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A 10-foot ladder is leaning against a vertical wall. The top

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A 10-foot ladder is leaning against a vertical wall. The top [#permalink] New post   21 Sep 2018, 09:01
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A 10-foot ladder is leaning against a vertical wall. The top of the ladder touches the wall at a point 8 feet above the ground. The base of the ladder slips 1 foot away from the wall.

Quantity A
Quantity B
1
The distance the top of the ladder slides down the wall


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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pranab223
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Re: A 10-foot ladder is leaning against a vertical wall. The top [#permalink] New post   22 Sep 2018, 04:33
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Carcass wrote:
Attachment:
wall.jpg


A 10-foot ladder is leaning against a vertical wall. The top of the ladder touches the wall at a point 8 feet above the ground. The base of the ladder slips 1 foot away from the wall.

Quantity A
Quantity B
1
The distance the top of the
ladder slides down the wall


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.



1st case:

Length of the ladder = 10 foot

Distance from the ground to point the ladder touches = 8 feet

So the distance of the base of the ladder from the wall =\(\sqrt{10^2 - 8^2} = 6\)

2nd case:

Now the distance of the base slips 1 foot away from the wall i.e = 6 + 1 =7

and the length of the ladder = 10 foot

SO the distance from the base to the point the ladder touches can be calculated by Pythagoras theorem = \(\sqrt{10^2 - 7^2} = \sqrt{51}\)

Now \(\sqrt{49} < \sqrt{51} < \sqrt {64}\)

or \(7 < \sqrt{51} < 8\)

so the new value of the base lies between 7 and 8

so the distance the ladder slides down = 8 - (number greater than 7 but less than 8) = less than 1

SO \(QTYA > QTY B\)
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Re: A 10-foot ladder is leaning against a vertical wall. The top [#permalink] New post   18 Jan 2019, 16:10
By Pythagoras theorem

Ans B
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Re: A 10-foot ladder is leaning against a vertical wall. The top [#permalink] New post   18 Jan 2019, 16:10
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pranab01 wrote:

SO the distance from the base to the point the ladder touches can be calculated by prognathous theorem = \(\sqrt{10^2 - 7^2} = \sqrt{51}\)


Please fix the typo
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Re: A 10-foot ladder is leaning against a vertical wall. The top [#permalink] New post   20 Feb 2020, 16:57
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considering the original condition, a 10 ft. ladder resting on a wall with its top 8 ft. above the ground. A right-angle triangle with a hypotenuse of 10 ft. (length of ladder) and height of 8 ft. can be drawn, by applying Pythagoras theorem we can derive the base of the triangle as 6 ft.
After the ladder has shifted by 1 ft. away from the wall (as shown), we can conclude that the top of the ladder has also slid down the wall (lets assume that distance as "d"). In the new triangle, the hypotenuse remains 10 ft., the base becomes (6+1) = 7 ft. and the height will be now (8-d) ft. Again on applying the Pythagoras theorem, we can calculate d=~0.86 ft., which is less than 1.

Hence, column (A) is greater than (B). Right answer is (A)
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Re: A 10-foot ladder is leaning against a vertical wall. The top [#permalink] New post   24 Feb 2020, 02:03
Thank you
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Re: A 10-foot ladder is leaning against a vertical wall. The top [#permalink] New post   11 Aug 2021, 06:14
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A 10-foot ladder is leaning against a vertical wall. The top [#permalink]
11 Aug 2021, 06:14
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