Math, asked by ayaanprimus, 10 months ago

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.

Answers

Answered by MukulCIL
3

Answer:

74

here it is not given that which digit is greater whether ones place or tens place , therefore answer may also be 47

Step-by-step explanation:

Let the original number be xy

value = 10x +y

reversed number = 10y +x

given that x-y = 3

now add the original number and reversed number

11x + 11y = 121

x+y = 11

and we know

x-y= 3

x= 7

y= 4

number is 74

Answered by Anonymous
0

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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