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CSAT Sample Paper Answers and Discussion

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I see that this formula works but i don't understand why because I thought the combination formula could only be used when the order doesn't matter. In this example order does matter so wouldn't it be a permutation?
Original post by Squiidsquid

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I see that this formula works but i don't understand why because I thought the combination formula could only be used when the order doesn't matter. In this example order does matter so wouldn't it be a permutation?
Original post by flamesofsnow
I see that this formula works but i don't understand why because I thought the combination formula could only be used when the order doesn't matter. In this example order does matter so wouldn't it be a permutation?


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Original post by Squiidsquid

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What do you mean 'arranged differently within themselves'? thanks. Also what was your method for sample paper 1 question 14?
has anyone done question 18 from paper 1
Original post by irishapplicant

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but I'm probably wrong


I got this in another way:

First we calculate all 3 point combinations, where we get: 560

Then, we subtract the number of "impossible" or degenerate triangles:

Horizontal Degenerates: [2xCombinations[3 points] per Horizontal Line ] *4 Lines = 8
Vertical Degenerates: [2xCombinations[3 points] per Vertical Line ] *4 Lines = 8
Obliques Degenerates: [2xCombinations[3 points] /(4 Point Oblique Line)] + [2xCombinations[3 points] /(3 Point Oblique Line)] *2 [Left Obliques and Right Obliques] = 8

So, 560 - (8+8+8) = 536

Any one cares to confirm??

Cheers
Original post by flamesofsnow
What do you mean 'arranged differently within themselves'? thanks. Also what was your method for sample paper 1 question 14?


as in technically there are 6 ways to arrange RRR but really its only one.
Original post by Squiidsquid
as in technically there are 6 ways to arrange RRR but really its only one.


ohh i get it thanks so much!
Original post by flamesofsnow
I see that this formula works but i don't understand why because I thought the combination formula could only be used when the order doesn't matter. In this example order does matter so wouldn't it be a permutation?


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Original post by uponthyhorse

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yeah that makes sense thanks. Have you done question 18?
Original post by Peractio
for strategy two

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Thanks, I realised this after I posted it, but couldn't change it on my phone because the LaTeX is really hard to do, so I only use my laptop. I'll amend it now :smile:
Original post by flamesofsnow
yeah that makes sense thanks. Have you done question 18?


Yes, this one is more like a riddle really

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Original post by Peractio
Mine was similar to yours, except

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Yes, you're right. Same method but I don't know how I got n^2.
Have answers for Sample Paper 2 Q: 13, 15, 17, 18 been posted?
Original post by popicecold99
Have answers for Sample Paper 2 Q: 13, 15, 17, 18 been posted?


I got these answers:

Q13 (not too sure about this)

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Q15

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Q17

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Q18 (long solution, I'll try and find a shorter way)

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(edited 5 years ago)
Original post by Integer123
I got these answers:

Q13 (not too sure about this)

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Q15

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Q17

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Q18 (long solution, I'll try and find a shorter way)

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Yes I got that for 13 and 15 too, for the others there seem to be some varying solutions. Not entirely sure how to resolve that, perhaps I will link to the different solutions.

Original post by popicecold99
Have answers for Sample Paper 2 Q: 13, 15, 17, 18 been posted?


I believe they have but I forgot to update the first post because of mock exams. Almost all of the questions have been covered at this point it's just a matter of me being less lazy :biggrin: I'll do it asap
Original post by popicecold99
Have answers for Sample Paper 2 Q: 13, 15, 17, 18 been posted?


Posted my solution to 15 on the previous page, got the same answer :smile: haven't done the others myself yet, there might be answers on previous pages though :smile:
Original post by flamesofsnow
has anyone done question 18 from paper 1


This question is tricky and I'm still not very sure what exactly it means (forgive me English is not my first language). Let's see if my solution is valid. :wink:

My claim is that the starting point has to be the North Pole.
By traveling east, the person will be at a different longitude (as he visits no point twice) After traveling n miles south and then n miles north, he will be at the same latitude. The only possible point where different meridians (lines of same longitude) meet is either the North pole or the South pole. Clearly it's not the South Pole.

Assume that "What is the closest you could have been to the South Pole when you started? " = "What it the closet you could have been to the South Pole during the journey?"Otherwise this is trivial. And I think "closet" means the "shortest 3D distance", because if it is spherical distance, the solution is trivial (pi*R-n)

The journey is symmetrical, so the person must reach the point closet to the South pole at least twice. WLOG, he reaches the point after going south for x miles.
Unparseable latex formula:

[br]let the distance to the South pole be d, and the angle he makes with the center and the axis is $ \theta $[br]Using cosine rule, $d^{2}= 2\times R^{2}-2\times R^{2} cos(\pi-\theta)$


As x<=n<R, the movement is restricted in the Northern Hemisphere, so
Unparseable latex formula:

$\theta<\frac{\pi}{2}$


Obviously, in this domain, d decreases as \theta increases. Thus, d achieves its minimum when x = n = R
Unparseable latex formula:

(\textbf{R can never be reached! Is this a tiny discrepancy or my entire argument is fallacious?}),[br] $$\theta = \frac{x}{R}=1 $$[br]Plug in, $$d^{2}=2\times R^{2}-2\times R^{2} cos(\pi-1)[br] \Rightarrow d= R\times \sqrt{2-2cos(\pi -1)}$$[br][br]

(edited 6 years ago)
Sample paper 2 Question 20

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Original post by Integer123
Sample paper 2 Question 20

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Nice proof! :biggrin:
How did you find the specific powers?
(I did the problem by dividing 2^33 and using the binomial expansion)

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