Question

A two-digit number is such that the ten's digit exceeds thrice the unit's digit by 3 and the number

obtained by interchanging the digits is 2 more than twice the sum of the digits. Find the number.​

Answer 1

Given :-

◉ A two digit number.

• The ten's digit exceeds thrice the unit's digit by 3.
• The number obtained by interchanging the digits is 2 more than twice the sum of the digits.

To Find :-

◉ The number

Solution :-

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

It is given that the ten's digit exceeds thrice the unit's digit by 3.

∴ x = 3y + 3 ...(1)

Now, The second case says that after interchanging the digits we get is 2 more than twice the sum of digits.

We know, A digit number is of the form , 10n + m

• n = Ten's digit
• m = unit's digit

⇒ 10y + x = 2(x + y) + 2

⇒ 10y + x = 2x + 2y + 2

⇒ 8y - x = 2

Substituting x = 3y + 2 from (1), we have

⇒ 8y - (3y + 3) = 2

⇒ 8y - 3y - 3 = 2

⇒ 5y = 5

⇒ y = 1

Put y = 1 in (1) , we get

⇒ x = 3×1 + 3

⇒ x = 6

We assumed this two-digit number to be 10x + y,

∴ Number = 61

Answer 2

Step-by-step explanation:

Assume that the ten's digit number be x and one's digit number be y.

A two-digit number is such that the ten's digit exceeds thrice the unit's digit by 3.

As per given condition,

→ x = 3y + 3 .................................(1)

The number obtained by interchanging the digits is 2 more than twice the sum of the digits.

• Original Number = 10x + y
• Interchanged Number = 10y + x

As per given condition,

→ 10y + x = 2(x + y) + 2

→ 10y + x = 2x + 2y + 2

→ 8y - x = 2

→ 8y - (3y + 3) = 2 [From (1)]

→ 8y - 3y - 3 = 2

→ 5y = 5

→ y = 1

Substitute value of y in (1)

→ x = 3(1) + 3

→ x = 6

Hence, the original number is 61 and interchanged number is 16.