Michael walks at the rate of 5 feet per second on a long straight path

Question

Michael walks at the rate of 5 feet per second on a long straight path. Trash pails are located every 200 feet along the path. A garbage truck traveling at 10 feet per second in the same direction as Michael stops for 30 seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck intersect?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

GMATinsight · Answer

Michael Speed = 5 ft / sec
Garbage truck speed = 10 ft/sec

Time that Michael takes to reach one pail from the other = 200/5 = 40 secs
Time that Garbage truck takes to reach one pail from the other = 200/10 = 20 secs

Garbage truck reached the second pails at the following time intervals - 20 sec to 50 seconds 
 Garbage truck at third Pail will be available at time interval 70 sec to 100 secs 
 Garbage truck at forth Pail will be available at time interval 120 sec to 150 secs 
 Garbage truck at fifth Pail will be available at time interval 170 sec to 200 secs and so on

Michael will reach at second pail at 40th sec i.e. It intersects with the truck 
 Michael will reach at third pail at 80th sec i.e. It intersects with the truck 
 Michael will reach at Forth pail at 120th sec i.e. It intersects with the truck 
 Michael will reach at Fifth pail at 160th sec i.e. It DOES NOT intersects with the truck

Since they also intersect at the first pail so total points of intersections = 4

Answer: Option A

Archit3110 · Answer

time taken by truck = 200/10 ; 20 sec time taken by Michae; 200/5; 40 seconds key catch of question +30 seconds halt at each pail so first stop ; 0 to 20 sec truck moves at 20+30 ; 50 seconds 2nd stop ; 70 sec ; move at 70+30 ; 100 sec 3rd stop; 120 sec ; move at 150 sec 4th stop; 170 ; move 200 sec 5th stop; 220 sec ; move 250 sec 6th; 270; move 300 sec for Michael 1st stop; 40 sec 2nd; 80 sec 3rd; 120 sec 4th; 160 sec 5th ; 200 sec 6th; 240 sec so we see that until 5th stop both truck and Michael are meeting each other; but at 6th stop Truck overtakes Michael and they dont meet any further IMO B; 5

Ritwick91 · Answer

IMO ans should be B (5 times).

Explanation:
Distance between two pails: 200 ft
Michael's speed: 5 ft/s ---> He will take 40 secs to reach from one pail to another
Truck's speed: 10 ft/s ---> It will take 20 secs to reach from one pail to another
Waiting time for garbage truck in each pail : 30 secs

Suppose Michael starts from pail A and as given the garbage truck starts from pail B. 
Michael will take 40 secs to reach B and in the same time Truck will reach C and wait for 20 secs (20+20 secs wait at C). 
In the next 40 secs, Michael will go from B to C, and Truck will go from C to D (10 secs wait at C+20 secs travel+10 secs wait at D).
In the next 40 secs, Michael will go from C to D, and Truck will just reach E from D (20 secs waiting at D+20 secs travel).
In the next 40 secs, Michael will go from D to E, and Truck will be midway from E to F (30 secs waiting at E +10 secs travel).
In the next 40 secs, Michael will go from E to F, and Truck will be just starting from F (10 secs travel +30 secs waiting at F) ----------1st intersect
In the next 40 secs, Michael will go from F to G, and Truck will go from F to G (20 secs travel+20 secs waiting at G) . -------------------2nd intersect
In the next 40 secs, Michael will go from G to H, and Truck will go from G to H (10 secs waiting at G + 20 secs travel+ 10 secs waiting at H) 
------Here there are two intersects: One while traveling from G to H, Truck crossed Michael and after reaching H, Michael intersected the truck. -------------------3rd and 4th intersect
In the next 40 secs, Michael will go from H to I, and Truck will go from H to I (20 secs waiting at H and 20 secs travel) -------5th intersect

And after this point the truck will always move ahead to Michael and never intersect. Please do let me know if the explanation is correct or not. Thanks.

effatara · Answer

A very interesting problem indeed!

P1____________P2___m1_________P3______m2______P4_________m3___P5____________ P6 ____________P7___m5___i3______P8______m6______P9
 (m4/i1) (i2) (i4) (i5)
P9_________m7___P10

Look at the above diagram carefully. P1/P2/P3… denotes the positions of the trash pails and i1/i2/i3… the points of intersection of Michael (M) and the garbage truck (GT). At the exact moment when M is crossing the first pail (P1), he sees GT leaving the second pail (P2). After 50 seconds (20 second to move from P2 to P3 and another 30 seconds to empty the pail), GT leaves P3. In these 50 seconds M has walked 250 feet i.e. he has crossed P2 and walked another 50 feet reaching ‘m1’ which denotes his position when GT leaves P3. In other words, he has shortened the distance between himself and GT by 50 feet. Similarly, ‘m2’ denotes his position when GT leaves P4, ‘m3’ when GT leaves P5, m4 when GT leaves P6 and so on. So, for every 50 second segment (after the point of time when M and GT left P1 and P2 respectively), the distance between M and GT reduces by 50 feet. Thus, four 50 second segments (200 seconds in total) would be needed for M to catch up with GT which means M catches up with GT at P6 just as it is driving off after loading the trash. This point is denoted by ‘m4/i1’ on the diagram since this is the first point of intersection and also M’s position when GT leaves P6. Thereon, M reaches P7 after 40 seconds at which point GT is still loading trash from P7 having reached there in 20 seconds and loaded for only 20 seconds with 10 seconds still to go. So the second point of intersection is P7 (marked ‘i2’ on the diagram). By the time GT leaves P7, M is already ahead by 50 feet (marked ‘m5’). Since GT gains 5 feet on M every second, it will need 10 seconds to cover M’s 50 feet lead. So it will overtake M 100 feet further on from P7 – also 100 feet short of P8. This is the third point of intersection (marked ‘i3’). From this point it takes M 20 seconds to reach P8 by which time GT had reached P8 and loaded for 10 seconds since it took 10 seconds to reach P8 from ‘i3’ and it has to load for 20 seconds more. So the fourth point of intersection is at P8 (‘i4’). By the time GT leaves P8, M has 100 feet (20x5) lead and it will take 20 seconds for GT to catch up with M. Which means GT will catch up with him at P9. This is the fifth and last point of intersection (‘i5’). By the time GT leaves P9 30 seconds later M will have a lead of 150 feet (marked ‘m7’) and just 50 feet short of reaching P10. In order to cover this 150 feet, GT needs 30 seconds but it doesn’t have because it will have to stop at P10 after reaching there in 20 seconds by which time M will have crossed P10 and walked a further 50 feet. By the time GT leaves P10, M will have reached P11 increasing his lead by another 50 feet to 200 feet. So M will continue to get further and further ahead of GT.

I had to make the explanation quite lengthy – maybe unnecessarily so – but I did so to make the concept absolutely clear to remove any confusion.

effatara · Answer

Sorry, the diagram got a bit messed up in the copy-paste process. I have modified it slightly; hope it is clear now.

P1____________P2___m1_________P3______m2______P4_________m3___P5____________ P6/(m4/i1) 
 ____________P7/i2___m5___i3______P8/i4______m6______P9/i5_________m7___P10

Michael walks at the rate of 5 feet per second on a long straight path

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