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Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

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Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

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#1 h Dec 23, 2024, 1:33 PM

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If the average of the set $5,x,10,10,10,10$ is $x,$ what is the value of $x$ ?

Compute the two-digit base $10$ number $n$ such that $n_9+n_7=n_{20}.$

William is ordering bottles. There are eight colors of bottles: White, Red, Blue, Green, Orange, Purple, Yellow, and Charteruse. What is the probability he puts the red bottle first and the white bottle last?

A paper towel roll is a cylinder with another cylinder in the middle cut out. Trying to save money, a CEO of a paper towel company makes the inside radius increase by $10\%.$ He is then sued, and forced to lower the price to match the original ratio of paper towel to cost. By what percentage does he lower the cost?

Eleven elves are making christmas presents. Each makes the same number of presents, and the sum of the digits of the total number of presents is $11.$ Also, after two elves steal all of the presents they made, the remaining number of presents ends in $5.$ How many presents did the two steal?

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#2 h Dec 23, 2024, 2:00 PM

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1
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NS0004

#3 h Dec 23, 2024, 5:02 PM

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1. (45+x)/6 = x so 45 +x = 6x so x = 9
3. You have 8! ways to order the bottles so the denominator will be 40320, Out of these you have 1 way to put the red bottle first and 1 way to put the white bottle last and 6! ways to order the middle 6 bottles. So the number of ways to put the red bottle first and the white bottle last is 6!= 720. So the probability will be 720/40320 which is 8!/6! which is just 1/56 because all the other terms cancel out. So the answer for number 3 is 1/56.
5. The restraints on the total number of presents, is that the digits must sum to 11 and the number must be divisible by 11. The only number that satisfies these constraints is 308 which 11 x 28. So each elf made 28 presents which means the presents that two of them stole is 2x28=56

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#4 h Dec 23, 2024, 5:07 PM

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$308$ is not the only multiple of $11$ with a digit sum of $11.$

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pingpongmerrily

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pingpongmerrily

#5 h Dec 23, 2024, 5:09 PM

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sol 1
sol 2
More coming later

This post has been edited 1 time. Last edited by pingpongmerrily, Dec 23, 2024, 5:17 PM

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#6 h Dec 23, 2024, 5:12 PM

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$28_7$ isn't possible, as $8$ is not a digit in bae $7.$

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pingpongmerrily

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pingpongmerrily

#7 h Dec 23, 2024, 5:15 PM

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fruitmonster97 wrote:

$28_7$ isn't possible, as $8$ is not a digit in bae $7.$

oh that's why there were two solutions which seemed off

ok let me fix that

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NS0004

#8 h Dec 23, 2024, 5:19 PM

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5. I just realized my mistake on problem 5, didnt read that after the presents of the two stealing elves were subtracted the number ended in 5. The correct answer is 110 because the total presents was 11x55 = 605. When 110 is subtracted from 605, you get 495 which ends in 5.

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#9 h Dec 23, 2024, 5:25 PM

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Also I just realized I made a small mistake in my solution for question 2 as I said, "So the probability will be 720/40320 which is 8!/6!" I meant to say 6!/8! instead.

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#10 h Dec 23, 2024, 5:29 PM

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P4 depends on the ratio of the radii of the two cylinders.
Its not a valid degree of freedom, I believe

This post has been edited 1 time. Last edited by Amkan2022, Dec 23, 2024, 5:29 PM

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