hard problem solving~~>< plz HELP

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September 8th 2006, 01:05 AM
fring

hard problem solving~~>< plz HELP

In a class students made a full cup of coffee with milk for themself. After finishing her cup, Kathy calculated that she drank 1/5 of the total amount of coffee and 1/7 of the total amount of milk used by all students. Assuming that everyone had cups of the same capacity, how many students were in the class? Find all possible anwsers.

>.< so difficult
any help appreciated
thank you!!
September 8th 2006, 01:42 AM
CaptainBlack

Quote:

Originally Posted by fring

In a class students made a full cup of coffee with milk for themself. After finishing her cup, Kathy calculated that she drank 1/5 of the total amount of coffee and 1/7 of the total amount of milk used by all students. Assuming that everyone had cups of the same capacity, how many students were in the class? Find all possible anwsers.

>.< so difficult
any help appreciated
thank you!!

Let the amount of coffee consumed by the class be $c$ cups

Let the amout of milk consumed by the class be $m$ cups.

We do not assume that $c$ and/or $m$ are integers,
just that they are greater than to zero (because Kathy drank some of both
neither can be zero).

Kathy drank a cup of coffee milk mixture, and from what we are told we
know that:

$ 1=\frac{c}{5}+\frac{m}{7} $ .

Rearranging this gives:

$ 35=7c+5m\ \ \ \dots(1) $ .

Also if there are $N$ students in the class, as they each drink a cup we have:

$ N=c+m\ \ \ \dots(2) $ .

From $(1)$ it is clear that $c\le 5$ cups and $m \le 7$ , so $N \le 12$ .

Now trial and error should allow you to find that the only solution consistent
with $(1)$ , $(2)$ , and the constraints on $c$ , $m$ and $N$ is:

$N=6,\ c=2.5,\ \mbox{and }m=3.5$ .

You will need to check this yourself.

RonL
September 8th 2006, 01:51 AM
malaygoel

Quote:

Originally Posted by CaptainBlack

Let the amount of coffee consumed by the class be $c$ cups

Let the amout of milk consumed by the class be $m$ cups.

We do not assume that $c$ and/or $m$ are integers,
just that they are greater or equal to zero.

Kathy drank a cup of coffee milk mixture, and from what we are told we
know that:

$ 1=\frac{c}{5}+\frac{m}{7} $ .

Rearranging this gives:

$ 35=7c+5m\ \ \ \dots(1) $ .

Also if there are $N$ students in the class, as they each drink a cup we have:

$ N=c+m\ \ \ \dots(2) $ .

From $(1)$ it is clear that $c\le 5$ cups and $m \le 7$ , so $N \le 12$

7c +5m=35
2c+5c+5m=35
2c+5n=35
N is an integer less than 7 otherwise c will be zero or negative.

Keep Smiling
Malay
September 8th 2006, 02:26 AM
CaptainBlack

Quote:

Originally Posted by malaygoel

7c +5m=35
2c+5c+5m=35
2c+5n=35
N is an integer less than 7 otherwise c will be zero or negative.

Keep Smiling
Malay

I saw no practical need to tighten up the constraint on $N$ as 12 is a small
enough number so that further reduction in computation in not worth the trouble
(and the lower constraint becomes evident when one does the search, my
search only went as far as $N=7$ as it was evident that no
greater value was viable).

My objective was to constrain the problem sufficiently so that a search for
feasible solutions would not be too arduous, and I stopped when this was
the case.

I expect with some effort we could constrain the problem sufficiently to
allow only a single feasible solution, and so need mo search at all.

RonL
September 8th 2006, 03:20 AM
malaygoel

Quote:

Originally Posted by CaptainBlack

I saw no practical need to tighten up the constraint on $N$ as 12 is a small
enough number so that further reduction in computation in not worth the trouble
(and the lower constraint becomes evident when one does the search, my
search only went as far as $N=7$ as it was evident that no
greater value was viable).

My objective was to constrain the problem sufficiently so that a search for
feasible solutions would not be too arduous, and I stopped when this was
the case.

I expect with some effort we could constrain the problem sufficiently to
allow only a single feasible solution, and so need mo search at all.

RonL

You are right.
2c+5n=35
2c<2n
2c+5n<7n
35<7n
n>5
the only solution is n=6, c=5/2 and m=7/2

Keep Smiling
Malay