A's speed is 20/13 times that of B. If A and B run a race

Let x be the percentage of the race which B runs

Then x = (13/20) * 0.8 = 52%

So B should get a lead of (100-52=) 48% of the race. Option (B)

Think of it this way: if we wanted the two to finish at the same time, we would give B would only run 13/20 of the race, or the inverse of 20/13 (20/13 x 13/20 = 1, meaning they finish at the same time).

B finishes 20% ahead of A, so we take the 13/20 and we multiply by 4/5, thereby shortening the amount B has to run by 4/5 or 80%. (4/5 x 13/20) = 52/100.

Then there is the final twist to the problem. Let’s think of it this way, B only has to run 52 meters out of one hundred whereas A has to run all 100 meters. However that doesn’t mean B gets a head start of 52 (then he would only have to run 48 meters). Because ‘B’ has to run 52 meters, he only gets a head start of 48 meters.

Therefore the answer is 48/100 or 48% (B).

Can you explain where the 25 comes from?

B beats A by 20% of the length of the race.
If the length of the race is 100 m, when A reaches 80m, B must be at the finish line i.e. at the 100m mark. In this case, B beats A by 20m i.e. 20% of the length of the race.
Similarly, if A reaches the 20 m mark, B must be at 25 m mark which must be at the finish line i.e. the length of the race must be 25m (20 is 80% of 25 so B beats A by 20% of the length of the race)
These values are assumed. All you need is the % of length which should be given as head start. So it doesn't matter what the actual length of the race is. You just take one scenario and find out the head start given in that. That will give you the required percentage. We know that in the time A covers 20m, B covers 13m so it is easy to assume the length of the race 25.

These are logical methods which don't require any algebra and you can pretty much do them in your head in a few seconds. But you need to be adept at handling numbers so you need to practice to get comfortable with them.

This is the problem I had difficultly with, redone, after Karishma explained it to me (thanks again!)

A's speed is 20/13 times that of B. If A and B run a race, what part of the length of the race should A give B as a head start, so that B beats A by 20% of the length of the race?

A. 44%
B. 48%
C. 52%
D. 42%
E. 46%

My single biggest problem was not realizing that A and B run for the same amount of time. If B beats A by 20% then A runs 80% of the entire distance of the race. Time = distance/rate. We know that the rate of A is 20/13ths that of B so if A runs 20 units/unit of time then B runs 13 units/unit of time:

(d/r)=(d/r) because the time they run is the same.
(.8*d/20) = (x*d/13) where x is the percentage of the distance B must run to win the race by 20%. Remember, we are looking for the percentage of distance A and B runs so we multiply A's distance by .8 (the distance of the race he will run, and the other distance by x, the percentage of the race B needs to run.
.8d/20 = xd/13
10.4d = 20xd
10.4 = 20x
10.4/20 = x
x = 52%.
Therefore B needs to run 52% of the race. B needs to start at the 48% mark.

Yeah, we know B is slower than A yet manages to beat A but 20% of the length of race. So B starts much ahead of A

So this is what the beginning of the race looks like. A at the start line and B somewhere in the middle. B has to run much less distance.
Start(A) _____________(B)____________________________ Finish


This is what happens at the end of the race:
Start ___________________________________(A)________Finish(B)

So A covers 80% of the length of the race (say, d) while B covers much less (say, x% of d). They do it in the same time so

\(\frac{(80/100)*d}{20} = \frac{x*d}{13}\)
So \(x = (80/100) * (13/20)\)
x = 52/100

B covered only 52% of d so he got a head start of 48% of d.

Solution:

<----------80%.........................>..<.. 20%.>
A........................B...............A............B
<--------x-------------->
<---------------------100---------------------.---------->


Let, Head start = x and velocity of B=13, so velocity of A=20
Here, (D=v t or, t=D/v, will use this formula)
Time requires for A to cover 80% distance = time requires for B to cover 100-x distance
or, 80/20 = (100-x)/13
or, x = 48%
(explained thoroughly)

Another way to visualize this..
It is given to us that
\(\frac{A_{speed}}{B_{speed}} = \frac{20}{13}\)

let the race be of 20 units.

Now..when A completes 80% of the distance of the race..B will complete 80% of his distance according to the ratio..
\(0.8*13 = 10.4\)

But what we want..is that when A completes 80% of the race..B should complete the entire race..

So the head start will be
\(20 - 10.4 = 9.6 units\)

In terms of percentage of 20 units..it is

48%(B)

Inference - A will covera distance of 0.80D and B will cover D-x distance in same time where x is the lead given to B.

Time to cover the distance is same.

So

0.80D/(20s/13)= (D-x)/s

10.40D=20D-20x

20x= 9.60D
X= 0.48D

So 48% lead distance is required.

Answer is B.

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