MHB How many weights of each type are needed at a gym to total 3180 pounds?

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The discussion centers on calculating the number of weights needed in a gym to total 3180 pounds, given specific relationships between different weight types. The initial approach incorrectly used the number of weights instead of their total weight in the equation. The correct formulation involves multiplying the number of each weight type by their respective weights: 100-pound, 50-pound, and 20-pound weights. This adjustment is necessary to ensure the solution yields whole numbers for the count of weights. The conversation highlights the importance of accurately representing the problem to achieve valid results.
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A gym offers a variety of weights for use by its mem-bers. If there are 6 more 50-pound weights than 100-pound weights and three times as many 20-pound
weights as 50-pound weights, for a total of 3180
pounds, how many of each weight are there?

this is howi solved it,

let
$x=$ # of 100 pound weights
$x+6=$ # of 50 pound weights
$3(x+6)=$ # of 20 pound weights

my equation,

$x+x+6+3(x+6)=3180=2x+3x+24=3180=5x+24=3180$ and then $x=631.2$

there are 631.2 (100 pound weights), 637.2(50 pound weeights), and 1911.6 (20 pound weights)

but my answers didn't make sense. because it's not a whole number. i expect to get a whole number.

can you help me with this. thanks.
 
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You are using the number of weights rather than the weight of each set in your equation. You want:

$$x(100\text{ lb})+(x+6)(50\text{ lb})+3(x+6)(20\text{ lb})=3180\text{ lb}$$
 
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