Answer: Option B
Answer: (b)Let the numbers be 2x, x and 4x respectively ∴ Average = ${2x+x+4x}/3$⇒ ${7x}/3$= 56⇒ x = ${3 × 56}/7$ = 24∴ First number = 2x = 2 × 24 = 48 Third number = 4x = 4 × 24 = 96 ∴ Required difference = 96 – 48 = 48 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 = $\text"3b"/ \text"1+b+ab"$ xHere, a = 2, b = $1/4$, x = 56First number = $\text"3ab"/ \text"1+b+ab"$x = ${3×2×{1/4}}/{1+{1/4}+2×{1/4}$×56= ${{3/2}×4}/{4+1+2}$×56 = 48Third number = $3/\text"1+b+ab"$ ×x = $3/{1+{1/4}+2×{1/4}$×56= ${3×4}/{4+4+2}$×56 = 96 Required difference = 96–48 = 48