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A ladder 25 feet long is leaning against a wall that is

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A ladder 25 feet long is leaning against a wall that is  [#permalink]

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New post 07 Apr 2012, 20:44
4
8
00:00
A
B
C
D
E

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  45% (medium)

Question Stats:

72% (02:18) correct 28% (02:35) wrong based on 332 sessions

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A ladder 25 feet long is leaning against a wall that is perpendicular to level ground. The bottom of the ladder is 7 feet from the base of the wall. If the top of the ladder slips down 4 feet, how many feet will the bottom of the ladder slip?

(A) 4
(B) 5
(C) 8
(D) 9
(E) 15

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Re: Ladder Issues  [#permalink]

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New post 07 Apr 2012, 22:18
25^2-7^2=576
it means that the height is equal to 24.

since the top of the ladder slips down 4 feet, then the height of the wall =24-4=20

the bottom =sqrt(25^2-20^)=sqrt(625-400)=15

ans is E , not C
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Re: Ladder Issues  [#permalink]

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New post 07 Apr 2012, 22:26
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LalaB wrote:
25^2-7^2=576
it means that the height is equal to 24.

since the top of the ladder slips down 4 feet, then the height of the wall =24-4=20

the bottom =sqrt(25^2-20^)=sqrt(625-400)=15

ans is E , not C


New,total distance between foot ladder and the wall = 15
slip = 15-7 = 8 hence C
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Re: A ladder 25 feet long is leaning against a wall that is  [#permalink]

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New post 08 Apr 2012, 00:51
1
boomtangboy wrote:
A ladder 25 feet long is leaning against a wall that is perpendicular to level ground. The bottom of the ladder is 7 feet from the base of the wall. If the top of the ladder slips down 4 feet, how many feet will the bottom of the ladder slip?

(A) 4
(B) 5
(C) 8
(D) 9
(E) 15


We have a right triangle with the hypotenuse of 25 feet and the base of 7 feet, hence its the height is \(\sqrt{25^2-7^2}=24\) feet;

Since the top of the ladder slips down 4 feet then the new height becomes 24-2=20 feet and the new base becomes \(\sqrt{25^2-20^2}=15\) feet;

Question basically asks about the difference between the old base and the new base, which is 15-7=8 feet.

Answer: C.
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Re: A ladder 25 feet long is leaning against a wall that is  [#permalink]

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New post 20 May 2014, 03:26
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Initially, base (Point E) of 25 feet long ladder (AE) is 7 meter away from the base of wall (Point B).

height of the top point (A) can be found using Pythagoras Theorem

AB = sqrt(AE^2-BE^2)
= sqrt (25^2-7^2) = sqrt (625-49) = sqrt(576) =24

When the top of the ladder slips 4 meter, height of top point reduces by 4 meter.
Now, as per diagram, DC is ladder and BD is height of top point and BC is the distance of base of ladder from wall.

we have , DC = 25, BD = 24-4 = 20,

By Pythagoras theorem, BC = sqrt (DC^2-BD^2) = sqrt (25^2-20^2) = sqrt(625-400)=sqrt (225) = 15

base point has slipped by a distance EC which can be found by subtracting BE from BC

EC = BC-BE = 15-7 = 8 .
Hence option C is the answer.
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Re: A ladder 25 feet long is leaning against a wall that is  [#permalink]

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New post 27 Nov 2016, 13:40
Ladder Before Movement
-------------
25^2 = 7^2 + x^2
625=49+x^2
x^2=24

Ladder After Movement
--------------------
25^2=20^2+x^2
625=400+x^2
225=x^2
x=15

Ladder moved 15-7 = 8 feet

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Re: A ladder 25 feet long is leaning against a wall that is  [#permalink]

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New post 28 Mar 2018, 09:07
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Re: A ladder 25 feet long is leaning against a wall that is &nbs [#permalink] 28 Mar 2018, 09:07
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