Question

the sum of digits of a two digit no is 10. the number formed by reversing the digits is 18 less than the original no
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Answer 1

$\large{\sf{\underline{\blue{ANSWER}}}}$

GIVEN:-

• Sum of a two digit no.is 10

• The number is reversed and the new number formed is 18 less than Original Number.

Now,

$\implies{Ones\:digit\:be=x}$

$\implies{Tens\:digit=(10-x)}$

$\implies{Original\:number=x+10(10-x)=100-9x}$

Again,

Atq

• After reversing the digit,

$\implies{Ones\:digit=(10-x)}$

$\implies{Tens\:digit=x}$

$\implies{New\:Number=10-x+10(x)}$

$\implies{New\:Number=9x+10}$

Now,

(New Number)-(Original Number)=18.

$(9x+10)-(100-9x)=18$

$9x+10-100+9x=18$

$18x-90=108$

$18x=108$

$x=\frac{108}{18}$

$x=6$

Hence,

Original Number is 64.

Answer 2

Let the two digit number xy i.e., 10x+y

A/c, " the sum of digits of a two digit no is 10 "

⇒ x + y = 10 ... (1)

A/c , " the number formed by reversing the digits is 18 less than the original no  "

Let reversed no. be " 10y+x "

⇒ 10x + y - ( 10y + x ) = 18

⇒ 9x - 9y = 18

⇒ x - y = 2 ... (2)

Now solve (1) + (2) , we get ,

⇒ ( x + y ) + ( x - y ) = 10 + 2

⇒ 2x = 12

⇒ x = 6

sub. x = 6 , in (2) , we get ,

⇒ y = 6 - 2

⇒ y = 4

So the original number is xy = 64.