Question

In a 2 digit number the sum of the digits is 9 islf the digits r teversed the number is increased by 9 find the number

Answer 1

Given :-

▪ The sum of the digits of a 2-digit number is 9. If the digits of the number is interchanged, the number is increased by 9.

To Find :-

▪ The Number

Solution :-

Let the one's and ten's digit be y and x respectively.

According to the question,

Case 1

Sum of the digits is 9.

⇒ x + y = 9

⇒ x = 9 - y ...(1)

Case 2

If the digits are reversed, the new number obtained is 9 more than the original number.

But, first we know, a 2-digit is of the form 10m + n where,

• m = ten's digit
• n = one's digit

⇒ 10x + y + 9 = 10y + x

⇒ 10x - x + y - 10y = -9

⇒ 9x - 9y = -9

⇒ x - y = -1

Substituting [x = 9 - y] from (1),

⇒ (9 - y) - y = -1

⇒ 9 - y - y = -1

⇒ -2y = -10

⇒ y = 5

Now, Put [y = 5] in (1) to get x

⇒ x = 9 - y

⇒ x = 9 - 5

⇒ x = 4

So, The number was,

⇒ 10x + y

⇒ 10 × 4 + 5

⇒ 40 + 5

⇒ 45

Hence, The required number is 45.

Answer 2

Given : In a two digit number the sum of the digits is 9. lf the digits are reversed the number is increased by 9.

　

To find : What is the number.

　

According to the question :

• Let "x" be the digit in tens place, "y" be the digit at one's place, let "10x + y" be the original number and "10y + x" be the number by reversing the digits.

　

→ x + y = 9

→ 10y - y + x - 10x = 9

→ 9y - 9x = 9 (Reversing we get -9)

• 9x - 9y = -9

→ x - y = -1

　

→ 2x = 8

→ x = 8/2

→ x = 4

　

→ 4 + y = 9

→ y = 5

　

→ 10 × 4 + 5

→ 40 + 5

→ 45

　

Therefore, 45 is the required number.