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Seven pieces of rope have an average (arithmetic mean) length of 68 ce

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Seven pieces of rope have an average (arithmetic mean) length of 68 ce  [#permalink]

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New post Updated on: 23 Nov 2018, 22:03
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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?


Hey guys,

Could you please help me understand the solution to the question below ? I don't see why "The maximum value of 4a + 14 will occur when the maximum value of a is used, and this will be the case only if the shortest 3 pieces all have the same length."

ANSWER:
Let a, b, c, d, e, f, and g be the lengths, in centimeters, of the pieces of rope, listed from least to greatest. From the given information it follows that d = 84 and g = 4a + 14. Therefore, listed from least to greatest, the lengths are a, b, c, 84, e, f, and 4a + 14. The maximum value of 4a + 14 will occur when the maximum value of a is used, and this will be the case only if the shortest 3 pieces all have the same length. Therefore, listed from least to greatest, the lengths are a, a, a, 84, e, f, and 4a + 14. The maximum value for 4a + 14 will occur when e and f are as small as possible. Since e and f are to the right of the median, they must be at least 84 and so 84 is the least possible value for each of e and f. Therefore, listed from least to greatest, the lengths are a, a, a, 84, 84, 84, and 4a + 14. Since the average length is 68, it follows that
a+a+a+84+84+84+(4a+14))/7=68,
or a= 30.
Hence, the maximum length of the longest piece is (4a + 14) = [4(30) + 14] = 134 centimeters.


MODERATOR NOTE : PLEASE USE SPOILERS TO HIDE THE OE,ALSO PLEASE FOLLOW THE RULES FOR POSTING IN THE FORUM AND POST THE ANSWER OPTIONS

Originally posted by onassih on 23 Nov 2018, 13:30.
Last edited by Abhishek009 on 23 Nov 2018, 22:03, edited 1 time in total.
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Re: Seven pieces of rope have an average (arithmetic mean) length of 68 ce  [#permalink]

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New post 24 Nov 2018, 00:18
onassih wrote:
Question:
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?


Hey guys,

Could you please help me understand the solution to the question below ? I don't see why "The maximum value of 4a + 14 will occur when the maximum value of a is used, and this will be the case only if the shortest 3 pieces all have the same length."

ANSWER:
Let a, b, c, d, e, f, and g be the lengths, in centimeters, of the pieces of rope, listed from least to greatest. From the given information it follows that d = 84 and g = 4a + 14. Therefore, listed from least to greatest, the lengths are a, b, c, 84, e, f, and 4a + 14. The maximum value of 4a + 14 will occur when the maximum value of a is used, and this will be the case only if the shortest 3 pieces all have the same length. Therefore, listed from least to greatest, the lengths are a, a, a, 84, e, f, and 4a + 14. The maximum value for 4a + 14 will occur when e and f are as small as possible. Since e and f are to the right of the median, they must be at least 84 and so 84 is the least possible value for each of e and f. Therefore, listed from least to greatest, the lengths are a, a, a, 84, 84, 84, and 4a + 14. Since the average length is 68, it follows that
a+a+a+84+84+84+(4a+14))/7=68,
or a= 30.
Hence, the maximum length of the longest piece is (4a + 14) = [4(30) + 14] = 134 centimeters.


MODERATOR NOTE : PLEASE USE SPOILERS TO HIDE THE OE,ALSO PLEASE FOLLOW THE RULES FOR POSTING IN THE FORUM AND POST THE ANSWER OPTIONS


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Re: Seven pieces of rope have an average (arithmetic mean) length of 68 ce &nbs [#permalink] 24 Nov 2018, 00:18
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