Kaplan Practice Test Q- 19

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priya_kapoor: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Wed Dec 16, 2009 8:04 pm; Location: New York, NY

Kaplan Practice Test Q- 19

by priya_kapoor » Sat Aug 07, 2010 4:40 pm

Hi,

So here is the question and the attached image for a reference. Here is my problem: I see the answer but when I tried to put in each answer into each triangle, I couldn't figure out the other triangle's lengths by doing the work. I guess what I would like to see if this wasn't a data sufficiency question, how would it be solved? Any and all help is welcomed!

In the figure above (or attached), line segment JK and LM represent two positions of the same board learning against the side of KN of a wall. The length of KN is how much greater than the length of MN?

1) The length of LN is square root of 2
2) The length of JN is 1
[/img]

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: Question from Kaplan Exam

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Source: Beat The GMAT — Data Sufficiency | Sat Aug 07, 2010 4:40 pm

selango: Legendary Member; Posts: 1460; Joined: Tue Dec 29, 2009 1:28 am; Thanked: 135 times; Followed by:7 members

by selango » Sat Aug 07, 2010 9:21 pm

LM=KJ[2 positions of same board]

stmt1,

LN=sqrt(2)

In triangle LMN,the sides are in ratio 1:1:sqrt(2)[45:45:90 traingle]

-->LN:MN:LM=1:1:sqrt(2)=sqrt(2):sqrt(2):2

LM=2

-->KJ=2

In traingle KJN,

KJ=2

JN:KN:KJ=1:sqrt(3):2[30:60:90 traingle]

-->JN:KN:KJ=1:sqrt(3):2[as KJ=2]

Now we got MN=sqrt(2) and KN=sqrt(3)

From this we can find the difference.

From option B we can find the lengths in the same way.

If this is DS question Pick D

Hope this clarify.

Last edited by selango on Sun Aug 08, 2010 12:56 am, edited 1 time in total.

--Anand--

Quote

kvcpk: Legendary Member; Posts: 1893; Joined: Sun May 30, 2010 11:48 pm; Thanked: 215 times; Followed by:7 members

by kvcpk » Sun Aug 08, 2010 12:52 am

I believe answer should be D.

Given an angle(not right angle) and a side length in right triangle, we can calculate other sides and angles.

stmt 1: LN is known
From Triangle LMN, MN can be calculated.
LM can also be calculated.

Now, LM = JK.
Hence from triangle JKN, KN can be calculated.
SUFF

stmt2: JN is known
From triangle JKN. JK can be found. and also KN can be found.

JK = LM.
Hence from triangle LMN, MN can be found.
SUFF

pick D

Hope this helps!!

Quote

kvcpk: Legendary Member; Posts: 1893; Joined: Sun May 30, 2010 11:48 pm; Thanked: 215 times; Followed by:7 members

by kvcpk » Sun Aug 08, 2010 12:52 am

selango wrote:LM=KJ[2 positions of same board]

stmt1,

LN=sqrt(2)

In triangle LMN,the sides are in ratio 1:1:sqrt(2)[45:45:90 traingle]

-->LN:MN:LM=1:1:sqrt(2)=sqrt(2):sqrt(2):2

LM=2

-->KJ=2

In traingle KJN,

KJ=2

JN:KN:KJ=1:sqrt(3):2[30:60:90 traingle]

-->JN:KN:KJ=1:sqrt(3):2[as KJ=2]

Now we got MN=sqrt(2) and KN=sqrt(3)

From this we can find the difference.

If this is DS question Pick A

Hope this clarify.

Why didnt you consider stmt2?

Quote

selango: Legendary Member; Posts: 1460; Joined: Tue Dec 29, 2009 1:28 am; Thanked: 135 times; Followed by:7 members

by selango » Sun Aug 08, 2010 12:58 am

Yes praveen Its D only..I typed it by mistake.

Now edited it..

--Anand--

Quote

KapTeacherEli: GMAT Instructor; Posts: 578; Joined: Tue Aug 25, 2009 6:00 pm; Thanked: 136 times; Followed by:62 members

by KapTeacherEli » Wed Aug 11, 2010 12:06 pm

priya_kapoor wrote: I guess what I would like to see if this wasn't a data sufficiency question, how would it be solved? Any and all help is welcomed!

Hi Priya,

GLad you saw the answer. Of course, on Test-day, we'd mark D and move on--if we have all the angles and at least one side of a triangle, that's always sufficient to answer for all the sides, and if we can find all the sides of one triangle then we have found the length of the board.

That being said, you're right--this type of problem could have shown up in the problem section. If it did, here's how we'd solve it:

The key deduction here is that we have two special right triangles. When the board is in the first position, it forms a 45 degree angle with the 90 degree wall--that means the last angle is 180 - (90 + 45) = 45 degrees. So, we have a right isosceles triangle, and the sides are always in the ratio of X:X:X root 2. Since one of the short sides is root 2, we know the other short side has the same measure and the long side (the board) is 2.

Similarly, when the board is in the second position, it forms a 30/60/90 triangle. This always has sides in the ratio of X : X root 3 : 2X. Given the shortest side has a measure of 1, the longest side, the board, has a measure of 2 (agreeing with our previous calculations!) and the middle side has a measure of root 3.

Thus, KN - MN = root 3 - root 2.

Eli Meyer
Kaplan GMAT Teacher
Cambridge, MA
www.kaptest.com/gmat