Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Sep 2107:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included!
Sep 2312:30 AM EDT
-01:30 AM EDT
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more
Sep 2401:30 AM EDT
-01:30 AM EDT
Tejas studied hard but kept repeating the same mistakes. After 3 attempts at 645, a focused 4-week strategy helped him identify and fix what was holding him back—leading to a 695 with just 12–15 hours of study per week.
Sep 2606:00 PM PDT
-07:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
19 Mar 2019, 21:48
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
41% (03:23) correct
59%
(03:31)
wrong
based on 34
sessions
History
Date
Time
Result
Not Attempted Yet
A large equilateral triangle is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have 3 rows of small congruent equilateral triangles, with 5 small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of 2003 small equilateral triangles?
(A) 1,004,004
(B) 1,005,006
(C) 1,507,509
(D) 3,015,018
(E) 6,021,018
Attachment:
16ca1f0305d713b2b9e5a0c53658d11829856ec7.png [ 15.51 KiB | Viewed 4532 times ]
20 Mar 2019, 02:17
Kudos
Bookmarks
Bunuel
As per counting of triangles shown in figure the triangle in first row = 1
triangles in second row = 3
triangles in third row = 5
Taringles in nth row = 2003
i.e. 1+2*(n-1) = 2003
i.e. n = 1002
Considering only triangles with peak vertex up
For only one equilateral triangle at the top the Toothpicks needed = 3*1
For two rows of Equilateral triangle, Toothpicks needed = 3*(1+2) where 1+2 represents triangles with peak vertex upside
For Three rows of Equilateral triangle, Toothpicks needed = 3*(1+2+3) where 1+2+3 represents triangles with peak vertex upside
i.e. triangles needed for 1002 rows = (1+2+3+----+1002)*3 = [(1/2)*1002*1003]*3 = 1,507,509
Answer: Option C
Attachments
File comment: www.GMATinsight.com
Screenshot 2019-03-20 at 2.41.32 PM.png [ 980.72 KiB | Viewed 3957 times ]
General Discussion
Debashis Roy
Joined: 23 Aug 2017
Last visit: 15 Dec 2019
Posts: 90
Own Kudos:
Given Kudos: 9
Schools: ISB '21 (A)
20 Mar 2019, 00:57
Kudos
Bookmarks
Bunuel VeritasKarishma chetan2u
The pattern is as follows:
1st row-1 triangle-3 sticks
2nd-3 trs-6 new sticks
3rd-5 trs-9 new
So for each nth row after 1st row containing r no of triangles..No of new sticks needed = 3*r-3*(r-n)...
**(This happens because the n no of inverted triangles share the same base from the above row and the sides from the adjacent upstand triangles)
The base row contains 2003 triangles..So total no of rows upto the base row= (2003-1)/2 +1=1002
So the series becomes
(3*1-3*0)+(3*3-3*1)+(3*5-3*2)....+(3*2003 - 3*1001).... 1001 since no exclusion from 1st row as mentioned above
=3*[(1+3+5+7+..+2003)-(1+2+3+...+1001)]=3*(S1-S2)
S1= 1+3+5+..+2003 = (1002/2)*(2004)=1002^2 = 1004004
S2= 1+2+3+...+1001=1001*1002/2=1001*501 = 501501
HEnce Answr = 3*(1004004-501501) = 3*502503 =1507509.... Ans (C)
The pattern is as follows:
1st row-1 triangle-3 sticks
2nd-3 trs-6 new sticks
3rd-5 trs-9 new
So for each nth row after 1st row containing r no of triangles..No of new sticks needed = 3*r-3*(r-n)...
**(This happens because the n no of inverted triangles share the same base from the above row and the sides from the adjacent upstand triangles)
The base row contains 2003 triangles..So total no of rows upto the base row= (2003-1)/2 +1=1002
So the series becomes
(3*1-3*0)+(3*3-3*1)+(3*5-3*2)....+(3*2003 - 3*1001).... 1001 since no exclusion from 1st row as mentioned above
=3*[(1+3+5+7+..+2003)-(1+2+3+...+1001)]=3*(S1-S2)
S1= 1+3+5+..+2003 = (1002/2)*(2004)=1002^2 = 1004004
S2= 1+2+3+...+1001=1001*1002/2=1001*501 = 501501
HEnce Answr = 3*(1004004-501501) = 3*502503 =1507509.... Ans (C)
