Middle School Math Quiz

by indophile Wed Jun 15, 2011 12:24 pm

I selected these 4 relatively easy questions from an actual quiz of 30 problems published in a math magazine. I also have the answers for these questions (with explanations) given in that magazine.

1. Each day Jane makes an open-faced sandwich using only one slice of bread. She uses 1 kind of meat or 1 kind of cheese or 1 kind of each. If she chooses from 4 kinds of bread, 5 kinds of meat, and 3 kinds of cheese, how many different sandwiches can she make?

2. Ten boys and girls write their names on slips of paper – one name per slip – to enter a prize drawing. Two of the names are drawn at random without replacement. If the probability that both winners are boys is 1/15, how many boys are in the group?

3. If Jack and Jill each randomly choose a whole number from 1 to 1000 inclusive, what is the probability that Jill’s number is larger than Jack’s?

4. Mental math – neither calculator nor pencil permitted. The sum of the first 50 positive odd integersis 50-squared. Find the sum of the first 50 positive even integers.

by charvaka Wed Jun 15, 2011 12:26 pm

1. 4x5 + 4x3 + 4x3x5 = 92

by Hellsangel Wed Jun 15, 2011 12:31 pm

2. 3 boys

by charvaka Wed Jun 15, 2011 12:32 pm

2. 3. Basically the denominator of the probability should be 9x10. Expressing it as a fraction of 90, we have 6/90. We also need teh numerator to be (n - 1) x n where n is the original number of boys. So n = 3.

by charvaka Wed Jun 15, 2011 12:40 pm

4. 50-squared + 50 = 2550. Basically each of the 50 even integers is 1 more than the corresponding odd number in the first series.

by charvaka Wed Jun 15, 2011 12:42 pm

3. My intuitive answer is 1/2. Neither of them have higher probability of picking the large number.

by indophile Wed Jun 15, 2011 12:47 pm

Incorrect.

by charvaka Wed Jun 15, 2011 12:49 pm

indophile wrote:Incorrect.

Which one? Or all of them?

by indophile Wed Jun 15, 2011 12:57 pm

charvaka wrote:
indophile wrote:Incorrect.
Which one? Or all of them?

Your answer to # 3 (i.e., 1/2) is incorrect.
Yours and HA's answers to 1,2,and 4 are correct.

by indophile Wed Jun 15, 2011 4:15 pm

indophile wrote:

1. 92. Jane can make 4*5 = 20 kinds of meat sandwiches, 4*3=12 kinds of cheese sandwiches, and 4*5*3 = 60 kinds of meat and cheese sandwiches. Thus, she can make a total of 20+12+60 = 92 different types of sandwiches.

2. 3 boys. Let b= the number of boys. The probability that the first winner is a boy is b/10, and the conditional probability that the second winner is a boy is (b-1)/9. The probability that both events occur is is the product of these two probabilities: b(b-1)/(10*9) = 1/15. In order for b > 0, b = 3. There are 3 boys in the group.

3. 999/2000. By symmetry, the probability that Jill’s number exceeds Jack’s number equals the probability that Jack’s number exceeds Jill’s number. Therefore, 2*P(Jill’s number > Jack’s number) + P (Jill’s number = Jack’s number) = 1. The probability that Jack and Jill independently choose the same number is 1/1000. Then, (1- 1/1000)/2 = 999/2000.

4. 2550. Sum the first 100 integers using Gauss method (Sum = n(n+1)/2). It’s 5050. Then subtract the sum of the first 50 odd integers: 5050-2500 = 2550 which is the sum of the first 50 even integers.
(Charvaka's solution is even easier than this)

by charvaka Wed Jun 15, 2011 4:39 pm

Very nice... I had the idea that neither of them has an "edge," but didn't take it to the logical conclusion.

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