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Question

(A) Let the three numbers be $x, y,$ and $z$.According to the problem:$x = 2y$ and $x = \frac{1}{2}z$,which implies $z = 2x = 2(2y) = 4y$.So,the numbe

Vedclass · Accepted Answer

$48, 24, 96$

Vedclass · Answer

(A) Let the three numbers be $x, y,$ and $z$.According to the problem:$x = 2y$ and $x = \frac{1}{2}z$,which implies $z = 2x = 2(2y) = 4y$.So,the numbers are $2y, y,$ and $4y$.The average of the three numbers is given as $56$.$\frac{2y + y + 4y}{3} = 56$$\frac{7y}{3} = 56$$7y = 56 	imes 3$$y = \frac{168}{7} = 24$.Now,calculating the numbers:First number $= 2y = 2 	imes 24 = 48$.Second number $= y = 24$.Third number $= 4y = 4 	imes 24 = 96$.Thus,the numbers in order are $48, 24, 96$.

Out of three numbers,the first is twice the second and is half of the third. If the average of the three numbers is $56$,the three numbers in order are

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