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A red box contains 7 ropes, with an average length of 60 cm per rope,

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A red box contains 7 ropes, with an average length of 60 cm per rope,  [#permalink]

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New post 07 May 2018, 22:47
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A red box contains 7 ropes, with an average length of 60 cm per rope, and median length of these 7 ropes is 45 cm. A blue box contains 5 ropes, with an average length of 50 cm per rope, and median length of these 5 ropes is 24 cm. Lengths of all ropes are in integers cm. If all these 12 ropes are to be put together in a separate empty box, what will be the median length of these 12 ropes?

(1) Out of 5 ropes in blue box, none of the ropes is longer than 89 cm.

(2) Out of 7 ropes in red box, none of the ropes is longer than 80 cm.
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A red box contains 7 ropes, with an average length of 60 cm per rope,  [#permalink]

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New post 09 May 2018, 21:56
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amanvermagmat wrote:
A red box contains 7 ropes, with an average length of 60 cm per rope, and median length of these 7 ropes is 45 cm. A blue box contains 5 ropes, with an average length of 50 cm per rope, and median length of these 5 ropes is 24 cm. Lengths of all ropes are in integers cm. If all these 12 ropes are to be put together in a separate empty box, what will be the median length of these 12 ropes?

(1) Out of 5 ropes in blue box, none of the ropes is longer than 89 cm.

(2) Out of 7 ropes in red box, none of the ropes is longer than 80 cm.


Red box. If we arrange the ropes in ascending order of their lengths, we will have:
_, _, _, 45, _, _, _

and the sum of these 7 values is 60*7 = 420. If we keep the smaller lengths to be very small, then the larger lengths can go very long. But we can determine the Minimum possible length of the longest rope here. For that, we will keep the first three values as '45' only. Now the sum of first four values = 45*4 = 180. And thus the sum of remaining three values = 420 - 180 = 240. Now to keep the longest value Minimum, the longest three ropes should be as close in length as possible, and since 240 is divisible by 3, the length of longest three ropes can each be = 240/3 = 80 cm.

So, we have determined that in red box, the Minimum length of the longest rope will be 80 cm, and that will happen in the following case:
45, 45, 45, 45, 80, 80, 80.

Same logic can be applied for blue box. Blue box has 5 ropes with a median 24 and average 50. So we have:
_, _, 24, _, _

and the sum = 50*5 = 250. To keep the length of longest rope in blue box minimum, we will keep length of first two ropes as 24 only. This gives sum of first three values = 24*3 = 72, and the sum of remaining two values = 250 - 72 = 178. And 178/2 = 89.

So we have determined that in blue box, Minimum length of longest rope will be 89 cm, and that will happen in the following case:
24, 24, 24, 89, 89.

Once all these 12 ropes are placed together and in ascending order of lengths, median length will be the average length of 6th and 7th ropes. So to answer this question we need to know the lengths of 6th and 7th ropes, once all these 12 ropes are arranged in ascending or descending order. Now lets move to the question statements.

(1) This tells us that the only possibility of lengths in blue box is 24, 24, 24, 89, 89.
But without knowing anything about lengths of 7 ropes in red box, we cant say which two rope lengths will be in the middle. Its possible that lengths of all 7 values lie between 24 and 89, but its also possible that they do not. Not sufficient.

(2) This tells us that the only possibility of lengths in red box is 45, 45, 45, 45, 80, 80, 80.
So we have the lengths of 7 ropes out of 12. And out of remaining 5 ropes (in red box), since the median is 24, we can be sure that the two smallest ropes will be either equal to or less than 24; which means the three smaller ropes of red box will definitely be less than 45. So now we have:

_, _, 24, 45, 45, 45, 45, 80, 80, 80 as the lengths of 10 ropes arranged in ascending order. What remains is the lengths of two longest ropes of blue box. Now at least one of those two ropes will be longer than 80 cm in length for sure (in order for sum of those 5 ropes of blue box to be 250) and whatever be the length of the second longest rope, the two ropes with middle lengths will be 45 cm only.

So to summarise this statement 2, once we combine the 12 ropes of the two boxes, smallest three ropes will be less than or equal to 24, longest four ropes will be 80 or more (three ropes of red box 80 each and one longest rope of blue box). What remains are the five ropes in between, out of which four ropes (from red box) are 45 in length, and no matter what the length of the fifth rope is (this is the second longest rope of blue box), the middle two ropes length wise will be 45 cm in length each only - no other possibility. Thus this statement is sufficient to answer the question.


Hence B answer
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Re: A red box contains 7 ropes, with an average length of 60 cm per rope,  [#permalink]

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New post 13 Sep 2018, 09:41
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While this question is solvable with the application of concepts, it seems too difficult to be featured on GMAT. Doesn't it?
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Re: A red box contains 7 ropes, with an average length of 60 cm per rope, &nbs [#permalink] 13 Sep 2018, 09:41
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A red box contains 7 ropes, with an average length of 60 cm per rope,

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