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

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.

Answer 2

Step-by-step explanation:

We Have :-

$a \: two \: digit \: number$

To Find :-

$number$

Solution :-

$let \: the \: first \: digit \: be \: x \\ \\ \\ second \: digit \: becomes \: x + 3 \\ \\ \\ acc \: to \: the \: question \: now \\ \\ \\ 10(x) + (x + 3) + 10(x + 3) + x = 121 \\ \\ \\ 10x + x + 3 + 10x + 30 + x = 121 \\ \\ \\ 11x + 3 + 11x + 30 = 121 \\ \\ \\ 22x + 33 = 121 \\ \\ \\ 11(2x + 3) = 121 \\ \\ \\ 2x + 3 = \dfrac{121}{11} \\ \\ \\ 2x + 3 = 11 \\ \\ \\ 2x = 8 \\ \\ \\ x = \dfrac{8}{2} \\ \\ \\ x = 4 \\ \\ \\ x + 3 = 4 + 3 = 7 \\ \\ \\ so \: the \: original \: number \: is \: 47$