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

  • Context: MHB 
  • Thread starter Thread starter bergausstein
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on calculating the number of weights needed at a gym to total 3180 pounds. The user initially set up the problem using variables for the number of 100-pound, 50-pound, and 20-pound weights but arrived at non-integer results. The correct approach involves setting up the equation as follows: 100x + 50(x + 6) + 60(x + 6) = 3180, which ensures that the total weight is accurately represented. This method leads to a solvable equation that provides whole number solutions for the weights.

PREREQUISITES
  • Understanding of algebraic equations and variables
  • Familiarity with weight conversions and calculations
  • Basic knowledge of gym equipment and weight types
  • Ability to solve linear equations
NEXT STEPS
  • Review algebraic equation solving techniques
  • Practice setting up equations based on word problems
  • Learn about weight distribution in fitness settings
  • Explore integer solutions in algebraic contexts
USEFUL FOR

Mathematicians, gym owners, fitness enthusiasts, and anyone involved in weight management or gym equipment logistics will benefit from this discussion.

bergausstein
Messages
191
Reaction score
0
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.
 
Mathematics news on Phys.org
You are using the number of weights rather than the weight of each set in your equation. You want:

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

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
7K
  • · Replies 5 ·
Replies
5
Views
2K