BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Triangles

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Legendary Member
Posts: 966
Joined: Sat Jan 02, 2010 8:06 am
Thanked: 230 times
Followed by:21 members

Triangles

by shankar.ashwin » Sat Jan 07, 2012 9:05 pm
A 25 ft long ladder is placed against the wall with its base 7 ft from the wall. The base of the ladder is drawn out so that the top comes down by half the distance that the base is drawn out. This distance is in the range:

A (5,8)
B (9,10)
C (2,7)
D (3,7)
E None of the above

[spoiler]Don't have a OA and don't exactly get what the questions asks.[/spoiler]

COuld someone help..Tough one I assume
Top
Source: — Problem Solving |
Legendary Member
Posts: 1084
Joined: Fri Apr 15, 2011 2:33 pm
Thanked: 158 times
Followed by:21 members

by pemdas » Sun Jan 08, 2012 12:26 am
Here we need to find within which range of values the distance of ladder jacked out (not 25 ft, but jacked out distance) is placed. The jacked out ladder's top has height of half the distance this ladder is jacked out.

The geometric figure will be a right triangle with the sides: base 7, height x/2, ladder's length x (jacked out). We should solve for x, and there's proportion such as 1:sqrt(3):2 (30-60-90 degrees). Hence, 7=sqrt(3)*(x/2) and 49=(3x^2)/4, 3x^2=196, x^2=~65, x is closer to 8 but is slightly greater, some 8.1.

let's see which answer choice contains 8.1

A (5,8) no
B (9,10) no
C (2,7) no
D (3,7) no
E None of the above :) Yes

e

[spoiler]on the second read after solving, i guess wording here is abra-cadabra (confusing), mainly relied on common sense about the object in q.
[/spoiler]
shankar.ashwin wrote:A 25 ft long ladder is placed against the wall with its base 7 ft from the wall. The base of the ladder is drawn out so that the top comes down by half the distance that the base is drawn out. This distance is in the range:

A (5,8)
B (9,10)
C (2,7)
D (3,7)
E None of the above

[spoiler]Don't have a OA and don't exactly get what the questions asks.[/spoiler]

COuld someone help..Tough one I assume
Success doesn't come overnight!
Top
User avatar
Community Manager
Posts: 1060
Joined: Fri May 13, 2011 6:46 am
Location: Utrecht, The Netherlands
Thanked: 318 times
Followed by:52 members

by neelgandham » Mon Jan 09, 2012 5:57 am
I might be wrong, but he is what I think.

Initial position : Right angled triangle ABC
7,24,25 is the pythagorean triple. So distance of the topmost point of the ladder to the ground is 24 ft.

Final position: Right angled triangle A'B'C
The base of the ladder is drawn out so that the top comes down by half the distance that the base is drawn out. So if the ladder is drawn out by x ft, then the top comes down BY 0.5x. Pythagorean theorem comes into picture now.

(24-0.5x)^2 + (7+x)^2 = 25^2
24^2 + 0.25x^2 - 24x + 49 + x^2 + 14x = 25^2
0.25x^2 - 24x x^2 + 14x
1.25x^2 - 10x = 0
1.25x(x -8) = 0
x = 8

Should be A if 8 is included in the set or E if it isn't.
Attachments
Beatthegmat_31.JPG
Anil Gandham
Welcome to BEATtheGMAT | Photography | Getting Started | BTG Community rules | MBA Watch
Check out GMAT Prep Now's online course at https://www.gmatprepnow.com/
Top
Master | Next Rank: 500 Posts
Posts: 382
Joined: Thu Mar 31, 2011 5:47 pm
Thanked: 15 times

by ArunangsuSahu » Mon Jan 09, 2012 1:28 pm
This is a problem of Rearrangement of Hypotenuse

Eqn 1: 25^2=24^2+7^2

Eqn 2: 25^2=(24-x)^2+(7+2x)^2

Find out 2x

2x = 8

This distance is in the range:

(A) should be the answer if the range here means inclusive else (E)
Top
Post new topic Post Reply
• Page 1 of 1