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DelSingh
Three straight metal rods have an average (arithmetic mean) length of 77 inches and the shortest rod has a length of 65 inches. What is the maximum possible value of the median length, in inches, of the three rods?

A. 71
B. 77
C. 80
D. 83
E. 89

If you know of any similar problems, please post.

Because it is mentioned that the average length of the three rods is 77 inches, we can assume there are three rods each of length 77 inches. Now we have been told that the shortest rod has a length of 65 inches. Thus, we can take 12 inches from one of the rods and will have to adjust it amongst the other two remaining rods. Now, if we divide 12 in any proportion other than 6 and 6, we will not have the maximum value for the median. Imagine, we redistribute 12 by giving 5 to one and 7 to another. This gives 65, 82 and 84.
Thus, giving 6 to both the rods, we have (77+6) = 83 inches for both the rods.

D.
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Smallest rod = 65
Average of all 3 rods = 77

Lets consider middle rod = 77, so the largest rod (x) calculation is:
65 + x = 77*2
x = 154 - 65 = 89.
Now, considering that middle rod = 77, the max lenght of the largest rod would be = 89.
So the max length of the middle rod possible = Avg of 89 & 77
= (89 +77)/2 = 83
Answer = D = 83
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DelSingh
Three straight metal rods have an average (arithmetic mean) length of 77 inches and the shortest rod has a length of 65 inches. What is the maximum possible value of the median length, in inches, of the three rods?

A. 71
B. 77
C. 80
D. 83
E. 89

Source: GMAT Prep Question Pack 1
Rated: Medium

If you know of any similar problems, please post.

Say the lengths of the rods in ascending order are \(x_1\), \(x_2\), and \(x_3\), where \(x_1\leq{x_2}\leq{x_3}\).

The median of a set with odd number of terms is just the middle term, when arranged in ascending/descending order, hence the median is \(x_2\).

Given that \(x_1+x_2+x_3=3*77\) --> \(65+x_2+x_3=3*77\) --> \(x_2+x_3=166\). We need to maximize \(x_2=median\), so we need to minimize \(x_3\).

The minimum value of \(x_3\) is \(x_2\) --> \(x_2+x_2=166\) --> \(x_2=median=83\).

Answer: D.

Similar topics:
gmat-diagnostic-test-question-79347.html
seven-pieces-of-rope-have-an-average-arithmetic-mean-lengt-144452.html
a-set-of-25-different-integers-has-a-median-of-50-and-a-129345.html
the-median-of-the-list-of-positive-integers-above-is-129639.html
in-a-certain-set-of-five-numbers-the-median-is-128514.html
given-distinct-positive-integers-1-11-3-x-2-and-9-whic-109801.html
set-s-contains-seven-distinct-integers-the-median-of-set-s-101331.html
three-boxes-have-an-average-weight-of-7kg-and-a-median-weigh-99642.html

Hope it helps.

Hi,

Is there a reason we cannot assume that the median is the mean in this case? Does that ONLY apply to evenly spaced sets?
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Bunuel
DelSingh
Three straight metal rods have an average (arithmetic mean) length of 77 inches and the shortest rod has a length of 65 inches. What is the maximum possible value of the median length, in inches, of the three rods?

A. 71
B. 77
C. 80
D. 83
E. 89

Source: GMAT Prep Question Pack 1
Rated: Medium

If you know of any similar problems, please post.

Say the lengths of the rods in ascending order are \(x_1\), \(x_2\), and \(x_3\), where \(x_1\leq{x_2}\leq{x_3}\).

The median of a set with odd number of terms is just the middle term, when arranged in ascending/descending order, hence the median is \(x_2\).

Given that \(x_1+x_2+x_3=3*77\) --> \(65+x_2+x_3=3*77\) --> \(x_2+x_3=166\). We need to maximize \(x_2=median\), so we need to minimize \(x_3\).

The minimum value of \(x_3\) is \(x_2\) --> \(x_2+x_2=166\) --> \(x_2=median=83\).

Answer: D.

Similar topics:
gmat-diagnostic-test-question-79347.html
seven-pieces-of-rope-have-an-average-arithmetic-mean-lengt-144452.html
a-set-of-25-different-integers-has-a-median-of-50-and-a-129345.html
the-median-of-the-list-of-positive-integers-above-is-129639.html
in-a-certain-set-of-five-numbers-the-median-is-128514.html
given-distinct-positive-integers-1-11-3-x-2-and-9-whic-109801.html
set-s-contains-seven-distinct-integers-the-median-of-set-s-101331.html
three-boxes-have-an-average-weight-of-7kg-and-a-median-weigh-99642.html

Hope it helps.

Hi,

Is there a reason we cannot assume that the median is the mean in this case? Does that ONLY apply to evenly spaced sets?

For an evenly spaced set (arithmetic progression), the median equals to the mean. Though the reverse is not necessarily true. Consider {0, 1, 1, 2} --> median = mean = 1 but the set is not evenly spaced.

So, for the original question we cannot assume that the mean and the median are the same.
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we can solve by eliminating the options.
avg is 77, lets assume the 3 rods are 77,77,77.

But the shortest is 65. In order to maximize median, the other 2 rods should have the same, highest possible value.

option a - if 2nd rod is 71, 3rd rod will be greater than 77 as the avg has to be 77. eliminate

option b - if 2nd rod is 77, 3rd rod will be greater than 77 as the smallest is 65 and avg has to be 77. eliminate

option c - if 2nd rod is 80, 3rd also has to be 80...but the sum (65 + 80 + 80) of 3 rods ends with 5(unit digit)...but we need the sum to end with 1(77*3). eliminate

option d can also be eliminated on similar lines (as mentioned in option c).
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Here is another way. We know that shortest rod is 65 so, 77-65 = 12 is the additional lenght to be distributed between other 2 rods.

There are 2 ways to do that.

1. distribute equally so data points will be 65, 83, 83 ( 6 given to each remaining data point )
2. distribute unequally 65, 65, 89 ( median will be minimized )

so, 83 is the max value of the median.
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Great Official Question.
Here is what i did in this one =>
Let the metal pieces be-->
W1
W2
W3

Using \(Mean = Sum/#\)


Sum(3)=77*3 = 231

Hence W1+W2+W3=231

W1=65
Hence W2+W3=166

To maximise the Median => We must minimise the value of W3

Hence W2=W3

Hence 2W2=166
Hence W2=83

Hence D
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Mean = 77
Rods = 3
Sum = 77*3
= 231
Shortest Rod's length = 65
Thus 231-65 = 166
Median max value = 166/2
=83
Value of 3 rods will be 65,83,83


:-D :-D
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­Classic optimization. In order to max the median, you minimize everything else:

­
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Average = 77
Total Length of all pieces combines = 77x3 = 231

65 _ _

i.e. sum of the other two rods = 231-65 = 166

For median to be maximum the third rod should have minimum length which can be eual to median

= 166/2 = 83

Answer: Option D

Related Hard Question:



DelSingh
Three straight metal rods have an average (arithmetic mean) length of 77 inches and the shortest rod has a length of 65 inches. What is the maximum possible value of the median length, in inches, of the three rods?

A. 71
B. 77
C. 80
D. 83
E. 89

PS01679

Source: GMAT Prep Question Pack 1
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