Question

the sum of the digits of a two digit number is 10 if the number formed by reversing the digit less than the original number by 36 then find the original number

Answer 1

let the 2 digit no. be 10x+y
ATQ, x+y = 10....... (i)
10y+x = (10x+y)-36
9y=9x-36
y=x-4...... (ii)

from (i) and (ii)
x+x-4=10
x=7
y=3

so no. is 73

Answer 2

$\circ\circ\circ\circ\underline{\fbox{\fbox{\fbox{\fbox{\fbox{\Large{\underline{\textbf{SOLUTION}}}}}}}}}\circ\circ\circ\circ$

$Let\:the\:unit\:place\:digit\:is\:'x'\\\\Then\:tens\:place\:digit=(10-x)\\\\Number=10(10-x)+x\\\\=100-10x+x\\\\=100-9x\\\\After\:reversing\:digit\:new\:number\\\\=10x+10-x\\\\=9x+10\\\\According\:to\:the\:question,\\\\\implies100-9x-9x-10=36\\\\\implies100-18x=46\\\\\implies100-46=18x\\\\\implies18x=54\\\\\implies x=3\\\\Number\\\\100-9\times3\\\\100-27\\\\=73\\\\=\boxed{\boxed{73\:answer}}$