Question

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Answer 1

Question

In a two digit number, the digit at ten's place is twice the digit at once place one's place. On reversing the digits, number became 36 less than the original number. What is the original number?

Solution

Let the digit at one's place be x

Digit at ten's place = Twice the digit at one's place = 2 * x = 2x

Number formed i.e original number = (10 * Digit at ten's place) + Digit at one's place

= 10(2x) + x

= 20x + x

= 21x

Number when digits are reversed =(10 * Digit at one's place) + Digit at ten's place

= 10(x) + 2x

= 10x + 2x

= 12x

Given

Original number - 36 = Number when digits are reversed

Substituting the values

⇒ 21x - 36 = 12x

⇒ 21x - 12x = 36

⇒ 9x = 36

⇒ x = 36/9

⇒ x = 4

Original number = 21x = 21 * 4 = 84

Hence, the iriginal number is 84.

Answer 2

Question :-

In a two digit number the digit at tens place is twice the digit at ones place .On reversing the digits number became 36 less then the original number .Find the original number ...?

Answer :-

→ Required number is 84 .

To Find :-

→ Original number

Step - by - step explanation :-

Let the digit at one's place is "a",

According to the question,

→digit at ten's place is equal to twice of digit at one's place = 2a ,

Original number formed ↓

→ 10× digit at ten's place + Digit at one's place

→ 10× 2a + 2a

→ 20a + a = 21a

According to the given condition ,

When digit are reversed ,

= 10× digit at one's place + digit at ten's place

= 10 ×a+2a

= 10a+2a

= 12a .

According to the question,

Original number - 36= reversed digit number ,

→ 21a - 36 = 12a

→ a = 4,

Hence original number after reversing the digits = 21a = 21×4= 84.

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