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If Susan takes 9 seconds to run y yards, how many minutes will it take her to run x yards at the same rate?

A)\(\frac{xy}{9}\)

B)\(\frac{9x}{60y}\)

C)\(\frac{60xy}{9}\)

D)\(\frac{xy}{540}\)

E)\(\frac{540x}{y}\)

Rate*Time=Distance.

Since, Susan takes 9 seconds to run \(y\) yards, then her rate is \(Rate=\frac{Distance}{Time}=\frac{y}{9}\) yards per second.

To run \(x\) yards at this rate she'll need \(Time=\frac{Distance}{Rate}=\frac{x}{(\frac{y}{9})}=\frac{9x}{y}\) seconds, which is \(\frac{9x}{60y}\) minutes.

Re: If Susan takes 9 seconds to run y yards, how many minutes [#permalink]

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09 Sep 2013, 08:27

Susan runs y yards in 9 seconds which means her rate over that particular distance is: r = y/9

We're told to look for how long it will take her to run x yards at the same rate (i.e. r=y/9)

Time taken: t = d/r t = x/r t = x/(y/9) t = 9x/y This gives us the time in seconds but we need to multiply by 60 (seconds) to get the time in minutes.

t = 9x/60y

I had a bit of trouble conceptualizing this problem at first, specifically why we would find the time of the new distance (in minutes) by dividing by the old rate. Well, I guess it was a silly problem to be caught up on. All we want to know is how long (t) it will take for x yards to be covered at the same rate we determined to be (y/9) Then, we just had to multiply by 60 seconds to get y in terms of minutes. I guess the final answer tripped me up because of the way it multiplied out (in other words, if t = d/r then why is d here equal to 9 seconds*x distance.) I'm not sure what I can do to help me better understand these kinds of problems.

ANSWER: b.) 9x/60y

P.S. Bunuel, I believe you got the correct answer (it's the same one I got) but you marked down the incorrect answer choice.

Re: If Susan takes 9 seconds to run y yards, how many minutes [#permalink]

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11 Nov 2014, 04:13

Solving these kinds of problems always causes trouble for me. Thats why i like to pick numbers.

Is it correct to do so as follows:

Let's say y equals 10 yards and x equals 100 yards. Which would mean that if those 10 yards are ran in 9 seconds, she runs 100 in 1,5. Then filling in this X and Y in the answer choices should result in 1,5 and if so that one is the correct answer.

9x/60y gives 900/600 which equals 1,5 so this is the right equation. Is this a correct way of solving this or did i get lucky?

Solving these kinds of problems always causes trouble for me. Thats why i like to pick numbers.

Is it correct to do so as follows:

Let's say y equals 10 yards and x equals 100 yards. Which would mean that if those 10 yards are ran in 9 seconds, she runs 100 in 1,5. Then filling in this X and Y in the answer choices should result in 1,5 and if so that one is the correct answer.

9x/60y gives 900/600 which equals 1,5 so this is the right equation. Is this a correct way of solving this or did i get lucky?

That's correct.

Note though, that for plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only.

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