Question

11. The sum of digits of a two-digit number is 9. If the digits are reversed, the number is increased by 63. Find the
number.
(a) 18
(b) 27
(C) 36
(d) 72
​

Answer 1

Answer : 18

Let x be the ones digit

⇒ The tens digit is 9−x

10(9−x)+x+63=10x+(9−x)

⇒90−10x+x+63=10x+9−x

⇒153−9x=9x+9

⇒153−9=18x

⇒144=18x

⇒x=8

⇒9−x=9−8=1

Therefore, the number we are looking for is

10(1)+8=18

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Answer 2

Given :-

• The sum of digits of a two-digit number is 9. If the digits are reversed, the number is increased by 63.

To find :-

• Required number

Solution :-

Let the tens digit be x then ones digit be y

• Original number = 10x + y

• According to the first condition

It is given that sum of two digit number is 9

→ x + y = 9

• According to the second condition

The digits are reversed, the number is increased by 63.

• Reversed number = 10y + x

→ 10x + y + 63 = 10y + x

→ 10x - x + y - 10y = - 63

→ 9x - 9y = - 63

→ 9(x - y) = - 63

→ x - y = 7

Add both the equations

→ x + y + x - y = 9 + 7

→ 2x = 16

→ x = 16/2

→ x = 8

Put the value of x in equation (ii)

→ x - y = 7

→ 8 - y = 7

→ y = 8 - 7

→ y = 1

Therefore,

• Tens digit = x = 8
• Ones digit = y = 1

Hence,

• Reversed number = 10y + x = 18
• Original number = 10x + y = 81