Of the three numbers, the first is twice the second and the second is thrice the third. If the average of the three numbers is 10, the largest number is :
Options:
A .  18
B .  30
C .  12
D .  15
Answer: Option A Answer: (a)Let the third number be x. ∴ Second number = 3x First number = 6x Now, $\text"x + 3x + 6x"/3$ = 10 ⇒ 10x = 30 ⇒ x = 3 ∴ The largest number = 6x = 6 × 3 = 18. 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 = 2, b = 3, x = 10 Largest number = $\text"3ab"/ \text"1+b+ab"$x = ${3×2×3}/{1+3+2×3}$×10= $18/10$ ×10 = 18
Submit Comment/FeedBack