Seema has three balls, $A,\ B,\ C$. If $A$ is twice as heavy as $B$ and $B$ weight one- third of $C$. Find the heaviest ball.

 Given: Seema has three balls, $A,\ B,\ C$. $A$ is twice as heavy as $B$ and $B$ weight one- third of $C$.

To do: To find the heaviest ball.

Solution:

Let $a,\ b\, c$ are the weight of the balls  $A,\ B,\ C$ respectively.

As given,

Weight of the ball $A=a=2b\ .....\ ( i)$     [$\because A$ is twice as heavy as $B$]

Weight of the ball $B=b=\frac{c}{3}\ .....\ ( ii)$      [$\because B$ weight one- third of $C$]

From $( i)$ and $( ii)$, 

$a=2b=\frac{2c}{3}$

Here we find,

$\frac{c}{3}<\frac{2c}{3}

Therefore, $weight( A)

Thus, $C$ is the heaviest ball.

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Updated on: 10-Oct-2022

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