Question

the digit at the tens place of a two-digit number is three times the digit at the ones place. If the sum of the number and the number formed by reversing it's digit is 88, find the number. ​

Answer 1

The digit at the tens place of a two digit number is three times the digit at ones place .

If the sum of this number and the number formed by reversing it's digits is 88 .

$\bf{\gray{\underbrace{\blue{TO\:FIND:-}}}} </p><p>$

The original number .

$\bf{\gray{\underbrace{\blue{SOLUTION:-}}}} </p><p></p><p>$

Let,

The tens digit be "y" .

And the ones digit be "x" .

★ Then, the original number is

$\bf\pink{(10y\:+\:x)}(10y+x) .</p><p>$

✞︎ According to the question,

CASE - 1 :-

y = 3 × x

$\bf{\green{\implies\:y\:=\:3x\:}}⟹ ----(1)$

CASE - 2 :-

☯︎ If we reserve the original number, then the new number is,

$\bf\pink{10x\:+\:y} \\ </p><p>\bf{\green{\implies\:10y\:+\:x\:+\:10x\:+\:y\:=\:88\:}}$

$➪ 11y + 11x = 88$

✨ Putting the value of “y = 3x” in the above equation,

➪ (11 × 3x) + 11x = 88

➪ 33x + 11x = 88

➪ 44x = 88

➪ x = 88/44

➪ x = 2

✨ Now, putting the value of “x = 2” in the equation (1),

➪ y = 3 × 2

➪ y = 6

✍️ Hence the original number is,

$\bf{\implies\:10y\:+\:x\:} \\ </p><p></p><p>\rm{\implies\:10\times{6}\:+\:2\:} \\ </p><p></p><p>\rm{\implies\:60\:+\:2\:} \\ </p><p></p><p>\bf\purple{\implies\:62}</p><p></p><p>\bf\red{\therefore}$

∴ The original number is "62" .

Hope its help u