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Intern  Joined: 29 Nov 2010
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A's speed is 20/13 times that of B. If A and B run a race  [#permalink]

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Question Stats: 63% (02:53) correct 37% (02:52) wrong based on 364 sessions

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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%
Math Expert V
Joined: 02 Sep 2009
Posts: 59721
Re: A's speed is 20/13 times that of B. If A and B run a race  [#permalink]

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anarchist112 wrote:
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%

You'll save time if you use plug-in method for this question.

Let the rate of A be 20 unit/time, then the rate of B will be 13 unit/time;
Also let the length of the race be 100 units;

Since A gives B a head start of x units and B beats A by 20 units, then A runs 100-20=80 units (A is 20 units behind when the race is over) and B runs 100-x units in the same time interval: $$\frac{80}{20}=\frac{100-x}{13}$$ --> $$x=48$$.

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GMAT 1: 780 Q51 V48 GRE 1: Q800 V740 Re: TIME AND SPEED QUESTION  [#permalink]

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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)
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Manager  Joined: 12 Oct 2011
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Re: TIME AND SPEED QUESTION  [#permalink]

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GyanOne wrote:
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)

Hi, can you please elaborate? I did not understand the logic used.
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Re: TIME AND SPEED QUESTION  [#permalink]

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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).
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GMAT 1: 610 Q43 V31 Re: TIME AND SPEED QUESTION  [#permalink]

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anarchist112 wrote:
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 approach is as follows:

First calculate the distance, B has covered with his speed in the time, in which A reached 80% of the race.
Then Add the remaining distance as head start for B to win the race.

Its best to apply Ratios concept here.
Since A's speed is 20/13 of B, therefore, B's speed is 13/20 of A
Distance covered by B = speed x time = (13/20) x (0.8) = 0.52%
(Which means B would have covered 0.52 of the race length during the time in which A has covered 0.8 of the race length.

Therefore to win, B needs a headstart of (1 - 0.52 = ) 0.48 of the race length.

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Re: A's speed is 20/13 times that of B. If A and B run a race  [#permalink]

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anarchist112 wrote:
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%

You can deduce the answer logically too.

A's speed : B's speed = 20:13
In the same time, A runs 20 units while B runs only 13 units. You want A to run 20 units but you want B to reach the end of the race which should be 25 units away (so that A is 20% behind). But B can run only 13 units. So he must have a head start of the rest of (25 - 13 = 12) units.

Head start will be 12/25 *100 = 48% of the race
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Re: A's speed is 20/13 times that of B. If A and B run a race  [#permalink]

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VeritasPrepKarishma wrote:
anarchist112 wrote:
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%

You can deduce the answer logically too.

A's speed : B's speed = 20:13
In the same time, A runs 20 units while B runs only 13 units. You want A to run 20 units but you want B to reach the end of the race which should be 25 units away (so that A is 20% behind). But B can run only 13 units. So he must have a head start of the rest of (25 - 13 = 12) units.

Head start will be 12/25 *100 = 48% of the race

Cool method. I wish I had started following you earlier. I have been on GMAT for a very long time but was not regular Intern  Joined: 23 Oct 2010
Posts: 2
Re: A's speed is 20/13 times that of B. If A and B run a race  [#permalink]

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Can you explain where the 25 comes from?
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Re: A's speed is 20/13 times that of B. If A and B run a race  [#permalink]

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sshin2 wrote:
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.
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Re: A's speed is 20/13 times that of B. If A and B run a race  [#permalink]

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

*New project from GMAT Club!!! Check HERE

All DS Distance/Rate Problems to practice: search.php?search_id=tag&tag_id=44
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Re: TIME AND SPEED QUESTION  [#permalink]

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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.

Originally posted by WholeLottaLove on 07 Aug 2013, 11:41.
Last edited by WholeLottaLove on 08 Aug 2013, 10:55, edited 1 time in total.
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Re: TIME AND SPEED QUESTION  [#permalink]

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WholeLottaLove wrote:
ChrisLele wrote:
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).

Ok, so we know that B runs slower than A yet still manages to beat A by a length of 20% of the total distance of the race. As stated in the question, B must start from a distance ahead of A that allows him to win my 20% even though A will be gaining on him the entire duration of the race. Why do we multiply 13/20 by 80%? I get that d=s*t and I see where why 13/20 is used but why is his distance 80% of A's? (I see that it comes from 100-20 but why???) Help!!!

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.
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Re: A's speed is 20/13 times that of B. If A and B run a race  [#permalink]

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anarchist112 wrote:
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%

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)
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Re: TIME AND SPEED QUESTION  [#permalink]

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Ahhhh! Because they both do it in the same time, we set (% of distance/rate) = (to be solved for distance/rate) This makes much more sense now. Thanks! VeritasPrepKarishma wrote:
WholeLottaLove wrote:
ChrisLele wrote:
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).

Ok, so we know that B runs slower than A yet still manages to beat A by a length of 20% of the total distance of the race. As stated in the question, B must start from a distance ahead of A that allows him to win my 20% even though A will be gaining on him the entire duration of the race. Why do we multiply 13/20 by 80%? I get that d=s*t and I see where why 13/20 is used but why is his distance 80% of A's? (I see that it comes from 100-20 but why???) Help!!!

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.
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Re: A's speed is 20/13 times that of B. If A and B run a race  [#permalink]

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Let A's speed be 20 and B's speed be 13 and let the total distance be 100
$$=>$$ Since B beats A by 20% of the length of the track, A only ran 80m.
$$=>$$ So using the formula $$Rate*Time=Distance$$, $$A's Time= 4s$$
similarly $$B's Time=$$ $$\frac{100}{13}$$
the difference between the Time of A and B is the time for which A has let B run ahead of him, i.e this is the headstart time
$$=>$$ $$\frac{100}{13}-4 = \frac{48}{13}s$$
To get the distance simply multiply $$rate*time$$ in case of B, which comes out to be $$13*\frac{48}{13} = 48$$
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Re: A's speed is 20/13 times that of B. If A and B run a race  [#permalink]

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mahendru1992 wrote:
Let A's speed be 20 and B's speed be 13 and let the total distance be 100
$$=>$$ Since B beats A by 20% of the length of the track, A only ran 80m.
$$=>$$ So using the formula $$Rate*Time=Distance$$, $$A's Time= 4s$$
similarly $$B's Time=$$ $$\frac{100}{13}$$
the difference between the Time of A and B is the time for which A has let B run ahead of him, i.e this is the headstart time
$$=>$$ $$\frac{100}{13}-4 = \frac{48}{13}s$$
To get the distance simply multiply $$rate*time$$ in case of B, which comes out to be $$13*\frac{48}{13} = 48$$

will B travel 100 unit ????
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Re: A's speed is 20/13 times that of B. If A and B run a race  [#permalink]

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Asifpirlo wrote:
mahendru1992 wrote:
Let A's speed be 20 and B's speed be 13 and let the total distance be 100
$$=>$$ Since B beats A by 20% of the length of the track, A only ran 80m.
$$=>$$ So using the formula $$Rate*Time=Distance$$, $$A's Time= 4s$$
similarly $$B's Time=$$ $$\frac{100}{13}$$
the difference between the Time of A and B is the time for which A has let B run ahead of him, i.e this is the headstart time
$$=>$$ $$\frac{100}{13}-4 = \frac{48}{13}s$$
To get the distance simply multiply $$rate*time$$ in case of B, which comes out to be $$13*\frac{48}{13} = 48$$

will B travel 100 unit ????

Yes B travels the whole 100 units.
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Re: A's speed is 20/13 times that of B. If A and B run a race  [#permalink]

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ChrisLele wrote:
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).

Hi Chris,

Thank you for the solution. I am having a hard time understanding this sentence "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." Could you please rephrase it?

TO
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A's speed is 20/13 times that of B. If A and B run a race  [#permalink]

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let race distance=100 yards
r=a's rate
(100-20)/r=(100-x)/(13/20*r)
x=48 yards
48/100=48%

Originally posted by gracie on 10 Oct 2015, 14:51.
Last edited by gracie on 02 Jul 2016, 11:05, edited 2 times in total. A's speed is 20/13 times that of B. If A and B run a race   [#permalink] 10 Oct 2015, 14:51

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