Fractions/ratios question I cant answer... any ideas??

jessica_r

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I have a word equation i cannot figure out! any one have any ideas?

There is a large bag of red and blue counters. One red and one blue counter are removed from the bag. This process is repeated until 1/4 of the red counter and 3/5 of the blue counters have been removed. What fraction of the original counters remain in the bag?

Im stumped!
I've written that:

Red - n = 3/4 Red
blue - n = 2/5 blue
n is the number of counters removed as it the same number removed from both.

therefore n = 1/4red and n = 3/5 blue. therefore 1/4 red = 3/5 blue. if you cross multiply that you get that 5 red = 12 blue.... I dont know how that helps!
 

tkhunny

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I have a word equation i cannot figure out! any one have any ideas?

There is a large bag of red and blue counters. One red and one blue counter are removed from the bag. This process is repeated until 1/4 of the red counter and 3/5 of the blue counters have been removed. What fraction of the original counters remain in the bag?

Im stumped!
I've written that:

Red - n = 3/4 Red
blue - n = 2/5 blue
n is the number of counters removed as it the same number removed from both.

therefore n = 1/4red and n = 3/5 blue. therefore 1/4 red = 3/5 blue. if you cross multiply that you get that 5 red = 12 blue.... I dont know how that helps!
Shall we also assume Red > 0 and Blue > 0?

It may be time for you to consider divisibility.

If Red = (12/5)*Blue, then Blue needs to be divisible by 5 and Red can't be any less than 12.
Likewise, if Blue = (5/12)*Red, then Red needs to be divisible by 12 and Blue can't be any less than 5.

Make any sense?

What can you do with that?
 

Dr.Peterson

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Suppose you started with r red and b blue. You've shown that 5r = 12b.

How many counters were in the bag originally? How many are in there now? (Express each in terms of r and b.)

Write a fraction representing the answer to the question, express it in terms of one variable (either r or b or something else), and simplify. You'll find that the variables disappear. You'll never even have to think about the actual numbers of r and b.
 

ksdhart2

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Your thinking is on the right track, but the conclusion that \(\displaystyle 5r = 12b\) is incorrect. We can see why this cannot possibly be the case by considering a simple example. Suppose we had r = 5 and b = 12. Since we can never remove a fractional number of counters, can you see why this means it's impossible to ever have exactly 1/4 of the original number of red counters? In order for there to be exactly 1/4 of the red counters left, we must have \(\displaystyle r = 20k, \: b = 48k\) for some integer k.

Using the simplest possible case where there's 20 red marbles, we see that in order for leave exactly 1/4 of the red counters behind, we have to remove 15 of each color counter. This means 33 blue counters are left, but \(\displaystyle 33/48 = 0.6875\) so it's impossible to have 25% of the red counters and 60% of the blue counters at the same time. A little bit of thinking can easily show that it's likewise impossible for any integer value of k.

Now, to see what went wrong and how to correct it. It should be obvious why our starting amount of counters must be 100% of the red counters and 100% of the blue counters. After removing some amount (n) of counters, we're left with 25% of the red counters and 60% of the blue counters. Can you see why that means the amount we removed must be 75% (3/4) of the red counters and 40% (2/5) of the blue counters? So we have \(\displaystyle n = \frac{3r}{4}\) and \(\displaystyle n = \frac{2b}{5}\).

Perform the same cross-multiplication and now that you've got the correct values here you'll get the correct ratio. Then use this ratio in the method Dr.Peterson described.
 

jessica_r

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Suppose you started with r red and b blue. You've shown that 5r = 12b.

How many counters were in the bag originally? How many are in there now? (Express each in terms of r and b.)

Write a fraction representing the answer to the question, express it in terms of one variable (either r or b or something else), and simplify. You'll find that the variables disappear. You'll never even have to think about the actual numbers of r and b.
Thanks Dr Peterson!

So we started with R and B initially and we're left with R-1/4R and B-1/4B... can i then substitute in r=12b/5?
 

jessica_r

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Your thinking is on the right track, but the conclusion that \(\displaystyle 5r = 12b\) is incorrect. We can see why this cannot possibly be the case by considering a simple example. Suppose we had r = 5 and b = 12. Since we can never remove a fractional number of counters, can you see why this means it's impossible to ever have exactly 1/4 of the original number of red counters? In order for there to be exactly 1/4 of the red counters left, we must have \(\displaystyle r = 20k, \: b = 48k\) for some integer k.

Using the simplest possible case where there's 20 red marbles, we see that in order for leave exactly 1/4 of the red counters behind, we have to remove 15 of each color counter. This means 33 blue counters are left, but \(\displaystyle 33/48 = 0.6875\) so it's impossible to have 25% of the red counters and 60% of the blue counters at the same time. A little bit of thinking can easily show that it's likewise impossible for any integer value of k.

Now, to see what went wrong and how to correct it. It should be obvious why our starting amount of counters must be 100% of the red counters and 100% of the blue counters. After removing some amount (n) of counters, we're left with 25% of the red counters and 60% of the blue counters. Can you see why that means the amount we removed must be 75% (3/4) of the red counters and 40% (2/5) of the blue counters? So we have \(\displaystyle n = \frac{3r}{4}\) and \(\displaystyle n = \frac{2b}{5}\).

Perform the same cross-multiplication and now that you've got the correct values here you'll get the correct ratio. Then use this ratio in the method Dr.Peterson described.
I am not following! well i understand most of it, but the question says 'this process is repeated until 1/4 of the red counters and 3/5s of the blue have been removed) so we're left with 75% red and 40% of the blue...

n=1/4r and n=3/5b and cross multiplying here gave me the 5r=12b
 

Dr.Peterson

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Thanks Dr Peterson!

So we started with R and B initially and we're left with R-1/4R and B-1/4B... can i then substitute in r=12b/5?
Yes. Keep going!
 

Dr.Peterson

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Your thinking is on the right track, but the conclusion that \(\displaystyle 5r = 12b\) is incorrect. We can see why this cannot possibly be the case by considering a simple example. Suppose we had r = 5 and b = 12. Since we can never remove a fractional number of counters, can you see why this means it's impossible to ever have exactly 1/4 of the original number of red counters? In order for there to be exactly 1/4 of the red counters left, we must have \(\displaystyle r = 20k, \: b = 48k\) for some integer k.

Using the simplest possible case where there's 20 red marbles, we see that in order for leave exactly 1/4 of the red counters behind, we have to remove 15 of each color counter. This means 33 blue counters are left, but \(\displaystyle 33/48 = 0.6875\) so it's impossible to have 25% of the red counters and 60% of the blue counters at the same time. A little bit of thinking can easily show that it's likewise impossible for any integer value of k.

Now, to see what went wrong and how to correct it. It should be obvious why our starting amount of counters must be 100% of the red counters and 100% of the blue counters. After removing some amount (n) of counters, we're left with 25% of the red counters and 60% of the blue counters. Can you see why that means the amount we removed must be 75% (3/4) of the red counters and 40% (2/5) of the blue counters? So we have \(\displaystyle n = \frac{3r}{4}\) and \(\displaystyle n = \frac{2b}{5}\).

Perform the same cross-multiplication and now that you've got the correct values here you'll get the correct ratio. Then use this ratio in the method Dr.Peterson described.
I think you misread a couple things. First, the 1/4 and 3/5 are the numbers removed, not the numbers remaining. Second, 5r = 12b means 5 times the initial number of red equals 12 times the initial number of blue (the definitions were not clearly stated at first). So your r=5 and b=12 would not satisfy that. Rather, you could have r=12 and b=5. After removal, you have 12-3 = 9 red, and 5=3 = 2 blue. Or it could have been r=60 and b=25 to start.
 

jessica_r

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thank you! i didnt know what do to when i got to 5R=12B but in this case (5)(r)=(12)(b) therefore r=12 and b=5! n=1/4r, n=3. thank you so much!!
 

ksdhart2

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I think you misread a couple things. First, the 1/4 and 3/5 are the numbers removed, not the numbers remaining. Second, 5r = 12b means 5 times the initial number of red equals 12 times the initial number of blue (the definitions were not clearly stated at first). So your r=5 and b=12 would not satisfy that. Rather, you could have r=12 and b=5. After removal, you have 12-3 = 9 red, and 5=3 = 2 blue. Or it could have been r=60 and b=25 to start.
Aw, farts! I'm apparently on a bit of a losing streak as far as reading comprehension goes. Sorry about that.
 

Dr.Peterson

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thank you! i didnt know what do to when i got to 5R=12B but in this case (5)(r)=(12)(b) therefore r=12 and b=5! n=1/4r, n=3. thank you so much!!
No, we don't know what numbers r and b actually are; we only know their ratio. We can say that for some k, r = 12k and b = 5k. As I said, they might be 12 and 5, or they might be 60 and 25, or whatever.

But how does what you've said answer the question? There's a lot to do that you haven't shown. Did you do what I suggested?
 
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