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Re: Set A consists of five consecutive positive integers. Set B consists [#permalink]
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Sonia0106 wrote:

Bunuel chetan2u @IanStewart- In statement 2, will we not use 5C2 =10 additional numbers in Set B? So in total 15 numbers? So it will mean 3 numbers are repeated and therefore discarded from Set B but overall it won't give us much so the statement 2 is insufficient. Just trying to understand whether there will be 12 values or 15 values based on calc.



It would be 12 as you have to discard numbers that are already there in B, that is repetition is not allowed.

But there will be times when the number could be less than 12 too.
a,a+1,a+2,a+3,a+4
If a+a+1<a+4 or a<3, the numbers would be even lesser.
3,4,5,6,7will not have 12.
Smallest 3 and largest 6+7 or 13. Total 11
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Re: Set A consists of five consecutive positive integers. Set B consists [#permalink]
chetan2u wrote:
Sonia0106 wrote:

Bunuel chetan2u @IanStewart- In statement 2, will we not use 5C2 =10 additional numbers in Set B? So in total 15 numbers? So it will mean 3 numbers are repeated and therefore discarded from Set B but overall it won't give us much so the statement 2 is insufficient. Just trying to understand whether there will be 12 values or 15 values based on calc.



It would be 12 as you have to discard numbers that are already there in B, that is repetition is not allowed.

But there will be times when the number could be less than 12 too.
a,a+1,a+2,a+3,a+4
If a+a+1<a+4 or a<3, the numbers would be even lesser.
3,4,5,6,7will not have 12.
Smallest 3 and largest 6+7 or 13. Total 11


Understood. Did the calculation with these variables now (a,a+1,a+2,a+3,a+4) and found that 3 numbers are always repeated (2a+3, 2a+4, 2a+5) when numbers are added to each other. So total will be 12 or less than that as you pointed out. Thank you.
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Re: Set A consists of five consecutive positive integers. Set B consists [#permalink]
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