Question

The digits of a two-digit number differ by 3. If digits are interchanged and the resulting number is added to the original number, we get 121. Find the original number

Answer 1

Answer:

let the one's place digit be y

let the ten's place digit be x

therefore the number=10x+y

x-y=3 (1)

interchanging the digits ,the new number =10y+x

So (10x+y)+(10y+x)=121

11x+11y=121

11(x+y)=121÷11

x + y=11 (2)

from(1) and (2)

x +y=11

+x -y=3

2x=14÷2

x=7

x - y=3

x+3=y

7-3=y

4=y

number=10x+y

10×7+4

74

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.