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?

Answer :

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 .