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Question 1 of 5
1. Question
How many rounds of matches does a knock-out tennis tournament have if it starts with 64 players and every player needs to win 1 match to move at the next round?
Correct
Ans-b
Since every player needs to win only 1 match to move to the next round, therefore the 1ST round would have 32 matches between 64 players out of which 32 will be knocked out of the tournament and 32 will be moved to the next round. Similarly in 2nd round 16 matches will be played, in the 3rd round 8 matches will be played, in 4th round 4 matches, in 5th round 2 matches and the 6th round will be the final match.
Hence total number of rounds will be 6 (2)6 = 64).
Hence, option (b) is the answer.
Incorrect
Ans-b
Since every player needs to win only 1 match to move to the next round, therefore the 1ST round would have 32 matches between 64 players out of which 32 will be knocked out of the tournament and 32 will be moved to the next round. Similarly in 2nd round 16 matches will be played, in the 3rd round 8 matches will be played, in 4th round 4 matches, in 5th round 2 matches and the 6th round will be the final match.
Hence total number of rounds will be 6 (2)6 = 64).
Hence, option (b) is the answer.
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Question 2 of 5
2. Question
There are N men sitting around a circular table at N distinct points. Every possible pair of men except the ones sitting adjacent to each other sings a 2 minute song one pair after other. If the total time taken is 88 minutes, then what is the value of N?
Correct
Ans-d
Total number of pairs of men that can be selected If the adjacent ones are also selected is NC2. Total number of pairs of men selected if only the adjacent ones are selected is N. Hence total number of pairs of men selected if the adjacent ones are not selected is
*C2- N.
Since the total time taken is 88 min, hence the number of pairs is 44.
Hence, NC2-N = 44 → N = 11.
Hence, Option (d) is the answer.
Incorrect
Ans-d
Total number of pairs of men that can be selected If the adjacent ones are also selected is NC2. Total number of pairs of men selected if only the adjacent ones are selected is N. Hence total number of pairs of men selected if the adjacent ones are not selected is
*C2- N.
Since the total time taken is 88 min, hence the number of pairs is 44.
Hence, NC2-N = 44 → N = 11.
Hence, Option (d) is the answer.
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Question 3 of 5
3. Question
In a class with boys and girls a chess competition was played wherein every student had to play 1 game with every other student. It was observed that in 36 matches both the players were boys and in 66 matche both were girls. What is the number of matches in which I boy and I girl play against each other?
Correct
Ans-A
Let the number of boys be B. Then Bc3 = 36 →B = 9
Let the number of girls be G. Then GC2 = 66 àG = 12.
Therefore total number of students in the class
= 12 + 9 = 21. Hence total number of matches
21C2 = 210.
Hence, number of matches between 1 boy and 1 girl = 210 – (36 +66) = 108.
Hence, Option (a) is the answer.
Incorrect
Ans-A
Let the number of boys be B. Then Bc3 = 36 →B = 9
Let the number of girls be G. Then GC2 = 66 àG = 12.
Therefore total number of students in the class
= 12 + 9 = 21. Hence total number of matches
21C2 = 210.
Hence, number of matches between 1 boy and 1 girl = 210 – (36 +66) = 108.
Hence, Option (a) is the answer.
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Question 4 of 5
4. Question
If x,y,z can only take the values 1,2,3,4,5,6,7 then find the number of solutions of the equation x+y+z=12
Correct
Ans-b
We will consider x = 7 to x=1
For X=7 ,y+z=6 No. of solutions = 4
For x = 6, y + z = 6. No. of solution =5
For x = 5, y + z = 7. No. of solutions = 6
For x = 4, y + z = 8. No. of solutions = 7
For x = 3, y = 2 = 9. No. of solutions = 6
For x = 2, y + z = 10. No. of solutions = 5
For x = 1, y + z = 11. No. of solutions = 4
Hence number of solutions = 37
Hence, Option (b) is the answer.
Incorrect
Ans-b
We will consider x = 7 to x=1
For X=7 ,y+z=6 No. of solutions = 4
For x = 6, y + z = 6. No. of solution =5
For x = 5, y + z = 7. No. of solutions = 6
For x = 4, y + z = 8. No. of solutions = 7
For x = 3, y = 2 = 9. No. of solutions = 6
For x = 2, y + z = 10. No. of solutions = 5
For x = 1, y + z = 11. No. of solutions = 4
Hence number of solutions = 37
Hence, Option (b) is the answer.
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Question 5 of 5
5. Question
If x,y are two digit no ,and u,v,x,y are digits then find the number of solutions of the equations:
(xy)2 =u!+v
Correct
Ans-b
(xy)2 = u! + v
Here xy is a two-digit number and maximum value of its square is 9801. 8! is a five-digit number =>u is less than 8 and 4! is 24 which when added to a single digit will never give the square of a two-digit number. Hence u is greater than 4. So, possible values of u can be 5, 6 and 7.
If u = 5, u!= 120 => (xy)2 = u! + v => (xy)2 = 120
+ v = 120 + 1 = 121 = 112
If u = 6, u! = 720 => (xy)2 = u! + y => (xy)2= 720
+ v = 720 + 9 = 729 = 272
If u = 7, u!= 5040 => (xy)2 = u! + y => (xv)2 =
5040 + v = 5040 + 1 = 5041 = 712
So there are three cases possible. Hence, 3 solutions exist for the given equation.
Hence, Option (b) is the correct answer
Incorrect
Ans-b
(xy)2 = u! + v
Here xy is a two-digit number and maximum value of its square is 9801. 8! is a five-digit number =>u is less than 8 and 4! is 24 which when added to a single digit will never give the square of a two-digit number. Hence u is greater than 4. So, possible values of u can be 5, 6 and 7.
If u = 5, u!= 120 => (xy)2 = u! + v => (xy)2 = 120
+ v = 120 + 1 = 121 = 112
If u = 6, u! = 720 => (xy)2 = u! + y => (xy)2= 720
+ v = 720 + 9 = 729 = 272
If u = 7, u!= 5040 => (xy)2 = u! + y => (xv)2 =
5040 + v = 5040 + 1 = 5041 = 712
So there are three cases possible. Hence, 3 solutions exist for the given equation.
Hence, Option (b) is the correct answer
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