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

[dewan_iitr]

Quantitative Question # 008
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Let S be the set of first 14 natural numbers. A special subset of S is a subset S' which satisfies the following three properties

a) S' has exactly 8 elements
b) If x belonging to S is even, then x is in S' if and only if x/2 is in S'
c) If y belonging to S is odd, then y is in S' if and only if (y+15)/2 is in S'

Let X denotes elements of S that cannot be the part of special subset. Then n(X) (i.e. number of elements in X) equals

(1)2 (2) 3 (3) 5 (4) 6 (5) none of these



S- 1-14
S' - 8 elemnets

x/2 =
2 - 1 ( ie if 2 is there 1 has to be there)
4 - 2
6 - 3
8 - 4
10 - 5
12 - 6
14 - 7

(y+15)/2 =
1 - 8
3 - 9
5 - 10
7 - 11
9 - 12
11 - 13
13 - 14


Now if we take
1 - 8 - 4 - 2 - 1 gives 4 elemenst
simillary 3- 9 - 12 - 6 - 3
or 14 -7 - 11 - 13 - 14

But 5-10-5
so 5 and 10 can never be part os S'

Answer (2)

[ravishah17]

1=> 8 is in S'
2=> 1 is in S'
3=> 9 is in S'
4=> 2 is in S'
5=> 10 is in S'
6=> 3 is in S'
7=> 11 is in S'
8=> 4 is in S'
9=> 12 is in S'
10=> 5 is in S'
11=> 13 is in S'
12=> 6 is in S'
13=> 14 is in S'
14=> 7 is in S'

Supp 1 is in S', 8,4,2
Supp 3 is in S', 9,12,6

5,10 are the only two elements which are dependent only on each other, rest come in pair of 4,

So n(X)=2

Answer is (1)2

[nakulnbhanage]

Condition for even natural numbers to be the part of S' is that (x/2) should be the part of the special set S'. And that for the odd numbers is that ((y+15)/2) should be the part of S'.

Accordingly the combinations work out to be ( ie if this is there, this has to be there)

9--12--6--3
8--4--2--1
7--11--13--14
5--10

So 5 and 10 cannot be chosen since the pre-requisite of S' is 8 elements. (If 5 and 10 are chosen, u cannot take other 6 numbers, since all appear in a group of 4)

So the answer is 2 (ie 5 and 10)

[implex]

Originally Posted by rajeev_hts

Let first "point of time" is x
Let 2nd "point of time" = y
Let 3rd "point of time" =z
Age of Anupam now = A
Age of Shrikant now = S
Age of Nilesh now = N

A+x=4 -> A=4-x --------(1)

N+x=3(S+x)

N=3S+2x ---- (A1)

S+y = 2(A+y)

S= 2A+y ------(A2)

From A1 and A2 and (1)

N=24-4x+3y -------(A3)

Also from prob statement N+y = 5(A+y)
N= 5A+4y --------(A4)

From A3 and A4 and (1)

y=4+x ---------(2)

From prob,
N+z=2(A+z)
N= 2A + z --> z=N-2A ----(A5)

From 1, 2, A4, A5

z=x+28 ---(3)

From Prob A+z = Y

Y=A+z ---(4)

From 1, 3, 4

Y= 32--- ANS

Looks like Y had a fix value.

you need to read the question again

there will be more than one y !

suppose I am 23 today you are 24 and someone else is 25

after 5 weeks
I may be 24 you 24 and someone else is 25

after 6 weeks
I may be 24 you 24 and he 26

[vedsameer]

If you classify first 14 natural numbers according to conditions .
we get 3 sets ...
i) 1,2,4,8
ii) 7,11,13,14
iii)5,10
As it says only 8 elements ... we cannot have 3rd set ..
So answer is 2 .. option 1.

[IIMA_2009]

Hey

The sets that can be formed..

--> 1 2 4 8
--> 3 6 9 12
--> 5 10
--> 7 11 13 14

Since only Eight Elements need to exist..

I can take any two sets at a time and get S'

Hence in all possible combination i will never have 5,10...

Hence the answer is [FONT='Georgia','serif']Option 1...[/font]

[pavanpadekal]

Originally Posted by Aarav

Overwhelming # of junta had asked to bring down QQAD's level when asked for their feedback, but interestingly very few of them are here on this thread.
If the problems are tough then people say QQAD is unrealistic and over the board. If problems are easy then good students of quant don't feel enough challenged. The root cause of this kinda scenario is -> after all we have just 1 problem in a day and it can't satisfy each and every person. Let me get a pulse of this year's batch, and I assure enough challenge to you all
Let me test you next week and take your feedback before deciding what kinda problems you would want in the future.

In the meanwhile, try this problem from last year's QQAD for some challenge

My Opinion/Suggestion (I hope i dont end up being groaned at)
1.As Vineet said, many people including myself(last year) do not / did not open the QQAD newsletters at all, most of the time fearing inability to solve the questions due to lack of concepts.
2.More people actively participating in the QQAD discussion doesnt exactly clutter the thread as we dont generally need to go back to older questions if we are regulars.If not regular most dont take the pain of looking for the questions they missed.Plus greater participation would let everyone benefit from QQAD.
3.Now does that mean we need to lower the level of QQAD? In a way yes.What I suggest is to begin with easier questions initially and then up the tempo each day.
4 Now another problem is that people see simple questions and solve/participate/benefit and the next day the question is from some topic they are not comfortable.Since its just 'one question' they tend to ignore it and wait for 'their topics'.Again the purpose of benefiting the majority of junta who are a little Quantophobic is lost.What to me seems a solution to this is to have sets of questions on a single topic maybe for a week.Initially easy sitters which everyone can solve.The next day take the level up a notch.So by the end of the week we must be having the most quantophobic junta also benefiting from the discussions and LEARNING.Even if the topic is beyond one's comfort zone people will venture into it and when they see they can solve a few among the tricky ones as well and interest levels will sustain.
5.At present very few people actually LEARN from QQAD.But I must say the few who are good at Quants already get a lot to hone their skills.Questions are beautifully crafted to enable them to learn.However it would benefit more if this learning curve didnt have any prerequisite knowledge of Quants.


Aarav and Vineet(though you aren't preparing the questions this year)! QQAD rocks!No question about it.

[implex]

Originally Posted by Aarav

A person is said to be n years old, where n is a non-negative integer, if the person has lived at least n years and has not lived n+1 years. At some point in time, Anupam is 4 years old and Nilesh is three times as old as Shrikant. At some other time, Shrikant is twice as old Anupam, and Nilsh is 5 times as old as Anupam. At yet another time, Nilesh is twice as old as Anupam, and Anupam is Y years old. There are different possibilities of what Y can be. What is the largest possible value of Y?

Scene1: Anu is 4 Nilesh is 3x and Shri is X
Scene 2 : Anu is Z Nilesh is 5z and Shri is 2z

Scene 3 Anu is Y and Nilesh is 2Y

now see as we move from Scene 1 to 2 ratio od nilesh to to shri declines
hence Scene 1 was before Scene 2

suppose in Scene 2 z is 4, then Nilesh is 20 and Shri is 8 wont work in scene 1

suppose z is 5 nilesh is 25 and shri is 10 won't work in scene 1

suppose z is 6 nilesh is 30 and shri is 12 won't wok
z=7, 35 and shri is 14
this can work !!
so z is 7

now the thing is at a particular age Y nilesh is 2 Y
2(z+x-1)=(5z+x) [ after X years we Nilesh will increase by X and and anu by x-1 for the max case}
x=3z+2 so clealry Y is x+z=4z+2
so we need to fidn the largest Z

if z is 8 nilesh is 40 and shri is 16 , this can work too

if z is 9 nilesh is 45 and shri is 18 this will work as well

if z is 10, nilesh is 50 and shri is 20
this will work as well
if z is 11, nilesj is 55 and shri is 22 work
if z is 12 nilesh is 60 and shri is 24 will work

if z is 13 nilesh is 65 and shri is 26 won't work

so z =12 max

so max Y =12.4+2 =48+2=50!!!

[D33pblue]

Quantitative Question A Day #008


For x(even) to be in S' x/2 should also be there in S'
and for x/2(odd) to be there in S' ((x/2)+ 15)/2 should also be present in S'

So for x to be present in S', ((x/2)+ 15)/2 should also be present and this is not possible if x and ((x/2)+ 15)/2 are same.
So equating them we get x = 10. Hence 10 cannot be there in S', which means that (x/2) i.e. 5 can also be not there in S'.

So the answer is 2 which is choice number (1).

[getneonow]

The first thing I like to do in this thread is simply put my solution without looking at others coz I'm sure there wud b better soln's here which will refrain me from posting mine

Todays qn is more of a grouping problem.

Even Set X E S ={2,4,6,8,10,12,14}
Then X E S' must be pairs of (2,1) ,(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)

Odd Set Y E S = {1,3,5,7,9,11,13}
Then Y E S' must be pairs of (3,9),(5,10),(7,11),(9,12),(11,13),(13,14)

We see that the pairs are in turn dependent on a different pair.

So if we group all the dependent pairs in different sets then we get

{1,2,4,8},{3,6,9,12},{5,10},{7,11,13,14}

Our S' must have some of these sets but we see that in 3 sets there are 4 elements each and in only one set there are 2 elements.

Since S' is a set of exactly 8 elements only two sets of 4 elements each will be in S' . Thus the odd set out here is the set {5,10} which cannot exist in S'.

=> # of elements which cannot belong in S' = 2 which are {5,10}

Ans Choice (1)

Neo