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Jim has arrived late for his golf tee time with his friends.
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Updated on: 13 Jul 2012, 01:46
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Jim has arrived late for his golf tee time with his friends. When he arrives at the course, he discovers that his friends have already walked 180 yards ahead on the course and that they are continuing to walk forward at a rate of 1.5 yards per second. A bystander with a golf cart takes pity on Jim and drives him to meet his friends at the rate of 4 yards per second. Once Jim meets his friends, he exits the cart and walks with them at their rate for the next 120 yards. What is Jim’s approximate average speed for his total trip? A. 1.8 yards per second B. 2.0 yards per second C. 2.4 yards per second D. 2.7 yards per second E. 3.0 yards per second This is a question from a Kaplan diagnostics test (free online). 1) How long can you solve this for? 2) How do you approach this type of questions? 3) Can you point me to other similar distance problems?
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Originally posted by stoy4o on 12 Jul 2012, 21:33.
Last edited by Bunuel on 13 Jul 2012, 01:46, edited 2 times in total.
Edited the question.




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Re: Jim has arrived late for his golf tee time with his friends.
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13 Jul 2012, 02:00
stoy4o wrote: Jim has arrived late for his golf tee time with his friends. When he arrives at the course, he discovers that his friends have already walked 180 yards ahead on the course and that they are continuing to walk forward at a rate of 1.5 yards per second. A bystander with a golf cart takes pity on Jim and drives him to meet his friends at the rate of 4 yards per second. Once Jim meets his friends, he exits the cart and walks with them at their rate for the next 120 yards. What is Jim’s approximate average speed for his total trip? A. 1.8 yards per second B. 2.0 yards per second C. 2.4 yards per second D. 2.7 yards per second E. 3.0 yards per second This is a question from a Kaplan diagnostics test (free online). 1) How long can you solve this for? 2) How do you approach this type of questions? 3) Can you point me to other similar distance problems? Jim's rate: 4 yards per second; Friends' rate: 1.5 yards per second; Their relative rate: 41.5=2.5 yards per second. To cover the distance of 180 yards between them \(\frac{180}{2.5}=72\) seconds are needed. Distance covered by Jim is \(4*72=288\) yards. To cover the next 120 yards Jim will need \(\frac{120}{1.5}=80\) seconds. \(Average \ rate=\frac{total \ distance}{total \ time}=\frac{288+120}{72+80}\approx{2.7}\). Answer: D. DS questions on Distance/Rate: search.php?search_id=tag&tag_id=44PS questions on Distance/Rate: search.php?search_id=tag&tag_id=64Hope it helps.
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Re: Jim has arrived late for his golf tee time with his friends
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12 Jul 2012, 23:28
stoy4o wrote: Jim has arrived late for his golf tee time with his friends. When he arrives at the course, he discovers that his friends have already walked 180 yards ahead on the course and that they are continuing to walk forward at a rate of 1.5 yards per second. A bystander with a golf cart takes pity on Jim and drives him to meet his friends at the rate of 4 yards per second. Once Jim meets his friends, he exits the cart and walks with them at their rate for the next 120 yards. What is Jim’s approximate average speed for his total trip?
1.8 yards per second 2.0 yards per second 2.4 yards per second 2.7 yards per second 3.0 yards per second
Hi, The question involves the concept of relative speed and average speed, To find the average speed, we have to find the total distance traveled and the total time taken: Jim covers the total trip in two parts: Part 1: Travelling on cart @ 4 yards/sec Time taken by Jim to meet his friends = \(\frac {distance}{rel. speed}\) = \(\frac{180}{41.5}\) = 72 sec Distance traveled = 4*72 = 288 yards Part 2: Jim travels 120 yards @ 1.5 yards/sec time taken = \(\frac {120}{1.5}\) = 80 secs Total distance travelled = 288 + 120 = 408 Total time taken = 72 + 80 = 152 Thus, average speed = \(\frac {408}{152}\) = 2.68 ~ 2.7 yards/sec Answer (D) It took me 2 minutes to solve & 10 mins to write here. Regards,



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Re: Jim has arrived late for his golf tee time with his friends.
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13 Jul 2012, 18:06
Thank you guys.
Bunnel, I'll take a look at your links. I appreciate it.



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23 Feb 2013, 18:52
can someone explain why relative speed is used here, ie 4  1.5 to get 2.5 yard per s



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Re: Jim has arrived late for his golf tee time with his friends.
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25 Feb 2013, 16:02
When Jim arrives he finds that his friends are 180 yards ahead of him and are still walking at a rate of 1.5 yards/sec. However, he gets a ride in a golf cart and is able to travel at a speed of 4 yards/sec. During this time, his friends are still walking at a speed of 1.5 yards/sec. Thus, the distance between them will get reduced at a rate of 41.5=2.5 yards/sec. The time taken to cover the distance of 180 yards is effectively increased and hence we use the relative speed to calculate the time. Consider this situation for example. You are walking with your friend when suddenly you meet someone and stop to talk to the person. Meanwhile your friend continues walking. After you end the conversation, you run to catch up with your friend. However, your friend continues to walk. Wouldn't this increase the time taken to reach your friend as opposed to the time taken if he waited for you when you start running? What is the reason for this increase in time? This is because even though you are running, your friend is also walking in the same direction and thus the distance between you is reducing at a slower rate than it would if he waited for you. Thus, you need to calculate the relative speed to account for this. Hope this helps! However, if you have any further concerns, please let me know. lemonhalls wrote: can someone explain why relative speed is used here, ie 4  1.5 to get 2.5 yard per s



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Re: Jim has arrived late for his golf tee time with his friends.
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25 Feb 2013, 21:03
lemonhalls wrote: can someone explain why relative speed is used here, ie 4  1.5 to get 2.5 yard per s The concept used here is relative speed. When two objects move in the same direction, one going faster than the other, the speed of the one going faster relative to the one going slower is the difference of the speeds. Check out this post I wrote sometime back on relative speed to get a hang of the basic concept. The answer explanation will make a lot of sense thereafter. http://www.veritasprep.com/blog/2012/07 ... elatively/
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Re: Jim has arrived late for his golf tee time with his friends.
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14 Mar 2013, 00:41
Hi karishma, I understand the concept of relative speed. However cant we solve this question with the approach outlined below??
The time when jim travels the 180 miles in cart is 45 seconds i.e. 180/4. In these 45 seconds his freinds travel a further 1.5*45 miles. So jim has to travel a total of 180+67.5 miles at a speed of 4 yards/mile to meet his friends. But by using this approach i am getting the answer as 2.59 approx. What am i missing.



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Re: Jim has arrived late for his golf tee time with his friends.
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14 Mar 2013, 01:08
sauravleo123 wrote: Hi karishma, I understand the concept of relative speed. However cant we solve this question with the approach outlined below??
The time when jim travels the 180 miles in cart is 45 seconds i.e. 180/4. In these 45 seconds his freinds travel a further 1.5*45 miles. So jim has to travel a total of 180+67.5 miles at a speed of 4 yards/mile to meet his friends. But by using this approach i am getting the answer as 2.59 approx. What am i missing. What happens in the time that Jim travels the 67.5 yards? During that time, his friends go further ahead at the speed of 1.5 yards/s and he still doesn't catch up with them. Time taken by Jim to cover 67.5 yards is 67.5/4 = 16.875 secs So the friends have traveled another 1.5*16.875 yards So it is best to use the relative speed concept. Jim has to cover 180 yards (relatively) at a relative speed of 4  1.5 = 2.5 yards/sec. Think of it like this: Out of his speed of 4 yards/sec, he uses 1.5 yards/sec to cover whatever distance his friends are covering now and the rest 2.5 yards/sec he uses to cover the gap of 180 yards between them. Time taken by Jim to meet his friends = 180/2.5 = 72 secs Time for which he travels at 1.5 yards/sec = 120/1.5 = 80 secs Average speed will be close to the middle of 1.5 and 4 but closer to 1.5 (using weighted averages). Middle of 1.5 and 4 is 2.75 so it will be 2.7 yards/sec
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14 Mar 2013, 01:22
Now i understand. Thank you so much karishma. Your explanations are the best I have encountered on this forum.



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Re: Jim has arrived late for his golf tee time with his friends
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08 Oct 2013, 00:23
saintforlife wrote: Jim has arrived late for his golf tee time with his friends. When he arrives at the course, he discovers that his friends have already walked 180 yards ahead on the course and that they are continuing to walk forward at a rate of 1.5 yards per second. A bystander with a golf cart takes pity on Jim and drives him to meet his friends at the rate of 4 yards per second. Once Jim meets his friends, he exits the cart and walks with them at their rate for the next 120 yards. What is Jim’s approximate average speed for his total trip?
1.8 yards per second 2.0 yards per second 2.4 yards per second 2.7 yards per second 3.0 yards per second Given Jim's friends has a lead of 180 yards and moving at 1.5 yards/sec speed Jim on golf cart with speed 4 yards/sec has to meet this friends, assuming he meets them after "t" secs So the total distance traveled by Jim and his friends should be same so as to meet them=> 180+1.5t=4t => 180=2.5t =>72secs, so Jim meets his friends after traveling 72 secs at 4 yards/sec speedFirst part of his journey (288 yards in 72 secs) Now, Jim travels 120 yards with speed 1.5 yards/sec in 120/1.5=80 secssecond part of his journey (120 yards in 80 secs) So avg speed of Jim= total distance traveled/ total time => (288+120)/(72+80)=408/152 =>2.7 yards/sec approx Ans D..hope its clear



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Re: Jim has arrived late for his golf tee time with his friends.
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31 Dec 2013, 09:40
stoy4o wrote: Jim has arrived late for his golf tee time with his friends. When he arrives at the course, he discovers that his friends have already walked 180 yards ahead on the course and that they are continuing to walk forward at a rate of 1.5 yards per second. A bystander with a golf cart takes pity on Jim and drives him to meet his friends at the rate of 4 yards per second. Once Jim meets his friends, he exits the cart and walks with them at their rate for the next 120 yards. What is Jim’s approximate average speed for his total trip? A. 1.8 yards per second B. 2.0 yards per second C. 2.4 yards per second D. 2.7 yards per second E. 3.0 yards per second This is a question from a Kaplan diagnostics test (free online). 1) How long can you solve this for? 2) How do you approach this type of questions? 3) Can you point me to other similar distance problems? Very nice question. Similar to Manhattan's GMAT 'Moving walkway' Let's get this started First, relative rate 41.5=2.5 yards per second to catch up with his buddies Therefore, time will be 180/2.5 =360/5=72 seconds Now in 72 seconds he traveled= 72*4 = 288 yards Now, he will travel the next 120 yards in = 120/1.5 = 240/3 in 80 seconds Hence average speed will be 288+120 / 80+ 72 = 2.7 Answer is D Kudos if it helps Cheers! J



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Jim has arrived late for his golf tee time with his friends.
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12 Dec 2015, 15:23
d=distance walked by J's friends from 180 yards to where J overtakes them (180+d)/4=d/1.5 d=108 total distance for J=180+108+120=408 yards total time for J=(180+108)/4+120/1.5=152 seconds 408/152=2.7 yards per second average speed



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12 Dec 2015, 18:40
Average Speed = Total Distance / Total Time
Distance 1 = 180 Distamce 2 = 120
Time 1 = 180/Relative Speed of Jim = 180/(4  1.5) = 180/2.5 = 72 Time 2 = 120/Speed of Group = 120/1.5 = 80 Total Distance = 300 Yards Total Time = 80+72 = 152 seconds Hence Average Speed apprx 2 / second
My 2 cents on problem solving Introduce numbers as late as possible into the calculation Solve it in equation format, use cancellation of terms etc



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Jim has arrived late for his golf tee time with his friends.
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05 Aug 2017, 17:21
stoy4o wrote: Jim has arrived late for his golf tee time with his friends. When he arrives at the course, he discovers that his friends have already walked 180 yards ahead on the course and that they are continuing to walk forward at a rate of 1.5 yards per second. A bystander with a golf cart takes pity on Jim and drives him to meet his friends at the rate of 4 yards per second. Once Jim meets his friends, he exits the cart and walks with them at their rate for the next 120 yards. What is Jim’s approximate average speed for his total trip?
A. 1.8 yards per second B. 2.0 yards per second C. 2.4 yards per second D. 2.7 yards per second E. 3.0 yards per second let d=distance from 180 yards to point where J meets friends (180+d)/d=4/1.5 d=108 yds 108+180+120=408 total yds (180+108)/4 yps=72 sec 120/1.5 yps=80 sec 72+80=152 total seconds 408 total yds/152 total sec≈2.7 yps average speed for total trip D



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Re: Jim has arrived late for his golf tee time with his friends.
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06 Nov 2018, 08:57
Bunuel wrote: stoy4o wrote: Jim has arrived late for his golf tee time with his friends. When he arrives at the course, he discovers that his friends have already walked 180 yards ahead on the course and that they are continuing to walk forward at a rate of 1.5 yards per second. A bystander with a golf cart takes pity on Jim and drives him to meet his friends at the rate of 4 yards per second. Once Jim meets his friends, he exits the cart and walks with them at their rate for the next 120 yards. What is Jim’s approximate average speed for his total trip? A. 1.8 yards per second B. 2.0 yards per second C. 2.4 yards per second D. 2.7 yards per second E. 3.0 yards per second This is a question from a Kaplan diagnostics test (free online). 1) How long can you solve this for? 2) How do you approach this type of questions? 3) Can you point me to other similar distance problems? Jim's rate: 4 yards per second; Friends' rate: 1.5 yards per second; Their relative rate: 41.5=2.5 yards per second. To cover the distance of 180 yards between them \(\frac{180}{2.5}=72\) seconds are needed. Distance covered by Jim is \(4*72=288\) yards. To cover the next 120 yards Jim will need \(\frac{120}{1.5}=80\) seconds. \(Average \ rate=\frac{total \ distance}{total \ time}=\frac{288+120}{72+80}\approx{2.7}\). Answer: D. DS questions on Distance/Rate: http://gmatclub.com/forum/search.php?se ... &tag_id=44PS questions on Distance/Rate: http://gmatclub.com/forum/search.php?se ... &tag_id=64Hope it helps. Hi, Please help me with where i'm thinking wrong. When he is covering the 180 yards that he is behind at 4 yards/second, his friends would not have stopped. In that time they would move some distance ahead, so he has to cover 180 yards + extra distance moved. Why have we not considered that? I'm really confused.



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Re: Jim has arrived late for his golf tee time with his friends.
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07 Nov 2018, 19:08
stoy4o wrote: Jim has arrived late for his golf tee time with his friends. When he arrives at the course, he discovers that his friends have already walked 180 yards ahead on the course and that they are continuing to walk forward at a rate of 1.5 yards per second. A bystander with a golf cart takes pity on Jim and drives him to meet his friends at the rate of 4 yards per second. Once Jim meets his friends, he exits the cart and walks with them at their rate for the next 120 yards. What is Jim’s approximate average speed for his total trip?
A. 1.8 yards per second B. 2.0 yards per second C. 2.4 yards per second D. 2.7 yards per second E. 3.0 yards per second Let’s let t = the time, in seconds, that Jim is in the golf cart. During these t seconds, he travels 4t yards. His friends have already walked 180 yards, plus 1.5t additional yards while Jim is in the golf cart catching up to them. Thus, we have:. 4t = 1.5t + 180 2.5t = 180 t = 72 seconds We know that Jim travels 4 x 72 = 288 yards in 72 seconds while he is in the golf cart. Next, let’s determine the time he walks with his friends after he catches up with them: 120/1.5 = 80 seconds So his total distance is 4 x 72 + 120 = 408 yards, and his total time is 72 + 80 = 152 seconds. Therefore, his average speed for the entire trip is: 408/152 ≈ 2.7 yards per second Answer: D
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