Bottle A contains more soda than Bottle B. Pour from Bottle A into Bottle B as much soda as B already contains. Now pour from B into A as much soda as A now contains. Then pour from A into B as much soda as B now contains. Each bottle now contains 64 ounces. How many more ounces of soda were in Bottle A than Bottle B at the beginning?

Bottle A started with 88 and bottle B started with 40

Step-by-step explanation:

To find out what they started with I got the ounces they finished with an the equation they did to get there.

Ended in 64 oz.

To get there:

A => B

B => A

A => B

Every time a pour is made the bottle to the right is doubled so I got 64 and divided it by 2 to see what B had before the pour of bottle A.

B had 32 oz. and A had 96 oz. before the last pour. I got 96 oz. because for them to end up with 64 oz. A has to have 32 oz. + 64 oz. since you are pouring 32 oz. into bottle B.

Now you have this equation:

A => B

B => A

96 oz. => 32 oz.

64 oz.

I then do the same thing I did to find 96 and 32. Since B is giving to A now it has to have more then A but it has to end up with 34 oz. for this is got 96 oz. and I divided it by 2 which got me 48. So on the second pour A started with 48 and B had to have given A 48 plus have 32 so it could end up with 34 therefore making it have 80 oz.

Now you have this equation:

A => B

80 oz. => 48 oz.

96 oz. => 32 oz.

64 oz.

For the last part to get the first ounces to answer the question you will do the same steps. Get half of 80 to see what B had before the 2nd pour which will be 40 and then you have to get 48 + 40 sine bottle A has to have the same amount of soda as B plus 48 oz. that it ends up with. This will have bottle A as 88 oz.

The

88 oz. => 40 oz.

80 oz. => 48 oz.

96 oz. => 32 oz.

Both will end up with 64 oz.

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step-by-step explanation:

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