Solution

The sum of two digits = 12

Let the ones digit of the number = x

then tens digit = 12 - x

and number = x + 10 (12 - x)

= x + 120 - 10x = 120 - 9x

Reversing the digits,

ones digit of new number = 12 - x

and tens digit = x

the number = 12 - x + 10 x = 12 + 9x

According to the condition,

12+9x=120−9x+54⇒ 9x+9x=120+54−12=174−12⇒ 18x=162⇒ x=16218=9

∴ Original number = 120 - 9x

= 120 - 9 × 9 = 120 - 81 = 39

Hence number = 39 Ans.

Check : Original number = 39

Sum of digits = 3 + 9 = 12

Now reversing its digit the new number will be = 93

and 93 - 39 = 54 which is given.

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Given:

The sum of a two-digit number is 12.

To Find:

The original number

Solution:

Let the two numbers be 'a' and 'b'

It is given that the sum of digits is a two-digit number 12.

Then,

⇒ a + b = 12 ..(i)

It is given that the new number formed by reversing the digits is greater than the original number by 54.

⇒ 54 + (10a + b) = (10b + a)

⇒ 54 + 10a + b = 10b + a

⇒ 9a - 9b = -54

⇒ a - b = -6 ..(ii)

Now, adding (i) and (ii)

⇒ a + b = 12

⇒ a - b = -6

⇒ 2a = 6

⇒ a = 3

So, b = 9

Now, to verify the solution,

39 + 54 = 93

So, the original number is 39 and when it is reversed is 93.