How Many People Went on the Cruise?
Date: 12/03/2001 at 17:26:34 From: Jessica Morris Subject: 7th grade word problem At the end of a special cruise, the employees could not remember the total number of people who were on board. However, they had the following data from the passenger list: 520 European females 340 North American women 340 North American males 580 girls 880 European citizens of which 580 were women 480 European adults Determine the total number of people on the cruise. I've spent a lot of time trying to figure this problem out. It seems as if the females would include the girls and women, and the adults would include women and men, but I am totally stuck. I have even had my parents try to help me and we are all confused. Please help if possible. Thank you!
Date: 12/03/2001 at 21:04:04 From: Doctor Peterson Subject: Re: 7th grade word problem Hi, Jessica. I'll just get you started by showing you how I approached the problem. If you think visually, this should work for you. If not, you can convert what I do to simultaneous equations and solve them by whatever means works best for you. First, I like to turn words into symbols, so I devised these: A = American B = British F = Female M = Male a = adults c = children There are two special terms: Women = Female adults = Fa Girls = Female children = Fc "Citizens" is a throw-away word, assuming that all the "British" people are British citizens. Now I can write the given numbers this way: BF = 26 AFa = 17 AM = 17 Fc = 29 B = 44 Fa = 29 Ba = 24 Now we can write these on a diagram. There are three categories being used to divide people up (A/B, F/M, and a/c), so we can use either the traditional three overlapping circles (which are hard to draw in text), or a cross-and-circle diagram: A | B | | | F AFa ********* BFa *** | *** ** | ** * AFc | BFc * * | * * | * ------------*-----------+-----------*------------ * | * * | c * * AMc | BMc * a ** | ** *** | *** M AMa ********* BMa | | | | The left side is Americans, the right side British; the top half is female, the bottom male; and the inside is children, the outside adults. Each region is labeled with a complete set of three letters; an incomplete set, such as Fa, means "all Female adults, British or American", so it is the union of two (or four) regions, in this case Fa = AFa + BFa. Now I can fill in the numbers, writing the number for a specific region (where all three categories are known) in the middle of the region, and the number for a combination of regions at their intersection: A | B Fa=29 | | F AFa=17 ********* BFa *** | *** ** Fc=29 ** * AFc | BFc * * | BF=26 * | * ------------*-----------+---------B=44---Ba=24--- * | * AM=17 | * * AMc | BMc * ** | c** *** | *** M AMa ********* a BMa | | | | For example, I wrote Fa=29 straddling the border of AFa and BFa, because Fa= AFa + BFa. I put B=44 where the line and circle cross, because B = BFc + BFa + BMa + BMc. (Actually, I just wrote the numbers, not the names, so the numbers stood out more.) Now you can look for numbers to subtract. We can start with the one specific number we are given, AFa=17. Look for a number on the border of that region: Fa=29. Subtract those to get BFa=12, since Fa = AFa + BFa. Keep going like this, and you'll find that you know every specific number except for one pair, for which you know the sum. (This makes sense, because you have 7 equations for 8 variables, so you wouldn't expect to be able to solve for all of them.) From this you can get the total number of people. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 12/04/2001 at 10:40:51 From: Jessica Morris Subject: Re: 7th grade word problem Dr. Peterson, Thank you for your help, but I am still struggling with part of this problem. I made symbols for all of the people in my problem, as you did, and I also drew it out on a graph, as you did. This is my problem: How can I have 580 European adult women, when the problem also says I only have 480 European adults? This doesn't make sense to me. Thank you! Jessica
Date: 12/04/2001 at 11:12:39 From: Doctor Peterson Subject: Re: 7th grade word problem Hi, Jessica. I knew I recalled doing a problem like this, and with the reminder that there was a bad number in there, I searched again and found this, an answer I sent before: ===================================================================== > Millennium Cruise > > Every week I give my students a problem of the week from a variety > of sources which I make note of in my plans, however I have > forgotten where this came from and believe I've made a typo > somewhere. If I haven't made a typo, then I guess(along with my > colleagues), I am stumped at the following problem: > > At the end of a special millenium cruise, the employees could not > remember the total number of people on board. They had the following > information from the passenger list: 520 European females, 340 North > American women, 340 North American males, 580 girls, 880 European > citizens of which 580 were women, and 480 European adults. Determine > the total number of people on board the Millennium Cruise. The 580 European women is too many; there are only 520 European females, a category that includes all the European women! Similarly, there are only 480 European adults, which would also include all 580 European women. Can we determine what the number should be? Here's a way to analyze the problem, and perhaps find a valid number for X, the number of European women. I find three categories, which I'll label by using capitals for the first and lower case for the second type in each: European (E) vs. North American (n); Female (F) vs. male (m); and Adult (A) vs. juvenile (j). There are eight combinations of these categories, and each of the numbers you give includes one or more of them: (a) 520 European females EFA+EFj (b) 340 North American women nFA (c) 340 North American males nmA+nmj (d) 580 girls EFj+nFj (e) 880 European citizens EFA+EFj+EmA+Emj (f) X European women EFA (g) 480 European adults EFA+EmA We're assuming that EFA = X. From (fg) we have (h) EmA = 480-X, and X = <480. From (af) we have (i) EFj = 520-X, and X = <520. From (di) we have (j) nFj = 580-EFj = 580-(520-X) = 60+X. From (ea) we have (k) EmA+Emj = 880-520 = 360. From (hk) we have (l) Emj = 360-(480-X) = X-120, and X >= 120. Now the total of all eight groups is EFA + EFj + EmA + Emj + nFA + nFj + (nmA+nmj) = X + (520-X) + (480-X) + (X-120) + 340 + (60+X) + 340 = 1640+X So you can replace the 580 with any value of X between 120 and 480, and the total number of people will be X+1640. I strongly suggest you try X = 360, which gives a nice solution to a Millennium cruise problem! (Or would you prefer X = 361 for the same reason?) On the other hand, what crew would have been unable to remember that number? There must have been too much to drink ... Sometimes figuring out what the problem should be can be more fun than actually solving the problem itself! ====================================================================== I think that should answer your question precisely. It's interesting that the word "millennium" was left out of your version. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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