the sum of digits of two digit number is 7. The number obtained by interchanging the digits exceed the original number by 27. Find the numbers
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the sum of digits of two digit number is 7. The number obtained by interchanging the digits exceed the original number by 27. Find the numbers
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Answer:
[tex]
let \: digit \: at \: ones \: place \: = \: x and \\
digit \: at \: tens \: place \: =y \\
So, \: the \: number \: will \: be \: 10y+x \\
x+y=7 ....(1) \\
reverse \: digit \: will \: be \: represented \: as, \: 10x+y \\
So, \\ \\
(10x+y)-(10y-x)=27 \\ \\
9x-9y=27 \\ \\
x-y=3 ....(2) \\ \\
Adding \: equation \: (1) \: and \: (2) \\ \\
2x=10 \\
x=5 \\
substituting \: in \: equation \: (1) \\ \\
y=7-5 \\
=2 \\
Thus, \: the \: number \: will \: be \: 25 \\
[/tex]
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Verified answer
Given the sum of digits of a two digit number, 7. And if the digits are reversed, the new number exceeds the original number by 27.
Let the digit at one's place be x and that of ten's place be y. Now, according to the question, equation will be-
=> x + y = 7 ...i)
Let's reverse the digits. So, the reversed new number will be 10x + y. Now that of new number, it will be exceeding the original by 27. So, the new equation will be-
=> 10x + y - (10y - x) = 27
Now solve this equation-
=> 10x + y - (10y - x) = 27
=> 10x + y - 10y - x = 27
=> 10x - x - 10y + y = 27
=> 9x - 9y = 27
=> 9(x - y) = 27
=> x - y = 3 ...ii)
After solving the equations i) and ii) we get-
=> x + y = 7
(+) x - y = 3
=> 2x = 7 + 3 ( y and -y got cancelled)
=> 2x = 10
=> x = 5
Now, put the value of x in eq(i)
=> x + y = 7
=> 5 + y = 7
=> y = 7 - 5
=> y = 2
∴ Hence, the original number is 25.