Question

The sum of a two-digit number and the number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the number.

Answer 1

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▶⏩ Let the ten's digit be x.
and the unit's digit be y.


$\bf \boxed{Real \: number = 10x + y.}$

$\bf \boxed{Interchange \: number = 10y + x.}$

▶⏩Now, A/Q.

↪➡ sum of real number and interchange number is 99.

$\bf{=> (10y + x) + (10x + y) = 99.}$

$\bf{=> 11x + 11y = 99.}$


$\bf{=> 11(x + y )= 99.}$

$\bf{ = > x + y = \frac{99}{11} .}$

$\huge { = > x + y = 9...........(1).}$


▶⏩ Now,

$\huge{ = > x - y =3.........(2).}$


$\boxed{Substract \: in \: equation (1) and (2).}$

x + y = 9.
x - y = 3.
(-)..(+)...(-).
________
=> 2y = 6.

$\huge \boxed{=> y = 3.}$

Put the value of ‘y’ in equation (1).

=> x + 3 = 9.
=> x = 9-3.

$\huge \boxed{=> x = 6.}$

▶⏩ The required number is

= 10x + y.

= 10×6 + 3.

$\huge{= 63.}$


✅✅ Hence, the number is finded.✔✔



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