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You go for a walk around the block with your daughter and. Two-thirds of the way, she stumbles, falls and gets a booboo. "Mommy, I have blood!" Some tears, some sobbing, some thumb sucking. You give your daughter a piggy-back ride the rest of the way and spend twice as much time on this little interval as you did while walking together. Father sees you approaching your home and says: "Hey, it looks like stuck on top of each other you are twice slower than walking side-by-side. I wonder how fast we'll go if I take both of you on my shoulders?" Is his estimation correct?

## 4 comments:

I think Father's estimation (twice slower) is different than what is given in the previous sentence. The earlier sentence says, I think, that the piggyback velocity was 1/4 the velocity of side-by-side. If I understand it right, his estimation is not correct.

And Dad's wondering how fast HE could carry Mom and Sissy is not relevant, and we have not information to guess.

I agree with Tom. Using distance = rate x time, we can see that the rate for the first leg of the walk is 2/3 d/t (because 2/3d = r x t). The rate for the second leg is 1/6 d/t (because 1/3d = r x 2t). 2/3 is 4 times 1/6 so walking side by side is 4 times as fast as the piggy back ride.

You both are right. Father mis-estimated the piggy-back speed. It is 4 times slower than walking side by side. And not 2 times as he suggested. Tom was first and deserves a puzzle point but it seems like he put "his signature" on some other puzzles as well. So, we'll give him a puzzle point for the other puzzles and this one will go to Jen. This is your second one, right? Or you are another Jen :)

No because they were two thirds of the way home, not one-half.

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