The sum of a two digit number and the number formed by interchanging the digit is 110. if 10 is subtracted from the original number, the new number is 4 more than 5 times the sum of the digits of the original number. Find the original number.
Answers
Given :
- the sum of a two digit number and the number formed by interchanging the digit is 110.
- 10 is subtracted from the original number
- The new number is 4 more than 5 times the sum of the digits of the original number.
To find :
- The original number =?
Step-by-step explanation:
Let the digit at ten's place be x and that at the unit's place be y.
Then, the required (original) number be 10x + y.
∴ The number obtained by interchanging its digit = 10y + x
Now,
According to the question :
➟ (10x + y) + (10y + x) = 110
➟ 11x + 11y = 110
➟ x + y = 10 ... (i)
Also,
➟ (10x + y) - 10 = 5(x + y) + 4
➟ 10x + y -10 = 5x + 5y + 4
➟ 5x - 4y = 14 .... (ii)
Multiplying (i) by 4, we get
➟ 4(x + y = 10)
➟ 4x + 4y = 40 ... (iii)
Adding (ii) and (iii), we get,
➟ 5x + 4x + 4y - 4y = 40 + 14
➟ 9x = 54
➟ x = 54/9
➟ x = 6
Putting x = 6 in (i), we get
➟ 6 + y = 10
➟ y = 10 - 6
➟ y = 4
So,
The original number is (10 x 6 + 4),i.e., 64.
Therefore, the original number = 64.
GIVEN :
- The sum of a two digit number and the number formed by interchanging the digit is 110.
- 10 is subtracted from the original number.
- The new number is 4 more than 5 times , the sum of digit of original number.
TO FIND :
- The original number = ?
STEP - BY - STEP EXPLANATION :
Let the digit at ten's places be 'X' and Let the digit as unit's place be 'Y'.
Then, the orignal number be 10x + y
Hence, the number obtained by interchanging the digits = 10y + x
Therefore,
AS PER THE QUESTION :-------
=> (10x +y) + (10y+x) = 110
=> (10x +y) + (10y+x) = 110 => 11 x + 11 y = 110
=> (10x +y) + (10y+x) = 110 => 11 x + 11 y = 110=> x + y = 110 ……… (i)
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=> (10 x + y ) - 10 = 5 (x+y) +4
=> (10 x + y ) - 10 = 5x + 5y + 4
=> 5x - 4x = 14 ……… (ii)
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Now , we will have to multiply (i) by 4
=> 4(x+y = 10)
=> 4(x+y = 10) => 4x + 4y = 40 ……… (iii)
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Add (ii) and (iii) we get here as,
=> 5x + 4x + 4y - 4y = 40 + 14
=> 5x + 4x + 4y - 4y = 40 + 14=> 40 + 14
=> 5x + 4x + 4y - 4y = 40 + 14
=> 40 + 14
=> 9x = 54
=> 5x + 4x + 4y - 4y = 40 + 14
=> 40 + 14
=> 9x = 54
=> x = 54/9
=> 5x + 4x + 4y - 4y = 40 + 14
=> 40 + 14
=> 9x = 54
=> x = 54/9
=> x = 6
Putting x = 6 in (i) we get ,
=> 6+y = 10
=> 6+y = 10=> y = 10-6
=> 6+y = 10=> y = 10-6 => y = 4
NOW,
The orignal number = (10×6+4) = 64