Another Algebra word problem :(?

Another Algebra word problem :(?

Postby filippo » March 21st, 2014, 10:31 pm

I can do normal problems well, but the word problems always get me. Here it is

Adam earns $37 mowing lawns. At his favorite store it costs $6.50 to rent a newly-released videogame and $4 to rent movies.

A. write an equation that gives the amount of money Adam has left, b, after renting 1 video game and m, movies.

B.How many movies can he rent if he rents no video games?

C. How many videogames can he rent if he rents no movies?

If you can show a bit on how you got the answers and not just give them to me that would be helpful. I am horrid at word problems, thanks guys.
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filippo
 
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Another Algebra word problem :(?

Postby blanco » March 21st, 2014, 10:36 pm

Rental cost is given by 6.5v + 4m.....so, money left would be:

b = 37 - 6.5v - 4m...........so, when v = 1 we have:

b = 37 - 6.5 - 4m

b = 30.5 - 4m.........(A)

When v = 0 we have:

b = 37 - 4m

so, m = 37/4 = 9.25......i.e. maximum 9 movies.......(B)

When m = 0 we have:

b = 37 - 6.5v

so, v = 37/6.5.......i.e. maximum 5 videogames.......(C)

:)>
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blanco
 
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Another Algebra word problem :(?

Postby beechy4 » March 21st, 2014, 10:42 pm

A. The money he spends is 6.5+4m. The money he has left is b=37-(6.5+4m)
B. 37/4=9.25 so he can rent 9 movies.
C. 37/6.5=5.69 so he can rent 5 games
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Another Algebra word problem :(?

Postby harlowe » March 21st, 2014, 10:54 pm

A. i dont really get it
B. 37 / 4 = 9.25. but he cant rent 9.25 movies so answer is 9
C.37 / 6.5 = 5.69 nut he cant rent 5.69 video games so the answer is 5
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harlowe
 
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Another Algebra word problem :(?

Postby galtero » March 21st, 2014, 10:58 pm

A. i dont really get it
B. 37 / 4 = 9.25. but he cant rent 9.25 movies so answer is 9
C.37 / 6.5 = 5.69 nut he cant rent 5.69 video games so the answer is 5
A b = 37 - 6.50 - 4m


B 9 movies ( 9 X 4 = 36 $ )

C 5 video games ( 5 X 6.5 = 32.50 $ )
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