CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

[kavita_iet]

Originally Posted by Aarav

I think I should write the answer now.

524 is the largest 3 digit perfect square in base 6 => x ranges from [√524] + 1 to 524 => x = 524 - 23 + 1 = 502 which is 1315 in base 7 and an even number in that base.

@ Aarav
Awesome question is round like jalebi
It will be good, if its possible to wait for ppl (like me ) who come late to the thread before posting your solution.

For all those like me who didnt understood here , x ranges from [√524] + 1 to 524
consider a number 196,
x(x-1) + x = 196
now for repn in two digits(99 to 10) we can start from base 15 to 196 ie sqrt(196) +1 to 196.

[dewan_iitr]

Originally Posted by sudeepdeb

hey satanica,

502 is the number of bases where the given condition satisfies. In my view 502(value of x) has nothing to do with base6.

nappy..
i feel satanica is correct here -
no. of bases u have taken out is in base 6
and
502 in base 6 = 182 in base 10 = 350 in base 7
u got 502 by 524 -22
in base 10 if u do the same thing it will be 196-14 = 182


got the mis

thus i think answer is 350....


got the mistake-
its got to be done in base 6 and no of bases is irrespective of that
still not able to completely digest this...

[Aarav]

Originally Posted by kavita_iet

@ Aarav
Awesome question is round like jalebi
It will be good, if its possible to wait for ppl (like me ) who come late to the thread before posting your solution.

Thanks Kavita, some Googling experts might be disappointed though robbed them of the chance to claim they know from where QQAD question is coming from.
Give me some points for ingenuity

Posted the solution early since people were going in all sorts of directions, hence it was better to restore things correct.

[sudeepdeb]

Originally Posted by dewan_iitr

nappy..
i feel satanica is correct here -
no. of bases u have taken out is in base 6
and
502 in base 6 = 182 in base 10 = 350 in base 7
u got 502 by 524 -22
in base 10 if u do the same thing it will be 196-14 = 182


got the mis

thus i think answer is 350....


got the mistake-
its got to be done in base 6 and no of bases is irrespective of that
still not able to completely digest this...

mandy,
the question says u need to take the number which is largest perfect square in base6 and find number of bases for that number.....
I think x = 502, (dhruv is also behind me, agar pange lene ho toh aajaa upar :P)

PS: I got confused with the assurance dewan posted.
Conclusion: Think less.

[implex]

Originally Posted by Aarav

Thanks Kavita, some Googling experts might be disappointed though robbed them of the chance to claim they know from where QQAD question is coming from.
Give me some points for ingenuity

Posted the solution early since people were going in all sorts of directions, hence it was better to restore things correct.

This was a good question, got it totally wrong!! bad day for me!

[fiction]

Largest 3 digit number is 555.
This when converted to base 10 is 215.
Now closest perfect square to this number is 196.(14^2)

So converting 196 back to base 6 we get 524. This is the largest 3 digit number in base 6 which is a perfect square.

Now lets consider a base 'x'

To get largest base where 524 can still be expressed in 2 digits the smallest 2 digit number in that base should be equated to 524.

=> x.1 + 0 = 524
=> x = 524.

So the largest base where 524 can still be represented as a two digit number is 524.

To get smallest base where 524 can still be expressed in 2 digits the largest 2 digit number in that base should be equated to 524.

=> (x).(x-1) + 1.(x-1) = 524
=> x = 22.something

So the smallest base where 524 can still be expressed as a 2 digit number will be 23.

So number of bases = 524-23+1
= 502

This converted to base 7 gives 1315 which is even in base 7.(thnx nappy)

so ans (4)

[adarshiitm]

Hello all....this is my first post in this forum.

and here is my solution---

Largest 3 digit number in base 6 is 555 = 215 on base 10

so maximum perfect square would be 196.

Now for representing 196 in 2 digits, the minimum base can be 15 as on base 14, 196 can be written as 100 which is 3 digits not 2.

the maximum base that can be used is 196 as for base more than 196, 196 can be written only in 1 digit

so number of bases that can be used = 196-15+1(for including 15) = 182

so the value of x is 182

which is even but not a perfect square irrespective of the base we use.

7 is a prime number which can be written as 10 on base 7, does that make 7 a non prime number? I don't think so.

[anubhutiv]

Largest 3 digit perfect square in base 6 = 524 in base 6 = 196 in base 10

For any base 'n' smallest 2 digit number = n and largest 2 digit number = (n-1)+n(n-1) = n*n -1

So, 196 base 10 is a two digit number from base 15 to base 196. x = 182

182 in base 10 = 350 in base 7 ... i think even number is one that is divisible by 2, so irrespective of base this will be even but not a perfect square.

[siddhartha.s]

Originally Posted by thebornattitude

------------------------------------------------------
Quantitative Question # 005
------------------------------------------------------


A particle moves around a circle (once) such that its displacement from the initial point in given time t is t(6-t) meters where t is the time in seconds after the start. The time in which it completes one-sixth of the distance is

(1) 0.60 s (2) 0.88 s (3) 1 s (4) 1.12 s (5) none of these

SOLUTION-

Max displacement occurs at diameter,and by looking at the displacement equation... we can infer that at t=3, max occurs.

Hence at t=3, max displacement= 9...means 2r=9(r=radius)

r=4.5...


Now at 1/6th of total distance, the angle in the arc will be=60deg. hence the displacement =r=4.5..

so to find t for it we put
t(6-t)= 4.5..solving we get

ANSWER=0.88 sec.

NOTE- people who are getting 1 as answer, I guess you are taking speed as constant for the equation, which is not the case,,the dependency is only between displacement and time,,and that is also not linear.

Sorry for bringing up an old question, but I am still confused as to how we can infer that at t=3, the max occurs?? would it be safe to assume this or is this logic base on some calculations, since nowhere it is given that the particle moves with constant velocity or speed? I can understand that the max displacement =2r, but how can we say that this occurs at t=3??
Somebody needs to clarify this in more detail to those of us who are not all that brainy

[avi_aks]

well siddhartha, ur doubt only seems to be how the max displacement occurs at t=3.

mathematically if u look at the equation:

lets assume the displacement to be 'x'.

then, x=t(6-t)
= 6t -t.t

hence, dx/dt = 6 - 2t (diffrenciating both sides wrt 't' )

At, x max .. dx/dt=0
hence, 6 - 2t = 0
or, t = 3.

hence, the max displacement occurs at 3.