Official Quant thread for CAT 2013

MBA CAT 2024
26608
@iLoveTorres said:
Sir please explain this part
Base length of the central parallelogram can be got by using pythagorous theorem on sides 1 and 1-1/n.

Now, side length of the square is the same as the height of this central parallelogram, which can be got by area of parallelogram / base length of parallelogram = 1/n / (root (1+(1-1/n)^2))

Team BV - Vineet
26609
@bodhi_vriksha @scrabbler seven different objects must be divided among three people. In how many ways can this be done if one or two of them can get no objects?
26610
@imrover said:
@bodhi_vriksha@scrabbler seven different objects must be divided among three people. In how many ways can this be done if one or two of them can get no objects?
a+b+c=7
9!/7!2!=36?
26611
@Subhashdec2 no no...things are different here, not identical.
26612
@imrover said:
@bodhi_vriksha@scrabbler seven different objects must be divided among three people. In how many ways can this be done if one or two of them can get no objects?
for first object we have 3 ways .. similarly for second we again have 3 ways and so on ... this is because one or two can be empty..
Hence no of ways are 3^7 :)

Team BV--Pratik Gauri
26613
@imrover said:
@bodhi_vriksha@scrabbler seven different objects must be divided among three people. In how many ways can this be done if one or two of them can get no objects?
3 cases.. correct me if i am wrong
case1: when everybody gets atleast 1.. this is equivalent to a+b+c=7 but since everybody gets atleast one a'+b'+c'=4 =>6C2
case 2: when one gets zero. this is equivalent to b+c=5 that is 6C1
case3: when two gets zero. this is equivalent to c=7 hence 1 case.
So totally 1+6+15=22 ways
26614
@imrover said:
@Subhashdec2 no no...things are different here, not identical.
are the people different as well?
26615
@albiesriram said:
lots of them. I know i suffered a lot. .last ques of the day from my side..
Alternate solution attached in the pic.

Team BV - Vineet
26616
@bodhi_vriksha thanks :)
26617
@Subhashdec2 lol.... i guess, they always are :)
26618
@iLoveTorres i reckon u too are taking the objects to be identical. they are given as different. refer to the soln by @bodhi_vriksha .
26619
@imrover said:
@Subhashdec2 lol.... i guess, they always are
hahahhah ryt
den see every thing has 3 options hence 3*3*3...7 times=3^7
26620
@bodhi_vriksha said:
What's the OA of this? Team BV--Pratik Gauri
ANS shd be D....
can you please solve it.

26621
@iLoveTorres said:
bhai yeh wala question aapne pehle bhi post kiya hua hai.. i guess the answer is 28.5% if my memory is good
nahi yar nahi post kia.....OA is b..getting it wrong

Total 5000
500 got wasted
2/3 means 3000 * 160 = 4,80,000 (20% discount)
rest 1500 * 200 = 3,00,000

total 7,80,000
profit = 280000/500000 * 100 = 56%


26622
@iLoveTorres said:
3 cases.. correct me if i am wrongcase1: when everybody gets atleast 1.. this is equivalent to a+b+c=7 but since everybody gets atleast one a'+b'+c'=4 =>6C2case 2: when one gets zero. this is equivalent to b+c=5 that is 6C1case3: when two gets zero. this is equivalent to c=7 hence 1 case. So totally 1+6+15=22 ways
suppose i have seven objects like this

_*_*_*_*_*_*_*_

i can select 2 spaces out of 8 to divide the objects in 3 groups.
so , 8C2 = 28...
may be some conceptual error..but what is it?
26623
@vbhvgupta said:
an orange seller make a profit of 20% by selling oranges at certain price. if he charges rs 1.2 higher per oranges he would gain 40%. Find the original price at which he sod an orange
Rs 7.2 ?
26624
@vbhvgupta said:
an orange seller make a profit of 20% by selling oranges at certain price. if he charges rs 1.2 higher per oranges he would gain 40%. Find the original price at which he sod an orange
OA 7.2
26625
@amresh_maverick said:
A is having Rs. 255 all in Re. 1 denominations, in how many minimum number of bags can he distribute this amount so that he can give any denomination from Re. 1 to Rs. 255 without opening any bag?
2^0 + 2^1 + 2^2 + .. .... + 2^7 = 255.

Hence 8 bags
26626

In how many ways can A,B,C,D, E, F, and G stand in queue such that D is always behind B and C is always behind D?


Team BV - Vineet
26627
@bodhi_vriksha said:
In how many ways can A,B,C,D, E, F, and G stand in queue such that D is always behind B and C is always behind D?Team BV - Vineet
A,B,C,D, E, F, G can stand in 7! ways .

requirement is : CDB - can be arranged in 3! ways but only one is acceptable

Hence total ways = 7!/3!

PS: galat hone ke pure assaar hain