A number consists of two digits, the difference of whose digits is 3. If 4 times the number equals 7 times the number obtained by reversing its digits, find the number.

[Hint. Original number is greater than the number obtained by reversing its digits. ∴ In original number, ten's digit greater than unit's digit.]

Let the tens and unit digits of required number be x and y. x > y (according to hint)

Given,

Difference of digits of the number is 3.

⇒ x - y = 3

⇒ x = y + 3     …..(1)

Original number = 10x + y

Number obtained by reversing the digits = (10y + x)

Given,

4 times the number equals 7 times the number obtained by reversing its digits.

⇒ 4(10x + y) = 7(10y + x)

⇒ 40x + 4y = 70y + 7x

⇒ 40x - 7x + 4y - 70y = 0

⇒ 33x - 66y = 0     …..(2)

Substituting the value of x from equation (1) in (2), we get :

⇒ 33(y + 3) - 66y = 0

⇒ 33y + 33 × 3 - 66y = 0

⇒ 99 - 33y = 0

⇒ 99 - 33y = 0

⇒ 33y = 99

⇒ y = $\dfrac{99}{33}$

⇒ y = 3.

Substituting value of y in equation (1), we get :

⇒ x = y + 3

⇒ x = 3 + 3

⇒ x = 6.

Original number = 10x + y

= 10 × 6 + 3

= 63.

Hence, the number is 63.