Question

in a two digit number sum of digits is 7. if the difference of 2 digit number and number obtained by reversing the digits is 9 then find the number​

Answer 1

let the digit at tens place be x and at unit place be y

sum of digit, x+y=7....(1)

original number=10x+y

number obtain by interchanging the digits=10y+x

difference,10x+y-10y-x=9

9x-9y=9

dividing by 9 on both sides

x-y=1....(2)

adding equation 1 and 2

x+y=7

-x-y=1

------------

2y=6

y=6/2

y=3

substituting y=3 in equation 2

x-3=1

x=1+3

x=4

the original number,10x+y=10×4+3

=40+3

=43

hope it helps you....

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Answer 2

Let digit in tens place=x

Let digit in ones place=y

Given,Sum of digits=7

A/Q

x+y=7

x=7-y   (equation 1)

Now original  Number

10(digit at tens place)+digit at ones place

Original Number=10x+y

Now Reverse Number

10(digit at ones place)+(digit at tens place)

Reverse number=10y+x

Now difference is 9

A/Q

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

10x-x+y-10y=9

9x-9y=9

9(x-y)=9        9 is common

x-y=9/9

x-y=1

we get

x-y=1

we know x=7-y

x-y=1

7-y-y=1

7-2y=1

-2y=1-7

-2y=-6

y=6/2

y=3

x=7-y

x=7-3

x=4

Original number=10x+y

=10*4+3

=40+3

=43

Original number=43