It takes Mike 1 hour and 30 minutes to commute from home to work at an average speed of 40 miles per hour. If Mike retur

[VJesus12]

It takes Mike 1 hour and 30 minutes to commute from home to work at an average speed of 40 miles per hour. If Mike returns home along the same route at an average speed of 45 miles per hour, how long does the return trip take?

A. 1 hour, 15 minutes

B. 1 hour, 20 minutes

C. 1 hour, 25 minutes

D. 1 hour, 30 minutes

E. 1 hour, 35 minutes

[spoiler]OA=B[/spoiler]

Source: Princeton Review

[deloitte247]

$$Average=\frac{dis\tan ce}{time\ taken}$$
Distance between Mike's home and workplace=>
$$=40\ miles\ per\ hour\ \cdot\ 1\frac{1}{2}=40\cdot1.5=60\ miles$$
Mike returns home along the same route at an average speed of 45 miles per hour.
$$Time\ taken\ for\ return\ trip=\frac{dis\tan ce}{speed}=\frac{60miles}{45miles\ per\ hour}$$ $$=\frac{4}{3}hours$$
$$=1\frac{1}{3}hours$$
=> 1 hour 20 minutes

Answer = option B

[marceltr]

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[GMATGuruNY]

It takes Mike 1 hour and 30 minutes to commute from home to work at an average speed of 40 miles per hour. If Mike returns home along the same route at an average speed of 45 miles per hour, how long does the return trip take?

A. 1 hour, 15 minutes

B. 1 hour, 20 minutes

C. 1 hour, 25 minutes
D. 1 hour, 30 minutes

E. 1 hour, 35 minute

(higher speed)/(lower speed) = 45/40 = 9/8
Rate and time have a RECIPROCAL RELATIONSHIP.
Since the speed home is 9/8 of the speed to work, the time home will be 8/9 of the time to work:
(8/9)(90 minutes) = 80 minutes = 1 hour 20 minutes

The correct answer is B.

[Scott@TargetTestPrep]

It takes Mike 1 hour and 30 minutes to commute from home to work at an average speed of 40 miles per hour. If Mike returns home along the same route at an average speed of 45 miles per hour, how long does the return trip take?

A. 1 hour, 15 minutes

B. 1 hour, 20 minutes

C. 1 hour, 25 minutes

D. 1 hour, 30 minutes

E. 1 hour, 35 minutes

[spoiler]OA=B[/spoiler]
Source: Princeton Review

Solution:

The distance between Mike’s home and his workplace is 40 x 1.5 = 60 miles. Therefore, the return trip takes 60/45 = 4/3 hours = 1 ⅓ hours = 1 hour 20 minutes.

Answer: B

[taniyuki]

$$Average=\frac{dis\tan ce}{time\ taken}$$
Distance between Mike's home and workplace=>
$$=40\ miles\ per\ hour\ \cdot\ 1\frac{1}{2}=40\cdot1.5=60\ miles$$
Mike returns home along the same route at an average speed of 45 miles per hour.
$$Time\ taken\ for\ return\ trip=\frac{dis\tan ce}{speed}=\frac{60miles}{45miles\ per\ hour}$$ $$=\frac{4}{3}hours$$
$$=1\frac{1}{3}hours$$
=> 1 hour 20 minutes

Answer = option B

Great, I agree with this solution