GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 17 Aug 2019, 07:55

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Manager
Manager
User avatar
Joined: 19 Aug 2010
Posts: 61
The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp  [#permalink]

Show Tags

New post 19 Feb 2011, 11:35
4
29
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

69% (02:26) correct 31% (02:37) wrong based on 226 sessions

HideShow timer Statistics

The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points respectively on them. How many triangles can be formed using these points as vertices.

A. 200
B. 205
C. 400
D. 410
E. 415
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 57022
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp  [#permalink]

Show Tags

New post 19 Feb 2011, 11:57
10
9
Most Helpful Community Reply
Retired Thread Master
User avatar
Joined: 27 Jan 2010
Posts: 137
Concentration: Strategy, Other
WE: Business Development (Consulting)
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp  [#permalink]

Show Tags

New post 25 Feb 2011, 10:55
5
1
medanova wrote:
The sides BC,CA,AB of triangle ABC have 3,4,5 interior points respectively on them. How many triangles can be formed using these points as vertices.
a.200
b.205
c.400
d.410


Using 1 point from each side of the triangle -
3C1 * 4C1 * 5C1 = 3 * 4 * 5 = 60

Using 1 point on 1 side and 2 points on the other side -
(3C1 * 4C2) + (3C1 * 5C2) + (4C1 * 3C2) + (4C1 * 5C2) + (5C1 * 4C2) + (5C1 * 3C2) = 145

Adding these we get = 60 + 145 = 205.

I was a little bit confused at first - I was not sure if we could use 2 of the interior points and 1 of the vertices of the existing triangle ABC (as the 3rd vertice) to form a new triangle. I needed to look at the question this way - How many triangles can be formed using only these points as vertices.
General Discussion
Manager
Manager
User avatar
Joined: 19 Aug 2010
Posts: 61
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp  [#permalink]

Show Tags

New post 19 Feb 2011, 12:01
Thank you ,

I initially included also the points A B C in the calculation but now I read the question again and found my mistake.
Manager
Manager
avatar
Joined: 26 Sep 2010
Posts: 123
Nationality: Indian
Concentration: Entrepreneurship, General Management
GMAT ToolKit User Reviews Badge
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp  [#permalink]

Show Tags

New post 25 Feb 2011, 23:20
Thanks for the question and great explanations.
_________________
You have to have a darkness...for the dawn to come.
Manager
Manager
User avatar
Joined: 19 Aug 2010
Posts: 61
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp  [#permalink]

Show Tags

New post 26 Feb 2011, 08:34
Thanks also from me.
Manager
Manager
User avatar
Joined: 20 Jul 2012
Posts: 115
Location: India
WE: Information Technology (Computer Software)
GMAT ToolKit User
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp  [#permalink]

Show Tags

New post 10 Sep 2013, 10:57
medanova wrote:
The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points respectively on them. How many triangles can be formed using these points as vertices.

A. 200
B. 205
C. 400
D. 410

[1*3*4]*5=60- when all the three points are on different sides..
[1*3+1*6]*5 +[1*10+1*3]*4+[1*10+1*6]*3=45+52+48-when two points on one side
total=60+45+52+48=205
Intern
Intern
User avatar
Joined: 17 Oct 2015
Posts: 17
Concentration: Technology, Leadership
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp  [#permalink]

Show Tags

New post 09 Jan 2016, 06:35
Bunuel, is there a easier way to do the math, instead of multiplying 12 x 11 x 10 x 9 x 8 x 7.....?

As far I understood your solution, I would need to do that at the C 3 12 combination.

Regards and have a great year!

Bunuel wrote:
Bumping for review and further discussion.
Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 7756
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp  [#permalink]

Show Tags

New post 09 Jan 2016, 06:51
4
mestrec wrote:
Bunuel, is there a easier way to do the math, instead of multiplying 12 x 11 x 10 x 9 x 8 x 7.....?

As far I understood your solution, I would need to do that at the C 3 12 combination.

Regards and have a great year!

Bunuel wrote:
Bumping for review and further discussion.


Hi,
Iwillhelp you out on that ..
the Q basically says there are three lines with 3,4,5 points respectively..
lets find ways triangle can be formed..
triangle requires three points,..
1) 2 points from line containing 3 points and one from remaining 4+5 points=3C2*(4+5)=27..
2) 2 points from line containing 4 points and one from remaining 3+5 points=4C2*(3+5)=48..
3) 2 points from line containing 5 points and one from remaining 4+3 points=5C2*(4+3)=70..
4) 1 point from each line containing 3,4,5 points =3*4*5=60..
total 27+48+70+60=205..


also you do not have to do the entire calculation in 12C3..
12C3=12!/3!9!= 12*11*10*9!/3!9!=12*11*10/3*2...
so you see you just have to multiply 12,11,and 10..
_________________
Board of Directors
User avatar
P
Joined: 17 Jul 2014
Posts: 2531
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
GMAT ToolKit User Reviews Badge
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp  [#permalink]

Show Tags

New post 06 Mar 2016, 19:56
very nice question...tried to analyze it first..then did not how to solve it..went with chetan2u 's method..
we can have 1 point on from 3, 1 point from 4, 1 point from 5: 3C1*4C1*5C1 = 60
we can have 1 point from 3, and 2 points from 5 = 3C1*5C2 = 30
we can have 1 point from 3, and 2 points from 4 = 3C1*4C2 = 18
we can have 1 point from 4, and 2 points from 3 = 4C1*3C2 = 12
we can have 1 point from 4, and 2 points from 5 = 4C1*5C2 = 40
we can have 1 point from 5, and 2 points from 3 = 5C1*3C2 = 15
we can have 1 point from 5, and 2 points from 4 = 5C1*4C2 = 30

now total: 60+30+18+12+40+15+30 = 205
_________________
Image
Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 7756
Re: The sides AB, BC & CA of a triangle ABC have 3, 4 and 5 interior point  [#permalink]

Show Tags

New post 11 May 2016, 09:55
1
2
techiesam wrote:
The sides AB, BC & CA of a triangle ABC have 3, 4 and 5 interior points respectively on them. Find the number of triangles that can be constructed using these points as vertices.

A. 220.
B. 205.
C.250
D.105.
E.225


hi,

different ways..
1) EASIEST approach..
lets see how many total can be made if none of the points were colliner that is NO three were in one line = 12C3 = 12!/9!3! = 220...
Due to collinear, extra triangles considered above = 3C3 + 5C3 +4C3 = 1+10+4 = 15..
total = 220-15 = 205

2) SECOND method
a) ONE point from each.
b) two points from one side and third from any of two other sides
_________________
Manager
Manager
avatar
Joined: 09 Jul 2013
Posts: 109
Re: The sides AB, BC & CA of a triangle ABC have 3, 4 and 5 interior point  [#permalink]

Show Tags

New post 11 May 2016, 12:14
2
To expand on Chetan4u's second method, each triangle can be formed with one point from each of the 3 lines AB (3 points), BC (4 points) and CA (5 points), or with two points from one line, and the third point from either of the other two lines.

The calculations look like this:
Scenario 1) One point from each line = \(3*4*5 = 60\)
Scenario 2) Two points from one line and one from the other two lines = \(3c2*9 + 4c2*8 + 5c2*7 = 27+48+70=145\)

(Quick explanation here in case I just lost some of you - Taking 2 points from the line AB, which has 3 points on it can be done in 3c2 ways - choosing two points from the 3 available on AB. Then we can choose the third point from any of the remaining points on the other two lines BC or CA, which have a total of 9 points among them. So the number of triangles that can be formed with two points on line AB is \(3c2*9 = 27\). Similarly for the other two lines.)

Add up the two scenarios: \(60+145 = 205\)

Answer: B
_________________
Dave de Koos
GMAT aficionado
Manager
Manager
User avatar
S
Joined: 01 Jan 2018
Posts: 78
CAT Tests
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp  [#permalink]

Show Tags

New post 04 Aug 2018, 21:42
Hi, Can someone please clear my doubt- If we are joining three points ..one point from each side of original triangle then 4 triangles formed. Why we dont consider those triangles as well?

Sent from my Redmi Note 3 using GMAT Club Forum mobile app
_________________
+1 Kudos if you find this post helpful
Manager
Manager
User avatar
G
Joined: 19 Nov 2017
Posts: 190
Location: India
Schools: ISB
GMAT 1: 670 Q49 V32
GPA: 4
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp  [#permalink]

Show Tags

New post 04 Aug 2018, 22:41
dvishal387 wrote:
Hi, Can someone please clear my doubt- If we are joining three points ..one point from each side of original triangle then 4 triangles formed. Why we dont consider those triangles as well?

Sent from my Redmi Note 3 using GMAT Club Forum mobile app


The triangles need to be formed from the interior points i.e. treating the interior points as vertices. This is mentioned in the question, just go back and have a look.
_________________
Regards,

Vaibhav



Sky is the limit. 800 is the limit.

~GMAC
Manager
Manager
User avatar
S
Joined: 01 Jan 2018
Posts: 78
CAT Tests
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp  [#permalink]

Show Tags

New post 04 Aug 2018, 23:45
vaibhav1221 wrote:
dvishal387 wrote:
Hi, Can someone please clear my doubt- If we are joining three points ..one point from each side of original triangle then 4 triangles formed. Why we dont consider those triangles as well?

Sent from my Redmi Note 3 using GMAT Club Forum mobile app


The triangles need to be formed from the interior points i.e. treating the interior points as vertices. This is mentioned in the question, just go back and have a look.
Okay..thanks

Sent from my Redmi Note 3 using GMAT Club Forum mobile app
_________________
+1 Kudos if you find this post helpful
Manager
Manager
avatar
B
Joined: 20 Apr 2019
Posts: 81
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp  [#permalink]

Show Tags

New post 10 Aug 2019, 02:15
Bunuel wrote:
medanova wrote:
The sides BC,CA,AB of triangle ABC have 3,4,5 interior points respectively on them. How many triangles can be formed using these points as vertices.
a.200
b.205
c.400
d.410


Total points: 3+4+5=12;

Any 3 points out of 12 but those which are collinear will form a triangle, so \(C^3_{12}-(C^3_{3}+C^3_{4}+C^3_{5})=220-(1+4+10)=205\).

Answer: B.

We substract (C^3_{3}+C^3_{4}+C^3_{5}) from C^3_{12} to exclude the triangles that would be possible by using the points from the same sides, correct?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 57022
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp  [#permalink]

Show Tags

New post 10 Aug 2019, 02:31
Luca1111111111111 wrote:
Bunuel wrote:
medanova wrote:
The sides BC,CA,AB of triangle ABC have 3,4,5 interior points respectively on them. How many triangles can be formed using these points as vertices.
a.200
b.205
c.400
d.410


Total points: 3+4+5=12;

Any 3 points out of 12 but those which are collinear will form a triangle, so \(C^3_{12}-(C^3_{3}+C^3_{4}+C^3_{5})=220-(1+4+10)=205\).

Answer: B.

We substract (C^3_{3}+C^3_{4}+C^3_{5}) from C^3_{12} to exclude the triangles that would be possible by using the points from the same sides, correct?


We subtract those combinations of points which are on a straight line (collinear) and thus do not form a triangle.
_________________
GMAT Club Bot
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp   [#permalink] 10 Aug 2019, 02:31
Display posts from previous: Sort by

The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne