Seven times a two digit number is equal to four times the number obtained by reversing the order of digits. Find the number, if the difference between its digits is 3.

Let the digit at tens place be x and the digit at unit place be y.

Number = 10x + y

Number on reversing the digits = 10y + x

And, the difference between the digits = x - y or y - x

Given:

⇒ 7(10x + y) = 4(10y + x)

⇒ 70x + 7y = 40y + 4x

⇒ 70x - 4x = 40y - 7y

⇒ 66x = 33y

⇒ $\dfrac{x}{y} = \dfrac{33}{66}$

⇒ y = 2x …………… (1)

x - y = 3 …………… (2)

Or

y - x = 3 …………… (3)

On solving equations (1) and (2), we get:

x - 2x = 3

x = -3

y = 2 * (-3) = -6

Number = 10x + y = 10 * (-3) + (-6) = -30 - 6 = -36

This case gives a result where the digits are negative (-3 and -6), which is not possible for a two-digit number.

On solving equations (1) and (3), we get:

2x - x = 3

x = 3

y = 2 * (3) = 6

Number = 10x + y = 10 * (3) + (6) = 30 + 6 = 36

Hence, the number = 36.