Dear minkathebestminkathebest wrote:Hello Gmatgury,
Why do you know that the following is true?
The minimum possible value for b and c is 30, implying 3 pieces with a length of 30.
The minimum possible value for e and f is 84, implying 3 pieces with a length of 84.
Best regards,
baBa
Hi, there. I'm happy to answer this.
Let's say the lengths of rope are
a, b, c, d, e, f, g
We know median = d = 84.
We know the average = 68.
Let' say, for the sake of argument, we have also figured out that a = 30.
Our job, in this question, is to make g, the max, as big as possible. Because the average = 68 is fixed, this is a zero-sum game. In other words, if I make g bigger, one or more of the other numbers have to be smaller. If my sole goal, within constraints, as it is in this question, is to make g as big as possible, then this implies that, within the constraint given, I have to make a + b + c + d + e + f as small as possible.
One constraint is that a = 30 is the minimum, so neither b nor c can less than 30, because no number can be smaller than the minimum. If another number were smaller than the minimum, then the minimum would be different. BUT, b and c can equal 30, equal to the minimum, and that won't change the minimum. That's the smallest we can make b & c, and we have to make them the smallest they can be to make g as large as it can be.
Similarly, d = 84, the median. How small can e & f be? Well, if either one, or both of them, is less than 84, then 84 would not be the median. BUT, both of them can equal 84, and that wouldn't not change the median. So given the constraint of the median, then e = f = 84 is the smallest we can make then. Again, we want to make e & f as small as they can be, so that g is as large as it can be.
Does all that make sense? Can you follow GMATGuruNY's superb analysis now?
Please let me know if you have any further question.
Mike
