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At a wedding reception, each of 11 guests brought a different individual as a date. If 5 people leave the reception early, what is the greatest possible number of guests remaining with their original dates?

a) 6
b) 7
c) 8
d) 9

User Sarim Sidd
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1 Answer

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Final answer:

The greatest possible number of guests remaining with their original dates after 5 people leave is 9, considering that the departing guests are part of the pairs.

Step-by-step explanation:

For the greatest number of guests remaining with their original dates after 5 people leave the wedding reception early, we should consider the worst-case scenario in which all 5 people who leave are from different pairs. In this case, 5 pairs would be affected. Since each guest brought a different individual as a date, there were initially 11 pairs, so if 5 pairs are affected, we have 11 - 5, leaving 6 pairs intact. However, those 5 people that left could also include dates from the intact pairs, which means the actual number leaving could be less than 5 pairs, resulting in more intact pairs. Therefore, the most pairs that can remain intact without their dates is actually 11 - (5/2), since each pair that loses a member loses two people. But, since you cannot have half a pair, we round down to the nearest whole number. So, the greatest possible number of guests remaining with their original dates is 11 - 2 = 9.

User Abhay
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