A Martian bat flies at 60 yards per second from its nest to a dry lake. When it arrives there, it immediately continues at 45 yards per second to a nearby cave. If the distance between the lake and the cave is half the distance between the nest and the lake, what is the average speed, in yards per second, of the bat during the whole flight?

A. 36
B. 40
C. 42
D. 54
E. 65
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Let the distance between nest to the dry lake be 180 ( LCM of 60 & 45 )
So, Time taken to reach the dry lake is 3 sec

The distance between the lake to the cave be 90
So, Time taken to reach the cave is 2 sec

Hence Avg speed = \(\frac{270}{5}\) = \(54\)

So, Answer will be (D) 54
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nest - lake = d
lake - cave = d/2


d + d/2/ d/60 + d/2*45
= 3d/2 / d/60 + d/90 = 3d/2 /3d + 2d/180 = 3d/2 / 5d/180 = 3/2 * 180 /5 = 54

D is answer

If the bat travels from the Nest to Lake at 60 yards per second and the nest to lake is double the distance from the Lake to the Cave you can use simple ballparking on this question to arrive at the answer.
With ballparking you can immediately remove choices A, B, C, and E by realizing that the bat is traveling a combination of 60 yards per second and 45 yards per second, so you can conclude that the average speed will be both above 45 and below 60. D is the only option.

Since we need to calculate average speed, we can use the equation:

Average speed = total distance/total time

We are given that a Martian bat flies at 60 yards per second from its nest to a dry lake and then immediately continues at 45 yards per second to a nearby cave. We also know that the distance between the lake and the cave is half the distance between the nest and the lake.

We can let d = the distance between the lake and the cave, and 2d = the distance between the nest and the lake.

Since time = distance/rate:

Time from nest to lake = 2d/60 =d/30

Time from lake to cave = d/45

Finally we can determine the average speed for the entire trip.

Average speed = total distance/total time

Average speed = (d +2d)/(d/30 + d/45)

Average speed = (3d)/(3d/90 + 2d/90)

Average speed = (3d)/(5d/90)

Average speed = 3/(5/90)

Average speed = 3 x (90/5)

Average speed = 3 x 18

Average speed = 54

Answer: D
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Let´s solve this problem without taking into account the "examiner´s generosity" (in terms of inviable alternative choices) nor exploring particular cases.

(Both approaches are important. We just want to avoid repetitions and present the UNITS CONTROL, one of the most powerful tools of our method!)

\(A \to B\,\,:\,\,\,\,\,\left\{ {\,\frac{{60\,\,{\text{yards}}}}{{1\,\,{\text{second}}}}\,\,\,\,\,;\,\,\,\,\,2x\,\,{\text{yards}}} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,B \to C\,\,:\,\,\,\,\left\{ {\,\frac{{45\,\,{\text{yards}}}}{{1\,\,{\text{second}}}}\,\,\,\,\,;\,\,\,\,\,\,x\,\,{\text{yards}}} \right.\)

\(? = \frac{{{\text{# }}\,\,{\text{total}}\,\,{\text{yards}}}}{{{\text{# }}\,\,{\text{total}}\,\,{\text{seconds}}}}\,\,\mathop = \limits^{\left( * \right)} \,\,\,\frac{{2x + x}}{{\frac{{2x \cdot \boxed3}}{{60 \cdot \boxed3}} + \frac{{x \cdot \boxed4}}{{45 \cdot \boxed4}}}}\,\, = \,\,\frac{{3 \cdot x \cdot 18 \cdot 10}}{{10 \cdot x}} = 54\,\,\,\,\left[ {\frac{{{\text{yards}}}}{{{\text{second}}}}} \right]\)

\(\left( * \right)\,\,\,\,\,\frac{{\left[ {\,{\text{yard}}\,} \right]}}{{\,\,\,\left[ {\,\frac{{{\text{yard}}}}{{{\text{second}}}}\,} \right]\,\,\,}} = \,\,\left[ {\,{\text{second}}\,} \right]\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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