Question

the sum of the digits of a 2 digit number is 7. If the digits are reversed, the number formed is 9 less than the original number. find the number

Answer 1

let the digit in ones place be x

and the digit in tens place be y

therefore, the two digit no. = 10y+x

the reversing no. = 10x+y

A/Q,

x+y=7

=>x=7-y

and,

10x+y=10y+x-9

=>10x-x+y-10y = -9

=> 9x-9y = -9

=> x-y = -1

=> 7-y-y=-1

=> -2y=-1-7

=> y = -8/-2= 4

x=7-4 = 3

therefore, the req. 2 digit no. = 10*4+3=43

Answer 2

Question:

The sum of a digit of a 2 digit number is 7. If the digits are reversed, the number formed is 9 less than the original number. Find the number.

Solution:

Take the no xy as 10x+y(expansion form)

Reverse the digits then th no is 10y+x

Difference

10x+y-(10y+x)

9x-9y

9(x-y)

ATQ

The difference is 9

9(x-y)=9

So,

x-y=9/9=1

x-y=1______(1)

x+y=7_____(2)

Adding (1) and (2)

2x=8

x=4

Putting in (2)

4+y=7

y=7-4

So,

x=4

y=3

Therefore

The final number is 43