A’s speed is 20/17 times that of B. If A and B run a race, what part

A will run 20*K units when B will run 17*K units .
20K - 17K = 3K units

3/20 answer

A and B will be running 20 km.

A can run 20 km/hour and B can run 17 km/hour. If B is allowed to start 3 km ahead, he will have traveled the same distance as A(20km). As such, 3/20 is the part of the length that B has to start at to make it a dead race.

D

Assume speed of B=x, so speed of A=20x/17; time taken by both A & B will be equal;
Speed =distance /time; we can write, Rate=distance/time
Equating both time A & B we get: LA*17/20X=LB/X; LA(length of A); LB(Length of B)
Therefore LB=17LA/20 => we need to calculate LA=LB= LA*(3/20 )

Hence answer is D

Thanks,

given :

Sa/Sb=20/17.

Ergo, Ta/Tb=17/20, Da/Db=20/17

which implies when A covers 100 , B would be at 85. Therefore, B needs to be given a head start of 15. ( so that both finish at 100 at the same time [ dead heat ].

Which is 3/20 of 100.

VERITAS PREP OFFICIAL SOLUTION:

We have the ratio of A’s speed and B’s speed. This means, we know how much distance A covers compared with B in the same time.

This is what the beginning of the race will look like:
(Start) A_________B______________________________

If A covers 20 meters, B covers 17 meters in that time. So if the race is 20 meters long, when A reaches the finish line, B would be 3 meters behind him. If we want the race to end in a dead heat, we want B to be at the finish line too at the same time. This means B should get a head start of 3 meters so that he doesn’t need to cover that. In that case, the time required by A (to cover 20 meters) would be the same as the time required by B (to cover 17 meters) to reach the finish line.

So B should get a head start of 3/20th of the race.

Answer (D)

Given: A's speed = (20/17)B

Ignoring distance for now, in order for A and B to finish at the same time, A's speed would need to be reduced by (17/20) so the fractions cancel out and A's speed = B's speed.

Now, s2 (adjusted speed) for A will be 17/20 out of 1 (s1 or original speed), so the ratio for s1/s2 = 1/(17/20) = 20/17

Since (s2/s1) = (t2/t1) = 20/17, we can see that 3/17 (or 1 - 20/17) additional time should be given to B for a head start to allow A and B to finish together.

So we have the additional time (3/17) and the adjusted speed for A (17/20). Using the d = s*t formula, we multiply them together and the 17 cancels out, leaving 3/20 (the answer choice).

let b speed = 17
a speed = 20

let the race be of 20 * 17 = 340 m long

time taken by A = 340/20 = 17 sec
time taken by B= 340/17= 20 sec

A defeats B by 3 sec

so B was short of 3*17 = 51 m

so A defeats B by 51m

give a head start of 51 m to B , he needs to cover 340-51=289 m which he covers in 17 seconds and hence we will get a tie ;

So answer is D (51/340 or 3/20)

Let's denote the speed of B as 1
Then the speed of A is \(\frac{20}{17}\)

Since the time is the same we can proceed with the following equation:

\(\frac{20}{17}(1-y)=1*1\) (where 1 is the full distance and 1-y is the abridged distance).

\(\frac{20}{17}-\frac{20}{17}y =1\)

\(\frac{20}{17}y=\frac{3}{17}\)

\(y=\frac{3}{20}\)

Answer D

We can let the race be 340 ft long and B runs 17 ft/s. So A runs 20 ft/s. So it will take A 17 seconds to run the race. However, in 17 seconds, B only runs 17 x 17 = 289 ft. So A has to give B a 340 - 289 = 51 ft head start, which is 51/340 = 3/20 of the length of the race.

Answer: D

Let the total distance = 1

Since A=\(\frac{20}{17}\)B, B=\(\frac{17}{20}\) A.
A & B both run the same total distance, so we can use the product of \(17*20= 340\) of A & B to represent the total distance: 1=\(\frac{340}{340}\)

We know that B covers \(289/340\) of the distance A runs, when running the total distance. Thus, A should give B a head start of \(\frac{340}{340}\) - \(\frac{289}{340}\) = \(\frac{51}{340}\)= \(\frac{3}{20}\).

Please hit Kudos if you liked this answer

let d=total length of race
x/d=B's head start as part of total length of race
d/(20/17)=(d-x)/1→
x/d=3/20
D

A’s speed is 20/17 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 the race ends in a dead heat?

Approaching this question logically would be the best start rather than thinking about formulas.

A’s speed is 20/17 times that of B.

From the above statement, we can assume that if B's speed is 17 km/hr then, A's speed would be 20 km/hr. In other words, if B covers 17 km , at the same time A will cover 20 km. Since the length of the race is not given, we have freedom to decide the race length.

What would be the best pick ?

100 km ? No way. This will end up in lots of unnecessary calculations.

20 km would be the right assumption to get the answer immediately or any multiple of 20 should also work.

Now, Why is 20 km the best pick?

As we already know that when A covers 20 km , B will cover only 17 km i.e 3 km behind A.

So, if we assume that the race length is 20 km, and to end the race in dead heat i.e both A and B should finish the race at the same time, B should have a head start of 3 km i.e 3/20 of the race length.

Option D is the answer.

Thanks,
Clifin J Francis,
GMAT QUANT SME

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