Login

The sum of the digits of a two-digit number is 7. when the digits are reversed the number is decreased by 9, find the number.

Let the digit at one's place is x and the digit at ten's place is y.
Then original number = 10y+x
Sum of the digits = x+y = 7...(i)
And according to the question we have;
10x+y = 10y+x-9⇒9y-9x = 9⇒y-x = 1 ...(ii)
Solving (i) and (ii) we get;
x+y+y-x = 7+1⇒2y = 8⇒y = 4
Then from (i) we have;
x+4 = 7⇒x = 7-4 = 3
Therefore the original number is 4×10+3 = 43.

  • 8
View Full Answer

see dost,

let one digit no. be x

and tens digit no. be y

then atq

x +y = 9 ...........{1}

original no. should be = 10y + x

when reversed , the no. becomes = 10x + y

so atq

(10x+ y)- ( 10y + x) = 9

x - y = 1.......{2}

taking {1} and {2}

x +y = 7

x - y = 1

2x = 8

x= 4

and since y = 7-x

hence y = 3

  • 5
What are you looking for?