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Five algebra problems

 
 
Levi
 
Reply Mon 24 Jan, 2005 05:38 pm
Show all your work for full credit!

1) Levi invested $15000. Some of this money was invested at 9¼ % simple interest and the remaining funds were invested at 11½ % simple interest. After 1 year, the total interest was $1623.75. How many dollars were invested at 9¼ %?

2) The sum of three positive numbers is 185. Find the first number if the third is four times the first and the second is 36 less than twice the third.

3) One plane leaves the airport at noon heading east at x mph. A second plane leaves 30 minutes later heading west at y mph. At 3:00 p.m. the 2 planes are 1800 miles apart. If at 5 p.m. the planes are 3100 miles apart, what is x-y?

4) The perimeter of a semicircular region (which includes the diameter as the base) is numerically equal to its area measured in cm². Find the number of cm in the radius of this semicircular region.

5) If the hands of a clock are correctly positioned, find the number of degrees in the acute angle formed by the hour hand and the minute hand at exactly 4:15 p.m.
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Type: Discussion • Score: 1 • Views: 2,382 • Replies: 31
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engineer
 
  1  
Reply Mon 24 Jan, 2005 06:57 pm
And you need this why?
These are all fairly straightforward. Is this homework? I won't just give you these answers, but if you have a specific question, I'd be happy to help.
0 Replies
 
Brandon9000
 
  1  
Reply Tue 25 Jan, 2005 01:53 pm
Re: Five algebra problems
Levi wrote:
Show all your work for full credit!

1) Levi invested $15000. Some of this money was invested at 9¼ % simple interest and the remaining funds were invested at 11½ % simple interest. After 1 year, the total interest was $1623.75. How many dollars were invested at 9¼ %?

Well, my work day is pretty boring, and I don't suppose it would injure your school career to do the first one for you:

Let's use D (for dollars) to represent the amount of money invested at 9 1/4%.
Let us use F (for final) to denote the amount of money in the account after one year.

F = D * 1.0925 + (15000 - D) * 1.1125
F = 1.0925D + 16687.50 - 1.1125D
F = 16687.50 - .02D
But we also know that F = 15000 + 1623.75 = 16623.75
Thus, 16687.50 - .02D = 16623.75
.02D = 16687.5 - 16623.75 = 63.75
D = 63.75/.02 = 3187.50 in dollars
0 Replies
 
Levi
 
  1  
Reply Tue 25 Jan, 2005 05:38 pm
These weren't for me. I thought you guys would like doing them.
0 Replies
 
engineer
 
  1  
Reply Wed 26 Jan, 2005 08:15 am
Up the difficulty
Levi wrote:
These weren't for me. I thought you guys would like doing them.


Ok, then you are going to have to up the difficulty. I'll modify number five for you. The hands of a clock are at a right angle at 3:00. What is the next time that that is true?
0 Replies
 
tex
 
  1  
Reply Tue 1 Feb, 2005 11:50 am
Time
3:32.7272727272...
0 Replies
 
engineer
 
  1  
Reply Tue 1 Feb, 2005 12:09 pm
Re: Time
tex wrote:
3:32.7272727272...


Correct. Let's try a physics one...

You are in a boat with a large rock on a lake. You can measure the level of the lake exactly. You throw the rock into the lake and it sinks. What happens to the level of the lake?
0 Replies
 
Einherjar
 
  1  
Reply Tue 1 Feb, 2005 12:17 pm
It sinks/falls/whatever the word is. When in the boat the rock adds to the boats displacement by a mass of water equal to its own mass. Once sunk it displaces a volume of water equal to its own volume. Given that the rodk is denser than the water, (it sinks) it will cause the boat to displace more additional water while in it than it will displace itself while at the bottom of the lake.
0 Replies
 
Bella Dea
 
  1  
Reply Tue 1 Feb, 2005 12:35 pm
Levi wrote:
These weren't for me. I thought you guys would like doing them.



It always amazes me that people actually like doing stuff like this. I was never good at math.
0 Replies
 
engineer
 
  1  
Reply Tue 1 Feb, 2005 12:55 pm
A conceptual challenge.
Einherjar wrote:
It sinks/falls


OK, you are too good. That one is generally good enough to fool a college student or two.

A conceptual one for you. No calculators allowed. Assume the Earth is a perfect sphere and a string is wrapped around the equator. One meter is added to the length of the string and the string is arranged so that it is off the Earth an equal distance everywhere. What type of instrument would you need to measure the distance from the surface? Microscope, micrometer, ruler, etc.
0 Replies
 
engineer
 
  1  
Reply Tue 1 Feb, 2005 01:07 pm
Math is easy!
Kristie wrote:

It always amazes me that people actually like doing stuff like this. I was never good at math.


Ah Kristie, you've been brainwashed by all those liberal arts folks. Math is much easier than English, for example. Think about it this way. In math, 123 means exactly one thing, one hundred twenty three. In English, bat could mean a wooden stick for hitting a ball or a flying mammal. A mathmatical sentence also has one meaning. English sentences can have layers of meaning with whole courses dedicated to understanding them. The difference is that you speak English everyday and probably do math rarely. If you did math regularly, it would be a snap. My son comes home with maybe 10 math problems a night. At his age, I would have to do sheets of 50. Relying on calculators means that math classes can move faster and cover more material, but it also means that people don't really get comfortable with the basics, learn to "speak" math, before moving on. My son is also doing work a couple of years earlier than I did.
0 Replies
 
Bella Dea
 
  1  
Reply Tue 1 Feb, 2005 01:11 pm
English was always easier for me than math. I majored in English in college. Barely passed math.

Guess it's that right brain/left brain thing.
0 Replies
 
Tryagain
 
  1  
Reply Tue 1 Feb, 2005 04:17 pm
Two pie's :wink:
0 Replies
 
Einherjar
 
  1  
Reply Wed 2 Feb, 2005 12:15 pm
Re: A conceptual challenge.
engineer wrote:
Einherjar wrote:
It sinks/falls


OK, you are too good. That one is generally good enough to fool a college student or two.

A conceptual one for you. No calculators allowed. Assume the Earth is a perfect sphere and a string is wrapped around the equator. One meter is added to the length of the string and the string is arranged so that it is off the Earth an equal distance everywhere. What type of instrument would you need to measure the distance from the surface? Microscope, micrometer, ruler, etc.


I would use a piece of string of the exact length of 1/(2pi) metres.
0 Replies
 
engineer
 
  1  
Reply Wed 2 Feb, 2005 06:57 pm
Correct again
Correct again. (Not that it was all that hard.)

What makes that one interesting is that the answer is independent of the size of the sphere.
0 Replies
 
markr
 
  1  
Reply Wed 2 Feb, 2005 10:55 pm
There's a circular statue in the middle of a circular room. You need to carpet the area between the circles. You don't know their diameters, but you know that the diameter of the room is three times the diameter of the fountain, and you know that when you lay your tape measure such that it touches the room's wall in two places and is tangent to the statue (largest possible chord of the larger circle that doesn't pass through the smaller circle) it measures 10 meters. How many square meters of carpet will be laid?
0 Replies
 
engineer
 
  1  
Reply Thu 3 Feb, 2005 07:14 am
markr wrote:
There's a circular statue in the middle of a circular room. You need to carpet the area between the circles. You don't know their diameters, but you know that the diameter of the room is three times the diameter of the fountain, and you know that when you lay your tape measure such that it touches the room's wall in two places and is tangent to the statue (largest possible chord of the larger circle that doesn't pass through the smaller circle) it measures 10 meters. How many square meters of carpet will be laid?


I got 25(pi) square meters. Interesting problem.
0 Replies
 
Einherjar
 
  1  
Reply Thu 3 Feb, 2005 07:54 am
markr wrote:
There's a circular statue in the middle of a circular room. You need to carpet the area between the circles. You don't know their diameters, but you know that the diameter of the room is three times the diameter of the fountain, and you know that when you lay your tape measure such that it touches the room's wall in two places and is tangent to the statue (largest possible chord of the larger circle that doesn't pass through the smaller circle) it measures 10 meters. How many square meters of carpet will be laid?


(10/2)^2 = R^2 - (R/3)^2

(8/9)R^2 = 5^2

R^2 = 25/(8/9) = 225/8

A = pi*R^2 - pi*(R^2)/3^2 = 8/9*pi*R^2 = pi*225*(8/8)/9 = (225/9)*pi = 25*pi

The area carpeted is 25*pi m^2.
0 Replies
 
engineer
 
  1  
Reply Thu 3 Feb, 2005 08:13 am
Next one from Norway
OK Einherjar, it's your turn to post one.
0 Replies
 
markr
 
  1  
Reply Thu 3 Feb, 2005 10:38 am
Einherjar wrote:
The area carpeted is 25*pi m^2.


This was another one where the sizes of the circles didn't matter. The only important piece of information was the 10 meter measurement. I threw the 3X stuff in there to throw you off, but I think it made the problem easier.

Good job, Einherjar!
0 Replies
 
 

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