RTD20

[yellowho]

Mary walks at an average speed of 8 kilometers per hour and runs at an average speed of
20 kilometers per hour. When he runs to work, he arrives 3.75 minutes sooner than when
he walks. What is the distance?

20(t-3.75/60)=8t
solve for t then multiply by 8. Is there a quicker way?

[Anurag@Gurome]

Mary walks at an average speed of 8 kilometers per hour and runs at an average speed of 20 kilometers per hour. When he runs to work, he arrives 3.75 minutes sooner than when he walks. What is the distance?

It can be a bit quicker if you directly form the equation on the distance. But I think they are almost same.

Say, the distance = d kilometers

Hence, (d/8) = (d/20) + (3.75/60)
=> (15d/120) = (6d/120) + (7.5/120)
=> 15d = 6d + 7.5
=> 9d = 7.5
=> d = (7.5/9) = (2.5/3) = 5/6

Therefore, the distance = (5/6) kilometers

[GMATGuruNY]

Mary walks at an average speed of 8 kilometers per hour and runs at an average speed of
20 kilometers per hour. When he runs to work, he arrives 3.75 minutes sooner than when
he walks. What is the distance?

20(t-3.75/60)=8t
solve for t then multiply by 8. Is there a quicker way?

We could plug in the answer choices, one of which would say that the distance = 5/6 kilometers.

Answer choice: d = 5/6 kilometers
Time to walk = d/r = (5/6)/8 = 5/48 hours.
Time to run = d/r = (5/6)/20 = 1/24 hours.
Difference = 5/48 - 1/24 = 3/48 = 1/16 hours = 3.75 minutes.