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The Official CAT 2007 Quant Thread
Quantitative Questions and Answers Discuss Quantitative and other Math related questions. Post your math doubts and get it solved by the smartest brains this side of the universe !

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sachin2807 sachin2807 is offline
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Re: The Official CAT 2007 Quant Thread - 30-10-2007, 04:05 PM

Quote:
Originally Posted by esh.nil View Post
the question is a+b+c = 4
so isnt the ans 6C2 = 15 ways??
out of this one is not possible... 004
so totally 14 ways..
they have given a totally diff solution.They divided it into four cases
CASE 1: All fruits are the same.
CASE 2: Three fruits same and fourth one is different.
CASE 3: Two fruits same while other two are different.
CASE 4: Two fruits are of one type while the other two are of another type

Totally,
2+6+3+3=14


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Re: The Official CAT 2007 Quant Thread - 30-10-2007, 04:10 PM

In triangle ABC,A(-8,5) B(-15,-19) C(1,-7).Then slope of angle bisector of angle A is ?
1) -12/7
2) 6/5
3) -11/2
4) 7/3


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Re: The Official CAT 2007 Quant Thread - 30-10-2007, 04:13 PM

Quote:
Originally Posted by sachin2807 View Post
they have given a totally diff solution.They divided it into four cases
CASE 1: All fruits are the same.
CASE 2: Three fruits same and fourth one is different.
CASE 3: Two fruits same while other two are different.
CASE 4: Two fruits are of one type while the other two are of another type

Totally,
2+6+3+3=14
@sachin.. my soln is also the same...
I used the formula isntead of using all the cases..
a+b+c = 4
(n+r-1)C(R-1) = (4+3-1)C2 = 6C2
in this one variety can be selected utmost 3,
so answer will be 6c2-1 =14

the usual way,
112--- 3 ways(3!/2!)
202--- 3 ways
301--- 6 ways
400 --- 2 ways(there is only 3 fruits of one type)..

@monil bhai.... why did u stop after finding the 755^1, why dint u proceed further to find the remainder?? I am not very clear with ur method, can u pls explain???


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Re: The Official CAT 2007 Quant Thread - 30-10-2007, 04:16 PM

Consider the fn f : Z+ --> Z+, where Z+ is the set of all positive integers.
If f satisfies
t^2 *f(s)=s{f(t)}^2,
find ratio of f(kx)/f(x)


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Re: The Official CAT 2007 Quant Thread - 30-10-2007, 04:20 PM

Quote:
Originally Posted by sharang View Post
In triangle ABC,A(-8,5) B(-15,-19) C(1,-7).Then slope of angle bisector of angle A is ?
1) -12/7
2) 6/5
3) -11/2
4) 7/3
I got the answer as (a) -12/7
but my approach is not the best one.... I am waiting for a better appraoch from puys..

What I did was to find the eqn of line BC and find a tentative point for the D(where ang bisector meets BC) from the ans and select it..


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Re: The Official CAT 2007 Quant Thread - 30-10-2007, 04:24 PM

Quote:
Originally Posted by esh.nil View Post
@monil bhai.... why did u stop after finding the 755^1, why dint u proceed further to find the remainder?? I am not very clear with ur method, can u pls explain???
Quote:
Originally Posted by reachmonil View Post
Expressing in base 3 means mod 3. By Euler's theorem, we have E(3)=2. So, what we have to find now is 755^1 in base 3 = 222. Hence the last 3 digits. Correct?
755^555 mod 3 would give the same remainder as 755^1 in base 10. So... I directly just convert 755^1 into base 3 and got the last 3 digits. Luckily, this answer is correct for the specific example. Though I need to standardise it for general approach, which I don't have enought time to do today. Maybe you can try it using 3-4 examples.


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Re: The Official CAT 2007 Quant Thread - 30-10-2007, 04:26 PM

hey hi.
this is shyam here from mumbai . have just joined in pagal guy and cant get the head or the tail of it. so if u can just help me out to get started with it . would highly appreciate a reply very soon as would not be interested in wasting time in looking for so as to get started it. thanakx man CHEERS!!!!!!!!!!!!!
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Re: The Official CAT 2007 Quant Thread - 30-10-2007, 04:30 PM

Quote:
Originally Posted by esh.nil View Post
I got the answer as (a) -12/7
but my approach is not the best one.... I am waiting for a better appraoch from puys..

What I did was to find the eqn of line BC and find a tentative point for the D(where ang bisector meets BC) from the ans and select it..
The ans is -11/2...anyways post ur approach


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Re: The Official CAT 2007 Quant Thread - 30-10-2007, 04:34 PM

Quote:
Originally Posted by esh.nil View Post
I got the answer as (a) -12/7
but my approach is not the best one.... I am waiting for a better appraoch from puys..

What I did was to find the eqn of line BC and find a tentative point for the D(where ang bisector meets BC) from the ans and select it..
i m getting -11/2 as the answer.
fst i used angle bisector theorem,
AB/AC=BD/DC
AB=25 and AC=15
so,BD/DC=5/3
so,corresponding point D will be (-5,-23/2)
nd then directly find the slope of the line joining pts A and D.


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Re: The Official CAT 2007 Quant Thread - 30-10-2007, 04:38 PM

Quote:
Originally Posted by reachmonil View Post
755^555 mod 3 would give the same remainder as 755^1 in base 10. So... I directly just convert 755^1 into base 3 and got the last 3 digits. Luckily, this answer is correct for the specific example. Though I need to standardise it for general approach, which I don't have enought time to do today. Maybe you can try it using 3-4 examples.
Monil bhai...The first example I took, there is a difference between ur ans and mine. :(
ur approach: consider (623)^7 so the remainder is same as 623,
calculating the last 3 digits
3|623
3|207--2
3|69--0
3|23--0
last 3 digits = 002
but acc to my method, the remainder of 623^7%27 = 2^7%27 = 20
20 in base 3 = 202...

So where is the mistake?? ??:??:


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