Math Games

Archive for April, 2008

Question: probability math question?

Electronic baseball games manufactured by Tempco Electronics are shipped in lots of 24. Before shipping, a quality-control inspector randomly selects a sample of 6 from each lot for testing. If the sample contains any defective games, the entire lot is rejected. What is the probability that a lot containing exactly 3 defective games will still be shipped?

Answer: this lot of 24 contains 21 ok & 3 defectives

ways of getting 6 ok = 21C6

total ways of getting 6 = 24C6

thus reqd. probability = 21C6/24C6 = 0.4032 or 40.32 %

MLB Predictions for 2010 by NJIT Professor Bruce Bukiet

Question: (Intercepts, Zeros, and Solutions)How do I solve these math problems?

1. Two students are traveling in two cars to a football game 135 miles away. The first car travels at an average speed of 45 miles per hour. The second car starts one half hour after the first and travels at an average speed of 55 mph. How long will it take the second car to catch up to the first car? Will the second car catch up to the first car before the first car arrives at the game?

2. Two cars start at one point and travel in the same direction at average speeds of 40 and 55 miles per hour. How much time must elapse before the two cars are 5 miles apart? (How do I solve this without graphing?)

3. An executive flew in a jet to a meeting in a city 1500 miles away. After traveling the same amount of time on the return flight, the pilot mentioned that they still had 300 miles to go. If the air speed of the plane was 600 miles per hour, how fast was the wind blowing? (Assume that the wind direction was parallel to the flight path and constant all day)

Answer: 1) When the second starts, the first car is 22.5 miles away. (45 mph times 1/2 hour)
The difference in speeds is 10 mph.
It will take 2.25 hours for the second car to catch up to the first.

In 2.25 hours at 45 miles/hour, the first car will travel 101.25 miles. The second car will overtake the first one before it reaches the destination.

2) I'm assuming the cars start at the same time.
One car travels 15 miles per hour faster than the other one. At the end of one hour, it will be 15 miles ahead of the other car. That means, it will be 5 miles ahead after one third of an hour = 20 minutes.

3) The time needed to fly to the meeting with the wind is
1500/(600+W)
The time needed to fly 1200 miles against the wind is
1200/(600-W)
I'll factor out 100
15/(6+W) = 12/(6-W)
Multiply both sides by (6+W)(6-W)
15(6+W)/(36-W^2) = 12(6+W)/(36-W^2)

Pull out the common factor of 1/(36-W^2) and multiply terms
90+15W = 72-12W
Subtract 72 from both sides; Add 15 to both sides
18 = 27W
W = 2/3
Since I pulled out the factor of 100, the wind speed is 200/3 = 66.666666 miles per hour.
Checking:
1500/666.6666 = 2.25.
1500/533.3333 = 2.8125
Difference in times = 0.5625
Multiply the difference in times by 533.3333
we get 300 miles, just as we should.

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