3 Problems

[Danforth]

#1: Al paints with watercolors on a sheet of paper 20 in. wide by 15 in. high. He then places this sheet on a mat so that a uniformly wide strip of the mat shows all around the picture. The perimeter of the mat is 102 in. How wide is the strip of the mat showing around the picture?


#2: A jeweler has five rings, each weighing 18 g, made of an alloy of 10% silver and 90% gold. He decides to melt down the rings and add enough silver to reduce the gold content to 75%. How much silver should he add?


#3: The radiator in a car is filled with a solution of 60% antifreeze and 40% water. The manufacturer of the antifreeze suggests that, for summer driving, optimal cooling of the engine is obtained with only 50% antifreeze. If the capacity of the radiator is 3.6 L , how much coolant should be drained and replaced with water to reduce the antifreeze concentration to the recommended level?




--I can't figure out how to solve these 3 problems taken from my homework. I do have the answers, as they are in the back of the book (I will list them below). We don't turn in our homework, but rather we are assigned this homework to help us practice and study for the tests. My problem is that I have a test tomorrow on this material, and am having trouble with 3 of the assigned problems. Any help for any of the problems is much appreciated. I've been trying for hours (unsuccessfully) to figure out how to get an eqation to solve these problems. So once again, thanks very much in advance to anyone who can (or atleast tries to) help.--

#1 - 4 in.
#2 - 18 g.
#3 - 0.6 L

 

[TchrQbic]

#1: Al paints with watercolors on a sheet of paper 20 in. wide by 15 in. high. He then places this sheet on a mat so that a uniformly wide strip of the mat shows all around the picture. The perimeter of the mat is 102 in. How wide is the strip of the mat showing around the picture?

You're given the measures of the sides of the painted sheet. A border x inches wide will surround it when it's placed on the larger mat. Note that the x inches will be added at each end of the sides, so that the width of the mat will be 15 + 2x and the length of the mat will be 20 + 2x. If you use the formula for the perimeter of a rectangle, you should be able to use those new measurements to set up an equation that's equal to the perimeter you are given for the mat.


#2: A jeweler has five rings, each weighing 18 g, made of an alloy of 10% silver and 90% gold. He decides to melt down the rings and add enough silver to reduce the gold content to 75%. How much silver should he add?

For this type of problem, you work with the amount and the percentage of ONE of the ingredients. Since we're asked to find the amount of silver he needs to add, we'll set it up based on the % of silver.

Altogether, five rings, each 18g, weigh a total of 90g. That amount is 10% silver. We will take that beginning amount and add x g of pure (100%) silver to it, getting a final amount of 90+x g that will be 25% silver. (If the result is 75% gold, it's 25% silver.)

Our equation, then, is (let x = the grams of silver)

.10(90g) + x = .25(90 + x)





#1 - 4 in.
#2 - 18 g.
#3 - 0.6 L

 

[Danforth]

Note that the x inches will be added at each end of the sides, so that the width of the mat will be 15 + 2x and the length of the mat will be 20 + 2x.

Thank you very much for the extremely fast response. I can't tell you how much I appreciate your help. I checked, as I'm sure you did as well, and both of your equations did work out perfectly.

I do have a question though. I apologize very much for bothering you again, and also for not understanding what you explained, but here is my question: I don't quite understand the part that I quoted above. Whenever I tried working out the problem myself, before posting here, I ended up with 15 + x and 20 + x. I know that you said that x inches will be added at each end of the sides, making it 15 + 2x and 20 + 2x. This does work out perfectly, but I can't seem to understand it. What exactly do you mean by "each end of the sides?" What makes it 2x instead of x?

Once again, I apologize for bothering you again, and for my ignorance, but I just can't seem to figure out how this works.

 

[Mrspi]

Note that the x inches will be added at each end of the sides, so that the width of the mat will be 15 + 2x and the length of the mat will be 20 + 2x.

Thank you very much for the extremely fast response. I can't tell you how much I appreciate your help. I checked, as I'm sure you did as well, and both of your equations did work out perfectly.

I do have a question though. I apologize very much for bothering you again, and also for not understanding what you explained, but here is my question: I don't quite understand the part that I quoted above. Whenever I tried working out the problem myself, before posting here, I ended up with 15 + x and 20 + x. I know that you said that x inches will be added at each end of the sides, making it 15 + 2x and 20 + 2x. This does work out perfectly, but I can't seem to understand it. What exactly do you mean by "each end of the sides?" What makes it 2x instead of x?

Once again, I apologize for bothering you again, and for my ignorance, but I just can't seem to figure out how this works.

If you draw a diagram, with the painting surrounded by a mat of width x, then what TchrQbic has suggested should be clear. You've got x inches of mat on the top and the bottom of the painting, and x inches of mat on the left and right sides of the painting. So, the total length of the picture with the mat should be 20 + 2x, and the total width should be 15 + 2x.

For your third problem,
let x = amount of solution drained from the radiator

Then, x will also be the amount of water ADDED to the radiator.

If you drain x L from the radiator, you'll have 3.6 - x L of 60% solution in the radiator. The water you add has NO antifreeze in it. You will end up with 3.6 L which is 50% antifreeze.

0.60(3.6- x) + 0(x) = 0.50(3.6)

Now, solve for x.....

 

[Danforth]

Thank you both very much. I understand each of these problems perfectly now. And drawing the diagram for problem #1 helped out very much also. It was much easier to understand.

Anyway, just saying thank you very much to both of you.