Question

1. 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.
number is 13. If the digits are interchanged, and the resulting​

Answer 1

Your Answer Is 74.

Given:-

• The digits of a two digit number differ by 3.

• If the digits interchange and the resulting number is added to the original number then we get 121.

To Find:-

• The Original Number.

so,

• Let the digit at ten's place be x.

• Let the digit at one's place be y.

• So the original number will be 10x+y.

• And the number when we interchange the digits would be 10y+x.

Therefore ATQ:-

↦ x - y = 3.

↦ x = 3 + y... eq(1).

Also,

↦ (10x + y) + (10y + x) = 121.

↦ 10x + y + 10y + x = 121.

↦ 11x + 11y = 121.

↦ 11 (x + y) = 121.

[Taking 11 as common].

↦ x + y = 121/11.

↦ x + y = 11.

↦ (3+y) + y = 11.

[Putting the value of x from eq(1)].

↦ 3 + y + y = 11.

↦ 3 + 2y = 11.

↦ 2y = 11 - 3.

↦ 2y = 8.

↦ y = 8/2.

↦ y = 4.

So by putting the value of y in eq(1) we get:-

↦ x = 3 + y.

↦ x = 3 + 4.

↦ x = 7.

Therefore the number is 10x + y.

= 10(7) + 4.

= 70 + 4.

= 74.

Therefore The Original Number Is 74.

Answer 2

answer

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.