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# In racing over a given distance d at uniform speed, A can be

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09 Jul 2012, 02:42
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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 09 Jul 2012, 03:26, edited 2 times in total.
Edited the question.
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09 Jul 2012, 03:24
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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 B is 8 yards ahead of C, 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.

Hope it's clear.
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Re: In racing over a given distance d at uniform speed, A can be [#permalink]

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09 Jul 2012, 06:14
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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.

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]

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20 Jul 2012, 19:39
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]

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22 Jul 2012, 02:19
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hardwoker1010 wrote:
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

Consider the case when A, B and C are racing each other. When A is on the finish line B must be 20 yards from the finish line and C must be 28 yards from the finish line. Hence when B is 20 yards from the finish line he's 28-20=8 yards ahead of C.

Hope it's clear.
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Re: In racing over a given distance d at uniform speed, A can be [#permalink]

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23 Jul 2012, 23:58
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.

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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27 Jun 2013, 01:25
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Re: In racing over a given distance d at uniform speed, A can be [#permalink]

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27 Jun 2013, 16:50
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1. When A touches d, B is at d-20 and C is at d-28
2. When B completes d, C completes d-10
3. Equate d-20 / d-28 = d / d-10 because the ratio of the speed of B to the speed of C is a constant.
4. d= 100
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Re: In racing over a given distance d at uniform speed, A can be [#permalink]

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04 Jul 2013, 09:44
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Think it into a other way.
1) When A completes, B is 20 Behind and C is 28 Behind. C is 8 behind B.
2) When B completes, C is 10 Behind.

What it refers is that in last 20 yards B has gained 2 yards, which refers that B is gaining 1 yard over C in every 10 yard it covers, so in last 80 yards it has gained 8 yards, so total is 80+20 = 100 :D
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03 Aug 2013, 11:16
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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

We are given nothing about time and nothing about rate. Solving using a traditional formula (i.e. distance = rate*time) might be difficult. Is that correct?

A must be 28 yards ahead of C. We know that A beats both B and C.

When A hits the finish line, it is 20 yards ahead of B and 28 yards ahead of C. This means at this time B is 8 yards ahead of C but it beats C by 10 yards. In 20 yards, it has to gain a two further yards between itself and C. Because A, B, C travel at a constant rate, B gains on C by 2 yards every 20 yards. This means that when B and C started out they were even and after 20 yards B was two ahead of C, after 40 yards B was 4 ahead of C, etc. B would have to travel 100 yards for it to gain 10 yards on C.

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Re: In racing over a given distance d at uniform speed, A can be [#permalink]

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16 Aug 2013, 02:13
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Another method to solve such questions : Let us consider the case when all three are racing each other. A is ahead of B by 20 yards at the finish and ahead of c by 28 yards, which means B is ahead of C by 8 yards at that instant. Now B defeats C by 10 yards, means the duration during which B covered 20, C covered 8+10=18. Hence, the ratio of their speeds is 10:9. That means in a race of 10yards B defeats c by 1 yard , so to defeat him by 10 yards the race should be of 10x10 =100 yards.
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Re: In racing over a given distance d at uniform speed, A can be [#permalink]

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26 Sep 2013, 00:49
Bunuel wrote:

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.

Can you explain this part with a diagram

<--------C---------------B-----------------A ( finish line)

So the distance between A and B is 20 and A and C is 28 and not b and c is this correct?
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Re: In racing over a given distance d at uniform speed, A can be [#permalink]

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26 Sep 2013, 00:56
fozzzy wrote:
Bunuel wrote:

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.

Can you explain this part with a diagram

<--------C-------(8)--------B-------(20)----------A ( finish line)

So the distance between A and B is 20 and A and C is 28 and not b and c is this correct?

Yes, that's correct: when A is on the finish line B is 20 yards back and C is 28 yards back.
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Re: In racing over a given distance d at uniform speed, A can be [#permalink]

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10 Jan 2014, 06:04
SravnaTestPrep wrote:
1. When A touches d, B is at d-20 and C is at d-28
2. When B completes d, C completes d-10
3. Equate d-20 / d-28 = d / d-10 because the ratio of the speed of B to the speed of C is a constant.
4. d= 100

Hi Sravna,

I dont follow how the ratio of the speeds are constant? what about the time factor while equating? as per the discussions in the various for this questions here it is re-iterated time and again that B gains 2 yards in the last 20 yards? if that is the case how can the speed ratios be constant?

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Re: In racing over a given distance d at uniform speed, A can be [#permalink]

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18 Jan 2014, 22:47
rawjetraw wrote:
SravnaTestPrep wrote:
1. When A touches d, B is at d-20 and C is at d-28
2. When B completes d, C completes d-10
3. Equate d-20 / d-28 = d / d-10 because the ratio of the speed of B to the speed of C is a constant.
4. d= 100

Hi Sravna,

I dont follow how the ratio of the speeds are constant? what about the time factor while equating? as per the discussions in the various for this questions here it is re-iterated time and again that B gains 2 yards in the last 20 yards? if that is the case how can the speed ratios be constant?

Hi Bunuel,

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Re: In racing over a given distance d at uniform speed, A can be [#permalink]

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13 May 2014, 08:28
narangvaibhav wrote:
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.

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..

You can'tresolve this..... 3equation and four variables....need one more equation here
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Re: In racing over a given distance d at uniform speed, A can be [#permalink]

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18 May 2014, 03:39
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rawjetraw wrote:
rawjetraw wrote:
SravnaTestPrep wrote:
1. When A touches d, B is at d-20 and C is at d-28
2. When B completes d, C completes d-10
3. Equate d-20 / d-28 = d / d-10 because the ratio of the speed of B to the speed of C is a constant.
4. d= 100

Hi Sravna,

I dont follow how the ratio of the speeds are constant? what about the time factor while equating? as per the discussions in the various for this questions here it is re-iterated time and again that B gains 2 yards in the last 20 yards? if that is the case how can the speed ratios be constant?

Hi Bunuel,

Lemme try and explain.

In a given times A runs d, B runs d-20, and C runs d-28. Since the time is same their speed a, b, c are in the ratio of the distance the travelled
Thus
a:b:c = d:d-20:d-28
Also, we know that b:c= d: d-10

Thus, equating b:c in both the cases, we get

d-20/d-28 = d/d-10

Solving this we get d= 100.

A more intuitive approach has been given by Bunuel in this tread. The approach works as the speeds are constant and are in direct proportion with the distance if time traveled is same.

Hope it helps!!!
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Re: In racing over a given distance d at uniform speed, A can be [#permalink]

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18 May 2014, 03:46
enders wrote:
narangvaibhav wrote:
Bunuel 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.

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..

You can'tresolve this..... 3equation and four variables....need one more equation here

You can resolve this very easily. Also, we don't need to find all the four variables. We can't find the speed of a, b and c individually as we don't know about the distance

The first one can be written as a/b = d/d-20
second one is b/c = d/d-10

Multiplying these two yeilds a/c = d^2/(d-10)(d-20)

Third one also says a/c = d/d-28

We equate these 2 and get the result as 100.
A more intuitive way has been already discusses. You should use that way.
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Re: In racing over a given distance d at uniform speed, A can be [#permalink]

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17 Jun 2014, 01:07
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rawjetraw wrote:
SravnaTestPrep wrote:
1. When A touches d, B is at d-20 and C is at d-28
2. When B completes d, C completes d-10
3. Equate d-20 / d-28 = d / d-10 because the ratio of the speed of B to the speed of C is a constant.
4. d= 100

Hi Sravna,

I dont follow how the ratio of the speeds are constant? what about the time factor while equating? as per the discussions in the various for this questions here it is re-iterated time and again that B gains 2 yards in the last 20 yards? if that is the case how can the speed ratios be constant?

Hi,

Very sorry for such a late reply.

Let us say that B travels a distance x and in the same time C travels distance y. The ratio of the distance = x/y. This is also the ratio of their speed since time is the same.

In the actual problem we have this ratio as d-20/d-28. This gives the ratio of their speed.

Similarly when B reaches d, c reaches d-10. The time taken by both to reach the respective distance is the same. This d/d-10 also gives the ratio of the speed of B and C. Thus we can equate the two and find d.
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Re: In racing over a given distance d at uniform speed, A can be [#permalink]

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18 Jun 2014, 05:27
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.

Hope it's clear.

Bunuel how can C be ahead of B unless I am missing something !!
Re: In racing over a given distance d at uniform speed, A can be   [#permalink] 18 Jun 2014, 05:27

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