Question

the sum of a two digit number and the number obtained by reversing its digit is 143.Find numbers if the tens place digit is greater than unit place digit by 3​

Answer 1

$\sf \bf \huge {\boxed {\mathbb {QUESTION}}}$

the sum of a two digit number and the number obtained by reversing its digit is 143.Find numbers if the tens place digit is greater than unit place digit by 3

$\sf \bf {\boxed {\mathbb {SOLUTION}}}$

• The number at the unit place digit be X.
• The number at the tens place digit be Y.

$\sf \bf \huge {\boxed {\mathbb {ANSWER}}}$

after interchanging the digits the number become 10 X+ Y.

$\sf \bf {\boxed {\mathbb {GIVEN}}}$

• The sum of a two digit number and the number obtained by reversing its digit is 143.

$\sf \bf {\boxed {\mathbb {TO\:FIND}}}$

• numbers if the tens place digit is greater than unit place digit by 3

$\huge \color{green} \boxed{\colorbox{lightgreen}{Answer}}$

So , 10 Y + X + 10 X + Y = 143

11 X + 11 Y= 143

X + Y = 13 ____(1)

Аlsо, in the given number the digit in unit’s рlасe is 3 mоre thаn the digit in the ten’s рlасe.

So X - Y = 3 ____(2)

after adding 1 and 2 we get ,

X = 8

substitute the value of X in (1) we get ,

Y = 5

$\huge\purple{\boxed{\red{\mathbb{\overbrace{\underbrace{\fcolorbox{blue}{white}{\underline{\purple{Answer : -}}}}}}}}}$

final answer is ,

Therefore the Number is

$58$

Answer 2

Answer:

ur final answer is 58...