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This is a difficult logical question
1) In 1st case whatever 8 numbers go in P, Q will have to take the left 2 from A, and then the rest 5 must be repeated to fulfill the conditions mentioned.
Sufficient
2) Here 65-50 would be =15, which could be made by 1+2+3+4+5 or 9+6 or 10+5 … this gives us the idea that the combination of numbers with sum 15 have been repeated but not the actual # of such numbers
Not sufficient
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kiran120680
List A ={1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

The integers from list A are used to form two distinct sets P and Q each consisting of distinct integers only. If all the integers from list A are used at least once and none of the integers are used more than twice, how many integers from list A are used more than once?

1) The numbers of integers in sets P and Q are, respectively, 2 and 3 less than the number of integers in list A.

2) The difference between the sum of the integers in sets P and Q and the sum of the integers in list A is 15.


There are total 10 distinct integers that have to be divided between P and Q. None of the integer can find a place in both P and Q. However each integer can be used twice.

Maximum possible integers including repetition in combined set of P and Q are 10+10=20
Whatever be the number of integers in combined set of P and Q, it will contain exactly 10 distinct integers 1-10.

1) The numbers of integers in sets P and Q are, respectively, 2 and 3 less than the number of integers in list A.
Number of integers in P=10-2=8
Number of integers in Q=10-3=7
Total in combined set of P and Q=8+7=15. But we know that exactly 10 of these are distinct.
So repetition =15-10=5
Sufficient

2) The difference between the sum of the integers in sets P and Q and the sum of the integers in list A is 15.
So Sum of integers in P and Q=(1+2…+10)+15
This 15 is surely due to repeated numbers.
15=5+10, so two could have been repeated.
15=1+2+3+4+5, so five could have been repeated.
15=4+5+6, so three could have been repeated.
Insufficient


A
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vinayakvaish
Hi please consider this:

Set A {1,2,3,4,5,6,7,8,9,10} from which two sets have to be created which contain distinct integers only. To find integers used twice, we have to look for overlaps in set p and set q

Statement 1: Set P has 8 elements whereas Set Q has 7

Set P can be {1,2,3,4,5,6,7,8} and Set Q can be {1,2,3,4,5,6,7}- overlap of 7 integers
Set P can also be {1,2,3,4,5,6,7,8} and Set Q {10,9,8,7,6,5,4} - overlap of 5 integers

Since it is given that each integer from Set A has to be used at least once hence we take the second case
Statement 1 is sufficient

Statement 2:

Sum of P+Q - Sum of A=15

15 can be the sum of 5 distinct integers at max ie. 1,2,3,4,5 but since we are not given how many elements to choose the elements may well be 4,5,6
Statement 2 is not Sufficient

In case the "must be used once" condition was absent in Set A, then Statement 1 would have gotten interesting.

Max overlap is 7 and Minimum overlap is 5 between two 7 and 8 element sets. Then Statement 1 would be insufficient and you would have to use the fact that sum of 15 can be achieved by a maximum of 5 distinct positive integers {1,2,3,4,5,}

Hence then the Answer would have been C

Here of course the Answer is A

vinayakvaish


The Statement states that EACH NUMBER IS USED ATLEAST ONCE. that being said , lets try and make a few use cases


P = {1,2,3,4,5,6,7,8}
now Q must have {9,10} given that each number must be used once . so we have only 5 spaces left, which can over lap . hence Suff

if again P = {10,9,8,7,6,5,4,3} Q will begin / have {1,2} leaving 5 places.
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