Answer: Option B
Answer: (b)Let the third number be x. ∴ Second number = 3x First number = 6x ∴ ${6x+3x+x}/3$ = 40⇒ 10x = 120 ⇒ x = 12 ∴ Required difference = 6x – x = 5x = 5 × 12 = 60 Aliter : Using Rule 15,From three numbers, first number is 'a’ times of 2nd number, 2nd number is 'b’ times of 3rd number and the average of all three numbers is x, then, First number = $\text"3ab"/ \text"1+b+ab"$ x ; Second number = $\text"3b"/ \text"1+b+ab"$ x ; Third number = $\text"3b"/ \text"1+b+ab"$ x Here, a = 2, b = 3, x = 40 Largest Number = First Number = $\text"3ab"/ \text"1+b+ab"$ ×x = ${3×2×3}/{1+3+2×3}$×40 = $18/10$×40 = 72 Smallest Number = Third Number = $3/\text"1+b+ab"$ ×x = $3/{1+3+2×3}$×40= $3/10$ ×40 = 12 Difference = 72 – 12 = 60