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Word problems (weights).

bergausstein

Active member
Jul 30, 2013
191
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.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You are using the number of weights rather than the weight of each set in your equation. You want:

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