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Lia,Phil, and Cam collected a total of $200.30 for a holiday fundraiser . Phil collects 12.80 more than Lia . Cam collects 3 times as much as Lia . How much does each person collect ?

 Dec 10, 2014

Best Answer 

 #1
avatar+270 
+8

This is an easy question on application of algebra.

First, you should write down all the facts you know.

    (P for Phil, L for Lia and C for Cam.)

    P + L + C = 200.30

    P = L + 12.80

    C = 3L

 

Second, you should find common letter in those equations.

    In this case, it's L.

    So, you should subtitute P = L + 12.80 and C = 3L to P + L + C= 200.30.

    Then you will get (L + 12.80) + L + 3L = 200.30

    

Third, then you can solve it using basic algebra.

    (L + 12.80) + L + 3L = 200.30

=  5L + 12.80 = 200.30

=  5L = 187.50

=  L = 37.5

 

Lia collected 37.5.

 

Fourth, You just subtitute L to other equations.

Then you will get C and P as well.

 

Hope it helped! 

 Dec 10, 2014
 #1
avatar+270 
+8
Best Answer

This is an easy question on application of algebra.

First, you should write down all the facts you know.

    (P for Phil, L for Lia and C for Cam.)

    P + L + C = 200.30

    P = L + 12.80

    C = 3L

 

Second, you should find common letter in those equations.

    In this case, it's L.

    So, you should subtitute P = L + 12.80 and C = 3L to P + L + C= 200.30.

    Then you will get (L + 12.80) + L + 3L = 200.30

    

Third, then you can solve it using basic algebra.

    (L + 12.80) + L + 3L = 200.30

=  5L + 12.80 = 200.30

=  5L = 187.50

=  L = 37.5

 

Lia collected 37.5.

 

Fourth, You just subtitute L to other equations.

Then you will get C and P as well.

 

Hope it helped! 

flflvm97 Dec 10, 2014
 #2
avatar+7188 
0

Thanks for answering that.....Zegroes and I were steering clear of this dead end

 Dec 10, 2014
 #3
avatar+4164 
0

the only dead end here is my lazyiness....Wait or maybe thats the beggining???

 Dec 10, 2014
 #4
avatar+7188 
0

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{l, c, p}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{L}}{\mathtt{\,\small\textbf+\,}}{\mathtt{C}}{\mathtt{\,\small\textbf+\,}}{\mathtt{P}}={\mathtt{200.3}}\\
{\mathtt{L}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12.8}}={\mathtt{P}}\\
{\mathtt{L}}{\mathtt{\,\times\,}}{\mathtt{3}}={\mathtt{C}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{l}} = {\frac{{\mathtt{75}}}{{\mathtt{2}}}}\\
{\mathtt{c}} = {\frac{{\mathtt{225}}}{{\mathtt{2}}}}\\
{\mathtt{p}} = {\frac{{\mathtt{503}}}{{\mathtt{10}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{l}} = {\mathtt{37.5}}\\
{\mathtt{c}} = {\mathtt{112.5}}\\
{\mathtt{p}} = {\mathtt{50.3}}\\
\end{array} \right\}$$

.
 Dec 11, 2014

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