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.)?

Math Problems, Number Theory Problems

Problem of the Day: 3/14/13

What is the sum of the terms in the sequence 35+34+36+37+33+32+38+39+ … +2+1+69+70?

Solution to yesterday’s problem:

We can assume that triangle PQR has integer sides because more information was not specified. This would mean that PQR is a Pythagorean triple with a short side of 9: a 9-12-15. So the hypotenuse of PQR is 15 cm.

Now we’re going to do some more estimating! You can sort of see that the small triangle QUT takes about a third of the hypotenuse. This goes along with our theory about integer lengths because it would make the hypotenuse 5, which leads us to believe that QUT is a 3-4-5 triangle. Also, the larger triangle RST takes up two-thirds of the hypotenuse, or 10 cm. This suggests that RST is a 6-8-10 triangle. So let’s show the triangle again with the lengths labeled:

20130314-091142.jpg

The length we actually want to find is US, conveniently the only segment we don’t know. We can use the Pythagorean theorem for triangle STU.

4^2+6^2=x^2

16+36=x^2

52=x^2

\sqrt{52}=x

Simplifying the radical,

\sqrt{4\times13}=x

\sqrt{4}\times\sqrt{13}=x

2\sqrt{13}=x

Math Problems, Number Theory Problems

Problem of the Day: 3/11/13

Two positive integers have a sum of 52. If their greatest common factor is 4, what is the least possible positive difference between the numbers?

Solution to yesterday’s problem:

If Carlo needs two cans of paint to coat three walls, then his rate of “paint usage” is \frac{2}{3} cans per wall. If he needs to paint eight 4-walled rooms, he has to paint 32 walls total. He also needs to paint two coats per wall, which is equivalent to painting 32 x 2 = 64 walls total. Using the rate mentioned above, we can write

\frac{2}{3}\times64

to represent the number of cans of paint he needs. This evaluates to \frac{128}{3}, or 42\frac{2}{3}. Remember that Carlo couldn’t very well buy two-thirds of a can of paint, so we round up to 43 cans.

Math Problems, Number Theory Problems, Uncategorized

Problem of the Day: 3/8/13

What is the greatest prime factor of 14! (factorial)?

Solution to yesterday’s problem:

In order to find the perimeter of this rectangle, we should first find the side lengths of the rectangle. Knowing that the area of the rectangle is 3x^2-11x+6, we can factor this to find the expressions that will multiply to get this area. On to factoring!

Factoring 3x^2-11x+6

Now, some students at my school like to use these modern techniques they call the ZBox, or some such thing, but on this blog we’re going to be old-fashioned and use plain old grouping. I guess the way the textbook explains grouping is a bit complicated and that’s why they don’t appreciate this wonderful technique!

Anyway, first we need to find factors of 18 (the first coefficient times the last coefficient) that will add up to the middle coefficient, 11. Factors of 18 are (1,18), (2,9),… Stop! We have our factors, 2 and 9. Let’s rewrite the -11x as the sum of those factors:

3x^2-9x-2x+6

Now we need to do the grouping part. Notice that the first two and the last two terms look like they could share some factors. (Sometimes these might be out of order, in which case you would rearrange them.) Note that I’ve put the negative INSIDE the grouping because otherwise it would cause some problems…

(3x^2-9x)+(-2x+6)

The terms in the first group have a common factor of 3x, so let’s pull that out and divide each term by 3x.

3x(x-3)+(-2x+6)

The second group has a common factor of 2, so we’ll pull that out as well.

3x(x-3)+2(-x+3)

At this point we would be able to pull out a common factor between these two remaining terms, but we can’t because there’s a subtle difference: (x-3) and (-x+3). We need to change the signs of the second term so that the two factors look the same. If we factor out a -1 from (-x+3), we’ll get

3x(x-3)-2(x-3)

Now let’s factor out an (x-3) from both terms. We’ll be left with

(3x-2)(x-3)

There are our factors and our side lengths! So to find the perimeter of the rectangle, we just take twice each term.

2(3x-2)+2(x-3)

6x-4+2x-6

8x-10

Math Problems, Number Theory Problems

Problem of the Day: 3/1/13

How many palindromic 10-digit phone numbers can exist if the first three digits must be different?

Solution to yesterday’s problem:

Let’s write two rates to represent how much water the tap can fill in one minute and how much the drain can empty in one minute.

The tap can fill the tub in 30 minutes, so it can fill 1/30 of the tub in one minute. The drain can empty the tub in 45 minutes, so it can drain 1/45 of the tub in one minute. We can write the amount that the tub fills in one minute as

\frac{1}{30}-\frac{1}{45}

Simplifying this fraction,

\frac{3}{90}-\frac{2}{90}

So the tub will fill by \frac{1}{90} in one minute. Logically, it will take 90 minutes to fill the entire tub.

Math Problems, Number Theory Problems

Problem of the Day: 2/21/13

How many consecutive zeros are there at the end of the number 105! (105 factorial)?

Solution to yesterday’s problem:

If three of Kelly’s blocks makes a half mile, then one of those blocks must be one-sixth of a mile. Kelly can walk one-sixth of a mile in twenty minutes (admittedly slow, maybe she socializes). Logically this proportion works out to one mile in 120 minutes, or two hours.

Two of her friend’s blocks make up a mile, so one of those blocks is a half mile. Using the proportion from above, 1 mile per 2 hours, we reason that 1/2 mile will take Kelly 1 hour to walk.