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July 6th, 2009, 10:48 AM  #1 
Senior Member Joined: Jul 2009 Posts: 136 Thanks: 0  I think this is an algebra problem...let me knowthanks!
Here is a logic problem that I tackled and because of the small numbers solved intuitively. 22 cents is distributed among ten students by means of nickels and pennies. How many nickels and how many pennies were distributed? I came up with three nickels and seven pennies. That's fine to do in my head because of the limited numbers, but isn't there an algebra equation for this involving "10," "22," if not also "5" and "1"? Let me know what you all think. Danke! 
July 6th, 2009, 01:01 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,472 Thanks: 2039 
If each student receives exactly one coin, the number of nickels is given by (22  10)/(5  1).

July 6th, 2009, 02:07 PM  #3 
Senior Member Joined: May 2008 From: Sacramento, California Posts: 299 Thanks: 0  Re: I think this is an algebra problem...let me knowthanks!
Were each of the students given a certain amount of coins?

July 7th, 2009, 08:04 AM  #4  
Senior Member Joined: Jul 2009 Posts: 136 Thanks: 0  Re:
OK, so would the equation be n = 22  10/ 51? And why subtract the number of students from the coin amount? Let me know, thanks. Quote:
 
July 7th, 2009, 08:07 AM  #5  
Senior Member Joined: Jul 2009 Posts: 136 Thanks: 0  Re: I think this is an algebra problem...let me knowthanks!
All I'm given in the logic problem is that there is the amount of 22 cents distributed to ten students through nickels and pennies. As noted previously, I can figure it out in my head as three nickels and seven pennies. So I'm figuring one coin per student. Anyway, I thought there must be an equation which would make this easier for large numbers too. Something as 22 cents = students with nickels + students with pennies (5 n + n ?), but then what to do with the information that there are ten students? Well, if you have anything let me know. It's interesting as a problem of distribution. Quote:
 
July 7th, 2009, 12:16 PM  #6 
Senior Member Joined: May 2008 From: Sacramento, California Posts: 299 Thanks: 0  Re: I think this is an algebra problem...let me knowthanks!
Then why can't the answer just be 22 pennies?

July 7th, 2009, 12:39 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,472 Thanks: 2039 
Each student is given one penny, leaving (22  10) pennies (worth of coins), which is distributed by replacing some pennies with nickels, which uses up (5  1) pennies = 4 pennies per exchange. Hence there are (22  10)/4 nickels. There are other possibilities if a student needn't receive exactly one coin. For example, one student could receive all the coins.

July 8th, 2009, 02:14 PM  #8  
Senior Member Joined: Jul 2009 Posts: 136 Thanks: 0  Re: I think this is an algebra problem...let me knowthanks!
Oh, that's the problem: distributed through nickels and pennies. Quote:
 
July 8th, 2009, 02:16 PM  #9  
Senior Member Joined: Jul 2009 Posts: 136 Thanks: 0  Re:
OK, thanks, this does help. There really are no other possibilities if the distribution is through nickels and pennies, one to each student. This is ood. I'm thinking as an equation it would have to be writen with a variable. Not sure what that would be. Quote:
 
July 10th, 2009, 09:55 AM  #10  
Senior Member Joined: Jul 2009 Posts: 136 Thanks: 0  this form may work....
Though does this work for other problems that are similar...? For instance, distributing $2 among 20 students through quarters and nickels. Again, in my head I can figure five quarters and 15 nickels = $2. Now using your formula, we have 200 (i.e., two hundred cents)  100 [i.e., one nickel per student] / 25  5 [i.e., quarters for nickels] if I follow your reasoning (which I'm not sure I do!). This leaves us with 100 / 20 = 5 which fits. So algebraically we write (I'm assuming): n = 200  100 / 25  5 Is that correct? Thanks for your help. Quote:
 

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