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.)

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.)?

Algebra Problems, Math Problems

Problem of the Day: 3/15/13

What is the probability that a term selected at random from the binomial expansion of (a+b)^{21} has a coefficient that is a multiple of 2? (edited)

Solution to yesterday’s problem:

The trick to sum of sequence problems is to find the pattern of the sums. In this problem, notice that the sum of the first and third terms, 71, is the same as the sum of the second and fourth terms. It’s also the sum of the third-last and last terms. So we can assume that the sum of two terms that have one term between them is 71.

The next step is to find the number of these pairs in the sequence. Well, we notice that the first two numbers starts counting down from 35, and each set of two numbers separated by two numbers continues this pattern down to 1 at the end. The numbers between the decreasing numbers start at 36 and increase to 70. It takes 35 numbers to count from 35 to 1, and 35 numbers to count from 36 to 70. Therefore there are 70 numbers and 35 pairs. Multiplying the number of pairs by the sum of each pair, 35 x 71 = 2485.

Another way this problem can be solved is to notice that the series essentially just contains the integers from 1 to 70. The sum of these integers is half the product of the biggest number and the next number after that.

frac{(70)(71)}{2}

frac{70}{2}times71

35times71

We wind up with the same answer, which is 2485.

Algebra Problems, Math Problems

Problem of the Day: 3/7/13

The area of a rectangle can be expressed as 3x^2-11x+6. What is the perimeter of this rectangle in terms of x?

Solution to yesterday’s problem:

If the frog has an equal probability of landing on any lily pad, and he can land on two possible pads (he cannot jump vertically and land on the same lily pad), then the probability of him landing on a particular pad is \frac{1}{2}. Therefore, the probability of any permutation of three jumps is \frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}.

Let’s make a tree diagram of all the possible ways he could jump. Say he started on lily pad 1. Here it is:

20130307-072009.jpg

Notice that there are two ends of the branches that have a one, which is the pad the frog started on. So the probability of any of these two permutations occurring is \frac{2}{8}, or \frac{1}{4}.

Algebra Problems, Math Problems

Problem of the Day: 2/27/13

Three positive integers have a product of 60. What is their smallest possible sum?

Solution to yesterday’s problem:

Let’s use the blanks method with probabilities for this problem. There are three blanks for three draws:

____ ____ ____

There are 12 possible marbles for the first draw, and 3 of them are green.

__3/12__ ____ ____

There are 11 possible marbles for the second draw (he has taken out one), and 2 of those marbles are green. Using the same principle for the third draw:

__3/12__ __2/11__ __1/10__

We need to multiply these probabilities to find out the probability of all three occurring. The product of the probabilities is 2/440, or 1/220.

Algebra Problems, Math Problems

Problem of the Day: 2/19/13

The company that makes Product X will make 100,000 units for every $1 increase in price. Two million people would buy the product if it were free, and 300,000 less people would buy it for every $1 increase in price. How should the Product X be priced so that there is no surplus or shortage?

Solution to yesterday’s problem:

First let’s find the rate at which each person works. Fred can mow a lawn in 3 hours; therefore, he can mow one-third of that lawn per hour. Following this, George can mow one-half of that lawn per hour. So their combined rates are

\frac{1}{3}+\frac{1}{2}

This simplifies to 5/6, or 5 lawns in 6 hours. We need to divide both numbers by 5 to find the time for one lawn: 5/5=1 and 6/5. So they will take 6/5 of an hour to mow the lawn; this simplifies to 1 1/5 hour, or 1 hour, 12 minutes.

Algebra Problems, Math Problems

Problem of the Day: 2/18/13

Fred can mow a certain lawn in 3 hours. His twin George can mow the same lawn in 2 hours. How long will it take for both of them to mow the lawn together?

Solution to yesterday’s problem:

We notice right away that there is a right triangle whose short leg is 3 inches and whose hypotenuse is 5 inches. Therefore the long leg, PT, is 4 inches. Since PQ is 3 inches as part of the square, the leftover segment QT is 1 inch. Now we notice two similar triangles: the small triangle outside the square and the large triangle, PRT. The ratio of their long legs will be equal to the ratio of their short legs. We can write this as a proportion:

\frac{4}{1}={3}{x}

Simplifying,

4={3}{x}

4x=3

x=\frac{3}{4}

We conclude that the length of the lower segment of QS is 3/4 inch.