Sample Question #110 (mathematical brainteaser)

My younger brother and I go to the same school. It takes me 20 minutes to walk to the school from home and it takes him 30 minutes. One morning, my brother takes off for school 5 minutes before I. After how many minutes of my walking will I catch up to my brother?

(Comment: according to an article in the *New York Times *a few months ago, a question similar to this one was asked of 8-year-old Terence Tao, a Chinese-Australian math prodigy who’s now a renowned mathematician at UCLA. He was able to figure out the answer *in his head *in no time! What a genius. Please try this on your own before peeking at my answer in the Comment section.)

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ANSWER

Before you start solving the problem, first state the assumption that the two brothers will be walking at their usual speeds.

Let D be the distance between school and home, and x be the time, in minutes, at which point "I" catch up to "my brother." Notice that by the time "I" walked for x minutes, the two brothers have travelled the same distance.

Then it’s easy.

x * D/20 = (5+x) * D/30

Solve for x. (x=10)

Note: I made up the numbers in the question; I’m not sure if the question Terence Tao was actually given when he was 8 years old involved numbers that made it easier to solve. Still, it’s absolutely amazing an eight-year-old could work out the equation very quickly in his head, without writing anything down. (Unless, of course, he’d already seen the question before!)

You only need suppose your brother walk 1 unit/min, and you are 1.5 unit/min. 5×1/(1.5-1)=10min. I guess more than half of Chinese children at 8 years old may solve this problem less than 1 minutes. Terence Tao is a genius, but not at this problem.

Hehe, when I was an eight-year-old boy in China I only wanted to play soccer, eat cotton candy every day, and chase girls older than myself. 🙂 I remember getting 60 (the passing grade) once in a while in my math classes.

-brett

How about calculating it this way? The brother needs extra 10 min to walk the same distance. He started 5 min earlier so he would need another 5 min after I finish. This is symmetric, so I must have caught him in the mid-way.