Three consecutive whole numbers are such that if they be divided by 5, 3 and 4 respectively, the sum of the quotients is 40. Find the numbers.

Let the three consecutive whole numbers be (x - 1), x, (x + 1).

According to the question, the sum of the quotients is 40.

$⇒ \dfrac{x - 1}{5} + \dfrac{x}{3} + \dfrac{x + 1}{4} = 40$

Since L.C.M. of denominators 5, 3 and 4 = 60, multiply each term with 60 to get:

$⇒ \dfrac{(x - 1) \times 60}{5} + \dfrac{x \times 60}{3} + \dfrac{(x + 1) \times 60}{4} = 40 \times 60$

⇒ (x - 1) $\times$ 12 + x $\times$ 20 + (x + 1) $\times$ 15 = 2400\\[1em]

⇒ (12x - 12) + 20x + (15x + 15) = 2400

⇒ 12x - 12 + 20x + 15x + 15 = 40

⇒ 47x + 3 = 2400

⇒ 47x = 2400 - 3

⇒ 47x = 2397

⇒ x = $\dfrac{2397}{47}$

⇒ x = 51

Other number = (x - 1), (x + 1)

= (51 - 1), (51 + 1)

= 50, 52

Hence, the numbers are 50, 51 and 52.