Algebra Word Question? - XP Math - Forums

 XP Math - Forums Algebra Word Question?

 07-27-2007 #1 Robert B Guest   Posts: n/a Algebra Word Question? I need help on a question that will be on my exam in my class. I don't understand the process to get the answer. The question is as follows:"Steve can mix 20 drinks in 5 minutes, Sue can mix 20 drinks in 10 minutes, and Jack can mix 20 drinks in 15 minutes. How much time will it take all 3 of them working together to mix the 20 drinks?"The answer I know is 2 minutes and 44 seconds, but I don't the process, like what I mentioned above. Help will be appreciated, asap.
 07-27-2007 #2 Chemist Chris Guest   Posts: n/a The answer is found in normalization. How many drinks can each make in 1 minute?Steve - 4 drinks/minute (20/5)Sue - 2 drinks/minute - (20/10)Jack - 1.33 drinks/minute - (20/15)The three of them can make 7.33 drinks per minuteUse ratio and proportion to set up:7.33 drinks/1 min = 20 drinks/ x minutesCross multiply to get 2.7397 minutes which equals 2 minutes 44 seconds.
 07-27-2007 #3 roynburton Guest   Posts: n/a For any problem like this, note what you need to find first. In this case, it is 20 drinks in a period of time with all three of them working on it.Then find the data that makes up the same value for each of them. In this case, again, it is some number of drinks in some period of time for each of the three. Put it in the form of 1 something in however-long-a-period of time for each. For instance, Steven's rate is 20 drinks / 5 minutes = 1 drink / 0.25 minutes.Then collect together the desired values. In this case, it is the simple rate you just located for each summed for the team of three people. In this case it is Steven + Sue + Jack = 20/5 (=1/0.25) + 20/10 (=1/0.5) + 20/15 (=1/0.75) = 220/30 = 1/0.1363636+.Then make a ratio comparison of that ratio, in either form, to the desired items divided by the unknown time. (Or it might be the number of items that is unknown and a known time. The comparison generates a simple algebra problem either way.) For instance: 220/30 = 20/x and x = 20*(30/220) = 2.727272+ minutes (which is 2 minutes 43.636363+ seconds or about 2 minutes 44 seconds).
 07-27-2007 #4 Matiego Guest   Posts: n/a First, find the rate of mixing for each person. Do this by dividing the drinks made by the time that they were made in.Steve mixes at 4 drinks per minute.Sue mixes at 2 drinks per minute.Jack mixes at 1 1/3 drinks per minute.Add the rates together:Together they mix at 7 1/3 drinks per minute.Since we need to determine time, we take the amount of drinks to be made and divide it by the combined rate of mixing.20 drinks divided by 7 1/3 drinks per minute equals 2 minutes and 72/100 of a minute, or 2 minutes and 43.6 seconds.
 07-27-2007 #5 Geoff L Guest   Posts: n/a Steve can mix 20/5 (or 4) drinks in 1 minute. Sue can mix 20/10 (or 2) drinks in 1 minute. Jack can mix 20/15 (or 4/3) drinks in 1 minute. So, in the course of 1 minute, the three of them can mix 4 + 2 + 4/3 (or 7 1/3) drinks. Since the end goal is to have 20 drinks mixed, to determine the number of minutes required you need to divide 20 by 7 1/3, and you will get approximately 2 minutes and 44 seconds.Hope this helps.

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