Question

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#Hard Question

A number has 2 digits whose sum is 9.The number formed by interchanging the digits is 9 more than the original number .Find the number

Answer 1

Answer:-

Given that, x + y = 9......(1)

( The sum of the digits = 9 )

Now, according to the question.

The digits of the number is: 10x + y.

The digits of the number when interchanged: 10y + x.

The digit of the number when interchanged, will will be nine more than original number.

So, 10x + y + 9 = 10y + x

➜ 10x - x + y - 10 y = -9

➜ 9x - 9y = -9

➜ 9(x - y) = -9

➜ x - y = -1.....(2)

Now, adding both the equations we got.

x + y = 9

x - y = -1

_______

2x = 8

➜ x = 4

So, 4 + y = 9

➜ y = 5

Now, Finding the number.

10x + y

➜ 10(4) + 5 = 45

$\rule{200}2$

Answer 2

Answer:

Let the Number be : (10a + b), where b is One's Digit and a is Ten's Digit.

★ We Have Given that :

↠ Sum of Digits = 9

↠ a + b = 9

↠ a = 9 – b ⠀— eq. ( I )

☯ According to the Question :

⇴ Original No. + 9 = Interchange No.

⇴ (10a + b) + 9 = (10b + a)

⇴ 9 = 10b + a – 10a – b

⇴ 9 = 9b – 9a

⇴ 9 = 9(b – a)

• Dividing both term by 9

⇴ b – a = 1

• putting the value of a from eq. ( I )

⇴ b – (9 – b) = 1

⇴ b – 9 + b = 1

⇴ 2b = 1 + 9

⇴ 2b = 10

• Dividing both term by 2

⇴ b = 5

☯ Putting value of b in eq. ( I ) :

⇴ a = 9 – b

⇴ a = 9 – 5

⇴ a = 4

∴ Number Formed will be : 10(4) + 5 = 45.