Answer: Option A
Answer: (a)Let second number be x ∴ The first number = 3x and the third number = 15x Now, x + 3x + 15x = 3 × 57 ⇒ 19x = 3 × 57 ⇒ x= ${3×57}/19$= 9 ∴ Required difference = 15x – x = 14x = 14 × 9 = 126 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 a = 3, b = 5, x = 57 First number = $\text"3ab"/ \text"1+b+ab"$x= ${3×3×5}/{1+3+15}$× 57 = $45/19$ × 57 = 135 Second number = $\text"3ab"/ \text"1+b+ab"$x = ${3×5}/{19}$× 57= $15/19$ × 57 = 45Third number = $3/ \text"1+b+ab"$x = $3/19$×57 = 9 Required result = 135 – 9 = 126