Date: 07/01/2002 at 14:06:29 From: Vickie Subject: Word Problem The problem is: a man has 20 coins. All of the coins are dimes and quarters. If the dimes were quarters and the quarters were dimes he would have 90 cents more than he has now. How many dimes and quarters does he have now? My solution was to start with two equations: #1 D + Q = 20 #2 .10D + .25Q = (I don't know what to put here)
Date: 07/03/2002 at 10:13:33 From: Doctor Ian Subject: Re: Word Problem Hi Vickie, You've made a good start. Can I make a suggestion? When working with money, it's often a good idea to express everything in terms of pennies, since that lets you work with integers. Your first equation lets you express the number of dimes in terms of the number of quarters, and vice versa. That is, all of the following equations say the same thing: d + q = 20 d = 20 - q q = 20 - d Setting up the second equation is a little trickier. If the guy HAS d dimes and q quarters, then the amount of money he has (expressed in pennies) is 10*d + 25*q Now, if the guy HAD q dimes and d quarters, the amount of money he would have would be 25*d + 10*q Does that make sense? What the second equation needs to express is that the second option would add up to 90 pennies more than the first option. In other words, you'd have to ADD 90 cents to what he actually HAS, to get what he WOULD have: (10*d + 25*q) + 90 = (25*d + 10*q) Here's where the first equation comes in handy! Since you can express q in terms of d, you can substitute (20-q) everywhere you see a d to get (10*(20-q) + 25*q) + 90 = (25*(20-q) + 10*q) Can you take it from here? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
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