Question

In a number consisting of 2 digits, the digit in ten's place is 1 more than thrice the digit in unit's place. If 45 be subtracted from the number, the digits interchange their places. Find the number
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Answer 1

Answer:

The required two digit number is 72.

Step-by-step-explanation:

Let the digit at tens place be x.

And the digit at units place be y.

Two digit number = xy

⇒ Original number = 10x + y

From the first condition,

Digit at tens place = Thrice of digit at units place + 1

⇒ x = 3y + 1 - - - ( 1 )

From the second condition,

Original number - 45 = Number with interchanged digits

⇒ xy - 45 = yx

⇒ 10x + y - 45 = 10y + x

⇒ 10x + y - 10y - x = 45

⇒ 10x - x - 9y = 45

⇒ 9x - 9y = 45

⇒ x - y = 5 - - - [ Dividing both sides by 9 ]

⇒ ( 3y + 1 ) - y = 5 - - - [ From ( 1 ) ]

⇒ 3y + 1 - y = 5

⇒ 3y - y = 5 - 1

⇒ 2y = 4

⇒ y = 4 ÷ 2

⇒ y = 2

By substituting y = 2 in equation ( 1 ), we get,

x = 3y + 1 - - - ( 1 )

⇒ x = 3 * 2 + 1

⇒ x = 6 + 1

⇒ x = 7

Now,

The original number = 10x + y

⇒ The original number = 10 * 7 + 2

⇒ The original number = 70 + 2

⇒ The original number = 72

∴ The required two digit number is 72.