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

Question

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.

(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.

amanvermagmat · Accepted Answer

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

sushforgmat · Answer

While this question is solvable with the application of concepts, it seems too difficult to be featured on GMAT. Doesn't it?

GmatPoint · Answer

Let us first consider the red box.
The average length of ropes = 60 cm
Therefore, the sum of the length of ropes in the red box = 7 x 60 = 420 cm
We have been given that the median value is 45 cm. Since the number of ropes in the red box is 7 (an odd number), the median represents the middle value when we arrange the ropes in either ascending or descending order of their lengths.
For instance, if we arrange the ropes in ascending order of their lengths, we get:
__ __ __ 45 __ __ __
Note: Here the spaces correspond to the lengths of the ropes.
 
We can follow the same steps as above for the blue box:
The average length of ropes = 50 cm
Sum of lengths = 5 x 50 = 250 cm
Median = 24 cm
Arranging the ropes in ascending order of their lengths:
__ __ 24 __ __

Let us now examine the statements individually.

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

If we observe the series, __ __ 24 __ __, we can say the two digits to the left of 24 cannot exceed 24 since they are in ascending order. If we assume their value to be maximum, i.e. 24 cm, we get the sum of the first three terms as 72 cm. Then, the sum of the last two terms = 250 - 72 = 178 cm. Hence, the sum of the last two terms can be greater than or equal to 178 cm, while the first two values can take values that are less than or equal to 24 cm.
However, if we notice closely, 178 = 89 x 2. Statement 1 explicitly mentions that the maximum length of a rope in the blue box cannot exceed 89 cm. Thus, we can have only one case such that the arrangement:
24 24 24 89 89.

However, when we combine the 12 ropes, we get(in ascending order):
24 24 24 45 89 89  .

The 3 values to the left of 45 in the original series will appear to the left of 45 in this series as well, and the 3 values to the right of 45 in the original series will appear to the right of 45 in this series as well. 
So, one thing that we are sure of is that 45 is the 7th number of this series. We need to find the 6th number as well in order to calculate the mean of the two middle values and get the median. However, we can have a variety of ways to distribute the remaining sum among the remaining six numbers. For example, the number to the immediate left of 45 might be 45 as well, and 44 as well, and so on, to state a few. Hence, we cannot conclusively reach the answer.
 Statement 1 in itself is not sufficient to answer the question.

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

If we observe the series, __ __ __ 45 __ __ __, we can say the two digits to the left of 45 cannot exceed 45 since they are in ascending order. If we assume their value to be 45, we get the sum of the first four terms as 180. Then, the last three terms = 420 - 180 = 240 cm. Hence, the sum of the last three terms can be greater than or equal to 240 cm since the first three terms can take values less than or equal to 45 cm.

However, on looking closely, 240 = 80 x 3, and since statement 2 says that the maximum length of a rope in the blue box cannot exceed 80 cm, we can have only one case:
45 45 45 45 80 80 80.

However, when we combine the 12 ropes, we get(in ascending order):
__ __ 24 45 45 45 45 80 80 80

On closely observing this series of 10 terms, we see that the four terms in the middle are all identical. So, whether we insert the remaining two terms to the left of 45 (value < 45) or the right of 45 (value > 45), the 6th and 7th terms will be 45 each in all cases. Even if one of the new numbers entered is 45, the 6th and 7th terms remain 45. Hence, the median is the average of the middle terms = 45 cm.  Hence, Statement 2 alone is sufficient to determine the median.

Therefore, we can say that Option (B) is the correct answer.

bumpbot · Answer

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards