Problem of the Day: 12/26/12

Dana bought a large holiday wreath for her front door, which is 9 feet tall and 5 feet wide. The radius of the wreath is 2 feet. Including the area containing the knob, how much space will be left on the door outside the wreath?

Solution to yesterday’s problem:
We can solve this problem with a simple proportion, knowing that each elf gets 1 cookie:
$\frac{\frac{3}{4}\, \textup{cups}}{20 \,\textup{cookies}}=\frac{x \,\textup{cups}}{400000 \,\textup{cookies}}$
Cross multiplying,
$\frac{3}{4}\times 400000=20x$
Simplifying,
$300000=20x$
$15000=x$
So Mrs. Claus will need 15,000 cups of sugar to give each elf a cookie. That’s more than 900 gallons of sugar!

Problem of the Day: 12/25/12

Merry Christmas!

Mrs. Claus is baking post-Christmas cookies to treat all the elves for their hard work. She needs 3/4 cup sugar for every 20 cookies she makes, and she has 400,000 elves working in the North Pole (they take shifts). If every elf should get a cookie, how much sugar will she need?

Solution to yesterday’s problem:
A good way to solve our elvish problem from yesterday is to use dimensional analysis. If we express each piece of information as a ratio, then we can multiply these ratios to get the value we want, in this case how many days it would take for the elves to make 3 billion gifts.

First, we know that each elf can make one gift in one minute. Let’s write this as a fraction.
$\frac{1 \textup{gift}}{1 \textup{minute}}$
Next, we know that there are 60 minutes in an hour, and 24 hours in a day.
$\frac{60 \textup{minutes}}{1 \textup{hour}}$, $\frac{24 \textup{hours}}{1 \textup{day}}$
We also have 200,000 elves, so we can add that as a constant on the left.
$200,000\times\frac{1 \textup{gift}}{1 \textup{minute}}\times\frac{60 \textup{minutes}}{1 \textup{hour}}\times\frac{24 \textup{hours}}{1 \textup{day}}$
The units (minutes, hours) cancel out, leaving the following:
$200,000 \textup{gifts}\times60\times\frac{24}{1 \textup{day}}$
Multiplying, we get $\frac{288 000 000 \textup{gifts}}{\textup{day}}$

Now we want to figure out how many days it would take the elves to make 3 billion gifts. To do this, simply divide 3 000 000 000 by 288 000 000, which gives us 10.416666…

Of course, we can’t leave the answer like that. The question is asking on what day in December will the elves have to start making gifts. They’ll have to work all of the 24th, 23rd, 22nd, 21st, 20th, 19th, 18th, 17th, 16th, and 15th, and they’ll also have to work part of the 14th. So the elves will have to start on December 14th.

Problem of the Day: 12/24/12

Santa’s elves are busy making Christmas gifts at the North Pole! They have to make 3 billion gifts in time for midnight on Christmas Eve – but luckily, they have 200,000 elves working at any given time. If each elf can make a gift in one minute (they’re highly trained), then on what day in December would the elves have to start making gifts?

Solution to yesterday’s problem:
To find the probability that five specific students are chosen out of 20, but the order of the students doesn’t matter, we can use combinations.

The number of possible ways that a group of five can be picked out of 20 is
$_{20}C_5=\frac{20!}{5!(20-5)!}$
To simplify this easily, we can rewrite the factorials as multiplications, then cancel:
$\frac{20\times19\times18\times17\times16\times15\times14\times13\times12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1}{5\times4\times3\times2\times1(15\times14\times13\times12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1)}$
Canceling,
$\frac{20\times19\times18\times17\times16}{5\times4\times3\times2\times1}$
Now we can do another round of dividing by common factors, like 20 and 5, 16 and 4 and 2, etc.
$4\times19\times6\times17\times2=15504$
So the chance that these 5 particular students are chosen is incredibly small: $\frac{1}{15504}$.

Problem of the Day: 12/23/12

Out of a classroom containing 20 people, each with a different name, what is the probability that Alfred, Camellia, Edward, Grace, and Ian will be chosen by the teacher to run errands before school today? Express your answer as a common fraction.

Solution to yesterday’s problem:
For this problem I can show you two different methods: a straightforward algebraic method, and a faster method.

Method 1 – Set up an equation. $x$ will be the lowest number, $x+1$ is the next one, and so on. Our equation looks like this:
$x+(x+1)+(x+2)+(x+3)=-118$
Simplifying,
$4x+6=-118$
$4x=-124$
$x=-31$
So the consecutive numbers are -31, -30, -29, and -28. The product of -31 and -28 is 868.

Method 2 – Knowing that -118 is the sum of four numbers, we can divide -118 by four to find the average of the four numbers. -118/4 is -29.5, which must be the average of our four numbers. Since they are consecutive, we can find the two nearest integers on either side of this decimal: -30 and -31, -29 and -28. These are our four integers, and the product of the least and greatest is 868.

Problem of the Day: 12/22/12

Four consecutive integers have a sum of -118. What is the product of the first and last of these integers?

Solution to yesterday’s problem:
In the figure from yesterday, the triangle $UVW$ is a right triangle, and the hypotenuse is 13 units long. Knowing that 5-12-13 is a Pythagorean triple, we can deduce that $UV=12$ and $WV=5$. We’re also given the fact that the ratio $TU:UV=\frac{3}{2}$, so we can solve a proportion to find the length of side $TU$.
$\frac{l}{12}=\frac{3}{2}$
Cross multiplying,
$2l=36$
$l=18$
So the length of side TU is 18 units.

Problem of the Day: 12/21/12

In the figure below, STUV is a rectangle. $UW = 13$ and the ratio $TU:UV=\frac{3}{2}$. What is the length of side TU?

Solution to yesterday’s problem:
To find the average speed for the entire flight, we just add the distances and the times together and solve with the formula $r=\frac{d}{t}$.

The distances are $100 + 600 + 200 = 900$ miles, and the times are $1 + 2 + 1\frac{1}{2} = 4\frac{1}{2}$.

Dividing the distance by the time,

$\frac{900}{4\frac{1}{2}}\times \frac{2}{2}=\frac{1800}{9}=200$

So the average flight speed is 200 miles per hour.

Problem of the Day: 12/20/12

A small plane travels 100 miles in one hour, 600 miles in two hours, then 200 miles in one and a half hours. What is the average speed of the plane throughout the trip?

Solution to yesterday’s problem:
Yay, I get to try out LaTeX!
The original length-to-width ratio is 2:1. So we can designate $x$ as the width and $2x$ as the height. When the length and width are both decreased by 2 inches, the ratio becomes 2.5:1. We’ll change 2.5 to 5/2 and write this equation to represent the change:
$\frac{2x-2}{x-2}=\frac{5}{2}$
Now let’s work this proportion! Cross multiplying,
$2(2x-2)=5(x-2)$
Distributing,
$4x-4=5x-10$
Rearranging by subtracting $4x$ and adding $10$,
$6=x$
So the width of the original board is 6 inches.