a number consists of two digit whose sum is 9 if 27 is subtract from the original number its digits are interchange find the original number
Answers
Question:
A number consists of two digits whose sum is 9. If 27 is subtracted from the original number its digits gets interchanged .Find the original number.
Solution:
Let the tens digit of the required number be x and the unit (once) digit of the number be y.
Then;
The original number = 10x + y
Also,
The new number formed after interchanging the digits = 10y + x
Now,
According to the question;
The sum of the digits of the required two-digits number is 9.
ie;
=> x + y = 9
=> y = 9 - x -----------(1)
Also,
It is said that ,
If 27 is subtracted from the original number (required number), then its digits gets interchanged.
ie;
=> 10x + y - 27 = 10y + x
=> 10x + y - 27 - 10y - x = 0
=> 9x - 9y - 27 = 0
=> 9(x - y - 3) = 0
=> x - y - 3 = 0
=> x - (9 - x) - 3 = 0 {using eq-(1)}
=> x - 9 + x - 3 = 0
=> 2x - 12 = 0
=> 2x = 12
=> x = 12/2
=> x = 6
Now,
Putting the x = 6 in eq-(1) , we get;
=> y = 9 - x
=> y = 9 - 6
=> y = 3
Thus,
Tens digit = x = 6
Unit digit = y = 3
Hence,
Required number = 10x + y
= 10•6 + 3
= 60 + 3
= 63.
Hence,
The required number is 63.
AnswEr :
Required Number is 63.
Explanation :
Let the Ones Digit of the Number be a and, Tens Digit be b.
• Required Number = (10b + a)
• After Interchanging = (10a + b)
As Per Question :
» a + b = 9
» a = 9 - b —(¡)
Given :
• If 27 is subtract from the original number its digits are interchange.
⇒ 10b + a - 27 = 10a + b
⇒ 10b + a - 27 - 10a - b = 0
⇒ 9b - 9a - 27 = 0
- Dividing Each term by 9
⇒ b - a - 3 = 0
⇒ b - (9 - b) - 3 = 0 —( from ¡ )
⇒ b - 9 + b - 3 = 0
⇒ 2b - 12 = 0
⇒ 2b = 12
- Dividing Each term by 2
⇒ b = 6
• Putting Value of b = 6 in (¡)
⇒ a = 9 - b
⇒ a = 9 - 6
⇒ a = 3
_________________________________
➙ Original Number
➙ (10b + a)
➙ (10 × 6 + 3)
➙ (60 + 3)
➙ 63
჻ Required Number is 63.