A bag contains some marbles. Half is given to A, one third of remaining to B and rest to C. If C has twice the number of marvels than B, how many marbels are there in all?

Let bag contains total marbles =

*x*

So, According to given information , Half is given to A

So,

A has total marbles = $\frac{x}{2}$

Remaining marbles after given half to A =

*x*- $\frac{x}{2}$ = $\frac{x}{2}$

From remaining one third of remaining to B , So

B has total marbles = $\frac{x}{2}$ $\times $ $\frac{1}{3}$ = $\frac{x}{6}$

So

C has total marbles = $\frac{x}{2}$ - $\frac{x}{6}$ = $\frac{3x-x}{6}$ = $\frac{x}{3}$

And also given C has twice the number of marbles than B , SO we get

C has total marbles = 2 ( B has total marbles )

$\frac{x}{3}$ = 2 ( $\frac{x}{6}$ )

$\frac{x}{3}$ = $\frac{x}{3}$

So here we can't find how many balls in the bag , there is some information is missing ,

As we take

*x*= 30

Than

A = 15

B = 5

And

C = 10

Or if we take

*x*= 60

Than

A = 30

B = 10

And

C = 20

So value of total balls in bag could be anything as here we have relationship between number of balls with A , B and C ,

So Please recheck your question again and get back to us , so we can help you .

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