Question

The difference between the digits of a two digit number is $3$ . If we reverse the digits of the number and multiply it by $4$ , we obtain $7$ times the original number. What is the original number?

Answer 1

Answer:

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Answer 2

The original number is 36.

Given: The difference between the digits of a two-digit number is 3. If we reverse the digits of the number and multiply it by 4, we obtain 7 times the original number.

To Find: The original number.

Solution:

Let the unit place of the original number be = y

and the ten's place of the original number be = x

We are given that the difference between the digits of a two-digit number is 3. So, assuming the unit's digit to be greater than the ten's digit, we can say

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

Now, the structure of the original number becomes = 10x + y

Now, to reverse the number means the unit's place becomes x and the ten's place becomes y. So, framing the reversed number we get,
the structure of the reversed number becomes = 10y + x

According to the conditions given,

    4 × ( 10y + x ) = 7 × ( 10x + y )

⇒ 40y + 4x = 70x + 7y

⇒ 33y = 66x

⇒ y = 2x

Putting y = 2 in (1), we get;

  y - x = 3    

⇒ 2x - x = 3

⇒ x = 3

So, y = 2×3 = 6

So, the original number is = 10x + y

Putting respective values, we get;

⇒ 10 × 3 + 6

⇒ 36

Hence, the original number is 36.

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