Official Quant thread for CAT 2013

Sorry for Spamming here,But I think the purpose of the thread is something I never get and that's why I refrained from Posting in 2012 thread. Think will do the same.


This is a thread which has the potential to be the best thread and by best, I don't mean the most posted one. But only if, we all stick to one agenda.

My question is, What is the Purpose of this thread. If this is the thread where we can show our mathematical prowess and impress people, think we are going right! But is this the purpose?

I highly doubt any newcomer can learn Quant from here. See, any of us who have prepared for 4-5 months will be able to solve most of the questions here, well, most of those who will ever come in CAT. It's the newbies who cannot be helped with this thread. It's been ONE, only ONE day and its difficult to follow and search this thread. I understand we all are eager to show what we can do, but let's not do it.

Lets Follow a structure, And then actually follow it so that Anyone who comes here, will know where to find what type of question. Can any of us really find anything in 2012 thread? Let's not do the same!

EDIT: Lets Chalk out a Plan, Maybe Topic Wise Questions First in Topic Wise Threads and then post only Relevant Question and More Importantly, ONLY Relevant Solutions Here. Any Mod can do this.Also, I am ready to Volunteer for the same.
@soumitrabengeri
@rkshtsurana said:
we have consider the case when Q will be in between x y z18! me yahi toh hgalike yaha 4! me ho rha he..x yz QxQzy , xzQy..and so on.


bhai ,both of u are absolutely right..

i made a small calculation mistake..(assumed the total things as 19)

now 24p7*18!*2!=24!*18!*2/17!=24!*36

i think this concludes it


@soumitrabengeri said:
In how many ways can the letters of the English alphabet be arranged so that there are 7 letters between the letters A and B?1) 31!*2!2) 24P7*18!*23) 36*24!4) 26P7*20!*2

A and B KO 2! ME SELECT KIYA
THEN 24 KO 7 ME SE CHOOSE KIYA AND THEN ARRANGE 24C7*7!

AND THEN 18! ME BAKI KO ARRANGE KIYA

SO 2!*24C7*7!*18!=36*24!
@soumitrabengeri said:
A locker at a famous bank can be opened by dialing a fixed 3 digit code(between 000 and 999). Dawood, a terrorist, only knows that the number is a 3 digit number and has only one six. Using this he tries to open the lock by dialing 3 digits at random. The probability that he succeeds in this is..1) 1/2432) 1/9003) 1/2164) 1/1000

xx6=9*9=81
x6x=81
6xx=81

total no. with one six=243

so probability=1/243
@soumitrabengeri said:
In how many ways can the letters of the English alphabet be arranged so that there are 7 letters between the letters A and B?1) 31!*2!2) 24P7*18!*23) 36*24!4) 26P7*20!*2
24C7*7!*18!*2
Option 2
@soumitrabengeri said:
2 women Katrina and Deepika are working on an embroidery design. If Deepika worked alone, she would need 8 hours more to complete the design than if they both worked together. Now, if Katrina worked alone, she would need 4.5 hours more to complete the design than if they were working together. What time would it take for Katrina alone to complete the design?1) 10.5 hours2) 12.5 hours3) 14.5 hours4) 18.5 hours

Say K and D are the hourly rates of work

=>Kt = 8D....................(i)
=>Dt = 4.5K.................(ii)

dividing (i) by (ii) we get (K/D) = 4/3

using (i) we get t = 8*(D/K) = 8*(3/4)= 6 hrs

so katrina working alone would take t+4.5 = 6+4.5 = 10.5 hrs

ATDH.
Mandeep and Jagdeep had gone to visit Ranpur, which is a seaside town and also known for the presence of the historical ruins of an ancient kingdom. They stayed in a hotel which is exactly 250 meters away from the railway station. At the hotel, Mandeep and Jagdeep learnt from a tourist information booklet that the distance between the sea-beach and the gate of the historical ruins is exactly 1 km. Next morning they visited the sea-beach to witness sunrise and afterwards decided to have a race from the beach to the gate of the ruins. Jagdeep defeated Mandeep in the race by 60 meters or 12 seconds. The following morning they had another round of race from the railway station to the hotel. How long did Jagdeep take to cover the distance on the second day?
A.53 seconds
B.47 seconds
C.51 seconds
D.45 seconds

@deedeedudu said:
Mandeep and Jagdeep had gone to visit Ranpur, which is a seaside town and also known for the presence of the historical ruins of an ancient kingdom. They stayed in a hotel which is exactly 250 meters away from the railway station. At the hotel, Mandeep and Jagdeep learnt from a tourist information booklet that the distance between the sea-beach and the gate of the historical ruins is exactly 1 km. Next morning they visited the sea-beach to witness sunrise and afterwards decided to have a race from the beach to the gate of the ruins. Jagdeep defeated Mandeep in the race by 60 meters or 12 seconds. The following morning they had another round of race from the railway station to the hotel. How long did Jagdeep take to cover the distance on the second day?A.53 seconds B.47 seconds C.51 seconds D.45 seconds
B. 47 seconds

Let speeds of Mandeep and Jagdeep be M and J respectively.
1000/J = 940/M
940/M - 1000/J = 0____(i)

1000/M -1000/J = 12_____(ii)

Subtract (i) from (ii):
60/M = 12
M = 5m/s
J = 1000/188 = 250/47 m/s

Time taken by Jagdeep to cover 250m = 250/(250/47) = 47s
@deedeedudu said:
Mandeep and Jagdeep had gone to visit Ranpur, which is a seaside town and also known for the presence of the historical ruins of an ancient kingdom. They stayed in a hotel which is exactly 250 meters away from the railway station. At the hotel, Mandeep and Jagdeep learnt from a tourist information booklet that the distance between the sea-beach and the gate of the historical ruins is exactly 1 km. Next morning they visited the sea-beach to witness sunrise and afterwards decided to have a race from the beach to the gate of the ruins. Jagdeep defeated Mandeep in the race by 60 meters or 12 seconds. The following morning they had another round of race from the railway station to the hotel. How long did Jagdeep take to cover the distance on the second day?A.53 seconds B.47 seconds C.51 seconds D.45 seconds
Ans is 47 secs..

Ratio of speed of jagdeep and mandeep = 1000/940=50/47

Speed of mandeep = Distance/Time = 60/12 = 5m/s
so speed of jagdeep = (50/47)*5=5.31m/s

Time taken by jagdeep to cover second distance = 250/5.31 = 47 secs..
@deedeedudu said:
Mandeep and Jagdeep had gone to visit Ranpur, which is a seaside town and also known for the presence of the historical ruins of an ancient kingdom. They stayed in a hotel which is exactly 250 meters away from the railway station. At the hotel, Mandeep and Jagdeep learnt from a tourist information booklet that the distance between the sea-beach and the gate of the historical ruins is exactly 1 km. Next morning they visited the sea-beach to witness sunrise and afterwards decided to have a race from the beach to the gate of the ruins. Jagdeep defeated Mandeep in the race by 60 meters or 12 seconds. The following morning they had another round of race from the railway station to the hotel. How long did Jagdeep take to cover the distance on the second day?A.53 seconds B.47 seconds C.51 seconds D.45 seconds
option B--47 seconds..

the ratio of speeds of J/M= 50/47...(1000/940)

and since the distance they have to cover is same (250m).. the ratio of the time taken by J/M=47/50..

hence Jagdeep takes 47 seconds.

P.S.== osrry for being so late in replying...but der ayr durust aye..cheers
5A(n+1) + 1 = 5*An + n. Given A5 = 55, find A55.
(n+1) is subscript.


Why this thread is so dead today..? 😞
S is the region described by the following:
1. |x|
2. |y|
3. x > y

Find the Perimeter of the region formed by the intersection of S and the unit circle with center at (1,1)

1. 8
2. 4
3. between 1 and 2.75
4. between 2.75 and 4
@vijay_chandola said:
5A(n+1) + 1 = 5*An + n. Given A5 = 55, find A55.
(n+1) is subscript.


Why this thread is so dead today..?
Is it 340.?
@sowmyanarayanan said:
S is the region described by the following:1. |x| 2. |y| 3. x > yFind the Perimeter of the region formed by the intersection of S and the unit circle with center at (1,1)1. 82. 43. between 1 and 2.754. between 2.75 and 4
option 3?
@vijay_chandola said:
5A(n+1) + 1 = 5*An + n. Given A5 = 55, find A55.(n+1) is subscript.Why this thread is so dead today..?
A(n+1) - An = (n-1)/5

=>A55 - A54 = 53/5
=>A54 - A53 = 52/5
=>A53 - A52 = 51/5
.
.
.
.=>A7-A6 = 5/5
=>A6 - A5 = 4/5

adding all we get

A55-A5 = (1/5){4+5+6+......53}
=>A55 = (1/5){(1+2+3+4....+52+53) - (1+2+3)} + A5.
=>A55 =(1/5)*{53*54/2 - 3*4/2} + 55 = 340.


ATDH.

@vijay_chandola said:
5A(n+1) + 1 = 5*An + n. Given A5 = 55, find A55.(n+1) is subscript.Why this thread is so dead today..?
A55=(5*55+5+6+7+..+54)/5
=(5*55+27*55-10)/5
=55+297-2
=350
@vijay_chandola said:
5A(n+1) + 1 = 5*An + n. Given A5 = 55, find A55.(n+1) is subscript.Why this thread is so dead today..?
350?
If x is the side of the largest equilateral triangle that can be drawn inside a square of side 1, what can be said about x?

1) 0.7
2) x = 1
3) 1
4) x > 1.1
@krum said:
A55=(5*55+5+6+7+..+54)/5=(5*55+27*55-10)/5=55+297-2=350
@pankaj1988 said:
350?

OA is 340.
refer to @anytomdickandhary sir's post
@vijay_chandola said:
If x is the side of the largest equilateral triangle that can be drawn inside a square of side 1, what can be said about x?1) 0.7 2) x = 13) 1 4) x > 1.1
x=1?