Math Problems, Proportions Problems

Problem of the Day: 2/28/13

A bathtub can fill to the top in 30 minutes. The drain can empty the tub in 45 minutes. If the tap is on and the drain open, how long will the bathtub take to fill?

Solution to yesterday’s problem:

Let’s list the factors of 60 to consider some of the possible integers.

1,60
2,30
3,20
4,15
5,12
6,10

We can say that two of the three integers must have a product of one of the factors because they are part of the product to make 60. We might reason that if the factors are closer together, then the sum will be smaller. So for (6,10), we can form (6,5,2) and (3,2,10) as two possible sets of factors. The sum of (6,5,2) is less: only 13. However, we’d better check some of the other factors because some of the larger numbers may have small factors as well. After a little trial and error, we find that (3,5,4) has a sum of 12. So the least possible sum of the factors is 12.

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.

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.

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.

Math Problems, Proportions Problems

Problem of the Day: 2/20/13

Kelly walks her dog around her block in 20 minutes. She knows that 3 of these blocks is half a mile. In how many minutes can she walk her dog around her friend’s block, two of which constitute a mile?

Solution to yesterday’s problem:

Let’s represent the problem with two linear equations. The first one, the supply equation/curve, has a slope of 100,000 because that’s how many is added for each $1 increase. So we can write that equation as

y=100000x

The demand equation has a y-intercept of 1,000,000 because that’s how many people would get the product at price zero. The slope is -300,000 because 300,000 less people would get it for each $1 increase. So the demand equation is

y=-300000x+1000000

The point where supply and demand are equal is the intersection of these two lines. We can find this point by setting the two equations equal to each other:

100000x=-300000x+1000000

Solving this equation,

400000x=1000000

(Dividing by 100,000 to make the numbers a bit easier)

4x=10

x=2.5

So Company X should theoretically price Product X at exactly $2.50.