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

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..

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.

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

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

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.

Answer: C. 100

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.

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?

Yes, that's correct: when A is on the finish line B is 20 yards back and C is 28 yards back.

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?

please help. confused

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

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!!!

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.

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.

(C) is the answer
I applied substitution method.
I considered the distance as 100 yards and A takes 1 hour to cover the distance. So his speed is 100 yards per hour. B's speed is 80 yards per hour.
Therefore B will take 75 minutes to cover 100 yards. In 75 minutes C covers 90 yards. Hence, in 60 minutes C covers 72 yards. That is his speed.

Let d=distance of race.
Between A's finish and B's finish, B runs 20 yards while C runs 18 yards.
B's rate to C's rate=10/9.
When B finishes, he is 10 yards ahead of C.
10/9=d/(d-10)
d=100 yards

The ratio of the speeds of two moving objects is equal to the ratio of the respective distances they cover within a given period of time.
Let SA, SB and SC be the speeds of A, B and C respectively. Then:

SA/SB=d/(d-20)......(i)
SB/SC=d/(d-10)......(ii)
SA/SC=d/(d-28)......(iii)

Multiplying Equations (i) and (ii), we get:
(SA/SB)*(SB/SC)=d^2/((d-20)(d-10)
SA/SC=d^2/(d^2-30d+200) But, from Eqn (iii), we know that SA/SC=d/(d-28). Therefore,
d^2/(d^2-30d+200)=d/(d-28)
d/(d^2-30d+200)=1/(d-28)
d^2-28d=d^2-30d+200
30d-28d=200
d=100

Ans: C