Posted by Daniel on Thursday, January 8, 2009 at 9:37pm.
My question:
A spinner looks like this: A disc divided into sections that might or might not be the same size, with a pointer that when spun will land on some part of the disc at random. The sections might be colours, or, as in these problems, have numbers in them.
There are two problems, not associated with each other. The first is to help you get the idea. The second is more challenging. You will be given a series of six clues for each problem that should enable you to draw the spinners.
Problem 3a:
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The four numbers on the spinner are equally likely. They are also all different.
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The smallest sum you can get in two spins of this spinner is 2; the largest is 14.
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It is impossible to get an odd number if you spin this spinner twice and add the results.
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If you spin this spinner twice, and add the two numbers, you are just as likely to get 10 as 6.
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The most likely sum of two spins of this spinner is eight.
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It is impossible to get an even number on this spinner if you spin three times and add.
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DRAW THE SPINNER.
Problem 3b:
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The spinner has three sections. While no two are the same size, one of the sections is half the size of another.
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You are more likely to get 2/3 than one if you sum two spins, but 5/6 is the most likely sum of all.
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The sum of the three numbers on the spinner is one.
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The largest number you can get in two spins of this spinner is one.
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If you spin the spinner twice and add, you get a sum of one about a quarter of the time.
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If you spun the spinner a hundred times and added up all the numbers, you'd probably get somewhere near 40.
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DRAW THE SPINNER.

Math  PsyDAG, Friday, January 9, 2009 at 10:55am
Although I cannot completely solve the problem (3a), I can give you some help.
If the highest number for two spins is 14 and the lowest is 2, one number must be 1 and the other 7.
If it is impossible to get an odd number from adding 2 spins, all the numbers have to be odd. That goes along with the final clue too.
With the other clues, it seems like the numbers are 1, 3, 5 and 7. Check to see if it fits.
I hope this helps you get the idea for solving the second problem, where the numbers are fractions. Thanks for asking.

Math  NIgel, Saturday, March 14, 2009 at 6:12am
Sorry for replying so late after you have done this but I only just came across the page.
for the first one the answer is 1,7,3,5. As the lowest number must be a 1 and the highest a 7, all numbers must be odd if they come up with an even number after two spins and odd after 3. the only two odds in between 1 and 7 are 3 & 5.
For the second one I have found an answer that explains all but the second part of 2 and number 5. I get 1/3; 1/6 & 1/2. the highest number must be 1/2 to equal to 1. As one number is half of another but not the same as any other we try to use half of 1/2 which is 1/4 but this will mean we have 2 x 1/4 and 1/2 which we can not do. 1/3 is 2x1/6 and comes up with 2/3 when adding 2 spins together on 50% of the spins.The equation we have to solve is 2x + x + y = 1 so x=1/6 and y = ½.
I hope this helps as I have never done these before.
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