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Time/speed/Distance problem

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by Joy Shaha » Tue Jan 24, 2017 10:39 am
Q. James started from his home and drove eastwards at a constant speed. Exactly 90 minutes after James stated from his home, his brother Patrick started from the same point and drove in the same direction as James did at a different constant speed. Patrick overtook James exactly 90 minutes after Patrick started his journey and then continued driving at the same speed for another 2 hours. By what percentage should Patrick reduce his speed so that James could catch up with Patrick in exactly 8 hours after Patrick overtook James?
A. 25%
B. 33%
C. 50%
D. 67%
E. 75%
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by DavidG@VeritasPrep » Tue Jan 24, 2017 11:36 am
Joy Shaha wrote:Q. James started from his home and drove eastwards at a constant speed. Exactly 90 minutes after James stated from his home, his brother Patrick started from the same point and drove in the same direction as James did at a different constant speed. Patrick overtook James exactly 90 minutes after Patrick started his journey and then continued driving at the same speed for another 2 hours. By what percentage should Patrick reduce his speed so that James could catch up with Patrick in exactly 8 hours after Patrick overtook James?
A. 25%
B. 33%
C. 50%
D. 67%
E. 75%
This couldn't be an official question. First, an official question would likely make it explicit that Patrick reduces his speed 2 hours after he overtakes James. More importantly, the question should make it clear that the percent change is an estimate.

Say James started at 10mph. After 90 minutes, or 1.5 hours, he'll have gone 15 miles. Patrick needs to make up those 15miles in 1.5 hours, so he needs to make makeup 15/1.5 = 10 miles each our. If James is going 10mph, and Patrick is making up 10 miles each hour, then Patrick would be traveling at 20mph.

If Patrick continues at the same rate for 2 more hours, and he's going 10mph faster, then he'll be 20 miles ahead at that point. We need James to catch him 8 hours after Patrick overtakes him, so this would be 6 hours after Patrick adjusts his speed.

James will need to make up 20 miles in 6 hours, or 20/6 = 10/3 = 3 1/3 miles each hour. If James is going 10mph and he needs to go 10/3 faster than Patrick, then Patrick would have to travel at 10 - 10/3 = 20/3 mph.

Patrick's old speed was 20 mph. We need him to travel at 20/3 mph so James can overtake him. Thus, he needs to reduce his speed by 20 - 20/3 = 40/3 mph. Thus his change will be [40/3]/20 = 40/60 = 2/3 = 66.666...%
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by GMATGuruNY » Tue Jan 24, 2017 11:47 am
Joy Shaha wrote:Q. James started from his home and drove eastwards at a constant speed. Exactly 90 minutes after James stated from his home, his brother Patrick started from the same point and drove in the same direction as James did at a different constant speed. Patrick overtook James exactly 90 minutes after Patrick started his journey and then continued driving at the same speed for another 2 hours. By what percentage should Patrick reduce his speed so that James could catch up with Patrick in exactly 8 hours after Patrick overtook James?
A. 25%
B. 33%
C. 50%
D. 67%
E. 75%
Let J's rate = 12 miles per hour.
P catches up to J after J has traveled for 180 minutes (3 hours).
In 3 hours, the distance traveled by J = rt = (12)(3) = 36 miles.
Since P starts 3/2 hours after J, P must travel these 36 miles in only 3/2 hours.
Thus, P's rate = d/t = 36/(3/2) = 24 miles per hour.

Over the next 2 hours, the difference between P's distance and J's distance = (P's rate - J's rate)(time) = (24-12)(2) = 24 miles.
Implication:
After 2 hours, P is 24 miles ahead of J.

Since J must catch up to P after another 6 hours -- 8 hours after P overtook J -- the required catch-up rate = (catch-up distance)/(catch-up time) = 24/6 = 4 miles per hour.
Implication:
For J to catch up to P in another 6 hours, J must travel 4 miles per hour FASTER than P.

Since J's rate = 12 miles per hour, P's rate must decrease to 8 miles per hour, with the result that J's rate will be 4 miles per hour faster than P's rate.
P's percent decrease from 24 miles per hour to 8 miles per hour = (24-8)/24 * 100 = (16/24)(100) = (2/3)(100) ≈ 67%.

The correct answer is D.
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by Matt@VeritasPrep » Wed Feb 01, 2017 5:47 pm
It's easiest to start without a rate equation, I think.

When Patrick catches James, James has been driving for (90 + 90) minutes and Patrick has been driving for 90 minutes. So Patrick must be going TWICE as fast: he's traveled the same distance in HALF the time.

From here, let's pick speeds. Say James = 20 and Patrick = 40. In the two hours after Patrick passes James, Patrick travels 2 * 40 = 80 miles, and James travels 2 * 20 = 40 miles.

From here, James has SIX hours to make up 40 miles, so he needs to go 40/6, or 20/3 miles per hour FASTER than Patrick. That means Patrick needs to go from 40mph to 20 - (20/3) mph, a change of 66.66666...%.
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