Official Quant thread for CAT 2013

@ScareCrow28 said:
Sir mat bolo mere bhai I distributed 1 chocolate to all of them, now only 3 are left. These 3 have 3 options. So 3*3*3 = 27. Highly circumspect @scrabbler Comments?
1 2 3
4 1 1 ---> 6c4*2c1*3!/2! = 90
3 2 1 ---> 6c3*3c2*3! = 360
2 2 2 ---> 1 way
total = 90 +360 + 1 = 451

yeh galat hai kya ?
@scrabbler said:
No clue....are the chocolates distinct or identical? When distributing 3 first you have assumed identical, then taken distinct for 3^3 so can't agree with your answer probably...sochna padegaAre there answer options? mailtoankitregardsscrabbler
no sir, options bhi nahi hai...
@mailtoankit said:
1 2 34 1 1 ---> 6c4*2c1*3!/2! = 903 2 1 ---> 6c3*3c2*3! = 3602 2 2 ---> 1 waytotal = 90 +360 + 1 = 451yeh galat hai kya ?
My solution is for identical chocolates
@mailtoankit said:
1 2 34 1 1 ---> 6c4*2c1*3!/2! = 903 2 1 ---> 6c3*3c2*3! = 3602 2 2 ---> 1 waytotal = 90 +360 + 1 = 451yeh galat hai kya ?
4 1 1 ---> 6!/(1!*1!*4!*2!) equal to 15
3 2 1 ---> 6!/(3!*2!*1!) equal to 60
2 2 2 ---> 6!/(2!*2!*2!*3!) equal to 15
total = 90

now there are 3 childern

so 90*3!

gives me 720

@scrabbler

720 hai option me

nahi toh mereko fir check karna padega


Abdullah, a trader had 3000 bananas with him and he had to take them to a market 1000 km away on his only camel, which can carry a maximum of 1000 bananas at a time. Unfortunately, the camel eats one banana for every 1 km, or a part thereof, that it travels while carrying the bananas. What is the maximum number of bananas that Abdullah can take to the market?
467
533
767
833
@scrabbler said:
No clue....are the chocolates distinct or identical? When distributing 3 first you have assumed identical, then taken distinct for 3^3 so can't agree with your answer probably...sochna padegaAre there answer options? mailtoankitregardsscrabbler
nai...After distributing 3 also, the chocolates have 3 options na? They are identical, right, but the children are diff na? Am I going wrong here?
@ScareCrow28 said:
Sir mat bolo mere bhai I distributed 1 chocolate to all of them, now only 3 are left. These 3 have 3 options. So 3*3*3 = 27. Highly circumspect @scrabbler Comments?
i think it should be 5c2 =10
@mailtoankit said:
1 2 34 1 1 ---> 6c4*2c1*3!/2! = 903 2 1 ---> 6c3*3c2*3! = 3602 2 2 ---> 1 waytotal = 90 +360 + 1 = 451yeh galat hai kya ?
If identical, then 4 1 1 for example is just 3 ways....6!/4! makes sense only if the chocs are distinct...

regards
scrabbler

@scrabbler said:
If identical, then 4 1 1 for example is just 3 ways....6!/4! makes sense only if the chocs are distinct...regardsscrabbler
haan sir, chocolates identical hai....my solution is wrong......ur approach ?
@ScareCrow28 said:
nai...After distributing 3 also, the chocolates have 3 options na? They are identical, right, but the children are diff na? Am I going wrong here?
Yes. Suppose the 1st of the remaining chocs goes to kid P and the 2nd to Q and 3rd to Q. Another case, 1st to Q 2nd to P 3rd to Q. You are counting them as two separate cases. But they are the same if chocs identical.

regards
scrabbler

@Spinelli said:
A water tank has steps inside it. A monkey is sitting on the topmost step(i.e. the first step). The water level is at the ninth step.(i) He jumps 3 steps down and then jump back 2 steps up. In how many jumps will he reach the water level?(ii) After drinking water, he wants to go back. For this, he jumps 4 steps up and then jumps back 2 steps down in every move. In how many jumps will he reach back the top step?
11 n 5
@scrabbler said:
If identical, then 4 1 1 for example is just 3 ways....6!/4! makes sense only if the chocs are distinct...regardsscrabbler
OA kya hai bhai?
@scrabbler said:
Yes. Suppose the 1st of the remaining chocs goes to kid P and the 2nd to Q and 3rd to Q. Another case, 1st to Q 2nd to P 3rd to Q. You are counting them as two separate cases. But they are the same if chocs identical.regardsscrabbler
Then it should be 5C2
@mailtoankit said:
haan sir, chocolates identical hai....my solution is wrong......ur approach ?
If the chocs are distinct:

Consider kids A, B, C.

If no conditions, the 6 chocs each have 3 ways to go, this gives 3*3*3*3*3*3 = 729 ways in all, some of which need to be subtracted (where only 1 or only 2 get chocs).

Now cases where A does not get a choc - 2^6 = 64. Which includes 2 cases where B and C respectivel also don't get a chos. So cases where only A doesn't get are 62. Similarly for only B and only C.

Cases where exactly 2 of them don't get a choc - 3 (AB, AC, BC).

so 729 - 3*62 -3 = 540.

If the chocs are identical, then we can use the theory of partitioning and say 6 in 3 grps such that no grp empty is 5C2 or 10 ways. Or else list them
411 141 114
321 312 231 213 132 123
222

regards
scrabbler

@joyjitpal said:
4 1 1 ---> 6!/(1!*1!*4!*2!) equal to 153 2 1 ---> 6!/(3!*2!*1!) equal to 602 2 2 ---> 6!/(2!*2!*2!*3!) equal to 15total = 90now there are 3 childernso 90*3!gives me 720@scrabbler720 hai option menahi toh mereko fir check karna padega
is solution mein burayi kya thi mere bhai
@joyjitpal said:
is solution mein burayi kya thi mere bhai
Burai ye hai = 90*3! = 540! :mg:
@pavimai said:
Abdullah, a trader had 3000 bananas with him and he had to take them to a market 1000 km away on his only camel, which can carry a maximum of 1000 bananas at a time. Unfortunately, the camel eats one banana for every 1 km, or a part thereof, that it travels while carrying the bananas. What is the maximum number of bananas that Abdullah can take to the market?467533767833
iska solution koi batao :)
@ScareCrow28 said:
Burai ye hai = 90*3! = 540!
@pavimai said:
Abdullah, a trader had 3000 bananas with him and he had to take them to a market 1000 km away on his only camel, which can carry a maximum of 1000 bananas at a time. Unfortunately, the camel eats one banana for every 1 km, or a part thereof, that it travels while carrying the bananas. What is the maximum number of bananas that Abdullah can take to the market?467533767833
It is 533.33

http://www.puzzle.dse.nl/math/bananas_us.html

refer to the link...bohot bada solution hai...
@Logrhythm said:
It is 533.33http://www.puzzle.dse.nl/math/bananas_us.htmlrefer to the link...bohot bada solution hai...
bhai mafh kar do ye solution merese nahi hoga