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Last updated date: 02nd Dec 2023
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MVSAT Dec 2023

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 143 .What can be the original number?

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Hint: - If there is a two digit number it means there are two places, ones and tens. For solving questions let assume the digit at any place be x and then apply the condition of the question to solve further.
Here in the given question there is a two digit number it means it has two places once place and tens place.
Now, let the digit at tens place be x.
Then the digit at once place be (x+3) as in the question it is given that the other digit is differ by 3.
Tens place=x
Once place= (3+x)
Therefore the two digit number forms by these two digits is
Number= 10x+1(x+3) as we know that for forming a two digit number we multiply the digit at tens place by 10 and then add it with the digit at once place by multiplying 1 at once place.
Therefore the required number is 11x+3
Now according to question we interchange the numbers,
i.e. tense place become= (3+x) and once place became=x
The new number formed is, Number= 10(x+3) +x
Therefore new number=11x+30
Now in the question it said if we add old number and new numbers it become 143
i.e. 11x+3+11x+30=143
   \Rightarrow 22x + 33 = 143 \\
   \Rightarrow 22x = 110 \\
   \Rightarrow x = \frac{{110}}{{22}} \\
  \therefore x = 5 \\
Now put the value of x in the required number,
$\therefore $Required number is $11x + 3 = 11 \times 5 + 3 = 58$

Note: - Whenever we face such a type of question the key concept for solving the question is first assume the digits at any place and then form the numbers by these digits according to the question. And then apply the statement of question to proceed a d then we will find the value of that digit. Now put the value of that digit to get the number.