official quant thread for cat08

[theGreenScar]

Since Convivals' conjectures are conniving, i'm posting a new question to keep the action burning:

here goes:

Laurel owes Hardy $10. Neither has any cash but Laurel has 14 DVD players which he values at $ 185 each. He suggests paying his debt with DVD players, Hardy making change by giving Laurel some CD players at $ 110 each. Is this possible? If so, how?

cheers,
TGS
PS: you have exactly 24 hours to solve this.

[IIM maniac]

Originally Posted by theGreenScar

Since Convivals' conjectures are conniving, i'm posting a new question to keep the action burning:

here goes:

Laurel owes Hardy $10. Neither has any cash but Laurel has 14 DVD players which he values at $ 185 each. He suggests paying his debt with DVD players, Hardy making change by giving Laurel some CD players at $ 110 each. Is this possible? If so, how?

cheers,
TGS
PS: you have exactly 24 hours to solve this.

Laurel will give 6 DVD player worth 1110 and in return Hardy will give 10 VCD players worth 1100, so 1110-1100 = 10.Regards.

[convivial]

Originally Posted by swati krishnan

Originally Posted by convivial
Amrita hosted a birthday party and invited all her friends and asked them to invite their friends.There are n people in the party.Only Satyam is not known to Amrita.Each pair that does not include Amrita or Satyam has exactly 2 common friends.Also,Satyam knows everyone except Amrita.If only 2 friends can dance at a time,how many dance numbers will be there at the party?

is it n-2?!!

swati!

sorry puys 4 the late reply
answer is 3n-7

lets take it logically......it is said that satyam is not known 2 amrita nor amrita 2 satyam and rest all frnds know satyam and amrita....also each pair has 2 comon friends,therefore those 2 common frnds r satyam and amrita....lets consider a pair of frnds X and Y and their common frnds i.e satyam and amrita....so total people in the party is n= 4.....now only 2 frnd can dance at a time...so both the frnds will dance with both the common frnds and once together.....
satyam with X
satyam with Y
amrita with X
amrita with Y
X with Y

so total no of dance nos. possible=5
when n=4. dance nos.=5

only 3n-7 satisfies the condition.....


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

[bubai376]

Originally Posted by theGreenScar

Since Convivals' conjectures are conniving, i'm posting a new question to keep the action burning:

here goes:

Laurel owes Hardy $10. Neither has any cash but Laurel has 14 DVD players which he values at $ 185 each. He suggests paying his debt with DVD players, Hardy making change by giving Laurel some CD players at $ 110 each. Is this possible? If so, how?

cheers,
TGS
PS: you have exactly 24 hours to solve this.

Laurel gives 6 DVD players.. which costs 1110$... then hardy gives 10 CD players.. it costs 1100$.. hence 1110-1100=10$ is the change... n laurel repays the debt....

[theGreenScar]

Originally Posted by theGreenScar

Since Convivals' conjectures are conniving, i'm posting a new question to keep the action burning:

here goes:

Laurel owes Hardy $10. Neither has any cash but Laurel has 14 DVD players which he values at $ 185 each. He suggests paying his debt with DVD players, Hardy making change by giving Laurel some CD players at $ 110 each. Is this possible? If so, how?

cheers,
TGS
PS: you have exactly 24 hours to solve this.

6 DVD players
10 VCD players. thats the exchange rate.

cheers,
TGS

[IIM maniac]

Interesting question : There is a circle of radius 1 cm. Each member of a sequence of regular polygon P1(n), n = 5,6..., Where n is the number of sides of polygon is circumscribing the circle and each member of a sequence of regular polygon P2(n), n = 4,5,6..., where n is the number of sides of a polygon, is incribed in the circle. Let L1(n) & L2(n) denote the perimeter of the corresponding polygons P1(n) & P2(n).
Then {L1(13)+2pi}/ L2(17) is :
1. Greater then pi/4 and less than 1
2. Greater then 1 and less than 2
3. Greater then 2
4. Less than 2.
Regards

[khanna_sumit]

Originally Posted by IIM maniac

Interesting question : There is a circle of radius 1 cm. Each member of a sequence of regular polygon P1(n), n = 5,6..., Where n is the number of sides of polygon is circumscribing the circle and each member of a sequence of regular polygon P2(n), n = 4,5,6..., where n is the number of sides of a polygon, is incribed in the circle. Let L1(n) & L2(n) denote the perimeter of the corresponding polygons P1(n) & P2(n).
Then {L1(13)+2pi}/ L2(17) is :
1. Greater then pi/4 and less than 1
2. Greater then 1 and less than 2
3. Greater then 2
4. Less than 2.
Regards

i think this question appeared in one of the CAT exams....
any L1(n) will have perimeter greater than the perimeter of the circle in question and L2(n) will have perimeter less than the circle. means L1(n) > 2*pie and L2(n) < 2*pie... which makes the given expression greater than 2.... hence option 3 is the answer.....

[convivial]

Originally Posted by khanna_sumit

i think this question appeared in one of the CAT exams....
any L1(n) will have perimeter greater than the perimeter of the circle in question and L2(n) will have perimeter less than the circle. means L1(n) > 2*pie and L2(n) < 2*pie... which makes the given expression greater than 2.... hence option 3 is the answer.....

hey puys...........just check out this site.........www.perplexus.info........i think it will be helpful 4 problem solving






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

[Cute.Sid]

Originally Posted by convivial

sorry puys 4 the late reply
answer is 3n-7

lets take it logically......it is said that satyam is not known 2 amrita nor amrita 2 satyam and rest all frnds know satyam and amrita....also each pair has 2 comon friends,therefore those 2 common frnds r satyam and amrita....lets consider a pair of frnds X and Y and their common frnds i.e satyam and amrita....so total people in the party is n= 4.....now only 2 frnd can dance at a time...so both the frnds will dance with both the common frnds and once together.....
satyam with X
satyam with Y
amrita with X
amrita with Y
X with Y

so total no of dance nos. possible=5
when n=4. dance nos.=5

only 3n-7 satisfies the condition.....


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

I thought it was like many pairs can dance at a time but the people forming the pair must be friends

[dashing]

Originally Posted by prashant225

hey guys

25^102/17

Remainder for the same?

Is there any short cut for this sum.Cauz it took me around 4 mins to solve this one today.

And find the numbers between 100 and 400 which are divisible by either 2,3,5 and 7.

regards
prashant

For "25^102/17"

102 is divisible by 17 so the remainder is zero

For "And find the numbers between 100 and 400 which are divisible by either 2,3,5 and 7."
havnt come across any shortcut till now