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Three straight metal rods have an average (arithmetic mean) [#permalink]

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27 Feb 2013, 21:23

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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?

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 \(x_2\) --> \(x_2+x_2=166\) --> \(x_2=median=83\).

Re: Three straight metal rods have an average (arithmetic mean) [#permalink]

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28 Feb 2013, 03:55

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DelSingh wrote:

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.

Re: Three straight metal rods have an average (arithmetic mean) [#permalink]

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27 Feb 2013, 22:32

We know that there are 3 metal rods with an average length of 77 inches. Total length=77*3=231 inches Shortest rod length=65 inches the sum of lengths of the two longer rods=231-65=166 Since, the longer rods have to be longer than 65, the smallest value one of these longer values could have is 66.

Now the question asks us for the longest median value. The median value is the middle value when the three rods are arranged in ascending order of lengths.

There are two ways of finding these values 1. Since we need to find the largest middle value, the two longer rods need to be of equal length. If both values are equal in length the middle value will be the largest possible value and the longest rod will be of the same length as the rod of median length. Hence, 2x =166 implies x=83.

Maximum length of the median rod=83

2. Testing the values given Let us test the values from the bottom

e=89 166-89=77. Since 77 is less than 89, it cannot be the median value. WRONG

d=83 166-83=83. Both the rods are of the same length. Hence, this is the largest possible middle value.

DelSingh wrote:

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?

Re: Three straight metal rods have an average (arithmetic mean) [#permalink]

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21 Aug 2013, 20:41

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
_________________

Re: Three straight metal rods have an average (arithmetic mean) [#permalink]

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07 Dec 2014, 12:04

Bunuel wrote:

DelSingh wrote:

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 \(x_2\) --> \(x_2+x_2=166\) --> \(x_2=median=83\).

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 \(x_2\) --> \(x_2+x_2=166\) --> \(x_2=median=83\).

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.
_________________

Three straight metal rods have an average (arithmetic mean) [#permalink]

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25 Aug 2015, 23:16

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).

Re: Three straight metal rods have an average (arithmetic mean) [#permalink]

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19 Sep 2015, 19:02

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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