You make one pitcher of iced tea and one pitcher of lemonade for a party. just before the party starts, you catch a little mischievous child mixing the two beverages. the child says that only one cup of the iced tea was taken and put in the lemonade. then, the child took only one cup of the lemonade-tea mixture that was just created and put it back into the pitcher of iced tea, hoping that no one would notice what had been done. is there now more lemonade in the iced tea or more iced tea in the lemonade?

  Each pitcher has the same fraction of the other drink.

Step-by-step explanation:

After 1 cup of tea is added to x cups of lemonade, the mix has the ratio 1:x of tea to lemonade. So, the fraction of mix that is tea is 1/(x+1).

The 1 cup of mix contains 1/(x+1) cups of tea and so x/(x+1) cups of lemonade. When that amount of lemonade is added to the tea, it brings the proportion of lemonade in the tea to (x/(x+1))/x = 1/(x+1), the same proportion as that of tea in the lemonade.

You can consider the degenerate case of one cup of drink in each pitcher. Then when the 1 cup of tea is removed from its pitcher and added to the lemonade, you have a 50-50 mix of tea and lemonade. Removing 1 cup of that mix and putting it back in the tea pitcher makes there be a 50-50 mix in both pitchers.

Increasing the quantity in each pitcher does nothing to change the fact that the mixes end up in the same ratio:

  tea:lemonade in Pitcher 1 = lemonade:tea in Pitcher 2

definitely a. the others have almost all data points on one side of the line.

step-by-step explanation: