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An irregular plot of land is surrounded by four straight segments of

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New post 11 Nov 2019, 03:56
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An irregular plot of land is surrounded by four straight segments of fence of meter lengths (5x + 5), (5x + 5), (3x + 10) and 2x. Is the total length of fence greater than 70 meters?

(1) The average (arithmetic mean) fence segment length is less than 25 meters.

(2) The largest fence segment is three times the length of the smallest fence segment.


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An irregular plot of land is surrounded by four straight segments of  [#permalink]

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New post 13 Nov 2019, 00:22
Bunuel wrote:
An irregular plot of land is surrounded by four straight segments of fence of meter lengths (5x + 5), (5x + 5), (3x + 10) and 2x. Is the total length of fence greater than 70 meters?

(1) The average (arithmetic mean) fence segment length is less than 25 meters.

(2) The largest fence segment is three times the length of the smallest fence segment.


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Total length =\((5x + 5)+ (5x + 5)+ (3x + 10)+ 2x=15x+20\)
So question is .. Is \(15x+20>{70}.....15x>{50}........x>{\frac{10}{3}}\)

1) The average (arithmetic mean) fence segment length is less than 25 meters.
\(\frac{15x+20}{4}\leq{25}.........15x+20\leq{100}........15x\leq{80}.......x\leq{\frac{16}{3}}\)
x=14/3..Yes
x=8/3...No

(2) The largest fence segment is three times the length of the smallest fence segment.
Since we do not know x, we cannot say if 5x+5 is greater or 3x+10, but 2x is surely the smallest
(a) 5x+5 is the greatest..
\(5x+5=3(2x)........x=5>\frac{10}{3}\)..Yes
(b) 3x+10 is the greatest..
\(3x+10=3(2x)........x=\frac{10}{3}\)..No
when \(x=\frac{10}{3}...3x+10=10+10=20....5x+5=\frac{50}{3}+5=\frac{65}{3}=21.66\)
So 5x+5 is always the greatest, and this option can be disregarded.
So x=5, and our answer is YES

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