Friday, May 31, 2019

POTW #34 & #35 - POTW 'til the End!

Remember to attempt BOTH POTWs regardless of grade and BOTH POTWs because I missed last week's, oops! Thanks Fiona for keeping me honest.

POTW #33 Solution Grade 8

POTW #33 Solution Grade 7

POTW #34 Question #1 Grade 8

POTW #34 Question #2 Grade 8

POTW #34 Question #1 Grade 7


 POTW #34 Question #2 Grade 7
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7 comments:

  1. EdwardJune 1, 2019 at 1:07 PM

    Grade 7 POTW #2:
    The only possible age of the father and son is 8 and 35, meaning that the son is 8 and the father is 35 years old. 35 + 8=43 and in one year, the sum has to be 45 years because both the son and father grow one year older. That would mean they would be 9 and 36 and 36/9=4.

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  2. Fiona BianJune 2, 2019 at 3:35 PM

    test

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  3. Fiona BianJune 2, 2019 at 3:36 PM

    Grade 7 POTW Question 1:

    Key Information:
    - 1/6 are seniors
    - 1/4 are teens or children
    - 3 times as many adults as teens
    - 138 children on board

    Let s represent the number of seniors, t represent the number of teens, c represent the number of children, a represent the number of adults and x represent the total number of passengers.

    Seniors = (1/6)x
    Children and teens combines = c + t = (1/4)n
    Teen = 138 + t = (1/4)n
    Teen = t = (1/4)n - 138
    Adults = a = 3t = 3((1/4)x - 138)

    x = (1/6)x + ((1/4)x - 138) + 3((1/4x) - 138) + 138
    x = (1/6)x + 1/4x - 138 + 3/4x - 414 + 138
    x = (1/6)x + x - 414
    x = (7/6)x - 414
    414 = (7/6)x - x
    414 = (1/6)x
    414 x 6 = x
    x = 2484

    There are 2484 passengers.

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  4. Fiona BianJune 2, 2019 at 3:38 PM

    Grade 8 POTW Question 2:

    Key Information:
    - 100 cards numbered 1 to 100
    - Starts red side up
    - Turns over multiples of 2, 3, 4, and then 5.
    - How many cards have red side facing up?

    First we need to account for the prime numbers that will be left untouched, meaning they will end up red because they started off red. There are 25 prime numbers, including 2, 3 and 5, but only 22 excluding them.
    Now we know that there is 22 + (some number) cards already that are red facing up.

    Now we need to consider all possible cases with numbers that may be multiples of one number, but not the others.

    Case 1: Multiple of 2 but nothing else.
    We know that there are 50 multiples of 2 from 1-100. But now we need to remove all the numbers that are multiples of 3,4 and 5. There are 25 numbers that are multiples of 4, leaving us 25 numbers left. Then there are 11 numbers that are multiples of 3 and 5, so that leaves us only 14 numbers.

    Case 2: Multiple of 3 but nothing else.
    There are 33 multiples of 3 from 1 to 100. But we need to remove the even numbered ones because they would be a multiple of 2 as well. There are 17 even numbers, leaving us only 16 numbers. Then we need to remove the numbers that are multiples of 5, which is only 2 numbers. This leaves us only 14 numbers.

    Case 3: Multiple of 4 but nothing else.
    This is impossible because a multiple of 4 must also be a multiple of 2. This case is irrelevant.

    Case 4: Multiple of 5 but nothing else.
    There are 20 numbers that are multiples of 5, but half of those end in 0, making them also a multiple fo 2. This leaves us 10 numbers. 3 of these would be a multiple of 3, which leaves us 7 numbers.

    Case 5: Multiple of 3, 4, 5 but not 2
    It is impossible for a number to be a multiple of 4 but not a multiple of 2. This case is impossible.

    Case 6: Multiple of 2, 4, 5 but not 3.
    If a number is a multiple of 2, 4 and 5, it must be a multiple of 20. There are only 5 numbers in tis cause. But 60 is a multiple of 3, so we need to ignore it. There are only 4 numbers.

    Case 7: Multiple of 2, 3, 5 but not 4.
    If a number is a multiple of 2, 3, 5, it needs to be a multiple of 30. There are 3 of these numbers, but 60 is a multiple of 4, so we ignore it. There are only 2 numbers.

    Case 8: Multiple of 2, 3, 4 but not 5.
    Multiples of 2, 3 and 4 are all multiples of 12. But we don’t need multiples of 5, also known as, 12 x 5, multiples of 60. There is only one multiple of 60 out of all the 8 multiples of 12. There are 7 numbers

    Now we need to subtract all these multiples to find those that aren’t touched or are flipped back to red.

    100 - 7 - 14 - 14 - 4 - 2
    = 52

    52 cards are red-side facing up

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  5. AnonymousJune 5, 2019 at 9:25 PM

    Grade 8 POTW #1
    Okay, so there are an x number of tokens. The average of the tokens are 56. Once the 68 is removed, the new average is 55 with x - 1 tokens. Therefore, we can find the value of x through this equation:
    (56x - 68)/(x - 1) = 55
    (56x - 68)/(x - 1) x (x - 1) = 55(x - 1)
    56x - 68 = 55x - 55
    56x - 68 + 68 - 55x = 55x - 55 + 68 - 55x
    x = -55 + 68
    x = 13
    Out of these 13 tokens, one is 68, and to maximize an integer that can appear, we have to minimize all other integers. In other words, there will be one 68, eleven 1s, and the remaining amount. This remaining amount will be:
    56(13) - 68 - 11 = 728 - 68 - 11 = 649
    Therefore, the largest possible integer that could appear on one of the tokens in the bag is 649.

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  6. AnonymousJune 5, 2019 at 9:25 PM

    Grade 8 POTW #2
    To start things off, there are 5 possible scenarios. They include whether a card is flipped four times, three times, two times, one time, or none. If it is flipped four times, it will remain red. If it is flipped three times, it will become yellow. If it is flipped two times, it will remain red. If it is flipped one time, it will become yellow. If it isn’t flipped, it will remain red. There are two ways to do this. We can either solve for the number of cards that will remain red, or solve for the number of cards that will become yellow, and subtract it from 100. I will do the first one.
    Okay, so we need to solve for each of these cases alone: flipped four times, two times, zero times.
    Flipped four times:
    The cards that will be flipped four times has to be a multiple of 4 x 3 x 5 = 60 (note that 2 wasn’t included as it is already a factor of 4). There is only 1 number that is a factor of 60 between 1 and 100.
    Flipped two times:
    If it is a multiple of 2 x 3 = 6, but not a multiple of 4 or 5, then it will remain red. There are 16 multiples of 6 between 1 and 100. 8 are multiples of 4, 3 are multiples of 5, but there is one overlap between 4 and 5. The overlap was already included so there are 16 - 7 - 2 = 7.
    If it is a multiple of 4 (2 and 4), but not a multiple of 3 or 5, then it will remain red. There are 25 multiples of 4 between 1 and 100. 8 are multiples of 3, 5 are multiples of 5, but there is one overlap between 4 and 5. The overlap was already included so there are 25 - 7 - 4 = 14.
    If it is a multiple of 2 x 5 = 10, but not a multiple of 4 or 3, then it will remain red. There are 10 multiple of 10 between 1 and 100. 5 are multiples of 4, 3 are multiples of 3, but there is one overlap between 3 and 4. The overlap was already included so there are 10 - 4 - 2 - 1 = 4.
    If it is a multiple of 3 x 5 = 15, but not a multiple of 2 or 4 (or just 2), then it will remain red. There are 6 multiples of 15 between 1 and 100. 3 are multiples of 2 so there are 6 - 3 = 3.
    In total, there are 7 + 14 + 4 + 3 = 28 cards flipped twice.
    Flipped zero times:
    A card isn’t flipped if it isn’t a multiple of 2, 3, 4, or 5, or in other words, if it is a one, or a prime number not equal to 2, 3, or 5. In total there are 25 prime numbers between 1 and 100. 25 - 3 = 22. 22 + 1 = 23 total numbers not flipped.
    In total there are 1 + 28 + 23 = 52 cards that will be red.
    Therefore, there are 52 cards that will be red faced up when flipped if it is a multiple of 2, 3, 4, or 5 if the cards are labelled from 1 to 100.

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