The sum of two digit number and the number obtained by reversing the order of its digits is 121

Solution:

Let us say that $\mathrm{x}$ be the ones digit and $\mathrm{y}$ be the tens digit. Therefore Two digit number before reversing $=10 \mathrm{y}+\mathrm{x}$ Two digit number after reversing $=10 \mathrm{x}+\mathrm{y}$ As per the question $\begin{array}{l} (10 \mathrm{y}+\mathrm{x})+(10 \mathrm{x}+\mathrm{y})=121 \\ \Rightarrow 11 \mathrm{x}+11 \mathrm{y}=121 \\ \Rightarrow \mathrm{x}+\mathrm{y}=11\dots \dots(i) \end{array}$ Since the digits differ by 3, so $x-y=3\dots \dots(ii)$ Adding equation(i) and equation(ii), we obtain $2 x=14 \Rightarrow x=7$ Putting $\mathrm{x}=7$ in equation(i), we obtain $7+y=11 \Rightarrow y=4$ Changing the role of $x$ and $y, x=4$ and $y=7$

As a result, the two-digit number is 74 or 47 .

Let x be the ones digit and y be the tens digit. ThenTwo digit number before reversing = 10y + xTwo digit number after reversing = 10x + yAs per the question(10y + x) + (10x + y) = 121⇒11x + 11y = 121⇒x + y = 11 …….(i)Since the digits differ by 3, sox – y = 3 ……….(ii)Adding (i) and (ii), we get2x = 14 ⇒ x = 7Putting x = 7 in (i), we get7 + y = 11 ⇒ y = 4Changing the role of x and y, x = 4 and y = 7

Hence, the two-digit number is 74 or 47.

>

The sum of a two digit number and the number obtained by reversing the order of its digits is 121 . If units and ten's digit of the number are x and y respectively, then write the linear equation representing the above statement.

Solution

Let unit digit = x

and tens digit = y

$\therefore$ Number = x + 10y

By reversing the digits,

units digit = y

and tens digit = x

$\therefore$ Number = y + 10x

Now sum of these two numbers = 121

$\therefore x+10y+y+10x=121\Rightarrow 11x+11y=121\Rightarrow x+y=11\left(\text{Dividing by 11}\right)\therefore x+y=11$

Mathematics

RD Sharma

Standard IX