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### 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.
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
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/
```
Associated Topics:
Middle School Word Problems

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