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

Redeem

Solve this

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 194
Joined: Mon Oct 15, 2012 7:14 pm
Location: India
Thanked: 47 times
Followed by:6 members

Solve this

by The Iceman » Thu Dec 06, 2012 1:34 am
In a triangle ABC, D is a point on the side BC and M, N are length of perpendicular dropped on line AD from the vertices B and C respectively. Is M > N?

(I) AB > AC

(II) BD < DC
Top
Source: — Data Sufficiency |
User avatar
Master | Next Rank: 500 Posts
Posts: 145
Joined: Tue Jan 31, 2012 6:50 am
Location: New Delhi
Thanked: 16 times
Followed by:2 members
GMAT Score:760

by nisagl750 » Thu Dec 06, 2012 5:14 am
The Iceman wrote:In a triangle ABC, D is a point on the side BC and M, N are length of perpendicular dropped on line AD from the vertices B and C respectively. Is M > N?

(I) AB > AC

(II) BD < DC
IMO C

I have one doubt though.
Since D is a point on side BC of the triangle, D will be somewhere between B & C
i.e. B & C will lie on either side of D (one to the left and one to the right.

If AD is perpendicular to BC, It is not possible to draw perpendiculars on AD from points B & C.
If ADC is an acute angle, It is not possible to draw a perpendicular on it from point B.
If ADC is an obtuse angle, It is not possible to draw a perpendicular on it from point C.

My question is , How can we have both M & N simultaneously if it is not possible to draw perpendiculars to AD from both B & C simultaneously?

Am I missing something?

Experts, Please help!!!
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 145
Joined: Tue Jan 31, 2012 6:50 am
Location: New Delhi
Thanked: 16 times
Followed by:2 members
GMAT Score:760

by nisagl750 » Thu Dec 06, 2012 5:45 am
nisagl750 wrote:
The Iceman wrote:In a triangle ABC, D is a point on the side BC and M, N are length of perpendicular dropped on line AD from the vertices B and C respectively. Is M > N?

(I) AB > AC

(II) BD < DC

If AD is perpendicular to BC, It is not possible to draw perpendiculars on AD from points B & C.

Am I missing something?
I guess I got my answer...
AD is INDEED the perpendicular to BC....This is the only possible way

So the question reduces to Is BD < CD? (a Yes/No question)

Statement 2 clearly answers the question.... Sufficient
Statement 1: Since AD is perpendicular to BC,
If AB>AC, It clearly means BD>CD
Sufficient

Answer is D

What is the OA?
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 194
Joined: Mon Oct 15, 2012 7:14 pm
Location: India
Thanked: 47 times
Followed by:6 members

by The Iceman » Thu Dec 06, 2012 8:38 am
nisagl750 wrote: I guess I got my answer...
AD is INDEED the perpendicular to BC....This is the only possible way
Think again. It indeed is possible even though AD were not a perpendicular to BC.
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 145
Joined: Tue Jan 31, 2012 6:50 am
Location: New Delhi
Thanked: 16 times
Followed by:2 members
GMAT Score:760

by nisagl750 » Thu Dec 06, 2012 8:48 am
The Iceman wrote:
Think again. It indeed is possible even though AD were not a perpendicular to BC.
How..?

My reasoning is...

Let AD is not Perpendicular to BC. then D will make an angle other than 90 degree..
So Either one of the angle ADC or angle ADB will be acute and the other will be obtuse.
Let BE & CF be the perpendicular to AD, BE=M, CF=N (according to question)

Case1:
Angle ADC is acute.
In this case Angle ADB will be obtuse....
In triangle BDE
Angle EDB is obtuse (E is on AD)
Angle BED is 90 degree....
Which is not possible (sum exceeds 180 and a triangle can have at most one right angle)

Case2:
Angle ADB is acute
In this case Angle ADC will be obtuse....
Same explanation as above for tiangle CDF

What am I missing?
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 194
Joined: Mon Oct 15, 2012 7:14 pm
Location: India
Thanked: 47 times
Followed by:6 members

by The Iceman » Thu Dec 06, 2012 9:03 am
nisagl750 wrote: How..?

What am I missing?
Even if <ADB is obtused, a perpendicular can be dropped from B to AD produced to E. In this case BE is the perpendicular and E lies outside the triangle.
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 145
Joined: Tue Jan 31, 2012 6:50 am
Location: New Delhi
Thanked: 16 times
Followed by:2 members
GMAT Score:760

by nisagl750 » Thu Dec 06, 2012 9:11 am
The Iceman wrote:
Even if <ADB is obtused, a perpendicular can be dropped from B to AD produced to E. In this case BE is the perpendicular and E lies outside the triangle.
Thats called thinking outside the Triangle.....

So, what is the OA?
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 194
Joined: Mon Oct 15, 2012 7:14 pm
Location: India
Thanked: 47 times
Followed by:6 members

by The Iceman » Thu Dec 06, 2012 9:51 am
nisagl750 wrote:
The Iceman wrote:
Even if <ADB is obtused, a perpendicular can be dropped from B to AD produced to E. In this case BE is the perpendicular and E lies outside the triangle.
Thats called thinking outside the Triangle.....

So, what is the OA?
OA is B.
Top
Senior | Next Rank: 100 Posts
Posts: 63
Joined: Tue Apr 06, 2010 10:01 pm
Thanked: 4 times
Followed by:1 members

by prat_agl » Sat Dec 08, 2012 1:46 pm
Can some one please explain how the OA is B?
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 194
Joined: Mon Oct 15, 2012 7:14 pm
Location: India
Thanked: 47 times
Followed by:6 members

by The Iceman » Tue Dec 11, 2012 5:02 am
prat_agl wrote:Can some one please explain how the OA is B?
Let D' be the midpoint of BC and let X' and Y' be the feet of the perpendicular from B and C to AD' respectively => X' = Y'. As D' shifts to right or left, we can know which of M or N is bigger. Thus, statement II answers

Statement I doesn't tell us anything

=> choice (B) is the right answer
Top
Post new topic Post Reply
• Page 1 of 1