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(C) Let the three numbers be $x, y,$ and $z.$According to the problem,the second number $y$ is twice the first number $x$,so $y = 2x$.The second numbe

Vedclass · Accepted Answer

$72$

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(C) Let the three numbers be $x, y,$ and $z.$According to the problem,the second number $y$ is twice the first number $x$,so $y = 2x$.The second number $y$ is also thrice the third number $z$,so $y = 3z$,which implies $z = y/3 = 2x/3$.The three numbers are $x, 2x,$ and $2x/3$.The average of these three numbers is given as $44$,so:$(x + 2x + 2x/3) / 3 = 44$$(3x + 2x/3) / 3 = 44$$(9x + 2x) / 9 = 44$$11x / 9 = 44$$x = (44 	imes 9) / 11 = 36$.The numbers are $x = 36$,$y = 2(36) = 72$,and $z = 72/3 = 24$.The largest number among $36, 72,$ and $24$ is $72.$

Of the three numbers,the second is twice the first and is also thrice the third. If the average of the three numbers is $44,$ the largest number is

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