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# A wire that weighs 20 pounds is cut into two pieces so that one of the

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Math Expert
Joined: 02 Sep 2009
Posts: 54376
A wire that weighs 20 pounds is cut into two pieces so that one of the  [#permalink]

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30 Jun 2017, 03:28
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Difficulty:

55% (hard)

Question Stats:

65% (02:06) correct 35% (01:59) wrong based on 63 sessions

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A wire that weighs 20 pounds is cut into two pieces so that one of the pieces weighs 16 pounds and is 36 feet long. If the weight of each piece is directly proportional to the square of its length, how many feet long is the other piece of wire?

(A) 9
(B) 12
(C) 18
(D) 24
(E) 27

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Re: A wire that weighs 20 pounds is cut into two pieces so that one of the  [#permalink]

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30 Jun 2017, 03:57
Bunuel wrote:
A wire that weighs 20 pounds is cut into two pieces so that one of the pieces weighs 16 pounds and is 36 feet long. If the weight of each piece is directly proportional to the square of its length, how many feet long is the other piece of wire?

(A) 9
(B) 12
(C) 18
(D) 24
(E) 27

Weight of 2nd piece = 4 pound

Since the weight is directly proportional to the square of its length., we may write
$$\frac{16}{36^2}$$ = $$\frac{4}{x^2}$$

Solving above, we get x = 18

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Re: A wire that weighs 20 pounds is cut into two pieces so that one of the  [#permalink]

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30 Jun 2017, 04:02
Bunuel wrote:
A wire that weighs 20 pounds is cut into two pieces so that one of the pieces weighs 16 pounds and is 36 feet long. If the weight of each piece is directly proportional to the square of its length, how many feet long is the other piece of wire?

(A) 9
(B) 12
(C) 18
(D) 24
(E) 27

$$20$$ pounds wire is cut into two pieces = $$16$$ pounds $$+$$ $$4$$ pounds

The piece weighs $$16$$ pounds is $$36$$ feet long.

Ratio of weight and length of $$16$$ pounds piece $$= \frac{16}{36^2} = \frac{4 * 4}{36 * 36} = \frac{1}{81}$$

Therefore required ratio of other part would also be $$\frac{1}{81}$$

Ratio $$= \frac{4}{x^2} =$$ $$\frac{1}{81}$$

$$x^2 = 81*4$$ $$=> x = \sqrt{81*4}$$

$$x = 9 * 2 = 18$$

Hence length of $$4$$ pounds wire $$= 18$$

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Re: A wire that weighs 20 pounds is cut into two pieces so that one of the  [#permalink]

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30 Jun 2017, 04:07
IMO (c)
Given, weight proportional to square (length)
=> w1 / w2 = square (l1/l2)
=> 16/4 = square (36/l2)
=> l2 = 18

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Re: A wire that weighs 20 pounds is cut into two pieces so that one of the  [#permalink]

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04 Jul 2017, 07:52
1
Bunuel wrote:
A wire that weighs 20 pounds is cut into two pieces so that one of the pieces weighs 16 pounds and is 36 feet long. If the weight of each piece is directly proportional to the square of its length, how many feet long is the other piece of wire?

(A) 9
(B) 12
(C) 18
(D) 24
(E) 27

Since the 20-lb wire is cut into two pieces and one of the pieces weighs 16 lbs, the other piece must weigh 4 lbs. Since the 16-lb piece has length of 36 ft and the weight of each piece is directly proportional to the square of its length, we can let x = the length in feet of the 4-lb piece and create the following proportion:

16/36^2 = 4/x^2

16x^2 = 4 * 36^2

4x^2 = 36^2

x^2 = 36^2/2^2

x^2 = 18^2

x = 18

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Re: A wire that weighs 20 pounds is cut into two pieces so that one of the   [#permalink] 04 Jul 2017, 07:52
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