Official Quant thread for CAT 2013

@veertamizhan said:
A train meets with an accident 150 km from the originating station. It completetd the remaining journey at half of the usual speed and reached an hour late at the destination. Had the accident took palce 30 km later, it would have been only half an hour late. Find the distance between the two stations.
let distance be = x and speed be = 2y

150/2y + (x - 150)/y = x/2y + 1
180/2y + (x - 180)/y = x/2y + 1/2

subtract both

=> 2y = 60 = speed

so , distance will be = 210
@EagleMenace said:
No need of any calculation.The answer is given in statement. Statement says it would have taken 30 mins less if accident had taken 30 km later, that means the distance he needs to travel to reach station is 60 km, hence he is taking 60 mins..So total distance is 150+60=210km
Best. Solution. :mg:

Solve the question in the attachement
@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 ?

@chillfactor said:
Solve the question in the attachement
2sqrt(221)?

Height of W from AB = h
1/2 * h * 20 = 1/2 * 16 * 12
h = 9.6

Vertical distance between W and K = 9.6 + 10 + 9.6 = 29.2

Let the perpendicular dropped from W to AB meet AB at E
WB/WA = WE/EA
16/12 = 9.6/EA
EA = 7.2

Horizontal distance = 20 - 7.2 - 7.2 = 5.6

WK = sqrt(29.2^2 + 5.6^2) = sqrt(852.64 + 31.36) = sqrt(884) = sqrt(4 * 221) = 2sqrt(221)
@chillfactor said:
Solve the question in the attachement
RT(884)
2 RT 221
@chillfactor said:
Solve the question in the attachement
2 rt (221) ??

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? ..



@chillfactor said:
Solve the question in the attachement
Let A be at origin
A(0,0) B(20,0) C(20,10) D(0,10) W(x,y)
Required distance=2WC where C is the center of rectangle
C=(20/2,10/2)=(10,5)
WC=20, WD=12
x^2+y^2-20y=12^2-10^2---1
x^2-40x+y^2-20y=16^2-20^2-10^2--2
Subtract 1 and 2
40x=288; x=7.2, y=19.6
Req distance=2*WC=2*sqrt((2.8)^2+(14.6)^2)=2sqrt(221)

@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? ..
23/144?
@grkkrg said:
23/144?
A-P-P-R-O-A-C-H :P
@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? ..
@ScareCrow28 said:
A-P-P-R-O-A-C-H
Got it from graph

Ashish is studying late into the night and id hungry . He opens his mother's snack cupboard without switching on the lights . Knowing that his mother has kept 10 packets of chips and biscuits in the cupboards . He pulls out 3 packets from the cupboards and all of them turn out to be chips . what is the probability that the snack cupboard contains 1 packets of biscuits and 9 packets of chips ?

i approached it like this.

number of possible cases

3chips packet 7biscuit
4chips 6 biscuit
5chips 5 biscuits
6chips 4biscuits
7chips 3 biscuits
8chips 2 biscuits
9chips 1biscuits
10 chips 0 biscuits(although i took this case also,because in forums everyone was taking it,but when it is given 10 packets of chips and biscuits,then i think ATLEAST 1 biscuits packet must be there.isn't it?? )

so my answer was 1/8,because above there are total 8 cases.

but the answer is Probability = C(9, 3)/{C(10, 3) + C(9, 3) + .... + C(3, 3)} = 84/330 = 14/55.

How??

This was IIFT 2012 question. can anyone explain why 1/8 is wrong answer and how we got 14/55 ?
@nole said:
Ashish is studying late into the night and id hungry . He opens his mother's snack cupboard without switching on the lights . Knowing that his mother has kept 10 packets of chips and biscuits in the cupboards . He pulls out 3 packets from the cupboards and all of them turn out to be chips . what is the probability that the snack cupboard contains 1 packets of biscuits and 9 packets of chips ?i approached it like this.number of possible cases3chips packet 7biscuit4chips 6 biscuit5chips 5 biscuits6chips 4biscuits7chips 3 biscuits8chips 2 biscuits9chips 1biscuits10 chips 0 biscuits(although i took this case also,because in forums everyone was taking it,but when it is given 10 packets of chips and biscuits,then i think ATLEAST 1 biscuits packet must be there.isn't it?? )so my answer was 1/8,because above there are total 8 cases.but the answer is Probability = C(9, 3)/{C(10, 3) + C(9, 3) + .... + C(3, 3)} = 84/330 = 14/55.How??This was IIFT 2012 question. can anyone explain why 1/8 is wrong answer and how we got 14/55 ?
Was discussed many times before
http://www.pagalguy.com/forums/quantitative-ability-and-di/official-quant-thread-c-t-88456/p-3611604/r-4108616?page=355
IFT 2012 question

The 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%increase
b)37%increase
c)45%increase
d)60%increase

How to approach this question?
@Torque024 yeah i have visited that link,but i didn't get the solution.Why the solution is linked to the 3 packet of chips ? that was my doubt.
@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

@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?
100 lekar kiya tha..45% aaya tha exam mein i guess
@nole said:
@Torque024 yeah i have visited that link,but i didn't get the solution.Why the solution is linked to the 3 packet of chips ? that was my doubt.
I mean why are we finding probability of finding 3 packet of chips for each of those 8 cases?
@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?
Let cement production in 2008 be 1
in 2012 it will be 1*1.18*1.18*.88*1.18
=1.445 ,so it increases
.445*100=44.5%