Question

1. If the number obtained by interchanging the
digits of a two-digit number is 18 more than
the original number and the sum of the digits
is 8, then what is the original number?​

Answer 1

Step-by-step explanation:

let x and y be the two digits respectively

then the number is 10y+x

x+y=8...... (1)

(10x+y)-(10y+x)=18

or, 9 (x-y)=18

or, x-y=2...... (2)

so x=5

and

y=3

so original number is 35

Answer 2

$\huge\sf{Answer}$

• Let the digit in the ones place = x

• Let the digit in the tens place = y

The two digit number is (10y + x)

Sum of the digits is 9

⇒ x + y = 9

⇒ x = 9 - y

The number obtained by interchanging the digit is 18 more than twice the original number.

⇒  (10x + y) = 2(10y + x) + 18

⇒ 10x + y = 20y + 2x + 18

⇒ 19y - 8x + 18 = 0

Substitute first equation into the second equation:

19y - 8(9 - y) + 18 = 0

19y - 72 + 8y + 18 = 0

27y - 54 = 0

27y = 54

y = 2

Substitute y = 2 .

x = 9 - 2

x = 7

Original number = 10y + x

Original number = 10(2) + 7

$\huge\boxed{27}$

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