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Statistics, Averages, Explanation, OG 183

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Statistics, Averages, Explanation, OG 183

by CarineK » Thu May 07, 2015 2:41 pm
I want to ask a question about a specific part of an answer explanation in OG 2015. I was only able to find the solution to #183 in an inefficient, guess and test sort of way.

Seven pieces of rope have an average length of 68 cm and a median length of 84 cm. If the length of the longest piece of rope is 14 cm more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in cm, of the longest piece of rope?

The explanation assigns the pieces a,b,c,d,e,f,g, where d=84, and g=4a+14. The maximum value for g will be when a is at maximum, which is only the case if a, b, and c are all of equal lengths. Here's the part I don't understand-- "The maximum value for 4a+14 will occur when e and f are as small as possible." From here, the explanation goes on to assign 84 as a value to e and f, and then solve for a using avg= sum/# of values. Why is the maximum value of g dependent on e and f/ how do we know to assign them 84? As long as g is higher than e and f, it's in the right "spot" in the ordered set, so why does it make any difference where e and f's values lie between 84 and (4a-14)-1? I understand it has something to do with knowing the avg of the entire set, but can't tie it all in.
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by GMATGuruNY » Thu May 07, 2015 2:46 pm
Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters . If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?

A) 82

B) 118

C) 120

D) 134

E) 152

OA: D

The book ( OG 13) posts a pretty confusing way to solve, would love to hear any other recommendations.

Thanks in advance for any guidance.
The sum of the lengths = 7*68 = 476.

Let the smallest piece = x.
Then the length of the longest piece = 4x+14.
Median piece = 84.
Let the remaining pieces be a, b, c, d.

Here are the 7 pieces, in ascending order:
x, a, b, 84, c, d, 4x+14.

To MAXIMIZE the value of 4x+14, we must MINIMIZE the values of a, b, c, and d.
The least possible value for a and b is x.
The least possible value for c and d is 84.
Here are the 7 pieces:
x, x, x, 84, 84, 84, 4x+14.

Since the sum of the lengths is 476, we get:
x + x + x + 84 + 84 + 84 + 4x+14 = 476
7x + 266 = 476
7x = 210
x = 30.

Thus:
Greatest possible value for the longest piece = 4x+14 = 4*30 + 14 = 134.

The correct answer is D.
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by GMATGuruNY » Thu May 07, 2015 2:48 pm
Alternate approach:

The sum of the lengths = 7*68 = 476.

We can PLUG IN THE ANSWERS, which represent the maximum length of the longest piece.
The longest piece is 14 centimeters more than 4 TIMES the length of the shortest piece. Thus, when 14 is subtracted from the correct answer choice, the result will be a multiple of 4.
Subtracting 14 from each answer choice, we get:
68, 104, 106, 120, 138.
106 and 138 are not multiples of 4.
Eliminate C and E.

Since the question asks for the MAXIMUM possible length, we should start with the GREATEST of the remaining answer choices.
To MAXIMIZE the length of the longest piece, we must MINIMIZE the lengths of all of the other pieces.

Answer choice D: 134
Shortest piece = 120/4 = 30.
The 7 pieces are:
30, b, c, 84, e, f, 134.
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.
Sum = 3(30) + 3(84) + 134 = 90+252+134 = 476.
Success!

The correct answer is D.
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by CarineK » Thu May 07, 2015 2:55 pm
Thank you so much, this clears up a lot!! Sorry for the doubled question post, I'm kinda new here and couldn't find that original.
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by Brent@GMATPrepNow » Thu May 07, 2015 4:31 pm
Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?

(A) 82
(B) 118
(C) 120
(D) 134
(E) 152

Answer is D
So, we have 7 rope lengths.
If the median length is 84, then the lengths (arranged in ascending order) look like this: {_, _, _, 84, _, _, _}

The length of the longest piece of rope is 14 cm more than 4 times the length of the shortest piece of rope.
Let x = length of shortest piece.
This means that 4x+14 = length of longest piece.
So, we now have: {x, _, _, 84, _, _, 4x+14}

Our task is the maximize the length of the longest piece.
To do this, we need to minimize the other lengths.
So, we'll make the 2nd and 3rd lengths have length x as well (since x is the shortest possible length)
We get: {x, x, x, 84, _, _, 4x+14}

Since 84 is the middle-most length, the 2 remaining lengths must be greater than or equal to 84.
So, the shortest lengths there are 84.
So, we get: {x, x, x, 84, 84, 84, 4x+14}

Now what?

At this point, we can use the fact that the average length is 68 cm.
There's a nice rule (that applies to MANY statistics questions) that says:
the sum of n numbers = (n)(mean of the numbers)
So, if the mean of the 7 numbers is 68, then the sum of the 7 numbers = (7)(68) = 476

So, we now now that x+x+x+84+84+84+(4x+14) = 476
Simplify to get: 7x + 266 = 476
7x = 210
x=30

If x=30, then 4x+14 = 134
So, the longest piece will be 134 cm long.

Answer = D

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by Scott@TargetTestPrep » Fri Jul 10, 2015 4:19 am
CarineK wrote:I want to ask a question about a specific part of an answer explanation in OG 2015. I was only able to find the solution to #183 in an inefficient, guess and test sort of way.

Seven pieces of rope have an average length of 68 cm and a median length of 84 cm. If the length of the longest piece of rope is 14 cm more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in cm, of the longest piece of rope?

The explanation assigns the pieces a,b,c,d,e,f,g, where d=84, and g=4a+14. The maximum value for g will be when a is at maximum, which is only the case if a, b, and c are all of equal lengths. Here's the part I don't understand-- "The maximum value for 4a+14 will occur when e and f are as small as possible." From here, the explanation goes on to assign 84 as a value to e and f, and then solve for a using avg= sum/# of values. Why is the maximum value of g dependent on e and f/ how do we know to assign them 84? As long as g is higher than e and f, it's in the right "spot" in the ordered set, so why does it make any difference where e and f's values lie between 84 and (4a-14)-1? I understand it has something to do with knowing the avg of the entire set, but can't tie it all in.
Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?

A) 82

B) 118

C) 120

D) 134

E) 152

Solution:

We need to first recognize that we are working with a maximum problem. Of the seven pieces of rope, we must make 6 of those pieces as small as we possibly can, within the confines of the given information, and in doing so we'll maximize the length of the 7th piece.

We are first given that seven pieces of rope have an average (arithmetic mean) length of 68 centimeters. From this information we can determine the sum of the lengths of the 7 pieces of rope.

average = sum/quantity

sum = average x quantity

sum = 68 x 7 = 476

Next we are given that the median length of a piece of rope is 84 centimeters, which means that when we arrange the pieces of rope from the shortest to the longest, the middle item (the 4th piece of rope) will have a length of 84 centimeters. We also must keep in mind that we can have pieces of rope of the same length. Let's first label our seven pieces of rope with variables or numbers, starting with the shortest piece and moving to the longest piece. We can let x equal the shortest piece of rope, and m equal the longest piece of rope.

piece 1: x

piece 2: x

piece 3: x

piece 4: 84

piece 5: 84

piece 6: 84

piece 7: m

Notice that the median (the 4th rope) is 84 cm long. In keeping with our goal of minimizing the length of the first 6 pieces, we will assign 84 to pieces 5 and 6 to make them as short as possible. Similarly, we have assigned a length of x to pieces 1, 2, and 3.

We can plug these variables into our sum equation:

x + x + x + 84 + 84 + 84 + m = 476

3x + 252 + m = 476

3x + m = 224

We also given that the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope. We can express this as:

m = 14 + 4x

We can now substitute (14 + 4x) for m into the equation 3x + m = 224. We obtain:

3x + 14 + 4x = 224

7x = 210

x = 30

Thus, the longest piece of rope is 4(30) + 14 = 134 centimeters.

Answer: D

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by Jim@StratusPrep » Fri Jul 10, 2015 4:29 am
The one thing that I would recommend that is different from the above solutions would be to not multiply 68 x 7. When you see the left side if the equation you have 7x + 266 an this equals 68(7), or:

7x + 266 = 68(7).

All the answers are integers, right? So it stands to reason that everything here is divisible by 7 and you get:

x + 38 = 68, or x = 30. Obviously the same answer, but I find that the fewer calculations you do, the smaller the chances you have to make a mistake. You might ask about 266/7, but you can use the nearest number calculation to se that 7 x 40 = 280 and then see the difference of 280 and 266 is 14.
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