Practice Test on Chapter 1 (One last problem)

[hannibolio]

Hi to all who will respond. I have another problem that has me stumped. My brother and I cannot seem to figure it out. We have tried many set ups for the problem. The problem is as follows:
A coin collection worth $3.45 contains nickels, dimes, and quarters. If there are twice as many dimes as quarters, and the number of nickels is 9 more than the number of quarters, find the number of each coin.

After many set ups, my brother came up with this one:

q + 2q + 9q = 3.45

12q = 3.45

q = .29

Not sure if it's right. I want to see what you guys think. I appreciate it.

 

[JeffM]

Hi to all who will respond. I have another problem that has me stumped. My brother and I cannot seem to figure it out. We have tried many set ups for the problem. The problem is as follows:
A coin collection worth $3.45 contains nickels, dimes, and quarters. If there are twice as many dimes as quarters, and the number of nickels is 9 more than the number of quarters, find the number of each coin.

After many set ups, my brother came up with this one:

q + 2q + 9q = 3.45

12q = 3.45

q = .29

Not sure if it's right. It cannot possibly be right. The answer has to be a non-negative whole number. I want to see what you guys think. I appreciate it.

I am going to give you what I believe is THE best path for solving any word problem

FIRST STEP Identify what are the unknowns to be found and assign a unique letter to each unknown IN WRITING.

This does three things. IT gets you started organizing the problem. It avoids having to keep too many details in your head. And it allows you to communicate efficiently with others about your problem. And it takes virtaully no time.

q = number of quarters

d = number of dimes

n = number of nickels

SECOND STEP Translate what the word problem says in English into math language using the letters assigned up in step 1. (You may also need to translate some general information that you are expected to know without being told into math language: example a quarter is worth 25 cents.)

25q + 10d + 5n = 345.

You can't add numbers of coins to get dollars or cents. But the number of cents that q quarters is worth is 25q. So do you see why the equation above makes sense? If you don't, don't be shy. Tell us and we shall explain it in more detail.

Now the problem gives you some other information. There are twice as many dimes as quarters. Set up an equation using your letters that says the same thing in math lingo. What do you get?

And the number of nickels is 9 more than the number of quarters. How do you say that in math lingo?

You have now transformed what was a word problem into a pure math problem. Some people are good at pure math problems but not so good at getting word problems into math form. These two steps show you how to do it.

STEP 3 Solve the pure math problem using any mathematical techniques you know. For this problem, after you have finished step 2 you have a system of three linear equations in three unknowns. Do you know how to solve such systems?

STEP 4 Check your answer. This habit will give you self reliance and will save your bacon on tests.

 

[HallsofIvy]

Hi to all who will respond. I have another problem that has me stumped. My brother and I cannot seem to figure it out. We have tried many set ups for the problem. The problem is as follows:
A coin collection worth $3.45 contains nickels, dimes, and quarters. If there are twice as many dimes as quarters, and the number of nickels is 9 more than the number of quarters, find the number of each coin.

After many set ups, my brother came up with this one:

q + 2q + 9q = 3.45

12q = 3.45

q = .29

Not sure if it's right. I want to see what you guys think. I appreciate it.

Unfortunately, I get the impression that neither you nor your brother has any idea what the question is asking.

What you have doesn't even make sense because you haven't said what "q" represents. I might be inclined to guess that q is the number of quarters, but

1) A quarter is worth $.25 and you haven't used that anywhere. I suspect the left side of the first equation is supposed to be the total number of coins but that surely is NOT equal to the monetary value. If you have 100 coins of various kinds that does NOT mean you have $1.00!

2) "9 more than" is NOT "9q" it is "q+ 9".

3) You can't have ".29" quarters! The number of quarters has to be a whole number. And the problem asks for the number of each kind of coin- your answer should have three whole numbers.

If you have q quarters, and there "are twice as many dimes as quarters" then there are 2q dimes. That you appear to have correct. If the number of nickels is "9 more than the number of quarters" then there are q+ 9 nickels. But now you have to use the fact that a quarter is worth $.25, a dime is worth $.10, and a nickel is worth $.05. If you have q quarters, then you have .25q dollars worth of quarters. If you have 2q dimes, then you have .10(2q)= .20q dollars worth of dimes. If you have q+9 nickels, then you have .05(q+ 9)= .05q+ .45 dollars worth of nickels. Add all of those together, set the sum equal to 3.45, and solve for q. After you have found the number of quarters use the fact that you have "twice as many dimes as quarters" and "9 more nickels than quarters" to determine the number of nickels.

 

[hannibolio]

Unfortunately, I get the impression that neither you nor your brother has any idea what the question is asking.

What you have doesn't even make sense because you haven't said what "q" represents. I might be inclined to guess that q is the number of quarters, but

1) A quarter is worth $.25 and you haven't used that anywhere. I suspect the left side of the first equation is supposed to be the total number of coins but that surely is NOT equal to the monetary value. If you have 100 coins of various kinds that does NOT mean you have $1.00!

2) "9 more than" is NOT "9q" it is "q+ 9".

3) You can't have ".39" quarters! The number of quarters has to be a whole number. And the problem asks for the number of each kind of coin- your answer should have three whole numbers.

If you have q quarters, and there "are twice as many dimes as quarters" then there are 2q dimes. That you appear to have correct. If the number of nickels is "9 more than the number of quarters" then there are q+ 9 nickels. But now you have to use the fact that a quarter is worth $.25, a dime is worth $.10, and a nickel is worth $.05. If you have q quarters, then you have .25q dollars worth of quarters. If you have 2q dimes, then you have .10(2q)= .20q dollars worth of dimes. If you have q+9 nickels, then you have .05(q+ 9)= .05q+ .45 dollars worth of nickels. Add all of those together, set the sum equal to 3.45, and solve for q. After you have found the number of quarters use the fact that you have "twice as many dies as quarters" and "9 more nickels than quarters" to determine the number of nickels.

Thank you so much. I must have been thinking about the problem way too much. It was a lot simpler than I originally thought. Everything makes sense now. Thanks again.