@ Zedai,
Do u think sufficient data is provided in the qn?
As per your Qn i can assume A has got 999 sweets and C got 1 sweet(to maximize a+b-c).
If B having 500 sweets then the total number of sweets that A and B having is 1498 numbers more than C...So as per the given data answer is 1498 according to me.:grin:
@ joethaliath
Ya we got to find out a+b-c
I agree with Raku
Hii frndz, Im Nihal.. Im new to PG and this thread..
just started preprtns.. want to take aimcats..
Happy to see a group helping eachother.. So lets do it.....:grin:
@Nihal..Welcome to the thread...!! Be active and let us work towards nailing our dream B School calls!!:biggrin:
@ Zedai,
Do u think sufficient data is provided in the qn?
As per your Qn i can assume A has got 999 sweets and C got 1 sweet(to maximize a+b-c).
If B having 500 sweets then the total number of sweets that A and B having is 1498 numbers more than C...So as per the given data answer is 1498 according to me.:grin:
@ joethaliath
Ya we got to find out a+b-c
Yep, the data is sufficient.. i managed to eliminate 2 options but then, i wanted a discreet ans..
My approach :
A+C => Even => (A, C) => (Even, Even) or (Odd, Odd) => (A - C) => Even
B => Even
=> A + B - C => (A - C) + B => Even
=> Cannot be 377 or 179
so, it has to be either of the remaining 2 options. :-/
Yep, the data is sufficient.. i managed to eliminate 2 options but then, i wanted a discreet ans..
My approach :
A+C => Even => (A, C) => (Even, Even) or (Odd, Odd) => (A - C) => Even
B => Even
=> A + B - C => (A - C) + B => Even
=> Cannot be 377 or 179
so, it has to be either of the remaining 2 options. :-/
@ zedai.
I guess u are going wrong here. There is no need of taking this odd, even approach for this question. Kindly see the solution i have provided. With the data given in the qn i am getting the answer as 1498. If u know the correct answer for this qn kindly do post.
Whatever u have posted is right. Possible values of a+b-c will always be even. Since a-c is even(a+c=1000) and b is also even.
But that approach wont lead you to a unique answer.
Sharing a question from CL PROC MOCK 1 :
What is the highest possible value of 'n' for which 3^(1024) - 1 is divisible by 2n?
a)13 b)10 c)11 d)12
Sharing a question from CL PROC MOCK 1 :
What is the highest possible value of 'n' for which 3^(1024) - 1 is divisible by 2n?
a)13 b)10 c)11 d)12
A=3^1024= (3^512+1)(3^512-1)
= (3^512+1)(3^256+1)(3^256-1)
Expanding like this we ll get
A=(3^512+1)(3^256+1)(3^128+1)(3^64+1)(3^32+1)(3^16+1)(3^8+1)(3^4+1)(3^2+1)(3+1)(3-1)
3^(even number)+1 will always be even
So it will be divided by 2
so in this expression of A there are 11 such factors.
Among those factors (3+1) = 4 will be having 2 times 2.
So total there are 12 factors which are multiples of 2.
So the ans should be n=12
(this soln is not my own creation. I had done similar kind of Qn somewhere last year.)
@Raku , Thanks a bunch 
I was not really happy with the solution that the CL guys put up , i liked your approach.. both are kinda the same , but this registered better !
A=3^1024= (3^512+1)(3^512-1)
= (3^512+1)(3^256+1)(3^256-1)
Expanding like this we ll get
A=(3^512+1)(3^256+1)(3^128+1)(3^64+1)(3^32+1)(3^16+1)(3^8+1)(3^4+1)(3^2+1)(3+1)(3-1)
3^(even number)+1 will always be even
So it will be divided by 2
so in this expression of A there are 11 such factors.
Among those factors (3+1) = 4 will be having 2 times 2.
So total there are 12 factors which are multiples of 2.
So the ans should be n=12
(this soln is not my own creation. I had done similar kind of Qn somewhere last year.)
friends one doubt, is it 2n or 2^n in the question?
Its 2^n.. 😃
Hi guys,
I am aneesh residing at tvm(kerala)
Completed my 2nd yr Btech in ECE on to thrd yr..
I am an mba aspirant and recently joined careerlauncher centre..
Actually i have no idea how to start my preps hopefully i can get the right advice from here !!
Puys..
Booking for Aimcat 1318 has started...
Is anyone confused about the option - "Your own venue for online mocks" ?
Just want to clarify whether its some sort of bug or whether T.I.M.E would provide us with the test code upfront ?
Will be taking up Aimcat 1319 this Saturday...
Cheers!!
P.S : Howz the prep goin for all??
Hi guys,
I am aneesh residing at tvm(kerala).................
Actually i have no idea how to start my preps hopefully i can get the right advice from here !!
First of all, welcome to the thread aneesh...
Looks like you have joined at the right time... the thread is just about to take off as most of the team are getting done with their exams...So stay active....:)
Regading prep... my advice would be to start off with the prep material Cl must have provided you with... You could even take up a couple of mocks..identify your weak areas & practice...practice... practice...
Unfortunately theirs no easy way...
You can always discuss your strategies/queries with the team & hopefully nail the CAT,this Oct-Nov...
Cheers!!!
My two cents for the day.. (courtesy vineet.nitd)
Though most are already aware of this funda...for the uninitiated...
Hope others will follow suit & chip in with more funda & interesting problems to discuss...:grin:
Number Systems (Concept 1)
1) Painted cube funda :
We assume the cube is divided into n^3 small cubes.
No. of small cubes with ONLY 3 sides painted : 8 ( all the corner cubes )
No. of small cubes with ONLY 2 sides painted :
A cube is painted on 2 sides means, it is on the edge of the bigger cube ,and we have 12 edges, each having n cubes. but since the corner cubes are painted on 3 sides, we need to neglect them. so in effect, for each side we will have (n-2) small cubes with only 2 sides painted.
thus, then number is, 12 * (n-2)
No of small cubes with ONLY 1 side painted :
for each face of the cube ( 6 faces ) we have (n-2)^2 small cubes with only one side painted. and we have 6 faces in total.
so th number is, 6*(n-2)^2
No of small cubes with NO sides painted :
if we remove the top layer of small cubes from the big cube we will end up a chunk of small cubes with no sides painted.
this number will be equal to, (n-2)^3.
2) Also, remember for Cuboids with all different sizes, the following are the results:
a x b x c (All lengths different)
Three faces - 8 (all the corner small cubes of the cuboid)
Two faces - There are two (a-2) units of small cubes on one face of the cuboid and there is a pair of such faces. Hence, number of such small cubes corresponding dimension a of the cuboid = 4(a-2).
Similarly, for others.
So, total with two faces painted = 4(a - 2) + 4(b - 2) + 4(c - 2)
One face - Since each face of the cuboid is a combination two different dimensions, hence for the face which is a combination of a and b dimensions, the number of small cubes is 2* (a-2)(b-2)
Similarly, for others.
So, total with one face painted = 2(a - 2)(b - 2) + 2(a - 2)(c - 2) + 2(b - 2)(c - 2)
Zero faces - The entire volume of small cubes except for two cubes in each of the rows and columns will not be painted at all. hence this is the simplest ...
(a - 2)(b - 2)(c - 2)
You can put different integer values for number of small cubes producing different edge lengths of cuboid to get varied results.
To verify for a cube, put a=b=c=L, you get
Three faces - 8
Two faces - 12(L - 2)
One face - 6(L - 2)^2
Zero faces - (L - 2)^3
Cheers!!
@Kevin , gr88 work... !! done with my btech (hopefully ) 2 days back..relaxing, enjoying the monsoons and wanna slowly start off one of these daaays...
turned the first few pages of the quant book i got from flipkart - QUANTUM CAT by Sarvesh Verma
If you guys havent noticed the below thread , plz do.. mainly deals with section 2 , check out Nishant88's post on LR , found it very useful!!
http://www.pagalguy.com/discussions/how-to-crack-section-ii-of-the-cat-25081933
Seems like everyone here is in deep prep mode...
Hence the low no of posts...:nono:
An update from my end...
Aimcat 1319 was a wake up call..
QA/DI : 93.5% (16C/2W)
VA/LR: 84.6% (17c/13w):banghead:
OVR : 93.44%
How abt the rest of you guys?
Seems like everyone here is in deep prep mode...
Hence the low no of posts...:nono:
An update from my end...
Aimcat 1319 was a wake up call..
QA/DI : 93.5% (16C/2W)
VA/LR: 84.6% (17c/13w):banghead:
OVR : 93.44%
How abt the rest of you guys?
QA/DI - 92.16% (44)
VA/LR - 94.52% (46)
OA - 95.74% (90)
My percentile took a dip from 97.xx in 1320 to 95.74 in 1319.Hope that the next mock will be a different case.
Qa-di : 99.4 - 66
va-lr : 99.74 - 62
oa : 99.92 - 128
@pugna that is an awesome percentile:clap:
will take mocks from july
hi all,
what about a pg meet in cochin on 24 th june(sunday)?please post your valuable suggestions.All other studygroups in pg are conducting meet on a regular basis,then why are we shying away from this:banghead:. come on puys,please suggest the venue,suitable date,time and all...............