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

[rajeev_hts]

@srikar2097

Simple any subset of 4 or more than 4 will have few pairs those will give summation 7.

[srikar2097]

Originally Posted by rajeev_hts

@srikar2097

Simple any subset of 4 or more than 4 will have few pairs those will give summation 7.

Agreed they might be few but if we need to consider them right?

Just because they are few, how can we neglect them

[rajeev_hts]

Originally Posted by srikar2097

Agreed they might be few but if we need to consider them right?

Just because they are few, how can we neglect them

Then we can't say equation x+y=7 has NO solution for that subset. It will have those "few" solutions.

I hope you got it.

[siddooba]

Originally Posted by srikar2097

Agreed they might be few but if we need to consider them right?

Just because they are few, how can we neglect them

No you have S1 = (1,6) ,S2= (2,5) ,S3 = (3,4)
At max you can choose 1 each form these 3 , so you cant have 4 element sets .

[selebratinglife]

Originally Posted by srikar2097

------------------------------------------------------
Quantitative Question # 024
------------------------------------------------------

The number of those subsets of {1, 2, 3, 4, 5, 6} such that the equation x+y = 7 has no solution in it is

(1) 18 (2) 21 (3) 27 (4) 36 (5) none of these

----------------------------------------------------------------

The cases for which there is no solution for the Eq x+y = 7 in the subsets are..
Number of subsets with single digit--(6)
Number of subsets with two digits--(12)[Total 6c2-3((1,6)(2,5)(3,4))
Number of subsets with 3 digits---( 8 ) [Can be calculated)
There is also null set which is a subset of the given set--(1)
No 4 digit or 5 digit subsets satisfy...
Adding--6+12+8+1=27
So option (3)

[slam]

Originally Posted by srikar2097

Slight correction, the no. of sets with 2 elements which will solve this equation is (x,y) = (1,6) (2,5) (3,4) (4,3) (5,2) (6,1)

You have missed out 3 sets. So the answer: 24

How is the set {2,1} different from the set {2,1}? When we talk of sets, the order in which we write the elements in the set should not matter. They will both be the same and contain the solution.

[selebratinglife]

Originally Posted by srikar2097

Slight correction, the no. of sets with 2 elements which will solve this equation is (x,y) = (1,6) (2,5) (3,4) (4,3) (5,2) (6,1)

You have missed out 3 sets. So the answer: 24

Nope..Here we are talking about subsets not about (x,y) so (1,6) or (6,1) doesn't matter. They are one and the same as far as subsets are concerned..Answer is still 27..

[slam]

Originally Posted by srikar2097

------------------------------------------------------
Quantitative Question # 024
------------------------------------------------------

The number of those subsets of {1, 2, 3, 4, 5, 6} such that the equation x+y = 7 has no solution in it is

(1) 18 (2) 21 (3) 27 (4) 36 (5) none of these

----------------------------------------------------------------

My solution is as below:
(1,6) (5,2) and (4,3) will be the three solution sets for x + y = 7

Nbr of sets with (1,6) = 4c0 + 4c1 + 4c2 + 4c3 + 4c4 = 16
Nbr of sets with (1,6) and (5,2) = 2c0 + 2c1 + 2c2 = 4
Nbr of sets with all three = 1

So, total = 16 * 3 - 4 * 3 + 1
= 48 - 12 + 1
= 37

Now, total number of sets are 2^6 = 64
So, the answer would be (64 - 37) = 27. => option (3)

[nsitaditya]

Null Set=1
Set with one element=6
Set with 2 elements= 6C2 - 3 = 12
Set with 3 elements {1,2,3},{1,3,6},{1,2,4},{1,4,6},{1,5,6},{2,3,6},{3 ,5,6},{4,5,6}
i.e 8 subsets

Hence total number of subsets= 1+6+12+8=27

Hence option is choice (3)

[rajeev_hts]

Originally Posted by selebratinglife

The cases for which there is no solution for the Eq x+y = 7 in the subsets are..
Number of subsets with single digit--(6)
Number of subsets with two digits--(12)[Total 6c2-3((1,6)(2,5)(3,4))
Number of subsets with 3 digits---( 8 ) [Can be calculated)
There is also null set which is a subset of the given set--(1)
No 4 digit or 5 digit subsets satisfy...
Adding--6+12+8+1=27
So option (3)

I think you are also including null set and {1} {2}...{6}, can we consider them to check equation x+y=7?