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

[gaggi_85]

(1)2IS ANSWER FOR QUESTION 008

12,6,3,9 ARE DEPENDENT ON EACH OTHER….IF ANY ONE IS THERE..THE OTHER 3 HAS TO BE THERE…
8,4,2,1 ARE DEPENDENT ON EACH OTHER….IF ANY ONE IS THERE..THE OTHER 3 HAS TO BE THERE…
7,11,13,14 ARE DEPENDENT ON EACH OTHER….IF ANY ONE IS THERE..THE OTHER 3 HAS TO BE THERE…

5 AND 10 ARE DEPENDENT ON EACH OTHER….IF ANY ONE IS THERE..THE OTHER 1 HAS TO BE THERE…

SO 5 AND 10 CANNOT BE THERE IN S’ AS AT A TIME ONLY 8 ELEMENTS CAN BE POSSIBLE….

[priyank12]

We have to start from 14...Now first considering 14..we should have 14,13,11,7
for 12 ----12,6,3,9
for 10 ---10,5
for 8 ----8,4,2,1
so in these all numbers till 14 are covered.......so the odd series is 10,5 which can't be placed with other numbers and adjusted in S'

so the answer is (1) 2

[rajatkapoor1986]

to satisfy condition 2, the set can be (2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)
to satisfy condition 3, the set can be
(1,,(3,9),(5,10),(7,11),(9,12),(11,13),(13,14)
s0 on combining above two,we can say that the numbers occur in following groups to satisfy above conditions,
(1,8,4,2),(3,9,12,6),(7,11,13,14)&(5,10) for presence of one member of group in s' means presence of other elements of pair.
now as per condition that no of elements in s' is 8,we can combine any 2 pf first 3 to form s'.But 5 & 10 special set s' will lead to failure of condition 1 as all other number occur in pair of 4.
so 5 &10 can't be member of s'
hence option 1.

[getneonow]

Originally Posted by 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).

I feel this solution is the beauty for today

Though its a simple problem, everyone(including me) have tried to group and find the set but this soln simply uses the algebraic expressions and finds the odd set out in the end by a little reasoning..awsome!!

Moreover this kind of soln comes handy if we are to deal with a large set of S..

I request the puys here to not only post their solns but also have a cursory look of other's solns too.. Assured 95% of them will post the same as your's but its only with the remaining 5% we get to learn. Atleast that's what I do in QQAD.

Neo

[ankitkothari.18]

Hey puys,

My first post to PagalGuy,
The answer is (1) 2.

The limit in the size of the set is 8. And, other than 5 and 10, picking any number in the set results in the addition of 3 more numbers to satisfy the conditions. Also, for any number, if a set is formed satisfying the constraints, the set size will always be a even number.

( ( y + 15 ) / 2 ) / 2 = y
or, y = 5
and, for y = 5, 10 has to be there.
This results in 2 numbers which cannot be there, 5 and 10.
(( y + 15 )/2)/2 = y y + 15 = 4y 3y = 15 y = 5

(( y + 15 )/2)/2 = y y + 15 = 4y 3y = 15 y = 5

[kavita_iet]

S = { 1,2,3,4,5,6,7,8,9,10,11,12,13,14}
x -> even
y -> odd
Checking dependencies
1 - 8, 3 - 9 , 5 - 10, 7 - 11, 9 - 12, 11 - 13, 13 - 14
2 - 1, 4 - 2 , 6 - 3 , 8 - 4 , 10 - 5 , 12 - 6, 14 - 7

Now satisfying the condition of max 8 elements
we get 3 chains of 4 elements - (1,2,4,8,) , (3,6,9,12) , (7,11,13,14)
and 1 chain of two elements - (5,10)
For getting 8 elemets in subset we cant use two numbers 5 and 10. ans (1)

This one was maggi noodles done in - 2 mins

[implex]

Originally Posted by 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).

I really doubt this
there is nothing wrong in x and ((x/2)+ 15)/2 being same.

The thing is (5,10) get disqualified because S' has 8 elements exactly

Here the thing to be aware of is
The subsets are either of 4 element or 2 element. Thats it !

[implex]

Originally Posted by kavita_iet

S = { 1,2,3,4,5,6,7,8,9,10,11,12,13,14}
x -> even
y -> odd
Checking dependencies
1 - 8, 3 - 9 , 5 - 10, 7 - 11, 9 - 12, 11 - 13, 13 - 14
2 - 1, 4 - 2 , 6 - 3 , 8 - 4 , 10 - 5 , 12 - 6, 14 - 7

Now satisfying the condition of max 8 elements
we get 3 chains of 4 elements - (1,2,4,8,) , (3,6,9,12) , (7,11,13,14)
and 1 chain of two elements - (5,10)
For getting 8 elemets in subset we cant use two numbers 5 and 10. ans (1)

This one was maggi noodles done in - 2 mins

solve the other problem Aarav posted !! it will take atleast 10 :P

[aditya_r]

Hey Puys,

Came to office late so replying late,

I too guess the answer is (1) 2.

my logic also being same 10,5 cant go in S' as 8 elements are needed and , there are 3 more sets of 4 elements each, hence its necessary to eliminate 5,10.

Lets see the explanation tomorrow of this question.

Bye Puys.

[kharesaurabh84]

------------------------------------------------------
Quantitative Question # 008
------------------------------------------------------


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


Soln.
All the possible subsets which follows point (b) & (c) are
1. (1,8,4,2)
2. (3,9,12,6)
3. (7,11,13,14)
4. (5,10)


by selecting any of two subsets from first 3 we can make a set S' of 8 element. But subset (5,10) can not be included in any of the case.

x=> (5,10)

n(x)=2

option 1 is right answer