Question

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 . Find the number .​

Answer 1

Answer:-

Let the digit at ten's place be x and digit at one's place be y.

The number = 10x + y or xy

Given:

Digit at the ten's place is three times the digit at ones place.

→ x = 3y -- equation (1)

Also,

The sum of the number and the number formed by reversing the digits = 88

Number formed by reversing the digits = 10y + x

Hence,

→ 10x + y + 10y + x = 88

→ 11x + 11y = 88

→ 11(x + y) = 88

→ x + y = 88/11

Substitute the value of x from equation (1).

→ 3y + y = 8

→ 4y = 8

→ y = 8/4

→ y = 2

Putting the value of y in equation (1) we get,

→ x = 3 * 2

→ x = 6

The number = xy = 62.

Therefore, the required two digit number is 62.

Answer 2

$\bf{\gray{\underbrace{\blue{GIVEN:-}}}}$

• 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 .

$\bf{\gray{\underbrace{\blue{TO\:FIND:-}}}}$

• The original number .

$\bf{\gray{\underbrace{\blue{SOLUTION:-}}}}$

Let,

• The tens digit be "y" .

• And the ones digit be "x" .

★ Then, the original number is $\bf\pink{(10y\:+\:x)}$ .

✞︎ 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,

• $\bf\pink{10x\:+\:y}$

$\bf{\green{\implies\:10y\:+\:x\:+\:10x\:+\:y\:=\:88\:}}$

➪ 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,

$\bf{\implies\:10y\:+\:x\:}$

$\rm{\implies\:10\times{6}\:+\:2\:}$

$\rm{\implies\:60\:+\:2\:}$

$\bf\purple{\implies\:62\:}$

$\bf\red{\therefore}$ The original number is "62" .