the sum of a 2 digit number is 11.if the number is obtained by reversing the digits is 9 less than the original number, find the number
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Let the tens digit be y and unirs digit x,so the number is (10y+x)
ATP... x+y=11=>x=11-y .....(i)
Another equation... 10y+x-9=10x+y =>10y-y-9=10x-x=>9x=9y-9=>x=9(y-1)/9 =>x=y-1
Putting the value of..... (i) in the equation
11-y=y-1=>-y-y=-11-1=>-2y=-12=>2y=12=>y=12/2 =>y=6
So the value of x is ........ x=y-1 =>x=6-1 =>x=5
SO THE TENS DIGIT(I.E. y) IS 6 AND THE UNITS DIGIT(I.E. x) IS 5
SO THE NUMBER(10y+x) IS (10×6+5=65) 65
HOPE IT HELPS
ATP... x+y=11=>x=11-y .....(i)
Another equation... 10y+x-9=10x+y =>10y-y-9=10x-x=>9x=9y-9=>x=9(y-1)/9 =>x=y-1
Putting the value of..... (i) in the equation
11-y=y-1=>-y-y=-11-1=>-2y=-12=>2y=12=>y=12/2 =>y=6
So the value of x is ........ x=y-1 =>x=6-1 =>x=5
SO THE TENS DIGIT(I.E. y) IS 6 AND THE UNITS DIGIT(I.E. x) IS 5
SO THE NUMBER(10y+x) IS (10×6+5=65) 65
HOPE IT HELPS
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