2005 June 12th, 11:05 PM
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#1 (permalink) |
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Within my grasp!
 Join Date: Jun 2005
Posts: 388
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Race problems from practice tests
1. In a 100m race, A can beat B by 25 m and B can beat C by 4m.In the same race, A can beat C by:
a 21 b 26 c 28 d 29 2. A and B take part in a 100m race. A runs at 5 Km/hr. A gives B a start of 8m and still beats him by 8 seconds. The speed of B is: a 5.15 kmph b 4.14 kmph c 4.25 kmph d 4.4 kmph
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Dinesh Last edited by Dkumar : 2005 June 13th at 12:58 AM. |
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2005 June 13th, 05:09 AM
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#2 (permalink) |
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Somebody Stop Me !!
Join Date: May 2005
Location: Waterloo, ON, Canada
Posts: 1,220
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Re: Race problems from practice tests
1. In a 100m race, A can beat B by 25 m and B can beat C by 4m.In the same race, A can beat C by:
Let, Speed of A=a Speed of B=b Speed of C=c Then (100/a)Xb=100-25=75...........(i) (100/b)Xc=100-4=96............(ii) (i)X(ii)....(10000/a)Xc=75X96 So, (100/a)Xc=75X96/100=72=100-28. So, answer=28m .......(C)
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2005 June 13th, 05:31 AM
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#3 (permalink) |
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Somebody Stop Me !!
Join Date: May 2005
Location: Waterloo, ON, Canada
Posts: 1,220
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Re: Race problems from practice tests
2. A and B take part in a 100m race. A runs at 5 Km/hr. A gives B a start of 8m and still beats him by 8 seconds. The speed of B is:
A takes 1 hr to go 5000m So, A takes 100/5000hr=.02hr=.02X3600sec=72sec to go 100m So, B takes (72+8) sec to go 100-8=92m So, Speed of B=92/80 m/s = (23/20)X(3600/1000)km/hr = 1.15X3.6km/hr = 4.14 km/hr...........................(Answer=B)
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2005 June 16th, 10:48 AM
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#6 (permalink) |
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Eager!
Join Date: Mar 2005
Posts: 51
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Race problem
Its given that A gives B a lead of 8 m. That means, as you must have seen in athletic matches, A and B dont have the same starting point. B stands 8 m ahead of A at the starting point. That means when A has to run 100 m, B has to run only 100 - 8, 92m.
HTH Chelsea |
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2005 June 16th, 10:52 AM
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#7 (permalink) |
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Eager!
Join Date: Mar 2005
Posts: 51
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I would solve the first problem this way
1st set of info Person distance A 100 B 75 2nd set of info Person distance B 100 C 96 We need to find how much distance is covered by C w.r.t A . i.e. when A covers 100 m. In other words when B covers 75m. Hence the ratio would be B C 100 96 75 x so x = 75 * 96/100 = 72. Or you can say this is what C covers when A covered 100 m. So difference between A and C is 28m i.e. A beats C by 28m Chelsea |
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2005 June 17th, 05:08 AM
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#9 (permalink) | |
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Eager!
Join Date: Jun 2005
Posts: 61
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shouldn't it be....?
Quote:
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2005 June 17th, 05:58 PM
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#10 (permalink) |
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Somebody Stop Me !!
Join Date: May 2005
Location: Waterloo, ON, Canada
Posts: 1,220
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No Kunal, that's the tricky thing about this problem.
See, the problem statement implies that : i)when A finishes 100m, B completes only 75m. ii)when B finishes 100m, C completes only 96m. This information actually give us the speed ratios of A,B,C.
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