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Among three numbers, the first is twice the second and thrice the third. If the average of the three numbers is 49.5, then the difference between the first and the third number is
Options:
A .  39.5
B .  41.5
C .  54
D .  28
Answer: Option C
Answer: (c)Let the second number be x. ∴ First number = 2x ∴ Third number =${2x}/3$∴ 2x+x+${2x}/3$ = 49.5×3⇒ 6x + 3x + 2x =49.5×9 = 445.5 ⇒ 11x = 445.5 ⇒ x = ${445.5}/11$ = 40.5∴ Required difference = 2x - ${2x}/3$ = ${4x}/3$= ${4×40.5}/3$ = 54Aliter : 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 = $3/2$, x = 49.5 First Number = $\text"3ab"/ \text"1+b+ab"$ 5x = ${3×2×{3/2}}/{1+{3/2}+2}×{3/2}$×49.5 = ${18/2}/{11/2}$×49.5= ${18×49.5}/11$ = 18×4.5= ${18×45}/10$ = 81Third Number = $3/\text"1+b+ab"$ ×x = $3/{1+{3/2}+2×{3/2}}$×49.5 = $3/{11/2}$ ×49 5. = 6 ×4.5 = 27 Difference = 81 – 27 = 54

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