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  • 1 Post By soroban
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July 25th, 2014, 08:05 PM   #1
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System of equations (practical problem)

It looks like a simple problem but I'm not sure how to going on.

In a pastry class each student prepare a dough of water and flour weighing 8 ounces. The amounts of flour and water for each student were varied but never 0. One student spent one quarter of the available water and 1/6 of the available flour. How many students were in the class?

My attempt:

$$
\frac{1}{4}mass_{water} + \frac{1}{6}mass_{flour} = 8 \\
\frac{3}{4}mass_{water} + \frac{5}{6}mass_{flour} = 8 \times(n-1),
$$

being $n$ the number of students. It seems that I need one equation more.

What do you think?

Last edited by dapias09; July 25th, 2014 at 08:07 PM.
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July 25th, 2014, 08:53 PM   #2
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Hello, dapias09!

You're off to a good start.


Quote:
In a pastry class, each student prepares a dough of water and flour
weighing 8 ounces. The amounts of flour and water for each student
were varied but never 0. One student spent one quarter of the available
water and 1/6 of the available flour. How many students were in the class?

I used your equations.
Let $n$ = number of students.

$\quad \frac{1}{4}W + \frac{1}{6}F \:=\:8$

$\quad \frac{3}{4}W + \frac{5}{6}F \:=\:8(n-1)$


I cleared out the fractions:

$\quad 3W + 2F \:=\:96$

$\quad 9W + 10F \:=\:96n-96$


I solved the system: $\begin{Bmatrix}W &=& 96 - 16n \\ F &=& 24n - 96\end{Bmatrix}$

Since $W > 0$, we have: $\:96-16n \:>\:0 \quad\Rightarrow\quad n \:<\:6$

Since $F > 0$, we have: $\: 24n - 96 \:>\:0 \quad\Rightarrow\quad n \:>\:4$


Therefore: $\:n \:=\:5.$

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July 26th, 2014, 08:11 AM   #3
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Hello soroban, thank you very much, your explanation makes sense for me
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