On a Race Track a Maximum of 5 Horses can Race Together at A Time GMAT Problem Solving

Question: On a race track a maximum of 5 horses can race together at a time. There are a total of 25 horses. There is no way of timing the races. What is the minimum number of races we need to conduct to get the top 3 fastest horses?

Solution and Explanation

Approach Solution (1):

First 5 races: All horses by five. We’ll have the five winners.
Race 6: The winners of previous five races. We’ll have the 3 winners.
Now it’s obvious that #1 is the fastest one (gold medal).
For the silver and bronze we’ll have the 5 pretenders:

(1) #2 from the last sixth race,
(2) #3 from the last sixth race,
(3) The second one from the race with the gold medal winner from the first five races,
(4) The third one from the race with the Gold medal winner from the first five races,
(5) The second one from the race with the one which took the silver in the sixth race.

Race 7: These five horse: first and second in this one will have the silver and bronze among all 25.

Correct option: B

Approach Solution (2):

Let the horses be numbered as H1, H2,…, H24, H25. Now First arrange 5 races among horses as:
Race 1: H1, H2, H3, H4, H5(let's say that H1 comes first in the race then H2 then H3 then H4 and then H5 and similarly happens for the next races)
Race 2: H6, H7, H8, H9, H10
Race 3: H11, H12, H13, H14, H15
Race 4: H16, H17, H18, H19, H20
Race 5: H21, H22, H23, H24, H25

Now arrange race between winner's of all the above races as :
Race 6: H1, H6, H11, H16, H21
So, Now we have H1 , H6, H11 as top 3 winner's of Race 6. So, there are some points that can be noticed through race 6, as:

1) As H16 comes fourth in the race, so we can be sure that horses (H17, H18, H19, H20) cannot acquire top 3 positions rather they can be at any position between fifth one to last position. So, in our final race we do not need to include H16, H17, H18, H19, and H20 as we are interested in only top 3 fastest horses.
2) As H21 comes fifth in the race, so we can be sure that horses (H22, H23, H24, H25) cannot acquire top 3 positions rather they can be at any position between sixth one to last position. So, in our final race we do not need to include H21, H22, H23, H24, and H25 as we are interested in only top 3 fastest horses.
3) As H11 comes third in the race, so we can be sure that horses (H12, H13, H14, H15) cannot acquire top 3 positions rather they can be at any position between fourth one to last position. So, in our final race only H11 will
participate and rest all (H12,H13, H14, and H15) need not to participate as we are interested in only top 3 fastest horses.
4) As H6 comes second in the race, so we can be sure that only H7 could have the possibility of third fastest horse and horses (H8, H9, H10) cannot acquire top 3 positions rather they can be at any position between fourth one to last position. So, in our final race only H6 and H7 will participate and rest all (H8, H9, and H10) need not to participate as we are interested in only top 3 fastest horses.
5) As H1 wins race 1 and also wins the race with the winner of all rest of the races, so, till now we have found that H1 is the fastest horse among all. So, we do not need to race H1 again with any other horse.
6) As H1 comes first in the race, so we can be sure that only H2 and H3 could have the possibility of second and third fastest horse respectively and horses (H4, H5)cannot acquire top 3 positions rather they can be at any position between fourth one to last position. So, in our final race only H4 and H5 will participate and rest all (H8, H9, and H10) need not to participate as we are interested in only top 3 fastest horses. So, for our final race, we have H2, H3, H6, H7, and H11 as the competitors of second and third position. So, we have,

Race 7: H2, H3, H6, H7, H11 Let's say H2 and H3 comes at first and second position respectively. So, finally we have top 3 fastest horses in just 7 races among 25 horses as:H1 > H2 > H3

Correct option: B

Approach Solution (3):

We want to get the top 3 fastest horses. The trick here is that though the horse #1 from the 6th race will be the fastest one, so gold medal owner but the horse #2 and #3 from this race many not be the 2nd and the 3rd fastest horses out of 25.

For the silver and bronze prize, we would have 5 pretenders:

#2 from the 6th race;
#3 from the 6th race;
The 2nd horse from the first round race with the Gold medal winner;
The 3rd horse from the 1st round race with the Gold medal winner;
The 2nd horse from the first round race with horse #2 in the 6th.

So total 7 races are needed.

Correct option: B

“When the numbers 5, 7, 11 divide a positive multiple of 17, the remainders left are respectively 4, 6, and 10. Which positive multiple of 17 gives the least number that satisfies the given condition?”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

Suggested GMAT Problem Solving Questions:

*The article might have information for the previous academic years, please refer the official website of the exam.

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On a Race Track a Maximum of 5 Horses can Race Together at A Time GMAT Problem Solving

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