y) + 3$
$\Rightarrow 10y + x - 4x - 4y = 3$
$\Rightarrow 6y - 3x = 3$
$\Rightarrow 2y - x = 1{\text{ }}...{\text{(1)}}$
Also given that when ${\text{18}}$ is added to the number the digits get interchanged.
Therefore, $(10y + x) + 18 = {\text{ }}10x + y$
$\Rightarrow 9x - 9y = 18$
$\Rightarrow x - y = 2{\text{ }}...{\text{(2)}}$
To find the values of${\text{ }}x$ and $y$ ,add ${\text{(1)}}$ and ${\text{(2)}}$ , we get
$\Rightarrow 2y - y - x + x = 1 + 2$
$\Rightarrow y = 3$
Put$y = 3$ in $x - y = 2$, we get
$\Rightarrow x - 3 = 2$
$\Rightarrow {\text{ }}x = 5$
As we know,
The original number is ${\text{(}}10y + x{\text{)}}$.
$\Rightarrow (10y + x) = 10(3) + 5 = 35{\text{ }}$
Putting the values of$x$and $y$ in original number, we get,
Hence, the number is ${\text{35}}$.

Note: To solve such types of questions, all the conditions must be considered carefully and then equated. Also, the numbers at one's place and ten’s place must not be mixed.