The sum of a two digit number and the number obtained by interchanging the digits is 132. If the two digits differ by 2. Let the ten's digit be x and unit digit be y then the two digit number in x and y is
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Let unit's digit =y and the ten's digit =x
So, the original number =10x+y
After interchanging the digits, New number =x+10y
The sum of the number =10x+y
The sum of the digit =x+y
According to the question, (10x+y)+(x+10y)=132
⇒11x+11y=132
⇒11(x+y)=132
⇒x+y=12…(i)
and 10x+y+12=5(x+y)
⇒10x+y+12=5x+5y
⇒10x−5x+y−5y=−12
⇒5x−4y=−12…(ii)
From Eq. (i), we get x=12−y…( iii )
On substituting the value of x=12−y in Eq. (ii), we get
5(12−y)−4y=−12
⇒60−5y−4y=−12
⇒−9y=−12−60
⇒−9y=−72
⇒y=8
On putting the value of y=8 in Eq. (iii), we get
x=12−8=4
So, the Original number =10x+y
=10×4+8
=48
Hence, the two digit number is 48.
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