12 Minute Test: Games and Tournaments - CAT MCQ

From the given information we can make the following conclusions:
The last 2 amongst the participants qualified (amongst top 6) were next round.
In R4 and R5 only one improvement happened, (which means the last participant cannot become first until he beats all the others).
Hence, R2 to R3 and R3 to R4 we have a double, which means P10 has to be 1st in R3 (double). His throw in R2 is > 87.2
Also P8 gets 6th rank in R3 and is the first to throw in R4. Since P8 has qualified, he beats P3 and P3 but not P1. So his throw in R3 is > 82.5 but < 82.9
In R4 P8 is the first one to throw. In R3 either P1 or P10 will be the best but P10 has not won a medal and there are only 3 improvements possible in the second phase. So he cannot throw more than 88.6. Hence, P1 is the best in R3.
So the results of the first three rounds can be tabulated as follows:

Given that the gold and bronze medalists improved their scores in the 5th and the 6th rounds respectively. One medalist improved in the 4th round. Hence, we can say that the three medal winners are out of P1, P5, P7 and P9. Only 3 improvements and every improvement is by the same difference (say x). We see that P9 (84.1) is way behind the other three, so he cannot improve and become a medal winner, which means the three medal winners are P1, P5 and P7.
Let us see their throws at the beginning of R4
P1 (88.6), P5 (86.4) and P7 (87.2)
We have to make 3 improvements and keep each improvement the same.
It is obvious that we cannot improve the largest number that corresponds to P1.
Let the lowest (P7) improve in the 4th round (87.2 + x)
Now if we look at the throws of P5, it is obvious that he cannot be the gold medalist.
So the difference between the throws of each of the medalists in the final tally is 1 m.
Let us assume P5 is the bronze medalist (one improvement in R6) and P1 is the silver medalist (no improvement).
Hence, P7 is the gold medalist.
So his score improves in R5.
We have (87.2 + 2x), 88.6 and (86.4 + x). For the difference of 1 m, the value of x has to be 1.2.
So the final lengths thrown by the gold, silver and bronze medalists are 89.6, 88.6 and 87.6.
(let us go back to R3 now, the length of throw of P10 has to be less than 87.6).
Let us represent all the data for the remaining 3 rounds in the form of a table:

From the given information we can make the following conclusions:
The last 2 amongst the participants qualified (amongst top 6) were next round.
In R4 and R5 only one improvement happened, (which means the last participant cannot become first until he beats all the others).
Hence, R2 to R3 and R3 to R4 we have a double, which means P10 has to be 1st in R3 (double). His throw in R2 is > 87.2
Also P8 gets 6th rank in R3 and is the first to throw in R4. Since P8 has qualified, he beats P3 and P3 but not P1. So his throw in R3 is > 82.5 but < 82.9
In R4 P8 is the first one to throw. In R3 either P1 or P10 will be the best but P10 has not won a medal and there are only 3 improvements possible in the second phase. So he cannot throw more than 88.6. Hence, P1 is the best in R3.
So the results of the first three rounds can be tabulated as follows:

Given that the gold and bronze medalists improved their scores in the 5th and the 6th rounds respectively. One medalist improved in the 4th round. Hence, we can say that the three medal winners are out of P1, P5, P7 and P9. Only 3 improvements and every improvement is by the same difference (say x). We see that P9 (84.1) is way behind the other three, so he cannot improve and become a medal winner, which means the three medal winners are P1, P5 and P7.
Let us see their throws at the beginning of R4
P1 (88.6), P5 (86.4) and P7 (87.2)
We have to make 3 improvements and keep each improvement the same.
It is obvious that we cannot improve the largest number that corresponds to P1.
Let the lowest (P7) improve in the 4th round (87.2 + x)
Now if we look at the throws of P5, it is obvious that he cannot be the gold medalist.
So the difference between the throws of each of the medalists in the final tally is 1 m.
Let us assume P5 is the bronze medalist (one improvement in R6) and P1 is the silver medalist (no improvement).
Hence, P7 is the gold medalist.
So his score improves in R5.
We have (87.2 + 2x), 88.6 and (86.4 + x). For the difference of 1 m, the value of x has to be 1.2.
So the final lengths thrown by the gold, silver and bronze medalists are 89.6, 88.6 and 87.6.
(let us go back to R3 now, the length of throw of P10 has to be less than 87.6).
Let us represent all the data for the remaining 3 rounds in the form of a table:

From the given information we can make the following conclusions:
The last 2 amongst the participants qualified (amongst top 6) were next round.
In R4 and R5 only one improvement happened, (which means the last participant cannot become first until he beats all the others).
Hence, R2 to R3 and R3 to R4 we have a double, which means P10 has to be 1st in R3 (double). His throw in R2 is > 87.2
Also P8 gets 6th rank in R3 and is the first to throw in R4. Since P8 has qualified, he beats P3 and P3 but not P1. So his throw in R3 is > 82.5 but < 82.9
In R4 P8 is the first one to throw. In R3 either P1 or P10 will be the best but P10 has not won a medal and there are only 3 improvements possible in the second phase. So he cannot throw more than 88.6. Hence, P1 is the best in R3.
So the results of the first three rounds can be tabulated as follows:

Given that the gold and bronze medalists improved their scores in the 5th and the 6th rounds respectively. One medalist improved in the 4th round. Hence, we can say that the three medal winners are out of P1, P5, P7 and P9. Only 3 improvements and every improvement is by the same difference (say x). We see that P9 (84.1) is way behind the other three, so he cannot improve and become a medal winner, which means the three medal winners are P1, P5 and P7.
Let us see their throws at the beginning of R4
P1 (88.6), P5 (86.4) and P7 (87.2)
We have to make 3 improvements and keep each improvement the same.
It is obvious that we cannot improve the largest number that corresponds to P1.
Let the lowest (P7) improve in the 4th round (87.2 + x)
Now if we look at the throws of P5, it is obvious that he cannot be the gold medalist.
So the difference between the throws of each of the medalists in the final tally is 1 m.
Let us assume P5 is the bronze medalist (one improvement in R6) and P1 is the silver medalist (no improvement).
Hence, P7 is the gold medalist.
So his score improves in R5.
We have (87.2 + 2x), 88.6 and (86.4 + x). For the difference of 1 m, the value of x has to be 1.2.
So the final lengths thrown by the gold, silver and bronze medalists are 89.6, 88.6 and 87.6.
(let us go back to R3 now, the length of throw of P10 has to be less than 87.6).
Let us represent all the data for the remaining 3 rounds in the form of a table:

From the given information we can make the following conclusions:
The last 2 amongst the participants qualified (amongst top 6) were next round.
In R4 and R5 only one improvement happened, (which means the last participant cannot become first until he beats all the others).
Hence, R2 to R3 and R3 to R4 we have a double, which means P10 has to be 1st in R3 (double). His throw in R2 is > 87.2
Also P8 gets 6th rank in R3 and is the first to throw in R4. Since P8 has qualified, he beats P3 and P3 but not P1. So his throw in R3 is > 82.5 but < 82.9
In R4 P8 is the first one to throw. In R3 either P1 or P10 will be the best but P10 has not won a medal and there are only 3 improvements possible in the second phase. So he cannot throw more than 88.6. Hence, P1 is the best in R3.
So the results of the first three rounds can be tabulated as follows:

Given that the gold and bronze medalists improved their scores in the 5th and the 6th rounds respectively. One medalist improved in the 4th round. Hence, we can say that the three medal winners are out of P1, P5, P7 and P9. Only 3 improvements and every improvement is by the same difference (say x). We see that P9 (84.1) is way behind the other three, so he cannot improve and become a medal winner, which means the three medal winners are P1, P5 and P7.
Let us see their throws at the beginning of R4
P1 (88.6), P5 (86.4) and P7 (87.2)
We have to make 3 improvements and keep each improvement the same.
It is obvious that we cannot improve the largest number that corresponds to P1.
Let the lowest (P7) improve in the 4th round (87.2 + x)
Now if we look at the throws of P5, it is obvious that he cannot be the gold medalist.
So the difference between the throws of each of the medalists in the final tally is 1 m.
Let us assume P5 is the bronze medalist (one improvement in R6) and P1 is the silver medalist (no improvement).
Hence, P7 is the gold medalist.
So his score improves in R5.
We have (87.2 + 2x), 88.6 and (86.4 + x). For the difference of 1 m, the value of x has to be 1.2.
So the final lengths thrown by the gold, silver and bronze medalists are 89.6, 88.6 and 87.6.
(let us go back to R3 now, the length of throw of P10 has to be less than 87.6).
Let us represent all the data for the remaining 3 rounds in the form of a table:

From the given information we can make the following conclusions:
The last 2 amongst the participants qualified (amongst top 6) were next round.
In R4 and R5 only one improvement happened, (which means the last participant cannot become first until he beats all the others).
Hence, R2 to R3 and R3 to R4 we have a double, which means P10 has to be 1st in R3 (double). His throw in R2 is > 87.2
Also P8 gets 6th rank in R3 and is the first to throw in R4. Since P8 has qualified, he beats P3 and P3 but not P1. So his throw in R3 is > 82.5 but < 82.9
In R4 P8 is the first one to throw. In R3 either P1 or P10 will be the best but P10 has not won a medal and there are only 3 improvements possible in the second phase. So he cannot throw more than 88.6. Hence, P1 is the best in R3.
So the results of the first three rounds can be tabulated as follows:

Given that the gold and bronze medalists improved their scores in the 5th and the 6th rounds respectively. One medalist improved in the 4th round. Hence, we can say that the three medal winners are out of P1, P5, P7 and P9. Only 3 improvements and every improvement is by the same difference (say x). We see that P9 (84.1) is way behind the other three, so he cannot improve and become a medal winner, which means the three medal winners are P1, P5 and P7.
Let us see their throws at the beginning of R4
P1 (88.6), P5 (86.4) and P7 (87.2)
We have to make 3 improvements and keep each improvement the same.
It is obvious that we cannot improve the largest number that corresponds to P1.
Let the lowest (P7) improve in the 4th round (87.2 + x)
Now if we look at the throws of P5, it is obvious that he cannot be the gold medalist.
So the difference between the throws of each of the medalists in the final tally is 1 m.
Let us assume P5 is the bronze medalist (one improvement in R6) and P1 is the silver medalist (no improvement).
Hence, P7 is the gold medalist.
So his score improves in R5.
We have (87.2 + 2x), 88.6 and (86.4 + x). For the difference of 1 m, the value of x has to be 1.2.
So the final lengths thrown by the gold, silver and bronze medalists are 89.6, 88.6 and 87.6.
(let us go back to R3 now, the length of throw of P10 has to be less than 87.6).
Let us represent all the data for the remaining 3 rounds in the form of a table: