Question

the sum of 2 digits number and number obtained by reversing the order of digits is 99 if the digits of the number differ by 3 find the number

Answer 1

♧♧HERE IS YOUR ANSWER♧♧

Let the digits are a and b.

Then the number is (10a + b).

When the digits are reversed, the number is (10b + a).

By the given condition :

a - b = 3 .....(i)

and

(10a + b) + (10b + a) = 99

=> 11a + 11b = 99

=> a + b = 9 .....(ii)

Now, adding (i) and (ii), we get :

2a = 12

=> a = 6

From (ii), we get :

6 + b = 9

=> b = 3

Therefore, the required number is 63.

♧♧HOPE THIS HELPS YOU♧♧

Answer 2

Hi ,

Let ten's place digit = x

Units place digit = y ,

The number = 10x + y ----( 1 )


If Reversing the digits then we get

the new number = 10y + x ---( 2 )

According to the problem given ,

10x + y + 10y + x = 99

11x + 11y = 99

Divide each term with ' 11 ' ,

x + y = 9 ---( 3 )

x - y = 3 ----( 4 ) [ given ]

Add ( 3 ) and ( 4 ) equations , we get

2x = 12

x = 12/2

x = 6

Put x = 6 in equation ( 3 ) we get ,

6 + y = 9

y = 3 ,

Therefore ,

Required number = 10x + y

= 10 × 6 + 3

= 60 + 3

= 63

I hope this helps you.

: )