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The Weekly Challenge: A different SAT problem every week. Each week we'll post a new question to test your SAT reasoning skills. Try the problem, check for hints, discuss the answer and check back for the solution. Watch for contests to win prizes for you and your school.

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  #1  
Old 01-22-2010, 10:50 PM
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donnalack donnalack is offline
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Default The Weekly Challenge is Back!

New cars roll off the assembly line in the following colors: red, gold, blue, and tan. If the cars come off the line repeatedly in that order, what color is the 1,067th car?
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Old 01-23-2010, 10:04 AM
dylanlack dylanlack is offline
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red, because when you try to reach the answer by multiplying nothing reaches but itself
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Old 01-24-2010, 08:52 PM
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donnalack donnalack is offline
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Dylan,

You are on the right track: The answer has to do with division. Try again!
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Old 02-03-2010, 12:46 PM
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Quote:
Originally Posted by donnalack View Post
New cars roll off the assembly line in the following colors: red, gold, blue, and tan. If the cars come off the line repeatedly in that order, what color is the 1,067th car?
There are four colors that repeat indefinitely. Divide 1,067 by four using long division. What you are interested in is the remainder. A decimal remainder that you would get using your calculator will not help. You need the whole number remainder.

1067 divided by four is 244 with a remainder of 3. The third car off the line is blue. It helps to see it as a chart:

Remainder: 1-------2-------3-------0
---------red------gold-----blue-----tan

Note that when you divide, the maximum remainder is always one less than the number you divide by.
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Last edited by donnalack; 02-03-2010 at 12:54 PM.
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Old 02-07-2010, 02:55 AM
adhityaen adhityaen is offline
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A single 6-sided die is rolled. What is the probability of each outcome? What is the probability of rolling an even number? of rolling an odd number?

The possible outcomes of this experiment are 1, 2, 3, 4, 5 and 6.

Probabilities:
P(1) = # of ways to choose a 1 = 1
----------------------------- ---
total # of numbers 5

P(2) = # of ways to choose a 2 = 1
---------------------------- ---
total # of numbers 5

P(3) = # of ways to choose a 3 = 1
---------------------------- ---
total # of numbers 5

P(4) = # of ways to choose a 4 = 1
--------------------------- ---
total # of numbers 5

P(5) = # of ways to choose a 5 = 1
--------------------------- ---
total # of numbers 5

P(even) = # of ways to choose an even number = 2
------------------------------------------- ---
total # of numbers 5

P(odd) = # of ways to choose an odd number = 3
----------------------------------------- ---
total # of numbers 5

The outcomes 1, 2, 3, 4 and 5 are equally likely to occur as a result of this experiment. However, the events even and odd are not equally likely to occur, since there are 3 odd numbers and only 2 even numbers from 1 to 5.
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Last edited by adhityaen; 02-07-2010 at 03:01 AM.
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