Official Quant thread for CAT 2013

MBA CAT 2024
12823
@bs0409 said:
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
i might sound dumb but, if S' have exactly 8 elements and X is the number of elements of S that are not part of S' then the answer should be 14-8 = 6
it is completely possible i'm not understanding the question at all.
12824
@bs0409 said:
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
Is it three elements?? (5,10 and 15)
12825
@somnathbhatta said:
i might sound dumb but, if S' have exactly 8 elements and X is the number of elements of S that are not part of S' then the answer should be 14-8 = 6 it is completely possible i'm not understanding the question at all.
IMO, what the question is asking for is if there are n1 to n14 elements in S, then how many of there cannot be a part of S'. So, basically the question is not about S, it is about S'.
But then I can be wrong. :)
12826
@pavimai what would 6/5 mean in the last equation?????
12827

A retailer buys a sewing machine at a discount of 15% and sells it for Rs.1955. Thus,he makes a profit of 15%. The discount is (in Rs.)______

12828
@cadmium said:
A retailer buys a sewing machine at a discount of 15% and sells it for Rs.1955. Thus,he makes a profit of 15%. The discount is (in Rs.)______
300
12829
@pyashraj said:
@bs0409Should be 2..5 n 10 are only the number that cannot be a part of S'..
@Logrhythm said:
IMO, what the question is asking for is if there are n1 to n14 elements in S, then how many of there cannot be a part of S'. So, basically the question is not about S, it is about S'. But then I can be wrong.
@Logrhythm said:
Is it three elements?? (5,10 and 15)
ANS is 3. 5,10 and 15 cannot be part of the subset S' in any case.

New Age Consultants have three consultants Gyani, Medha and Buddhi. The sum of the number of projects handled by Gyani and Buddhi individually is equal to the number of projects in which Medha is involved. All three consultants are involved together in 6 projects. Gyani works with Medha in 14 projects. Buddhi has 2 projects with Medha but without Gyani, and 3 projects with Gyani but without Medha. The total number of projects for New Age Consultants is one less than twice the number of projects in which more than one consultant is involved.
3. What is the number of projects in which Medha alone is involved?
[1] Uniquely equal to zero
[2] Uniquely equal to 1
[3] Uniquely equal to 4
[4] Cannot be determined uniquely
[5] None of the above.
4. What is the number of projects in which Gyani alone is involved?
[1] Uniquely equal to zero [2] Uniquely equal to 1
[3] Uniquely equal to 4 [4] Cannot be determined uniquely [5] None of the above.
12830
@cadmium said:
A retailer buys a sewing machine at a discount of 15% and sells it for Rs.1955. Thus,he makes a profit of 15%. The discount is (in Rs.)______
CP = 1955/1.15 = 1700
MP = 100/85*1700 = 2000.
Discount = 15/100*2000 = 300.
12831
@cadmium said:
A retailer buys a sewing machine at a discount of 15% and sells it for Rs.1955. Thus,he makes a profit of 15%. The discount is (in Rs.)______
Price before discount = 2000
Dis = 300
12832
@bs0409 said:
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
please give the explanation .. couldn't get it at all
12833
@bs0409 said:
ANS is 3. 5,10 and 15 cannot be part of the subset S' in any case.New Age Consultants have three consultants Gyani, Medha and Buddhi. The sum of the number of projects handled by Gyani and Buddhi individually is equal to the number of projects in which Medha is involved. All three consultants are involved together in 6 projects. Gyani works with Medha in 14 projects. Buddhi has 2 projects with Medha but without Gyani, and 3 projects with Gyani but without Medha. The total number of projects for New Age Consultants is one less than twice the number of projects in which more than one consultant is involved. 3. What is the number of projects in which Medha alone is involved? [1] Uniquely equal to zero [2] Uniquely equal to 1 [3] Uniquely equal to 4 [4] Cannot be determined uniquely [5] None of the above. 4. What is the number of projects in which Gyani alone is involved? [1] Uniquely equal to zero [2] Uniquely equal to 1 [3] Uniquely equal to 4 [4] Cannot be determined uniquely [5] None of the above.
Refer to the attached figure
12834
1,1,1,2,2,2,3,4,4,4,5,6,?....wt is the next term??
12835
@sumit99 said:
1,1,1,2,2,2,3,4,4,4,5,6,?....wt is the next term??
6 .

Solution:
1,1,1,2,2,2,3,4,4,4,5,6,?
club the given terms in groups of 2 :
(1,1), (1,2), (2,2), (3,4), (4,4), (5,6), {6,6}
and the sequence continues as
(1,1), (1,2), (2,2), (3,4), (4,4), (5,6), (6,6), (7,8), (8,8), ....
no mathematical formula used. simple logic. please share the answer. different approach if any .. please share that too.
12836
@bs0409 said:
ANS is 3. 5,10 and 15 cannot be part of the subset S' in any case.

New Age Consultants have three consultants Gyani, Medha and Buddhi. The sum of the number of projects handled by Gyani and Buddhi individually is equal to the number of projects in which Medha is involved. All three consultants are involved together in 6 projects. Gyani works with Medha in 14 projects. Buddhi has 2 projects with Medha but without Gyani, and 3 projects with Gyani but without Medha. The total number of projects for New Age Consultants is one less than twice the number of projects in which more than one consultant is involved. 3. What is the number of projects in which Medha alone is involved? [1] Uniquely equal to zero [2] Uniquely equal to 1 [3] Uniquely equal to 4 [4] Cannot be determined uniquely [5] None of the above. 4. What is the number of projects in which Gyani alone is involved? [1] Uniquely equal to zero [2] Uniquely equal to 1 [3] Uniquely equal to 4 [4] Cannot be determined uniquely [5] None of the above.

q1)1
q2)4
12837

New Age Consultants have three consultants Gyani, Medha and Buddhi. The sum of the number of projects handled by Gyani and Buddhi individually is equal to the number of projects in which Medha is involved. All three consultants are involved together in 6 projects. Gyani works with Medha in 14 projects. Buddhi has 2 projects with Medha but without Gyani, and 3 projects with Gyani but without Medha. The total number of projects for New Age Consultants is one less than twice the number of projects in which more than one consultant is involved. 3. What is the number of projects in which Medha alone is involved? [1] Uniquely equal to zero [2] Uniquely equal to 1 [3] Uniquely equal to 4 [4] Cannot be determined uniquely [5] None of the above. 4. What is the number of projects in which Gyani alone is involved? [1] Uniquely equal to zero [2] Uniquely equal to 1 [3] Uniquely equal to 4 [4] Cannot be determined uniquely [5] None of the above.
1.5 2.4
12838

hi puys,


Any shorter/faster method of calculating HCF and LCM?


Thanks

12839

Find the unit digit:
1^2! + 2^3! + 3^4! + 4^5! +.........+150^151! +...153^154!
a) 1
b) 0
c) 5
d) 7

12840
@ravi.theja said:
Find the unit digit:1^2! + 2^3! + 3^4! + 4^5! +.........+150^151! +...153^154!a) 1b) 0c) 5d) 7

Lets first not consider 1^2! +2^3+3^4!

Of the remaining 150 numbers, 75 are even and 75 are odd to the power 4n

even no^4n = 6
odd no^4n =1

So (75*6)+(75*1)=unit digit = 5

Unit digit of 1^2! +2^3+3^4! = 0

So Overall unit digit =5

Am I correct?
12841
@nramachandran said:
Lets first not consider 1^2! +2^3+3^4!Of the remaining 150 numbers, 75 are even and 75 are odd to the power 4neven no^4n = 6odd no^4n =1So (75*6)+(75*1)=unit digit = 5Unit digit of 1^2! +2^3+3^4! = 0So Overall unit digit =5Am I correct?
nope 😞
12842
@ravi.theja said:
Find the unit digit:1^2! + 2^3! + 3^4! + 4^5! +.........+150^151! +...153^154!a) 1b) 0c) 5d) 7

till 150 , we will have 15 blocks(1,6,1,6,5,6,1,6,1,0) .. but catch is 2^3! ends in 4 where as other 2 series will end in 6

so till 150 the number ends in 5 - 2 = 3

now add rest of three numbers

3 + 1+ 6 + 1 = 1