This past week I gave you two SAT problems. You can find the first one here. The second one required a diagram, and you can view that one here.
I also gave you a riddle of sorts, though it involved a lot of math. I appreciated all the complaints about the wording, but the fact remains that I got in one try and so did Eric.
Finally, this brain teaser comes to you compliments of my friend, Aithan. You'll find another question from Victor in the comments. I personally agree with Eric's answer to that one. Maybe Victor will tell us if he's right.
And what good are all these fun questions without answers?
SAT question one:
Answer: 105
For this problem there are a number of strategies you can use. One is to recognize that if the average of ten numbers is 100, then the numbers must add up to 1000. If we then remove the numbers 90 and 70 from the group we have eight numbers that add up to 840. Dividing this by eight gives us a new average of 105.
SAT question two:
Answer: A) 21
This problem simply requires you to know the pythagorean theorem which states that in a right right traingle with legs of length a and b and hypotenuse of length c, a^2 + b^2 = c^2. Using this on all the triangles eventually yields x^2 + y^2 = 21.
Mathy riddle question:
Answer: 3 minutes fast
The initial trip was suppose to take an hour, but was five minutes late, so it actually took 65 minutes. The fact that Mr. Moody's watch shows the train to be three minutes early indicates that he has no idea the train is actually late. Therefore, when he sets his watch forward he is actually setting to match an hour-long trip when the trip was actually 65 minutes. At this point his watch is off by five minutes. The return trip is 25% faster. This refers to the speed, not the time. To figure out what this means for the speed we can use the face that distance = speed x time. For both trips the distance remains the same, so speedtrip1 x timetrip1 = speedtrip2 x timetrip2. We also know that speedtrip2 = 1.25 x speedtrip1. This means that speedtrip1 x 65 = 1.25 x speedtrip1 x timetrip2. Dividing both sides of the equation by 1.25 and speedtrip1 gives us timetrip2 = 52. This is a difference of eight minutes on a trip that should take one hour. However, because Mr. Moody's watch is off by five minutes, the station is actually only fast by three minutes.
People on a bus question:
Answer: 4 people get off the bus
There is a pattern to this problem. One way to figure it out is to consider a group of four people. Let's assume one has a blue hat. The person with this hat will see all white hats on the other people. Since they know someone has to have a blue hat, this person will immediately know he has the blue hat and get off at the first stop. Now consider the case where two people have blue hats. They will both see one blue hat and two white hats. If the person that they see with the blue hat was the only one wearing a blue hat, that person would immediately know it and would have gotten off at the first stop. Because he did not get off, then there must be another blue hat. Since these people can only see one blue hat, they know they must be wearing the other one, and they will then get off at the second stop. Assume now that there are three blue hats. These three people see two blue hats and one white hat. They know that if there were only the two blue hats they see then those two people would have gotten off at the second stop. When they don't get off they know there must be another blue hat. Since they only see two, they must be the one's wearing it. They will then get off at the third stop. Now think back and realize it doesn't matter how many white hats there are in any of these situations, so we don't need to start with four people for this to work. The pattern will continue, so if people get off at the fourth stop, there must be four blue hats.
PS. I advise you to watch this.
This is just going to turn into a forum for riddles, I think.
ReplyDeleteHi Jon!
ReplyDeleteOMG HI JUDY!!!!
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