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Saturday, August 19 2017 @ 07:23 PM CDT
 Forum Index >  Test Paper Related >  Primary 6 Matters
 Prelims Practice! Help solve! Patterns Question! p6
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By: kesterasd (offline)  Thursday, September 12 2013 @ 02:29 AM CDT (Read 1847 times)  
kesterasd

Help me solve this question asap! Solve (b) and (c) please. Very urgent. (a) is done already. Thank you!

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By: kesterasd (offline)  Thursday, September 12 2013 @ 02:31 AM CDT  
kesterasd

Help me solve please, anyone? I will be grateful to you for that if you do this before 15/9/13, Sunday. Thank you!

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By: echeewh (offline)  Thursday, September 12 2013 @ 03:15 AM CDT  
echeewh

Hi <kesterasd>,

I suppose you have forgotten to attach / post your question here.

Best regards,
Edward

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By: kesterasd (offline)  Thursday, September 12 2013 @ 10:11 AM CDT  
kesterasd

Is it attached now? Thanks for reminding.

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By: echeewh (offline)  Thursday, September 12 2013 @ 08:10 PM CDT  
echeewh

Hey <kesterasd>,

//updated 26/9/13, 09 40
part (c) - re-edited solution
//


Out of curiousity, I am wondering why are you not able to solve it or have the solution ready when you are the setter (as noted in your attachment). Smile . Anyway, my worked solution is as follows:

Let n = Pattern number

Number of W triangles = [n(n+1)] / 2
Number of B triangles = [(n-1)n] / 2
Total number of triangles = n × n
Number of triangles (1 stick) = [(n-2)(n-1)] / 2
Number of triangles (3 sticks) = 1

(a)
Pattern 5
Total triangles = 5 × 5 = 25
Number of triangles (1 stick) = (3 × 4) / 2 = 6
Number of triangles (2 sticks) = 25 - 6 - 1 = 18

Number of Sticks = (18 × 2) + 6 + 3 = 36 + 9 = 45

(b)
Pattern 100
Total triangles = 100 × 100 = 10000
Number of triangles (1 stick) = (98 × 99) / 2 = 4851
Number of triangles (2 sticks) = 10000 - 4851 - 1 = 5148

Number of Sticks = (5148 × 2) + 4851 + 3 = 10296 + 4854 = 15150

(c)
Shaded triangles = B triangles
Number of B triangles = [(n-1)n] / 2

[(n-1)n] / 2 = 210
(n-1) × n = 420
20 × 21 = 420
n = 21

Hence, it is Pattern 21

=========

Trust this helps.

Do let me know if this is different from your <Answerkey> or if there's further clarification.

Cheers,
Edward

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By: kesterasd (offline)  Thursday, September 12 2013 @ 08:56 PM CDT  
kesterasd

<echeewh>, the setter is not me, it is the person I share this account with. And he don't want to give me the answer key.
Anyway, thank you for helping me solve! Razz Big Grin

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By: echeewh (offline)  Friday, September 13 2013 @ 06:51 PM CDT  
echeewh

Hey <kesterasd>,

Following is an <Alternative method> on the pattern for calculating the Number of Sticks. So this applies to (a) and (b) only.

Pattern 1 2 3 4 5
Triangles 1 4 9 16 25
Sticks 3 9 18 30

Total number of triangles = n × n
Number of W triangles = [n(n+1)] / 2
Number of B triangles = [n(n-1)] / 2

Number of Sticks = Pattern + Triangles + [n(n+1)] / 2
= n + (n × n) + {[n(n+1)] / 2}

(a)
Pattern 5

Number of Sticks = 5 + (5 × 5) + [(5 × 6) / 2]
= 5 + 25 + 15 = 45

(b)
Pattern 100

Number of Sticks = 100 × (100 × 100) + [(100 × 101) / 2]
= 100 + 10000 + 5050
= 15150

(c)
Same as previous method.

=========

Trust this helps too... Big Grin

Cheers,
Edward

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By: echeewh (offline)  Wednesday, September 25 2013 @ 08:50 PM CDT  
echeewh

Hey <kesterasd>,

I have re-edited the solution to part (c) of your question on 26/9/13, 09 40. My apologies for the mistake. Kindly review (c) on top.

Additionally I have found another alternative method (pattern) to calculate the Number of Sticks for (a), (b) as follows:

*** - for alignment purpose

Pattern ************1 2 3 4 5
Triangles **********1 4 9 16 25
White Triangles ** 1 3 6 10
Sticks ************* 3 9 18 30

Number of Sticks = 3 × (White Triangles)
Number of White Triangles
= n(n+1) / 2

(a)
Hence, in Pattern 5,

White Triangles = (5 × 6) / 2 = 15
Number of Sticks = 15 × 3 = 45

(b)
in Pattern 100,

White Triangles = (100 × 101) / 2 = 5050
Number of Sticks = 5050 × 3 = 15150

==========

Trust this helps and any inconvenience caused is regretted.

Do let me know again if there are further clarifications.

All the best in your PSLE,
Edward


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