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19 Mar 2019, 21:48
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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41% (03:23) correct
59%
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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 4395 times ]
20 Mar 2019, 02:17
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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
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General Discussion
Debashis Roy
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20 Mar 2019, 00:57
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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)
