Official Quant thread for CAT 2013

@nole said:
IFT 2012 questionThe annual production in cement industry is subjected to business cycles.The production increases for two consecutive years consistently by 18% and decreases by 12% in the third year.Again in the next two years,it increases by 18% each year and decreases by 12% in the third year.Taking 2008 as the base year,what will be the approximate effect on cement production in 2012 ?a)24%increaseb)37%increasec)45%increased)60%increaseHow to approach this question?
c) 45% increase

1.18*1.18*0.88*1.18
This comes close to 1.44
So 45% increase
@ScareCrow28 said:
Duels in the town of discretion are rarely fatal. There, each contestant comes at a random moment between 5 a.m. and 6 a.m. on the appointed day and leaves exactly 5 minutes later, honor served, unless his opponent arrives within the time interval then they fight. What fraction of duels lead to violence? ..

1 - 11^2/12^2

= 23/144
@grkkrg How do we know that the increase or decrease would be applied keeping the FIRST year in mind and not the LATEST year?? I did keeping latest year and scrwed the question.
@grkkrg said:
2sqrt(221)?Height of W from AB = h1/2 * h * 20 = 1/2 * 16 * 12
WK = sqrt(29.2^2 + 5.6^2) = sqrt(852.64 + 31.36) = sqrt(884) = sqrt(4 * 221) = 2sqrt(221)
@rkshtsurana said:
RT(884)2 RT 221
@Brooklyn said:
2 rt (221) ??
Yup, 2 ˆš221 is correct.
Check the attachment for another approach

@nole said:
@chillfactor thanks.I understood your solution,it may sound silly but i want to know how did u realise that c(6,4) approach would lead to redundant cases ? I mean is there any way to avoid such approach in exam to save time ?
I have already explained that why your approach was incorrect.

Solve this question, it will help you.

In how many ways we can form a team of five from 9 persons, three persons each from three different countries, such that at least one person from each country is selected.
@nole said:
@grkkrg How do we know that the increase or decrease would be applied keeping the FIRST year in mind and not the LATEST year?? I did keeping latest year and scrwed the question.
It is mentioned that 2008 is the base year. :)
@chillfactor said:
Yup, 2 ˆš221 is correct.Check the attachment for another approachI have already explained that why your approach was incorrect.Solve this question, it will help you.In how many ways we can form a team of five from 9 persons, three persons each from three different countries, such that at least one person from each country is selected.

Cases -> 3,1,1 : 3c1*3c1*3c3 *3

2,2,1 -> 3c1*3c2*3c2 *3

add all above-> 108??

Edited
@chillfactor said:
In how many ways we can form a team of five from 9 persons, three persons each from three different countries, such that at least one person from each country is selected.
108?

A1,A2,A3,B1,B2,B3,C1,C2,C3

Arrangements
2,2,1:
3C1 * 3C1 * 3C2 * 3C2 = 3 * 3 * 3 * 3 = 81

3,1,1:
3C1 * 3C1 * 3C1 = 27

Total = 108
@grkkrg said:
108?A1,A2,A3,B1,B2,B3,C1,C2,C3Arrangements 2,2,1: 3C1 * 3C1 * 3C2 * 3C2 = 3 * 3 * 3 * 3 = 813,1,1: 3C1 * 3C1 * 3C1 = 27Total = 108
u forgot to permute last case
@grkkrg ok got it.thanks :)
@Brooklyn said:
u forgot to permute last case

3 1 1:
Select which 3 people from 1 country: 3C1
Select 1 person from 2nd country = 3C1
Select 1 person from 3rd country = 3C1
Total = 27

What am I missing?
@grkkrg said:
3 1 1:Select which 3 people from 1 country: 3C1Select 1 person from 2nd country = 3C1Select 1 person from 3rd country = 3C1Total = 27What am I missing?
there are 3 countries so u can choose 311 as -> 3!/2! ->3 ways

311 or 131 or 113
@Brooklyn said:
there are 3 countries so u can choose 311 as -> 3!/2! ->3 ways 311 or 131 or 113
Yes.. that's what the first 3C1 is for..
@chillfactor said:
Yup, 2 ˆš221 is correct.Check the attachment for another approachI have already explained that why your approach was incorrect.Solve this question, it will help you.In how many ways we can form a team of five from 9 persons, three persons each from three different countries, such that at least one person from each country is selected.
hmm 108 it wud be..

@chillfactor said:
Yup, 2 ˆš221 is correct.Check the attachment for another approachI have already explained that why your approach was incorrect.Solve this question, it will help you.In how many ways we can form a team of five from 9 persons, three persons each from three different countries, such that at least one person from each country is selected.

total = 9c5 = 126 ways

when all 5 selected persons are from only 2 countries = 3c2*6c5 = 18 ways

required ways = 126 - 18 = 108 ways
@chillfactor 1)person from same country thrice,then 2 persons from different countries(aaabc)- 3*3*3*5!/3!
2)(aabbc)-3*3*3*5!/2!2!


is 1350 answer correct ?
@chillfactor said:

In how many ways we can form a team of five from 9 persons, three persons each from three different countries, such that at least one person from each country is selected.
x1 x2 x3 / y1 y2 y3 / z1 z2 z3

3 1 1 = 3C1*3C1*3C1 = 27
2 2 1 = 3C1*3C1*3C2*3C2 = 81

total = 108

sir jii ___/\___
@nole said:
@chillfactor1)person from same country thrice,then 2 persons from different countries(aaabc)- 3*3*3*5!/3!2)(aabbc)-3*3*3*5!/2!2!is 1350 answer correct ?
you are permuting people also, q is only about selecting team
@Brooklyn ya ya sorry 😃 got it.
@chillfactor sirji kuch aur Qs hein kya ? mein toh offc me hu cant post..
@chillfactor said:
In how many ways we can form a team of five from 9 persons, three persons each from three different countries, such that at least one person from each country is selected.
311---3C1*3C1*3C1=27
221---3C2*3C2*3C2*3C1=81
Total=108