Question

The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. What can be the original number? please answer it urgent

Answer 1

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ original \ number \ is \ 85.}$

$\sf\orange{Given:}$

$\sf{\implies{A \ two \ digit \ number \ differ \ by \ 3}}$

$\sf{\implies{If \ the \ digits \ are \ interchanged}}$

$\sf{and \ the \ resulting \ number \ is \ added \ to}$

$\sf{the \ original \ number, \ we \ get \ 143.}$

$\sf\pink{To \ find:}$

$\sf{Original \ number.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{Let \ the \ ten's \ place \ of \ the \ digit \ be \ x}$

$\sf{and \ unit's \ place \ be \ y.}$

$\sf{According \ to \ the \ first \ condition.}$

$\sf{x-y=3...(1)}$

$\sf{\implies{Original \ number=10x+y}}$

$\sf{\implies{Number \ with \ reversed \ digit's=10y+x}}$

$\sf{According \ to \ the \ second \ condition.}$

$\sf{(10x+y)+(10y+x)=143}$

$\sf{\therefore{11x+11y=143}}$

$\sf{\therefore{11(x+y)=143}}$

$\sf{\therefore{x+y=\frac{143}{11}}}$

$\sf{\therefore{x+y=13...(2)}}$

$\sf{Add \ equations (1) \ and \ (2)}$

$\sf{x-y=3}$

$\sf{+}$

$\sf{x+y=13}$

__________________

$\sf{2x=16}$

$\sf{\therefore{x=\frac{16}{2}}}$

$\boxed{\sf{\therefore{x=8}}}$

$\sf{Substitute \ x=8 \ in \ equation (2)}$

$\sf{8+y=13}$

$\sf{\therefore{y=13-8}}$

$\boxed{\sf{\therefore{y=5}}}$

$\sf{Original \ number=10(8)+5=80+5}$

$\sf{\therefore{Original \ number=85}}$

$\sf\purple{\tt{\therefore{The \ original \ number \ is \ 85.}}}$

Answer 2

here is your answer:

let the no. at ones place be x

the no. at tens place be x+3

so, the no. is 10(x+3)+x

number whose digits are interchanged = 10x+(x+3)

ATQ

10(x+3)+x-10x+(x+3)=143

= 10x+30+x-10x+x+3=143

= 10x+x+10x+x= 143-30-3

= 22x= 110

=x=¹¹0/22

=x= 5

so. the no. is 10(5+3)+5

= 50+30+5

= 80+5

= 85

so. the ans is 85

hope it helps

please mark me as brainiest