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500 ds

This topic has 6 member replies
dunkin77 Master | Next Rank: 500 Posts Default Avatar
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Posted:
269 messages

500 ds

Post Wed Apr 04, 2007 2:00 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    Hi,

    Pls see the attached.

    I thought the answer was C) but turned out to be B).

    Can anyone pls explain? thanks!
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    jayhawk2001 Community Manager
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    Post Wed Apr 04, 2007 9:12 pm
    We have to use properties of similar triangles here. Draw lines
    from B and C to the ground. Lets call these points D and E respectively
    on the ground. BD and CE are parallel

    We are given that AB = BC.

    (2) says that BD = 5. We can use properties of similar triangles now.
    BD / CE = AB / AC. We can hence compute CE

    So B is the correct answer.

    dunkin77 Master | Next Rank: 500 Posts Default Avatar
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    Post Thu Apr 05, 2007 7:50 am
    Thanks for explanation Jay.

    I am still a bit confused cause we only know AB=AC and BD=5,
    so, BD / CE = AB / AC (let's say AB=AC=3)

    5 / CE = 3 / 3

    5/CE=1

    CE=5..........?? Crying or Very sad

    I would appreciate it if you could explain a bit more... thanks again for your help!!

    gabriel Legendary Member
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    Post Thu Apr 05, 2007 10:01 am
    dunkin77 wrote:
    Thanks for explanation Jay.

    I am still a bit confused cause we only know AB=AC and BD=5,
    so, BD / CE = AB / AC (let's say AB=AC=3)

    5 / CE = 3 / 3

    5/CE=1

    CE=5..........?? Crying or Very sad

    I would appreciate it if you could explain a bit more... thanks again for your help!!
    dunkin u have got evrything right except for the part that AB= AC ... AB is actually = BC ... therefore AC=2AB ... so we have 5/CE = AB/AC .... 5/CE = 1/2 .. therefore CE = 10 .. hence the answer is B

    vk.neni Junior | Next Rank: 30 Posts Default Avatar
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    12 messages
    Post Thu Apr 05, 2007 10:52 am
    Hi,
    Couldn't we solve this using information in (i) x=30? With this info the small triangle (ABD) becomes a 30-60-90 and we know the length of one of the sides, so we can figure the other two sides. Once, we've this info, we can use the similar triangle method to solve it. So, the answer would be D.

    myprepgmat Newbie | Next Rank: 10 Posts Default Avatar
    Joined
    10 Mar 2007
    Posted:
    2 messages
    Post Sat Apr 07, 2007 5:06 am
    Ans is B

    I'll explain...

    Let the height of point B is h and the point is H. Let the length of seesaw ie AC is 2d..

    Now draw perpendicular lines to ground from point B and C. Let they intersect at ground at D ( from Point B) & E(point C).

    Now we will get two similar triangles.. ie ABD and ACE..

    we have BD/AB = CE/AC

    ie h/d= H/2d given AB=BC we assumed AC=2d

    i.e. h=H/2

    given h=5ft

    therefore H=10ft

    case 1 doesnot give any unique value as the length of AC or height of the point B is not given.

    gabriel Legendary Member
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    Post Sat Apr 07, 2007 9:51 am
    vk.neni wrote:
    Hi,
    Couldn't we solve this using information in (i) x=30? With this info the small triangle (ABD) becomes a 30-60-90 and we know the length of one of the sides, so we can figure the other two sides. Once, we've this info, we can use the similar triangle method to solve it. So, the answer would be D.
    dude one of the most basic mistake for a DS... carrying information from one statement to the other .. the length of the side is mentioned only in the second statement.. so cant use it for the first ..

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