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:

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Answer 2

$\pink{Given\::}$

The sum of digits of a two-digit

number is 9. If the digits are

reversed, the number is increased

by 63.

$\red{To find \::}$

Required number

$\purple{Solution \::}$

Let the tens digit be x then ones digit be y

Original number = 10x + y

According to the first condition

$\orange{ According \: to \: the \: first \: condition\:}$

It is given that sum of two digit number is 9

→ x + y = 9

According to the second condition

$\blue{ According \: to \: the \: second \: condition\:}$

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

$\purple{ Hence, \</p><p></p><p>Reversed \: number = 10y + x = 18:</p><p>Original \: number = 10x + y = 81\:}$

$\blue{Here \: is \: your \: answer \: be \: happy \::}$