Logical Reasoning Set 7
|
P, Q, R and S were distributed 12 lottery tickets each bearing a different number among 1 to 12. No two persons got the same number
of lottery tickets. Both the lottery tickets owned by P bear even numbers. Also, the number on any of the lottery tickets that R got is
not the square of a prime number. Neither the lottery ticket numbered 1 nor the lottery ticket numbered 9 is with Q. R received lottery
ticket number 12. The sum of the numbers on two lottery tickets of R is the same as the sum of the numbers on all the lottery tickets of
P. All the prime numbered lottery tickets are owned by Q.
|
Questions Set 7
Q1) If the lottery ticket numbered 14 is given to anyone except to the person having the maximum number of lottery tickets without violating any of the earlier conditions and the sum of numbers on all the lottery tickets of R is more than that of Q, then, what is the sum of all the numbers on the lottery tickets of P?1) 12
2) 16
3) 18
4) Can't be Determined
Q2) What is the minimum possible difference between the sum of numbers on all the lottery tickets of any two persons?
1) 1
2) 3
3) 4
4) 5
Q3) Which of the following conditions is required to get the unique solution for the distribution?
1) Twice the total of numbers on all lottery tickets of P is equal to the total of numbers on all lottery tickets of Q.
2) The sum of all the numbers on all the lottery tickets of one of the persons is a perfect cube.
3) The sum of numbers on all lottery tickets of R is 19.
4) Any one of the above.
Q4) What is the maximum possible difference between the sum of numbers on all the lottery tickets of any two persons?
1) 21
2) 22
3) 24
4) 23
SOLUTIONS:
12 lottery tickets can be distributed among 4 persons such that no two persons get the same number of lottery tickets in 2 ways.
Either 1, 2, 3, and 6 or 1, 2, 4, and 5.
Consider the distribution 1, 2, 4, and 5.
Q can have lottery tickets 2, 3, 5, 7 and 11 only. As both P and R have at least two lottery tickets, S must have only one lottery ticket.
It must be numbered 9 as R cannot have a square of a prime number and P cannot have an odd one. Also, P should have 4. Since R cannot have square of a prime number.
Distribution 1:
S: 9
P: 4, 10
R: 1, 6, 8, 12
Q: 2, 3, 5, 7, 11
Now, consider the distribution 1, 2, 3, and 6.
Distribution 2:
S: 9
P: 8, 10
R: 1, 6, 12
Q: 2, 3, 5, 7, 11, 4
S: 9
P: 8, 10
R: 1, 6, 12, 14
Q: 2, 3, 5, 7, 11, 4
Sum of the numbers on the lottery tickets of P = 8 + 10 = 18
2) Minimum possible difference:
In Distribution 1 = 28 - 27 = 1
In Distribution 2 = 19 - 18 = 1
3) Choice 1, refer to 'Distribution 1' where 2P = Q.
Choice 2, refer to 'Distribution 1' where R = 27.
Choice 3, refer to 'Distribution 2' where R = 19.
4) Maximum possible difference:
In Distribution 1 = 28 - 9 = 19
In Distribution 2 = 32 - 9 = 23