Sum of the digit of two digit number is 9. the number obtained by interchanging the digits exceeds the given number by 27.find the original number.
Answers
Given that the sum of the digits is 9
x + y = 9 (equation 1)
Given that the number obtained by interchanging the digits exceeds the given number by 27
10y + x = 10x + y + 27
9x - 9y = - 27
taking 9 as common
x - y = - 3 (equation 2)
Adding equation 1 and 2
x + y = 9
x - y = - 3
2x = 6
x = 3
3 + y = 9
y = 6
The number is 10x + y is 36.
Step by step explanation:
(Hint:While writing a number in expanded form, its digits are multiplied by their values,i.e,
36=3×10+1×6)
Solution: Let the units digit of the original number be x and the tens digit of the original number be 9-x.
The original number in expanded notation
=10(9-x)+1×x
=90-10x+x
=90-9x
On interchanging the digits, the units digit is now =9-x and the tens digit is now x.
The number obtained by interchanging the digits in expanded notation=10x+1×(9-x)
=10x+9-x
=9x+9
Applying the given condition,
The number obtained on interchanging the digits-The original number =27
9x+9-(90-9x)=27
9x+9-90+9x=27
18x-81=27
18x=27+81
18x=108
x=108/18
x=6
Hence, the units digit=x=6
The tens digit=9-x=9-6=3.
So, the original number is 36.
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