Question

The sum of a two digit 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

Step-by-step explanation:

Let the digit at ten's place = x

Let the digit at one's place = y

original number = 10x + y

According to given :

x - y = 3 .............(1)

y - x = 3 .............(2)

When the digits are interchanged, then new number = 10y + x

according to the given :

(10x + y) + (10y + x) = 99

⇒ x + y = 9 .........(3)

first Case :

adding (1) and (3), we get

2x = 12 ⇒ x = 6

Put x = 6 in (1), we get y = 3

So, original number = 10 × 6 + 3 = 63

Second Case :

adding (2) and (3) , we get

2y = 12 ⇒ y = 6

from (2), we get x = 3

so original number = 36

Answer 2

Step-by-step explanation:

Let the digit at ten's place = x

Let the digit at one's place = y

original number = 10x + y

According to given :

x - y = 3 .............(1)

y - x = 3 .............(2)

When the digits are interchanged, then new number = 10y + x

according to the given :

(10x + y) + (10y + x) = 99

⇒ x + y = 9 .........(3)

first Case :

adding (1) and (3), we get

2x = 12 ⇒ x = 6

Put x = 6 in (1), we get y = 3

So, original number = 10 × 6 + 3 = 63

Second Case :

adding (2) and (3) , we get

2y = 12 ⇒ y = 6

from (2), we get x = 3

so original number = 36