Question

The tens digit of a two digit number exceed the unit digits by 5. If the digits are reversed, the new number is less by 45. If the sum of digits is 9. Find the numbers.

Answer 1

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 question to solve further.

Complete step-by-step answer:

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 unit place be x.
Then the digit at tens place be (x+5) as in the question it is given that the tense digit is exceeded by 5.
Once place =x
Tens place = (5+x)
Therefore the two digit number forms by these two digits is
Number= 10(5+x)+1(x) 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+50
Now according to question we interchange the numbers,
I.e. once place became = (5+x) and tens place become =x
The new number formed is, Number= 10(x) +1(x+5)
Therefore new number=11x+5
In the question it is given that the sum of the digit is 9 i.e.
\[
   \Rightarrow \left( {x + 5} \right) + x = 9 \\
   \Rightarrow 2x = 9 - 5 \\
   \Rightarrow x = \dfrac{4}{2} \\
  \therefore x = 2 \\
 \]
Now we put the value of x to find the value of numbers.
Old number=11x+50=72
And the new number=11x+5=27.
Hence the required number is 72.

Note: - Whenever we face such type of question the key concept for solving the question is first assume the digits at any place and then formed the numbers as we know that the rule of making two digit numbers i.e. 10p+q, where p is digit at tens place and q is the digit at once place. And then apply the statement of question to proceed and then we will find the value of that digit. Now put the value of that digit to get the number.