Question

13. Sum of the digits of a two-digit number is 9. The number obtained by interchanging the
digits exceeds the given number by 27. Find the original number.​

Answer 1

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Let the two digit number be 10x + y

Given that the sum of the digits is 9

x + y = 9 (equation 1)

Given that the number obtained by interchanging the digits exceeds the given number by 27.

10y + x = 10x + y + 27

9x - 9y = - 27

taking 9 as common

x - y = - 3 (equation 2)

Adding equation 1 and 2

x + y = 9

x - y = - 3

2x = 6

x = 3

3 + y = 9

y = 6

• The number is 10x + y is 36.

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Let the unit place digit be x.

And tens place digit be y.

Original Number = 10y + x

Reversed Number = 10x + y

According to the Question,

1st Part,

Sum of digits = 9

⇒ x + y = 9

⇒ x = 9 - y ...... (i)

2nd Part,

⇒ (10x + y) - (10y + x) = 27

⇒ 10x + y - 10y - x = 27

⇒ 9x - 9y = 27

⇒ x - y = 3

⇒ 9 - y - y = 3     [From (i)]

⇒ -2y = -6

⇒ y = 6/2

⇒ y = 3

Putting y's value in Eq (i), we get

⇒ x = 9 - y

⇒ x = 9 - 3

⇒ x = 6

Original number = 10y + x = 10 × 3 + 6 = 30 + 6 = 36.

Hence, the required original number is 36.