A computer store sold a total of 300 items in a month, and sold six times as many hard drives as they did CD-ROM drives, and half as many floppy drives as hard drives, how many of each item were sold?
Let H = number hard drives
C = number CD-Roms
F = number floppy drives.
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H = 6C
(1/2)H = F
F + H + C = 300
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Three equations.
Solve for H, C, F.
Post your work if you get stuck.
To solve this problem, let's assign variables to the unknown quantities. Let's say the number of CD-ROM drives sold is represented by "x," the number of floppy drives sold is represented by "y," and the number of hard drives sold is represented by "z".
From the given information, we can create a system of equations:
1) Total items sold: x + y + z = 300
2) Hard drives sold: z = 6x (six times as many hard drives as CD-ROM drives)
3) Floppy drives sold: y = 0.5z (half as many floppy drives as hard drives)
Now, let's solve the system of equations:
Substituting equation 2 into equation 3:
y = 0.5(6x)
y = 3x
Substituting equations 2 and 3 into equation 1:
x + 3x + 6x = 300
10x = 300
x = 30
Substituting x = 30 into equations 2 and 3:
z = 6(30)
z = 180
y = 3(30)
y = 90
Therefore, the number of CD-ROM drives sold is 30, the number of floppy drives sold is 90, and the number of hard drives sold is 180.