Question

The digit of a two digit number differ by 5 if the digits are interchanged and resulting number is added to the original number we get 121 find the original number

Answer 1

Let us take the two digit number such that the digit in the units place is x. The digit

in the tens place differs from x by 3. Let us take it as x + 3. So the two-digit number is

10 (x + 3) + x = 10x + 30 + x = 11x + 30.

With interchange of digits, the resulting two-digit number will be 10x + (x + 3) = 11x + 3

If we add these two two-digit numbers, their sum is

(11x + 30) + (11x + 3) = 11x + 11x + 30 + 3 = 22x + 33

It is given that the sum is 121.

Therefore, 22x + 33 = 121

22x = 121 – 33

22x = 88

x=4

The units digit is 4 and therefore the tens digit is 4 + 3 = 7.

Hence, the number is 74 or 47.

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Answer 2

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

The digit of a two digit number differ by 5, if the digits are Interchanged and resulting number is added to the original number we get 121.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The original number.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

Let the ten's place digit be r

Let the one's place digit be m

$\boxed{\bf{The\:original\:number=10r+m}}}}}$

$\boxed{\bf{The\:reversed\:number=10m+r}}}}}$

A/q

$\longrightarrow\sf{r-m=5}\\\\\longrightarrow\sf{r=5+m.....................(1)}$

&

$\longrightarrow\sf{10r+m+10m+r=121}\\\\\longrightarrow\sf{11r+11m=121}\\\\\longrightarrow\sf{11(r+m)=121}\\\\\longrightarrow\sf{r+m=\cancel{\dfrac{121}{11} }}\\\\\longrightarrow\sf{r+m=11}\\\\\longrightarrow\sf{5+m+m=11\:\:\:[from(1)]}\\\\\longrightarrow\sf{5+2m=11}\\\\\longrightarrow\sf{2m=11-5}\\\\\longrightarrow\sf{2m=6}\\\\\longrightarrow\sf{m=\cancel{\dfrac{6}{2} }}\\\\\longrightarrow\sf{\pink{m=3}}$

Putting the value of m in equation (1),we get;

$\longrightarrow\sf{r=5+3}\\\\\longrightarrow\sf{\pink{r=8}}$

Thus;

$\boxed{\bf{The\:original\:number=10(8)+3=80+3=\boxed{83}}}}}}$