Can anyone help me?

**angelmt** · July 9th 2009, 11:31 PM

A mug is 1/6 full of milk. If 565ml are added, the mug will be full. How much does the mug hold? Can someone help me find the solution?

**Chris L T521** · July 9th 2009, 11:36 PM

Originally Posted by angelmt

A mug is 1/6 full of milk. If 565ml are added, the mug will be full. How much does the mug hold? Can someone help me find the solution?

Let $M$ represent the amount of milk the mug can hold.

So we can set up the following equation:

$\underbrace{\frac{1}{6}M}_{\text{amount originally in mug}}+\underbrace{565}_{\text{amount added to mug}}=\underbrace{M}_{\text{total amount mug can hold}}$

Now, solving for $M$ , we have $565=\frac{5}{6}M\implies M=\dots$

Can you finish the problem?

**Soroban** · July 10th 2009, 06:38 AM

Hello, angelmt!

I don't suppose you made a sketch . . .

A mug is 1/6 full of milk.
If 565 ml are added, the mug will be full.
How much does the mug hold?

Code:

  -   *-------*   -
  :   |       |   :
  :   |       |   :
  :   |       |   :
  :   |  565  |   :
  C   |       | (5/6)C
  :   |       |   :
  :   |       |   :
  :   |       |   :
  :   * - - - *   -
  :   |///////| (1/6)C
  -   *-------*   -

Let $C$ = capacity of the mug (in ml).

The bottom $\tfrac{1}{6}C$ is filled with milk.

The top $\tfrac{5}{6}C$ contains 565 ml of milk.

There is our equation: . $\tfrac{5}{6}C \:=\:565$

Now solve for $C$ . . .

**Raoh** · July 10th 2009, 10:20 AM

hello

let $x$ be what you're looking for

$x = \frac{5}{6} +\frac{1}{6}$ ,when you're mug is full.
at the beginning the mug is $\frac{1}{6}$ full,which means $x$ equal only $\frac{1}{6}$ ,and when you added $565 ml$ , $x$ is $\frac{1}{6} + 565ml$ ,that is $565 = \frac{5}{6} = 5\times\frac{1}{6}$ and so $\frac{1}{6} = \frac{565}{5} =113 ml$ .
back to your $x$
$x = 113 + 565 = 678 ml$

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