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DS

This topic has 1 expert reply and 2 member replies
jet1445 Newbie | Next Rank: 10 Posts Default Avatar
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DS

Post Wed Jan 24, 2007 9:50 pm
Anyone explains to me, Please.

1. On level farmland, two runners leave at the same time from the intersection of two country roads. One runner jogs due north at a constant rate of 8 miles per hour while the second runner jogs due east at a constant rate that is 4 miles per hour faster than the first runner's rate. How far apart, to the nearest mile, will they be after hour ?
(A) 6
(B) 7
(C) 8
(D) 12
(E) 14

Ans. B

2. The rear wheels of a car crossed a certain line 0.5 second after the front wheels crossed the same line. If the centers of the front and rear wheels are 20 feet apart and the car traveled in a straight line at a constant speed, which of the following gives the speed of the car in miles per hour? (5,280 feet = 1 mile)

(A) (20/5280)*{(60^2)/0.5}
(B) (20/5280)*{60/0.5}
(C) (20/5280)*{0.5/(60^2)}
(D) (20*5280)/{(60^2)*0.5}
(E) (20*5280)/{60*0.5}

Ans.A

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Post Fri Oct 06, 2017 5:47 pm
jet1445 wrote:
2. The rear wheels of a car crossed a certain line 0.5 second after the front wheels crossed the same line. If the centers of the front and rear wheels are 20 feet apart and the car traveled in a straight line at a constant speed, which of the following gives the speed of the car in miles per hour? (5,280 feet = 1 mile)
(A) (20/5280)*{(60^2)/0.5}
(B) (20/5280)*{60/0.5}
(C) (20/5280)*{0.5/(60^2)}
(D) (20*5280)/{(60^2)*0.5}
(E) (20*5280)/{60*0.5}
Ans.A
average speed = distance covered / time taken
distance covered between the rear and front wheels are 20 ft.
since 5280 ft = 1 mile
20 ft = (20 ft * 1 mile )/ 5280 ft = (20/5280) miles

Time taken while crossing the line = 0.5 sec

since 1 hr = 60 minutes = 60 * 60 sec
therefore, 0.5 sec = (0.5 sec * 1 hr) / 60^2 sec
= 0.5 / 60^2

speed of the car = (20/5280) miles / (0.5 / 60^2) hr
= (20/5280) miles * (60^2 / 0.5) hr
= (20/5280) (60^2 / 0.5) miles/hr

hence, A is correct Shocked Mr. Green

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Stacey Koprince GMAT Instructor
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Post Fri Jan 26, 2007 11:40 pm
Can you please post the sources for your questions?

Also can you please check the text for your first question? Something's not quite right. At the end, it says "how far apart... will they be after hour?" Is that supposed to be an hour or some other time? (If so, something else is wrong - check the source or throw it out.)

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g2000 Junior | Next Rank: 30 Posts Default Avatar
Joined
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Post Thu Jan 25, 2007 10:57 am
For the question 1, I get square root of 208. I don't know why.... It seems like a triangle...

For Question 2.
Displacement = speed * time

Let s be speed,
s * 0.5 sec = 20 ft
s = 20/0.5 ft/sec

20/0.5 ft/sec
= (20 * 1/5280) / 0.5 mi/sec
= (20 * 1/5280) / 0.5 * 60*60 mi/hr
= (20 / 5280) * (60^2) / 0.5 mi/hr

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