Question

the digits of a two digit number differ by 3 if the digits are interchanged and the resulting number is added to the original number we get 121 find the original number​

Answer 1

Step-by-step explanation:

let us take the two digit no. such that the digit in the unit place is x

the digit in the tens place is x+3

so the two digit no. =10(x+3)+x=11x+30

with interchanging the digits the resulting no.=10x+(x+3=11x+3

add both these no.=11x+30+11x+3=22x+33

sum that is given if we add the no. is 121

so 22x+33=121

22x=121-33

22x=88

x=44

Answer 2

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.