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03 Dec 2024, 10:50
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AfifaAhana888
Correct answer should be D i.e. 1.15... If you put 0.35 in 1 value in avg, you get (sum of remaining 4) = 2.15, now if you put 1.15, you get (remaining 3 sum) = 1. So, even if all are equal, they would be 0.33, 0.33, 0.34 (but it breaks the question stem which states that lightest sample is 0.35). So, if the heaviest is 1.15, then it will break the question stem and so, it can not be the weight.
03 Dec 2024, 10:52
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PriyankaGmatIMD
Can someone help me with this
03 Dec 2024, 11:20
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The best approach for this situation is to find the total number of combinations for a group using 12 people (6 couples) and subtract the number of possibilities in which the opposite of our objective happens (the number of groups where couples are together).
The total ways are 12c4: 12 people to place in 4 spots. So 495.
Now calculate how many groups have only ONE couple together and how many groups have TWO couples together.
It is easier to select couples than people for this part so: for the 2 spots available (each spot can hold one couple) we have that 6c1 is selecting one couple, and 10c2-5 is randomly selecting 10 people for the two remaining single person spots but removing the 5 combinations where the other two remaining people selected would be a couple.
And 6c2 is selecting 2 couples of the 6 available couples.
Then (total- combinations where there is a couple in a group) thus: 495-240-15= 240. So (e)
The total ways are 12c4: 12 people to place in 4 spots. So 495.
Now calculate how many groups have only ONE couple together and how many groups have TWO couples together.
It is easier to select couples than people for this part so: for the 2 spots available (each spot can hold one couple) we have that 6c1 is selecting one couple, and 10c2-5 is randomly selecting 10 people for the two remaining single person spots but removing the 5 combinations where the other two remaining people selected would be a couple.
And 6c2 is selecting 2 couples of the 6 available couples.
Then (total- combinations where there is a couple in a group) thus: 495-240-15= 240. So (e)
