Question

The sum of a hvo digit number and number obtained on reversing the digits is
99. If number obtained on reversing the digits is 9 more than the original
number, find the number.

Answer 1

Step-by-step explanation:

10x+y+10y+x=99

11x+11y=99

x+y=99/11

x+y =9

10x+y+9= 10y+x

9x-9y=-9

y-x=1

y+x=9

2y=10

y=5

x4

number = 45

Answer 2

Correct Question :

The sum of a two digit number and number obtained on reversing the digits is 99. If number obtained on reversing the digits is 9 more than the original number, find the number.

Solution :

• Let the tense digit be 'x' and ones digit be 'y',

According to the first condition,

• ∴ Original number = 10x + y

Reversing the number from the given number,

⟹ (10x + y) + (10y + x) = 99

⟹ ( 10x - x) + (10y - y) = 99

⟹ 11x + 11y = 99

⟹ x + y = 9 -------- (1)

According to the second condition,

On reversing the digits, the number obtained is 9 more than the original number.

Reversed number = 9 + Original number

⟹ 10y + x = 9 + 10x + y

⟹ (10x - x) - (10y - y) = -9

⟹ 9x - 9y = -9

⟹ x - y = -1 ------- (2)

Subtracting both (1) & (2) we get,

x - y = -1

x + y = 9

_______

2y = 10

_______

[ While subtracting the equation, signs of Eq (2) will change, And so both 'x' will get cancelled. ]

⟹ 2y = 10

⟹ y = 10 / 2

⟹ y = 5

Substituting 'y' value in Eq (2) :

⟹ x + y = 9

⟹ x + 5 = 9

⟹ x = 9 - 5

⟹ x = 4

• ∴ The number is 45.