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BTGmoderatorLU
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A car traveled from Los Angeles to San Francisco in 6 hours at an average rate of x miles per hour. If the car returned along the same route at an average rate of y miles per hour, how long did it take for the car to make the entire round trip, in minutes?
$$A.\ \left(6+\frac{6x}{y}\right)\cdot60$$
$$B.\ \left(6+\frac{6y}{x}\right)\cdot60$$
$$C.\ 30\cdot\left(x+y\right)$$
$$D.\ 10\cdot\left(x+y\right)$$
$$E.\ \frac{\left(x+y\right)}{360}$$
The OA is A.
Can any expert please explain to me what I'm doing wrong:
I know that the car traveled 6x one way, so it must travel back 6x the other way, so total round trip 12x. Te rates (in minutes) are x/60 for the first leg, and y/60 for the second leg.
Thus, isn't the time simply the total distance over the combined rates? Or
$$\frac{12x}{\frac{\left(x+y\right)}{60}}\ \Rightarrow \frac{720}{\left(x+y\right)}?$$
Thanks in advance.
$$A.\ \left(6+\frac{6x}{y}\right)\cdot60$$
$$B.\ \left(6+\frac{6y}{x}\right)\cdot60$$
$$C.\ 30\cdot\left(x+y\right)$$
$$D.\ 10\cdot\left(x+y\right)$$
$$E.\ \frac{\left(x+y\right)}{360}$$
The OA is A.
Can any expert please explain to me what I'm doing wrong:
I know that the car traveled 6x one way, so it must travel back 6x the other way, so total round trip 12x. Te rates (in minutes) are x/60 for the first leg, and y/60 for the second leg.
Thus, isn't the time simply the total distance over the combined rates? Or
$$\frac{12x}{\frac{\left(x+y\right)}{60}}\ \Rightarrow \frac{720}{\left(x+y\right)}?$$
Thanks in advance.
