SAT Ratios, Rates & Proportions Question 36 | Free SAT Prep | Aniko

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Question 36·Hard·Ratios, Rates, Proportional Relationships, and Units

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Pump A removes water from a pool at a constant rate. Pump B removes water at a rate that is 60 percent faster than the rate of Pump A. Working together, the two pumps remove 18,000 gallons of water in 2 hours. How many hours would Pump A, working alone, take to remove 12,000 gallons of water?

For work-rate problems, first assign a variable to one worker’s rate and express all other rates in terms of that variable, especially when you see phrases like “60 percent faster,” which means multiplying the base rate by $1.6$ . Next, convert any “together they complete X in Y hours” information into a combined rate (amount ÷ time) and set that equal to the algebraic combined rate to solve for the variable. Finally, answer the question being asked—often a different amount of work for one worker—using $\text{time} = \dfrac{\text{amount}}{\text{rate}}$ , and simplify carefully without unnecessary rounding so you can match an exact answer choice quickly.

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Introduce a variable for a rate

Try letting Pump A’s rate be $r$ gallons per hour. How can you express Pump B’s rate in terms of $r$ if it is 60% faster than Pump A?

Turn the “together” information into an equation

If Pump A’s rate is $r$ and Pump B’s rate is written in terms of $r$ , what is their combined rate? Use the fact that together they remove 18,000 gallons in 2 hours to find a numerical value for this combined rate.

Solve for Pump A’s rate, then use time = amount ÷ rate

Once you set the combined rate in terms of $r$ equal to the numerical combined rate, solve for $r$ . Then, use $\text{time} = \dfrac{\text{amount}}{\text{rate}}$ for Pump A removing 12,000 gallons alone.

Be careful with “60 percent faster”

Remember that “60 percent faster” means adding 60% of the original rate to the original rate (a factor of $1.6$ ), not just $0.6$ times the original rate.

Desmos Guide

Compute Pump A’s rate from the combined rate

From the word problem, their combined rate is $18{,}000 \div 2 = 9{,}000$ gallons per hour, and that equals $2.6$ times Pump A’s rate. In Desmos, in one expression line type 9000/2.6 and note the decimal output; this is Pump A’s rate in gallons per hour.

Compute the time for Pump A alone

In a new expression line, type 12000 / (9000/2.6) (time = amount ÷ rate, using Pump A’s rate from step 1). The value Desmos displays is the number of hours Pump A would need alone. Compare this decimal to the fraction choices by typing each choice (for example, 52/15, 13/4, etc.) into Desmos and seeing which fraction matches the decimal time.

Step-by-step Explanation

Represent the pump rates with a variable

Let Pump A’s pumping rate be $r$ gallons per hour.

“60 percent faster” means Pump B’s rate is 60% more than Pump A’s rate:

100% of $r$ is $r$
60% of $r$ is $0.6r$
So Pump B’s rate is $r + 0.6r = 1.6r$ gallons per hour.

Together, the two pumps have a combined rate of

r + 1.6r = 2.6r \text{ gallons per hour.}

Use the information about working together

Together, the pumps remove 18,000 gallons in 2 hours.

Rate is amount ÷ time, so their combined rate is

\frac{18{,}000}{2} = 9{,}000 \text{ gallons per hour.}

Set this equal to the expression for their combined rate in terms of $r$ :

2.6r = 9000.

Now solve for $r$ .

First rewrite $2.6$ as a fraction: $2.6 = \dfrac{26}{10} = \dfrac{13}{5}$ . So

\frac{13}{5}r = 9000.

Multiply both sides by $\dfrac{5}{13}$ :

r = 9000 \cdot \frac{5}{13} = \frac{45{,}000}{13} \text{ gallons per hour.}

This is Pump A’s rate.

Find Pump A’s time to pump 12,000 gallons alone

Time is given by

\text{time} = \frac{\text{amount}}{\text{rate}}.

For Pump A alone to remove 12,000 gallons,

\text{time} = \frac{12{,}000}{\dfrac{45{,}000}{13}}.

Dividing by a fraction is the same as multiplying by its reciprocal:

\text{time} = 12{,}000 \cdot \frac{13}{45{,}000}.

Simplify $12{,}000/45{,}000$ :

\frac{12{,}000}{45{,}000} = \frac{12}{45} = \frac{4}{15}.

\text{time} = \frac{4}{15} \cdot 13 = \frac{52}{15} \text{ hours}.

Thus, Pump A alone would take $\boxed{\dfrac{52}{15}}$ hours to remove 12,000 gallons.

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation