How Many Men, Women, and Children at the Circus?
Date: 10/18/2000 at 23:41:58 From: Morgan Subject: Word Problem A circus performance is witnessed by 120 people who have paid $120. The men paid $5, the women $2, and the children $.10 each. How many of each went to the circus? a) Solve this problem using any method and show all work. b) Explain your logic in writing and describe why your answer is correct.
Date: 10/19/2000 at 17:59:00 From: Doctor Greenie Subject: Re: Word Problem Hi, Morgan. Thanks for sending your question to us here at Dr. Math. This is quite an ambitious problem for an 11-year-old. There are ways to solve problems like this using algebra, which you might learn about in high school. I will assume that at your age you have had only a very slight introduction to the ideas of algebra. There are ways for a younger student like you to solve the problem, too - but to do so without doing a huge amount of work takes a lot of thinking and analysis. The basic problem in solving a problem like this by elementary methods is that there are so many possibilities to consider. For instance, one approach to this problem would be to try every combination of number of children, number of women, and number of men that totals 120; however, if my calculations are correct, there are 7,381 different ways to make a total of 120 people using different numbers of men, women, and children. Solving the problem by trying 7,381 different cases does not look like an easy way. So we need to analyze the problem to see if we can logically reduce the number of cases we need to look at. Let's show a list of all the possible cases we need to consider. As we use logic to eliminate some of these possible cases, we will display this list again to see how we are cutting down on the number of cases we need to consider. At this point we have 7,381 cases to consider: children women men -------- ----- --- 0 0 120 0 1 119 : : : (No, I'm NOT going to list all 7,381 of them!) : : : 119 0 1 119 1 0 120 0 0 So, what can we do to cut down on the number of cases we need to consider? Well, one thing to notice is that the men's and women's ticket prices are whole numbers of dollars, while the child's ticket price is a fraction of a dollar (ten cents). Since the total amount of money from ticket sales is exactly $120.00, what do we know about the number of children? Could it be 55, or 72? No! If there were 55 children, the total paid by those 55 children would be $5.50, and the overall total could not come out to be exactly $120.00; similarly, if there were 72 children, you could also not get a total of exactly $120.00. (1) IN ORDER FOR THE TOTAL AMOUNT FROM ALL TICKET SALES TO BE EXACTLY $120.00, THE NUMBER OF CHILDREN MUST BE A MULTIPLE OF 10. So now our list looks something like this: children women men -------- ----- --- 10 0 110 10 : : 10 110 0 20 0 100 20 : : 20 100 0 : : : 110 0 10 110 : : 110 10 0 120 0 0 Well, that was a big improvement. We've cut the possible choices for the number of children down from (0, 1, 2, ..., 120) to (0, 10, 20, ..., 120). What more can we do to cut down on the number of cases we need to look at? Each woman paid $2 and each man paid $5. What is the largest number of men and women together that could have bought tickets? If there were no men, then the largest number of women there could be would be 60 (60 women at $2 each is $120, which is the total for ALL tickets sold). Since a man's ticket costs more than a woman's ticket, we know that (2) THE NUMBER OF MEN AND WOMEN TOGETHER MUST BE LESS THAN 60; THEREFORE, SINCE THE TOTAL NUMBER OF TICKETS IS 120, THE NUMBER OF CHILDREN MUST BE MORE THAN 60. By logical analysis, we have set a lower limit on the number of children. We know it must be (a) a multiple of 10, and (b) greater than 60, so it must be at least 70. Now, by a similar analysis, can we find an upper limit on the number of children? Yes, we can. We can quickly eliminate the case of 120 children and no men or women, because that case has a total ticket sales amount of only $12. What about the cases for 110 children or 100 children? If there are 110 children, then ticket sales to children totals $11; we need another $109 from ticket sales to a total of 10 men and women. But a man's ticket costs $5, so even if all 10 of the adults were men we could not get the other $109 we need. And if there were 100 children, then ticket sales to children would total $10, and even with the 20 adults being all men, the total ticket sales amount would only be $110. If there are 90 children, then ticket sales to children will total $9, and we would need $111 from ticket sales for a total of 30 men and women. Since 30 men's tickets would bring in $150, there is a chance that there is a combination of 30 men and women for which the ticket sales would bring in the required remaining $111. (3) WITH MEN'S AND WOMEN'S TICKET PRICES BEING $5 AND $2, THERE MUST BE MORE THAN 20 ADULTS TO RESULT IN A TOTAL OF $120 IN TICKET SALES; THEREFORE THE NUMBER OF CHILDREN IS LESS THAN 100. Combining results (1), (2), and (3), we now know the number of children can only be 70, 80, or 90; so the list of combinations we need to consider now looks like this: children women men -------- ----- --- 70 0 50 70 : : 70 50 0 80 0 40 80 : : 80 40 0 90 0 30 90 : : 90 30 0 At this point, let's stop and see what we have accomplished by analyzing the problem. We started out with the possible number of children being anywhere from 0 to 120 (121 choices); we have cut that down to the three choices 70, 80, or 90. So we have cut the amount of work we might have to do by a factor of about 40. Now let's go ahead and see how we can finish this problem. Let's look first at the possible solutions to the problem in which there are 70 children. The 70 children's tickets cost $7, so the total cost of the combined 50 adult tickets must be $113. Each man's ticket costs $5 and each woman's ticket costs $2. What combinations of 50 adult tickets can cost a total of $113? Well, the $113 is an odd number, while the $5 is odd and the $2 is even. The cost of any number of women's tickets is going to be an even number; on the other hand, the cost of any even number of men's tickets will be an even number, while the cost of any odd number of men's tickets will be an odd number. To get a total of $113 from the sales of adult tickets, then, there must be an odd number of men's tickets. So let's make a table of the total ticket sales amounts if there are 70 children and 50 adults, with an odd number of men: number of number of number of total cost children men women of tickets --------- --------- --------- ---------- 70 1 49 $110 70 3 47 $116 70 5 45 $122 70 : : : Now, going down in this table from one entry to the next, each time we try a combination with 2 more men and 2 fewer women, the total ticket sales amount goes up by $6 (2 fewer $2 tickets were sold; and 2 more $5 tickets were sold, for a net increase of $6). So once we see that the combination of 70 children, 3 men, and 47 women brings in less than $120, and the combination of 70 children, 5 men, and 45 women brings in more than $120, we know that there is no answer to the problem with 70 children. So we don't need to check any more of the combinations with 70 children and 50 adults. Let's do the same analysis for the possible cases with 80 children. Now the ticket sales from the children's tickets is $8. So now, using the same logic as in the case with 70 children, we can see that to get a total of $112 from the adult tickets, we have to have an even number of men. So now we make a table of the total ticket sales amounts if there are 80 children and 40 adults, with an even number of men: number of number of number of total cost children men women of tickets --------- --------- ---------- ---------- 80 0 40 $ 88 80 2 38 $ 94 80 4 36 $100 80 6 34 $106 80 8 32 $112 80 10 30 $118 80 12 28 $124 80 : : : As before, we find that there is no combination of 120 people that results in $120 in total ticket sales if the number of children is 80. Lastly we perform the same analysis for the possible combinations where there are 90 children. We are back now to the case where the number of men must be an odd number, because the total sales of children's tickets amounts to $9. So now we make a table of the total ticket sales amounts if there are 90 children and 30 adults, with again an odd number of men: number of number of number of total cost children men women of tickets --------- --------- ---------- ---------- 90 1 29 $ 72 90 3 27 $ 78 90 5 25 $ 84 90 7 23 $ 90 90 9 21 $ 96 90 11 19 $102 90 13 17 $108 90 15 15 $114 90 17 13 $120 At long last we have found the one combination that satisfies the conditions of the problem - 90 children, 17 men, and 13 women. This is a huge amount of work for an 11-year-old student. While there is a lot of work required, the work isn't particularly difficult. But the analytical abilities and thought processes that will help cut down on the amount of work are not yet well-developed in most students at that age. There are other refinements we could have made in our method to cut down a bit more on the amount of work we had to do, but I wanted to keep my solution as elementary as possible to give younger students a better chance of being able to follow it. There are algebraic methods for solving problems like this also, but as the question was submitted by an 11-year-old student, I think those methods are not appropriate in this response. If you are interested in these methods and have the knowledge of algebra required to understand them, you can find information about them in the archives by going to the Dr. Math search engine at http://mathforum.org/mathgrepform.html and entering the phrase "Diophantine equation" (without quotes). I hope you didn't get tired of reading before you reached the end of this response. There is just no short way to give a complete answer to a question like this.... - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum