CAT 2007: Quantitative Questions a Day 101 to 133 - The Discussions

KOOL030

Quote:

Originally Posted by atshubha

dont get this part (b+c) = -a right (b+c)+c = -a+c ??

b+2c
=(b+c)+c
=a+c(since b+c=a)
=b(since a+c=b)

I HOPE YOU GET IT NOW

sunit_chotu

Quote:

Originally Posted by Aarav

Take any point A. A is joined to at most three points, call them B, C, D. Any other point P must be joined to one of B, C, D. Each of B, C, D is joined to at most two other points. So there are at most 1 + 3 + 3·2 = 10 points in all. But is 10 possible?

Take points A, B, C, D, E, F, G, H, I, J and edges AB, AC, AD, BE, BF, CG, CH, DI, DJ, EH, EJ, FG, FI, GJ, HI. It is easy to check that every point has degree 3. Now take any two points not joined by an edge. For A,E or A,F both are joined to B. Similarly, both A,G and A,H are joined to C, and both A,I and A,J are joined to D. Both B,C and B,D are joined to A. B,G and B,I are joined to F. B,H and B,J are joined to E. Similarly, C,D are joined to A. C,E and C,I are joined to H. C,F and C,J are joined to G. Similarly, D,F and D,H are joined to I. D,E and D,G are joined to J. E,F are joined to B. E,G are joined to J. E,I are joined to H. F,H are joined to I. F,J are joined to G. G, I are joined to F. H, J are joined to E.

here is the figure attached Sir. Hope I have done justice to the figure u had in mind Aarav.

Rohit .Sinha

Quote:

Originally Posted by sunit_chotu

here is the figure attached Sir. Hope I have done justice to the figure u had in mind Aarav.

This is beautiful.

PS: Shouldn't the question read that evry node has a degree 3 rather than atmost.
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Rohit

rahul.maverick

Isnt it possible to have 10 points ...

rahul.maverick

Quote:

Originally Posted by sunit_chotu

here is the figure attached Sir. Hope I have done justice to the figure u had in mind Aarav.

just check sunit whether i am right or not ....

sunit_chotu

Quote:

Originally Posted by rahul.maverick

Isnt it possible to have 10 points ...
just check sunit whether i am right or not ....

Ya That is the right answer which Aarav has mentioned in the previous page. it is possible to have 10 points and that is the maximum of what we can have considering the given condition. . Ya Rahul ur figure is correct.

sunit_chotu

Quote:

Originally Posted by Rohit .Sinha

This is beautiful.

PS: Shouldn't the question read that evry node has a degree 3 rather than atmost.
-------------
Rohit

If the degree is atmost 3 then only u will get the maximum value of p that is 10 other wise not. That is u can have a maximum degree of 3 and to maximize the value of p u need to take the max value of degree that is 3.

Aarav

Quote:

Originally Posted by sunit_chotu

here is the figure attached Sir. Hope I have done justice to the figure u had in mind Aarav.

You are too good

@All ... Tomorrow's both the questions are going to be sitters. Get the answers in the very 1st attempt.

Aarav

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Quantitative Question # 131
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Part (A)

Which one of the following is the remainder when x + x^7 + x^16 + x^37 is divided by x^4 - x?

(1) 4x (2) 2x(x-1) (3) x^2 + 2x (4) x(x-1) (5) none of these

Part (B)

Reema tries to prove her conjecture that f(a)/(a-b) is an integer for two distinct integers a and b for a given polynomial f(x) which satisfies the following properties I and II
(I) Every coefficient in f(x) is an integer (II) f(a)f(b) = -(a-b)^2

Find out the mistake in her proof if there is any

Proof

(a) For a natural number n, f(a) - f(b) is divisible by a-b since a^n-b^n is divisible by a - b (from I)

(b) Thus, [f(a) - f(b)]/(a-b) is an integer. The quadratic equation that has 2 solutions f(a)/(a-b) and -f(b)/(a-b) is x^2 - x[f(a) - f(b)]/(a-b) + 1 = 0 (from II)

(c) Since, f(a)/(a-b) is rational, (from I) and [f(a) - f(b)]/(a-b) is integer => f(a)/(a-b) is an integer

(1) (a) (2) (b) (3) (c) (4) exactly 2 of the foregoing (5) none of these

ballz

1st part

5)

2nd part

3)