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# A rod 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: 58340
A rod that weighs 20 pounds is cut into two pieces so that one of the  [#permalink]

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25 Sep 2018, 05:12
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Difficulty:

55% (hard)

Question Stats:

57% (01:41) correct 43% (01:44) wrong based on 94 sessions

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A rod 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 rod?

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

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

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25 Sep 2018, 05:33
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Bunuel wrote:
A rod 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 rod?

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

$$Weight = k*(Length)^2$$
16 = k (36)^2
i.e. k = 16/1296

Remaining weight = 20-6 = 4

Now, $$4 = (16/1296)*L^2$$
i.e. $$L^2 = (4*1296)/16$$

i.e. $$L = (2*36)/4$$

i.e. $$L = 18$$

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

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17 Jun 2019, 18:09
Bunuel wrote:
A rod 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 rod?

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

We have to find x,

16/4 = (36/x)^2

x= 18
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Re: A rod that weighs 20 pounds is cut into two pieces so that one of the  [#permalink]

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17 Jun 2019, 18:33
Given:
Weight of first-rod piece = 16 pounds, Length of first-rod piece = 36 feet
Weight of each piece (W) ∝ L^2
Therefore, W = K.L^2 (K is a constant)
and, K = W/L^2 = 16/36^2

The Weight of the 2nd piece = 4 pounds
Therefore, 4 = K.L^2 = 16/36^2.L^2
L^2 = 9*36
L=3*6 = 18

Therefore, Length of the second-rod piece = 18 feet
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Re: A rod that weighs 20 pounds is cut into two pieces so that one of the  [#permalink]

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01 Oct 2019, 08:30
Bunuel wrote:
A rod 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 rod?

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

given
weight = k*(l)^2
16=k*36^2
k=1/81
weight of other piece = 4
4=1/81*L^2
√4*81 =L
L=18
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Re: A rod that weighs 20 pounds is cut into two pieces so that one of the  [#permalink]

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07 Oct 2019, 20:02
Bunuel wrote:
A rod 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 rod?

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

We can create the proportion:

16/36^2 = 4/n^2

16n^2 = 4 x 36^2

4n^2 = 36^2

n^2 = (36 x 36)/4

n^2 = 9 x 36

n = 3 x 6

n = 18

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Re: A rod that weighs 20 pounds is cut into two pieces so that one of the   [#permalink] 07 Oct 2019, 20:02
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