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Math Challenge IB

 
 
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Math Challenge IB
by Amber Lin - Friday, June 21, 2019, 9:13 PM
 

I need help on these following problems in Chapter 9:

9.3(c)

9.22

9.23

9.24


 
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Re: Math Challenge IB
by Areteem Professor - Monday, June 24, 2019, 7:50 PM
 

Hi Amber,

Please remember to tell us from which subject you need help. (The last few times you asked for help you mentioned Counting and Probability, but now I see you are taking Number Theory as well. We just want to be sure we are helping you with the correct problems.)

Here are some hints for Counting and Probability. If you needed help with the Number Theory problems, please let us know and we'll reply to this post again.

9.3(c): Since we are picking the integers in order, and it is possible to pick repeats, they can be picked in $\underbrace{10 \times \dots\times 10}_{5\text{ times}} = 10^5$ different ways. For the mean of the list of numbers to be $2$, the numbers must add to $2 \times 5 = 10$. So, we want to count how many different integer solutions are there to the equation $a + b + c + d + e = 10$, where each number is at least one. Note we are picking the numbers in order: first $a$, then $b$, and so on. We can count these using the positive version of stars and bars: there are $\displaystyle \binom{10 - 1}{5 - 1} = \binom{9}{4} = 126$ solutions. Therefore, the probability that the mean of the list of numbers is $2$ is $\dfrac{126}{100000} = \dfrac{63}{500000}$.

9.22: Say $n$ is the number of additional quizzes she takes. What is the maximum possible sum of the scores of her current and additional quizzes? (Recall each quiz is worth at most $100$ points). How many quizzes would she have taken in total? Using these, find an expression for the average of all of her quizzes and use it to set up an inequality to find an average of at least $90$.

9.23: Since the list of numbers has an even number of numbers, the median will be the average of the two numbers in the middle of the ordered list. If all numbers were $1$, clearly the median would be $1$. If all numbers were $5$, clearly the median would be $10$. What numbers in between $1$ and $10$ are also possible? For  example, if the numbers (in order) were $1, 1, 2, 6$, the median would be $(1 + 2)\div 2 = 1.5$. 

9.24: To yield the largest possible difference between the median and the mean, we want to make the median as small as possible while making the mean as big as possible. Clearly the smallest possible median is $0$. To get this median, what is the least number of $0$'s you would need in your list? What should the remaining values be so that the mean is as big as possible? What is the mean in this case?

Let us know if you need additional hints. Remember it is also really helpful to us if you tell us what you have tried so far, and/or where you got stuck so we can provide you with better hints.