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 numbers and the
numbere formed by reversing its digit
il 88. Find the number,​

Answer 1

Answer:-

Original number = 62

$\rule{200}{1}$

Solution:-

Let the digit at unit's place be x so,digit at ten's place is 3x

Therefore,

→ Original number = x + 10(3x)

→ Original number = x + 30x

→ Original number = 31x

And now,

→ Interchanged number = 3x + 10x

→ Interchanged number = 13x

A/q,

→ 31x + 13x = 88

→ 44x = 88

→ x = 88/44

→ x = 2

∴ Digit at unit's place = 2

★ Digit at ten's place —

→ Digit at ten's place = 3x

→ Digit at ten's place = 3(2)

→ Digit at ten's place = 6

Therefore,

→ Original number = 31x

→ Original number = 31(2)

→ Original number = 62

And, Interchanged number (If required!)

→ Interchanged number = 13x

→ Interchanged number = 13(2)

→ Interchanged number = 26

Therefore,

$\therefore\underline{\boxed{\textsf{Original two-digit number = {\textbf{62 }}}}}$

Answer 2

Answer:

Step-by-step explanation:

Given :-

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 numbers and the  number formed by reversing its digit  is 88.

To Find :-

The Number.

Solution :-

let the digit at one's place be x.

And the digit at ten's place be 3x.

Original number = 10 (3x) + x  

Interchanged Number = 10x + 3x

According to the Question,

⇒ 10x + 3x + (10(3x) + x) = 88

⇒ 13x + 30x + x = 88

⇒ 44x = 88

⇒ x =  88/44

⇒ x = 2

Unit's place Digit = 2

Ten's place Digit = 3x = 3 × 2 = 6

Original number = 62

Hence, the required number is 62.