Question

Difficulty: HardBasic Algebraic Expressions and One-Step Equations

A municipal water authority pumps water into a storage reservoir. During a drought conservation phase, the authority reduces its standard daily pumping volume by 35%35\%. If the resulting reduced daily pumping volume is WW acre-feet, which of the following equations correctly expresses the standard daily pumping volume, SS, in acre-feet, in terms of WW?

  1. S=W0.65S = \frac{W}{0.65}Answer
  2. B
    S=0.65WS = 0.65W
  3. C
    S=W0.35S = \frac{W}{0.35}
  4. D
    S=0.35WS = 0.35W
  5. E
    S=W+0.35S = W + 0.35

Answer

The standard daily pumping volume is given by S=W0.65S = \frac{W}{0.65}.
Reducing the standard volume SS by 35%35\% leaves 65%65\% of SS, which can be written algebraically as 0.65S0.65S. Since this reduced amount equals WW, the relationship is 0.65S=W0.65S = W. Dividing both sides by 0.650.65 isolates SS to give the one-step equation solution S=W0.65S = \frac{W}{0.65}.

Step-by-Step Solution

1
Express the reduced volume in terms of the standard volume SS
Reduced volume =S0.35S=0.65S= S - 0.35S = 0.65S
A 35%35\% reduction leaves 100%35%=65%100\% - 35\% = 65\% of the original standard volume SS.
2
Set up the one-step algebraic equation matching the given variable WW
0.65S=W0.65S = W
The reduced volume is defined as WW acre-feet.
3
Solve the one-step equation for SS
S=W0.65S = \frac{W}{0.65}
Divide both sides of the equation by 0.650.65 to isolate SS.

Key Concept

Formulating and solving a one-step algebraic equation involving a percentage decrease.

Alternative Method

Convert 65%65\% to the fraction 1320\frac{13}{20}. Then 1320S=W\frac{13}{20}S = W, which yields S=2013W=W0.65S = \frac{20}{13}W = \frac{W}{0.65}.
Estimated Time:1m 30s
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