Seven pieces of rope have an average (arithmetic mean) lengt

Because e, f, and g cannot be less than the median, which is d=84 (\(a\leq{b}\leq{c}\leq{d}\leq{e}\leq{f}\leq{g}\)).

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Hope it helps.

They're telling us that:\(L = 4S + 14\), and they want us to maximize L, thus we plug in options to solve for S such that we maximize S (this maximizes L). So our answer is the option that gives us: \(S* = \frac{(L -14)}{4}\), where * stands for the maximum value of S given the options

The answer needs to be divisible by four after subtraction with 14. The only of the options that are capable of this are B and D (104/4 = 26 and 120/4 = 30). A is obviously not a viable option since the middle value is 84.

Of course, they're asking for "LONGEST possible value" so naturally you take the option that has the highest value of the two: So our answer is D.

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Thanks Bunnel..
Could you please elaborate more about WHY "We need to maximize g. Now, to maximize g, we need to minimize all other terms." I just want to understand how does it work.

Because the question asks to find the maximum possible length of the longest piece of rope, which we denoted as g.

to minimize y we took 84, we cud have valued all the rope lengths to our minimum=a.

We can save a little bit of time by cancelling out the "7" before multiplying the mean by it:

(a+a+a+84+84+84+(14+4a))/7=68

=> (7a + 266)/7=68
=> a + 38 = 68
=> a=30
=> g=4(30)+14=134

pls help me here! Why cant we just take all the values as 84 itself.

We are told that the average length is 64, a median length is 84 centimeters and the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope. How can all the pieces be 84 centimeters from this?

Complete solution is here: seven-pieces-of-rope-have-an-average-arithmetic-mean-lengt-144452.html#p1159013

Also, check similar questions here: seven-pieces-of-rope-have-an-average-arithmetic-mean-lengt-144452.html#p1211023

Hope this helps.

I tried to solve this question by this approach but I got wrong answer choice..Can you please explain why I cannot use this approach? why I am wrong here..

Mean = 1st and last term/2 (smallest x + largest y/2)
Given Mean = x+y/2 = 68
x+y = 136
now replace y= 4x+14
5x+14= 136
x= 24.5
which gives Ans choice E.....
PLS explain...

The lengths of the pieces of the rope does not form an evenly spaced set to use (mean)=(first+last)/2.

Does this make sense?

It's a little confusing....
Since we have the equation \(g=4a+14\), to maximize g, we also need to maximize a... isn't this the case?

No. Because we have fixed total length of the rope: 7*68 centimeters. If you increase a, you'd be decreasing g.

For the solution we don't even need an arithmetic mean. The median has a property that (first number + last number)/2 = median (in case of odd number of values) --> we have smallest number=x and largest number = 4x + 14

(4x+14+x)/2=84 --> x ≈ 30 --> 4x+14=134 (D)

Im unable to understand that how minimizing all other terms would maximize value of g, can you please explain?

Answer this: the sum of two positive integers is 10. What is the maximum possible value of the largest of the integers?

We need to first recognize that we are working with a maximum problem. This means that 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 doing so will 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 we can determine the sum.

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. Thus when we arrange the pieces of rope from least length to greatest, the middle length (the 4th piece) 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. Thus, pieces 5 and 6 are either equal to the median, or they are greater than the median. 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. So we can say:

m = 14 + 4x

We can now plug 14 + 4x in for m into the equation 3x + m = 224. So we have:

3x + 14 + 4x = 224

7x = 210

x = 30

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

Answer is D.