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Middle School Math Quiz

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Post by indophile Fri Jul 22, 2011 12:32 pm

Relatively easy Middle School quiz problems from a math magazine. Please explain your answers (no need for a big narrative, just a short explanation will do).


1. John picked a chrysanthemum with a Fibonacci number of petals. He plucked one petal: “She loves me.” He plucked a second petal: “She loves me not.” He plucked a third petal: “She loves me.” He continued in this way until he reached the last petal. What is the probability that the last petal was a “she loves me” petal?

2. Angie rolls one ordinary six-sided die repeatedly, keeping track of each number she rolls, and stopping as soon as she rolls any number three times. If Angie stops after her twelfth roll and the sum of these rolls is 47, which number has she rolled three times?

3. A prime-prime is a prime number that yields a prime when its units digit is omitted. For example, 317 is a prime-prime (317 is a prime, and 31 is also a prime). How many two-digit prime-primes are there? Explain how you got to your answer.

4. Nine points are placed in a square array, as shown in the figure below. If 3 points are randomly selected, what is the probability that they are verticies of a triangle?

0 0 0
0 0 0
0 0 0


Last edited by indophile on Fri Jul 22, 2011 12:34 pm; edited 1 time in total (Reason for editing : improve font)

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Post by Guest Fri Jul 22, 2011 1:22 pm

Do they teach anything other than probability in middle school. Sad

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Post by indophile Fri Jul 22, 2011 1:23 pm

Tracy Whitney wrote:Do they teach anything other than probability in middle school. Sad

Yes, they do, but not much fun.

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Post by charvaka Fri Jul 22, 2011 3:26 pm

4. 19/21. Total number of combinations possible = 9C3 = 84. Of these, 8 combinations are straight lines (3 horizontal, 3 vertical, 2 diagonal). So combos that give you triangles = 84-8=76. Probability of getting a triangle = 76/84 = 19/21
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Post by charvaka Fri Jul 22, 2011 3:31 pm

3. 9. Only those two-digit primes beginning with 2, 3, 5, 7 count. That is: 23, 29, 31, 37, 53, 59, 71, 73, 79.
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Post by Hellsangel Fri Jul 22, 2011 3:35 pm

4. 8 possibilities yield straight lines out of 9 C 3
so 1- (8/9 C 3) = 76/84 = 19/21


Last edited by Hellsangel on Fri Jul 22, 2011 3:39 pm; edited 1 time in total
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Post by charvaka Fri Jul 22, 2011 3:37 pm

2. 6. Any number can occur at most twice in the first 11 throws. 2*(1+2+3+4+5+6) = 42. If anything other than 6 came thrice, the number would add up to less than 47.
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Post by indophile Fri Jul 22, 2011 3:38 pm

charvaka wrote:4. 19/21. Total number of combinations possible = 9C3 = 84. Of these, 8 combinations are straight lines (3 horizontal, 3 vertical, 2 diagonal). So combos that give you triangles = 84-8=76. Probability of getting a triangle = 76/84 = 19/21

Correct.

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Post by indophile Fri Jul 22, 2011 3:39 pm

charvaka wrote:3. 9. Only those two-digit primes beginning with 2, 3, 5, 7 count. That is: 23, 29, 31, 37, 53, 59, 71, 73, 79.

Correct.

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Post by indophile Fri Jul 22, 2011 3:40 pm

Hellsangel wrote:4. 8 possibilities yield straight lines out of 9 C 3
so 1- (8/9 C 3) = 76/84 = 19/21

Yes.

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Post by charvaka Fri Jul 22, 2011 3:42 pm

For question 1, I am struggling with reasonableness of assumption. What's the largest number of petals a chrysanthemum flower can possibly have? The smallest number? Once we establish the bounds, the answer is the probability that the Fibonacci number is an odd one.
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Post by indophile Fri Jul 22, 2011 3:45 pm

charvaka wrote:2. 6. Any number can occur at most twice in the first 11 throws. 2*(1+2+3+4+5+6) = 42. If anything other than 6 came thrice, the number would add up to less than 47.

Correct.

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Post by indophile Fri Jul 22, 2011 3:52 pm

charvaka wrote:For question 1, I am struggling with reasonableness of assumption. What's the largest number of petals a chrysanthemum flower can possibly have? The smallest number? Once we establish the bounds, the answer is the probability that the Fibonacci number is an odd one.

The answer is independent of how large a number of petals exist in that flower. All you need to know is that the number of petals is a Fibonacci number, and that you do not normally have a chysanthemum with a small number of petals like a jasmine (malle) or a crossandra (kanakaambaram).

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Post by charvaka Fri Jul 22, 2011 3:54 pm

Then the answer is 2/3. Because of the nature of the Fibonacci sequence, where there are two odd numbers and an even number, then two odd numbers and an even number.

Edit: We don't even need the assumption on the lower end. It holds right from the beginning: 1,1,2,3,5,8,13,21,34, etc.
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Post by indophile Fri Jul 22, 2011 4:02 pm

charvaka wrote:Then the answer is 2/3. Because of the nature of the Fibonacci sequence, where there are two odd numbers and an even number, then two odd numbers and an even number.

Yes. John's first petal was a "she loves me" petal. Hence all odd petals will be "she loves me" petals. The Fibonacci sequence begins 1,1,2,3,5,8,.... We observe that every third number is even and that other numbers are odd. Therefore, the probability that the chrysanthemum has an odd number of petals is 2/3.

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