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In racing over a given distance d at uniform speed, A can be [#permalink]
 09 Jul 2012, 03:42
Question Stats:
31% (03:01) correct
68% (01:42) wrong based on 1 sessions
In racing over a given distance d at uniform speed, A can beat B by 20 yards, B can beat C by 10 yards, and A can beat C by 28 yards. Then d, in yards, equal A. not determined by the give information B. 58 C. 100 D. 116 E. 120
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Last edited by Bunuel on 09 Jul 2012, 04:26, edited 2 times in total.
Edited the question.
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Re: In racing over a given distance d at uniform speed, A can be [#permalink]
09 Jul 2012, 04:24
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carcass wrote: In racing over a given distance d at uniform speed, A can be beat B by 20 yards, B can beat C by 10 yards, and A can beat C by 28 yards. Then d, in yards, equal
A. not determined by the give information
B. 58
C. 100
D. 116
E. 120 Given: A can beat B by 20 yards, A can beat C by 28 yards, B can beat C by 10 yards, So, when A is on the finish line B is 20 yards back and C is 28 yards back. Hence, 20 yards before the finish line C is 8 yards ahead of B, and since the final difference between B and C is 10 yards, then B gains 2 yards every 20 yard. To gain the final difference of 10=2*5 yards B should run total of 20*5=100 yards. Answer: C. Hope it's clear.
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Re: In racing over a given distance d at uniform speed, A can be [#permalink]
09 Jul 2012, 07:14
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Bunuel wrote: carcass wrote: In racing over a given distance d at uniform speed, A can be beat B by 20 yards, B can beat C by 10 yards, and A can beat C by 28 yards. Then d, in yards, equal
A. not determined by the give information
B. 58
C. 100
D. 116
E. 120 Given: A can beat B by 20 yards, A can beat C by 28 yards, B can beat C by 10 yards, So, when A is on the finish line B is 20 yards back and C is 28 yards back. Hence, 20 yards before the finish line C is 8 yards ahead of B, and since the final difference between B and C is 10 yards, then B gains 2 yards every 20 yard. To gain the final difference of 10=2*5 yards B should run total of 20*5=100 yards. Answer: C. Hope it's clear. Hi Can we also resolve by comparing time taken, i.e. d/a=(d-20)/b d/b=(d-10)/c d/a=(d-28)/c we can resolve d=100 from these eqn..
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Re: In racing over a given distance d at uniform speed, A can be [#permalink]
20 Jul 2012, 20:39
I want to ask bunnel how did he get this
" 20 yards before the finish line C is 8 yards ahead of B, and since the final difference between B and C is 10 yards.."
can you pls explain a bit in detail.
thanks
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Re: In racing over a given distance d at uniform speed, A can be [#permalink]
22 Jul 2012, 03:19
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Re: In racing over a given distance d at uniform speed, A can be [#permalink]
24 Jul 2012, 00:58
Bunuel wrote: carcass wrote: In racing over a given distance d at uniform speed, A can be beat B by 20 yards, B can beat C by 10 yards, and A can beat C by 28 yards. Then d, in yards, equal
A. not determined by the give information
B. 58
C. 100
D. 116
E. 120 Given: A can beat B by 20 yards, A can beat C by 28 yards, B can beat C by 10 yards, So, when A is on the finish line B is 20 yards back and C is 28 yards back. Hence, 20 yards before the finish line C is 8 yards ahead of B, and since the final difference between B and C is 10 yards, then B gains 2 yards every 20 yard. To gain the final difference of 10=2*5 yards B should run total of 20*5=100 yards. Answer: C. Hope it's clear. Hi Bunnel, Can you please explain this part? since the final difference between B and C is 10 yards, then B gains 2 yards every 20 yard. To gain the final difference of 10=2*5 yards B should run total of 20*5=100 yards.
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Re: In racing over a given distance d at uniform speed, A can be
[#permalink]
24 Jul 2012, 00:58
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