Question

sum of digits of two numbers is 12. the number formed by reversing the digits is 54 greater than the original number. find the original number.

Answer 1

Answer:

39

Step-by-step explanation:

Let the digits be a and b.

∴ The original number = 10a + b.

Number formed by reversing the digits = a + 10b.

(i) Sum of two numbers is 12:

a + b = 12

(ii) Number formed by reversing the digits is 54 greater than original.

⇒ (a + 10b) = 10a + b + 54

⇒ a + 10b - 10a - b = 54

⇒ -9a + 9b = 54

⇒ a - b = -6

On solving (i) & (ii), we get

⇒ a + b = 12

⇒ a - b = -6

   --------------

   2a = 6

      a = 3

Substitute a = 3 in (ii), we get

⇒ a - b = -6

⇒ 3 - b = -6

⇒ b = 9.

Therefore, the original number is 39.

Hope it helps!

Answer 2

Step-by-step explanation:

let a and b the digit

a+b=12 ........(1)

54+(10a+b)=(10b+a)

54+10a+b=10b+a

9a-9b=-54

a-b=-6........(2)

adding (1) and (2)

a+b=12

a-b=-6

----------

2a=6

a=3

so b=9

39+54=93 (verified)

orig no.=39 and  No. when digits reversed=93