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

[D33pblue]

Originally Posted by implex

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 !

Consider this
Suppose lets say that the question is 1)S' has only 2 elements, rest of things remain same.

Then you would reason that (5, 10) should be in S'

But 10 can be in S' iff 5 is in S'
And 5 can be in S' iff 10 is in S'

How would you resolve this deadlock then ?

In this case(of S' having only 2 elements): You would reason that (5,10) is in, where as I would reason that S' cannot have any elements !

[twinkle2]

option 1
is d ans acc 2 me
1,2,8, 4 occur in one pair
7 11 13 14 occur in anodr pair
3,9,12,6 in anothr pair
and 5,10 in one pair

and for s` to have 8 elements 5, 10 will always b excluded so answer is 2 nos can never b included..
plzz tell if my logic is ryt. didnt c nybodys else`s reasong

[D33pblue]

Oh! I got where I am going wrong.

The solution I posted is not correct, but coincidently leads to a correct answer

(5, 10) are not included purely for the reason that S' has only 8 elements
and including (5,10) would prevent us from achieving that

Thanks Implexfor pointing that out !

[implex]

Originally Posted by D33pblue

Consider this
Suppose lets say that the question is 1)S' has only 2 elements, rest of things remain same.

Then you would reason that (5, 10) should be in S'

But 10 can be in S' iff 5 is in S'
And 5 can be in S' iff 10 is in S'

How would you resolve this deadlock then ?

In this case(of S' having only 2 elements): You would reason that (5,10) is in, where as I would reason that S' cannot have any elements !

10 can be in S' iff 5 is in S'
And 5 can be in S' iff 10 is in S'

aren't these two equivalent?? Am I missing something? the thing is if and only if. so if 5 is there 10 is there

if 10 is there 5 is there

the proposition is true and the converse as well !!
so I really am not sure what you wanted to say there!!!

[implex]

Originally Posted by getneonow

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

this solution is beautifully wrong!! the thing is all that glitters is not gold

[anubhutiv]

As everyone has indicated:
Quadraplet 1: (1,2,4,8 )
Quadraplet 2: (3,6,9,12)
Quadraplet 3: (7,11,13,14)

Duplet left : (5,10)

Since, at any time S' has 8 elements, so at any time X shall have rest of the six elements which are not in S'.

Answer should be choice (4). That is '6'.

[implex]

Originally Posted by anubhutiv

As everyone has indicated:
Quadraplet 1: (1,2,4,8 )
Quadraplet 2: (3,6,9,12)
Quadraplet 3: (7,11,13,14)

Duplet left : (5,10)

Since, at any time S' has 8 elements, so at any time X shall have rest of the six elements which are not in S'.

Answer should be choice (4). That is '6'.

read the question again !!

[thebornattitude]

QQAD no.008

I always miss on doing the question early due to this !!@@@## office neways

SOLUTION-

The numbers that reapeat in a cycle ,to occur in the subset S' are

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


As all except 5-10 have 4 elements,,,so choosing any 2 among the three will give us a subset S'.. Just 5-10 will not be apart of S'.

Answer- option 1- 2

[convivial]

possible set of nos. can b

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

the set S contains 8 elements.........so any of the two pairs of 4 elements except 5,10 can form set S

X={5,10}
n(X)=2

option 1 holds true......



cogito,ergo,sum................

[Aarav]

Solution to last year's QQAD.

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 Nilesh is 5 times as old as Anupam. At yet another time, Nilesh is twice as old as Shrikant and Anupam is Y years old. There are different possibilities of what Y can be. The largest possible Y is in the range

(1) [15, 18] (2) [21, 25] (3) [27, 32] (4) [35, 39] (5) none of these

Solution:

Let each f1 lies in [0, 1) and each y1 lies in (0, 1).
At some point, Anupam has lived 4 + f1 years, Nilesh has lived n + f2 years and Shrikant has lived s + f3 years where n and s are integers satisying n = 3s.
At another time, say a + f4 years later where a is an integer, Shrikant is twice as old as Anuapm .... => Anupam would have lived 4 + a + f1 + f4 years, he will be 4 + a + y1 years old. Similarly, Nilesh would be n + a + y2 years old and Shrikant would be s + a + y3 years old.

=> s + a + y3 = 2(4 + a + y1) and n + a + y2 = 5(4 + a + y1)
At yet another time, say b + f5 years later after Anupam lived 4 + f1 years, Anupam is 4 + b + y4 years old and n + b + y5 = 2(s + b + y6). This equation and n = 3s imply b = n - 2s + y5 - 2y6 = s + y5 - 2y6.

=> s = a + 8 + 2y1 - y3 and 3s = 4a + 20 + 5y1 - y2.
=> m = 4m - 3m = 12 + 3y1 - 4y3 + y2
Combining the above we get Y = 16 + 3y1 - 4y3 + y2 + y5 - 2y6 + y4, since each y(i) lies between 0 to 1 => T <= 16 + 3 + 1 + 1 + 1 = 22.

Thus, Anupam's age is 4.6, Shrikanth's is 16 and Nilesh's is 48.6
Next: Anupam's age is 12, Shrikant's is 22.4 and Nilesh's is 55
Next: Anupams age is 22.1, Shrikant's is 33.5 and Nilesh's is 66.1