SOLUTION: I am stumped! I have been at this math problem for what seems like hours. I would really like it if you can help me out. The answer may be so easy to get, but I just can't figure i

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Question 178134: I am stumped! I have been at this math problem for what seems like hours. I would really like it if you can help me out. The answer may be so easy to get, but I just can't figure it out. Please help.

The word problem is as followed:
A computer store sold a total of 300 items last month. The store sold six times as many hard drives as they did CD ROM drives, and half as many floppy drives as hard drives. Based on this information how many of each item was sold?

Thanks so much .. your help is greatly appreciated!

Found 2 solutions by gonzo, Mathtut:
Answer by gonzo(654) (Show Source):
You can put this solution on YOUR website!
store sold 300 items total.
let x = number of hard drives sold
let y = number of cd drives sold
let z = number of floppy drives sold
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x + y + z = 300
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The store sold six times as many hard drives as they did CD ROM drives.
hard drives = 6 * cd drives meaning:
x = 6y
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store sold half as many floppy drives as hard drives.
floppy drives = 1/2 * hard drives meaning:
z = x/2.
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with this information you should be able to reduce your equation to one unknown.
you have z = x/2.
this is your first substitution.
---
you have x = 6y.
if you divide both sides of this equation by 6, you get:
y = x/6.
this becomes your second substitution.
---
since all variables are now in terms of x, your equation becomes:
x + x/6 + x/2 = 300
multiply both sides of this equation by 6 to get:
6x + x + 3x = 1800
which becomes:
10x = 1800
which becomes:
x = 180
y = 180/6 = 30
z = 180/2 = 90
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your answer should be:
x = 180
y = 30
z = 90
---
x + y + z = 300 becomes:
180 + 30 + 90 = 300 becomes:
300 = 300
which is true so the values for x,y,z are good.
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Answer by Mathtut(3670) (Show Source):
You can put this solution on YOUR website!
let x, y and z be the number of hard drive CD roms, and floppy drive sold, respectively
:
x+y+z=300.......eq 1
x=6y............eq 2
z=x/2..........eq 3
:
re write eq 2 to y=x/6. Now take that value of y which is x/6 and z's value from eq 3 which is x/2 and plug them into eq 1
:
x+(x/6)+(x/2)=300
:
6x+x+3x=1800........mulitiplied each term by 6 to get rid of fractions
:
10x=1800
:
hard drives sold
:
cd roms sold
:
floppies sold