Algebra Problems, Geometry Problems, Math Problems, Number Theory Problems, Probability Problems

Daily Quiz #2: March 23, 2014

As of writing this post, it looks like many of you have not seen the contest yet. I’m hoping those of you that do know about it will tell the others, and I’ll give you a day to get points for the questions you answer on Quiz #1. Remember, I’m preparing you for your competition this weekend!

6. Willy Wonka’s famous ice cream shop has three types of ice cream cones: the Rocky Road Regular, the Tiger Stripe Tall, and the White Chocolate Wide. Yum. They all have the same dimensions, except the tall’s height is three times the regular’s, and the wide’s diameter is twice the regular’s. Does the tall cone or the wide cone hold more ice cream, and how much more? Express your answer as a common fraction (ratio of the bigger cone to the smaller one). (10 pts.)

7. Given a standard, fair coin, what is the probability of flipping the coin three times and getting exactly two heads? (10 pts.)

8. If the sum of 10 consecutive numbers is 395, what is the sum of the odd numbers among them? (10 pts.)

9. Bonus. Joh can mow half an acre of lawn in 1 hour. His good friend Num can mow a full acre in one and a half hours. If they start mowing a 10.5-acre lawn at 10:00 AM together, at what time will they finish? (20 pts.)

10. Bonus. The height of a falling object in feet as time passes is given by h = -16t2vts, where v is the initial velocity of the object and s is the initial height. If a projectile is launched off a 24-foot building at an initial velocity of 40 feet per second, how many seconds later will the projectile hit the ground? (20 pts.)

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Algebra Problems, Geometry Problems, Math Problems, Number Theory Problems, Probability Problems

Daily Quiz #1: March 22, 2014

The state competition is coming up! We’re already making great progress towards doing well at state – now we just need to keep up the skills until next weekend. So, here’s what I have in mind: every day from now until Thursday, I’ll post five problems on this blog. You, the Mathlete, should solve everything you can and post your answers (along with your name) in the comments. The next day, the answers will be posted so you can check your work.

Here’s the game, though: you will receive points for each question you answer correctly. Every day, I’ll post each person’s rankings along with the answers. Who will be the best Mathlete of all? We shall see….

1. I bought a one-year membership to the Abscissa Nature Park the other day for $99. The membership lets me visit the park for only $5, while non-members must pay $18. How many visits do I have to make this year to make buying the membership worth my money? (10 pts.)

2. Srikar is looking for books at the Broadmoor Library, and there are 70 fiction books and 80 nonfiction books on the nearest shelf. If no two books are the same, what is the probability that he will randomly choose one book of each kind without replacement? (10 pts.)

3. Sunjay is thinking of three positive integers. He tells his friend Robert that the sum of the first and second numbers is 8 less than the sum of the first and third, and 14 less than the sum of the second and third. If the second number is three times the first number, what is the sum of all three numbers? (10 pts.)

4. Bonus. What is the sum of 17268 and 3246? Express your answer in base 10. (20 pts.)

5. Bonus. The plot of land for Caroline’s new house is a trapezoid with two right angles, two sides of length 300 m, and a diagonal of length 500 m. What is the length of the other diagonal? Express your answer as a decimal to the nearest tenth of a meter. (20 pts.)?

Math Problems, Probability Problems

Problem of the Day: 3/12/13

In how many different ways can 10 people be seated at a round table? Note that arranging the people the same way at different positions around the table does not count.

Solution to yesterday’s problem:

Since the common factor of the two numbers is 4, both numbers are divisible by 4. We can therefore say that the first number is 4 times a number, or 4m. The second number is 4 times another number, or 4n. The problem tells us that

4m+4n=52

Factoring out a 4 on the left side,

4(m+n)=52

We can now divide by 4:

m+n=13

So the range of possible values for m and n is such that their sum is 13: 1 and 12, 2 and 11. However, logic tells us that since we want the smallest possible difference between the numbers, we should choose numbers close to 13/2; namely, 6 and 7. Using 6 and 7, our original numbers would be 24 and 28. These do have a sum of 52 and a GCF of 4, and their difference is 4.

Math Problems, Probability Problems

Problem of the Day: 3/9/13

In music, a single beat consists of one quarter note, which can be subdivided into two eighth notes. Each eighth note can be further subdivided into two sixteenth notes. How many four-beat rhythms can be constructed from quarter notes, eighth notes, and sixteenth notes?

Solution to yesterday’s problem:

All you need to do in this problem is write the factorial out as its entire product, like this:

14\times13\times12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1

Then split each factor into its prime factors:

7\times2\times13\times3\times2\times2\times11\times...

We don’t need to go any further because we already have the greatest prime factor we’re going to find here: 13.

Math Problems, Probability Problems

Problem of the Day: 3/6/13

A frog can jump between three lily pads with equal probability of landing on any one. If the frog can only jump to a different lily pad, what is the probability that the frog will start and end on the same lily pad after three jumps?

Solution to yesterday’s problem:

For every red flower in a garden, there are three blue flowers. For every two blue flowers, there are five purple flowers. For every flower of these colors, there are six yellow flowers. If there are fifteen purple flowers, how many flowers are there in all?

If there are fifteen purple flowers, then there are six blue flowers by the second statement. Then there are two red flowers by the first statement. Then there are 23 total flowers, and 23 x 6 = 138 yellow flowers. Adding these up together, we find that there are 161 flowers in this (apparently) large garden.

Math Problems, Probability Problems

Problem of the Day: 2/26/13

A bowl contains three red, green, blue, and purple marbles. If Stefan draws out three marbles at random without replacement, what is the probability that all three marbles will be green?

Solution to yesterday’s problem:

If you were given this problem in, say, a Countdown Round at Mathcounts, you would probably realize that time is of the essence. So we need a way to quickly find the ratio of the two areas. Well, we could use an important principle in geometry: if the dimensions of any 2D shape are cut in half, then the area is 1/4 the original area. Notice that the diameter of a small circle is half the diameter of the large circle. So one of the smaller circles is 1/4 the area of the bigger circle! Combining the two smaller circles, we find the ratio 1/2 to be our final answer.

Math Problems, Probability Problems

Problem of the Day: 2/22/13

A display table at a store has 20 ties. A equal number contain red and blue, but 8 ties are striped red and blue. What is the probability of picking a tie with only red?

Solution to yesterday’s problem:

This is one of those problems that you just have to use your logic to solve. Math competitions like AMC and Mathcounts love these problems because they would require titanic effort to compute manually (calculating 105 factorial). In this case, you’ll need to find the number of spots in the expansion of the factorial (105 x 104 x …) where a zero is introduced.

Which digit would likely introduce a zero? Of course, any multiple of ten will add a zero to the end of the product. However, we also know that multiples of five can add a zero if the other number is even. Luckily, we know that the number will be even because we definitely multiply by an even number somewhere. So every multiple of five adds a zero – that’s 105/5 = 21 zeros.

However, we’re not finished yet! We need to account for the factor of 100. We know multiplying something by 100 adds 2 zeros to the product, and we only accounted for 1 of the zeros. So let’s just tack a zero on the end of our number, and we have a final answer of 22 zeros.