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Help with an Algebra 1 word problem


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Hi.  I am using Anita Harnadek's Algebra I Word Problems book from Critical Thinking Co. with my dd.  We came across this problem last night, and I nor my husband could figure it out.  It is as follows:

 

A camera store owner makes a set profit on each roll of film she sells.  Yesterday she sold a certain number of film.  If she had sold 5 fewer but made 1c more on each, her profit on them would have been $2.80 less.  If she had sold 10 more but made 1c les on each, her profit on them would have been $6.65 more.  How many did she sell?  What was the profit on each one?  What was the total profit on them?

 

The answer key lists the formulas and the final answers, but not how to get from the formulas to the final answers. 

 

f = the number of rolls of film sold yesterday; p = the profit ($) made on each roll of film sold yesterday; t = the total profit ($) made on the rolls of film sold yesterday;  pf = t; (f-5)(p+.01)=t-2.80; (f+10)(p-.01)=t+6.65; p=.80, f =125, t=100.

 

This problem is in a section labeled "Mixtures" with the heading" "Use Your Common Sense".  :~

 

If any of you could help me out, I would be very appreciative!  Thanks!

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 pf = t
(f-5)(p+.01) = t - 2.80
(f+10)(p-.01) = t + 6.65

 

(f-5)(p+.01) =  t - 2.80
fp - 5p + 0.01f - 0.05 = t - 2.80

Since fp = t,

t - 5p + 0.01f - 0.05 = t - 2.80

Subtract t from both sides,
- 5p + 0.01f - 0.05 = - 2.80
-500p + f = -275                                 -> equation (1)

 

(f+10)(p-.01)= t + 6.65
fp + 10p - 0.01f -0.1 = t + 6.65

Since fp = t,

t + 10p - 0.01f -0.1 = t + 6.65

Subtract t from both sides,
10p - 0.01f -0.1 = 6.65
1000p - f = 675                                      -> equation (2)

 

equation (1) +  equation (2)
500p = 400
p = 0.80

 

substitute p = 0.8 into equation (1)
-400  + f = -275
f = 125

 

substitute p = 0.8, f = 125 into pf = t
t = 100

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