Quant by Arun Sharma

There Are 4 letters and 4 envelopes. in how many ways can wrong choices be made?

a)4^3 b)4! - 1 c) 16 d) 4^4 - 1

this sum looks like a sum on de-arrangement but the answer as per the book does not match if de-arr is followed.

It is not a question of dearrangement because dearrangement requires that all the elements go into wrong places. The question here also asks for cases where even 1 wrong choice will do. Total number of ways = 4! . All correct choices = 1 way. So, at least 1 incorrect = 4! - 1
A passenger train left town Alpha for town Beta. At the same time, a goods train left Beta for Alpha. The speed of each train is constant throughout the whole trip.Two hours after the trains met they were 450km apart.The passenger train arrived at the place of destination 16hrs after their meeting and the goods train 25hrs after the meeting.How long did it take the passenger train to make the wholetripp

a)21hrs
b)28hrs
c)14hrs
d)None of these
A passenger train left town Alpha for town Beta. At the same time, a goods train left Beta for Alpha. The speed of each train is constant throughout the whole trip.Two hours after the trains met they were 450km apart.The passenger train arrived at the place of destination 16hrs after their meeting and the goods train 25hrs after the meeting.How long did it take the passenger train to make the wholetripp

a)21hrs
b)28hrs
c)14hrs
d)None of these

boss it not possible to caculate
A passenger train left town Alpha for town Beta. At the same time, a goods train left Beta for Alpha. The speed of each train is constant throughout the whole trip.Two hours after the trains met they were 450km apart.The passenger train arrived at the place of destination 16hrs after their meeting and the goods train 25hrs after the meeting.How long did it take the passenger train to make the wholetripp

a)21hrs
b)28hrs
c)14hrs
d)None of these


I am getting 36 hrs as answer. What is the OA?

I solved this problem by set theory looking for a better method:
Quantitative Aptitude_Arun Sharma, Chapter 11, Geometry and Mensuration.
LOD II Q.No 43.

Circles are drawn with 4 vertices as the centre and radius equal to the side of a square. If the square is formed by joining the mid-points of sides of another square of side 2*(6^(1/2)) find the area common to all four circles.

hi Puys,

I want the detailed step by step solution for this problem in Arun Sharma..P&C..LOD3; problem no 13..pg no 461

Q. In an exam, the maximum marks for each of the 3 papers is 50 each. The maximum marks for 4th paper is 100. Find the number of ways with which a student can score 60% marks in aggregate.

Option : a) 330850 b) 223551 c) 110551 d) 220800 e) none of these

Ans : Option c) 110551..

No idea how such problems are to be solved..there is one more similar problem in LOD3..problem no 50..pg 464..that is also similar..

Puys..pls help..with 6 days to CAT I am freakin out with such problems..

wish all the puys all the best for CAT 09..

this question seems really difficult....
60% of 250 = 150
now you have to find number of ways in which 150 can be written as sum of 4 poistive numbers(x,y,zit comprises of lengthy steps..so better to leave them as CAT wants time management and want us to leave such type of questions...i don't think there is ant trick to solve this question in 1-2 minutes...

Please note that this is not a Time Management thread. It's primary purpose is to help fellow puys solve some questions that they could not. If you don't know how to solve it then you don't need to post (just like I didn't :)), there would be someone who knows how to approach these types of problems and he'll reply with the solution.
this question seems really difficult....
60% of 250 = 150
now you have to find number of ways in which 150 can be written as sum of 4 poistive numbers(x,y,zit comprises of lengthy steps..so better to leave them as CAT wants time management and want us to leave such type of questions...i don't think there is ant trick to solve this question in 1-2 minutes...

chk this out buddy!
http://www.pagalguy.com/discussions/the-shout-boxers-team-09-25043743

what is the last 2 digits of
a)(101*102*103*197*198*199)
b)(65*29*37*63*71*87)
plz do post the full method..!!!!!

remainder when 7^99/2400
a.1 b.343 c.49 d.7

remainder when 7^99/2400
a.1 b.343 c.49 d.7


7^99 can be written as ((2400+1)^24)*343
=>(((7n+1)^24)*343)/2400
so the remainder comes out to be 343
option b..
remainder when 7^99/2400
a.1 b.343 c.49 d.7

7^4 = 2401


7^99 => (7^3) mod 2400 => 343.

Hello everyone,


I have grt amount of difficulty :w00t: while i am solving DI section especially bcos of long divison. CAn u pls help me with different methods of divison you are using??

Lets take fr Eg:: 3218/5439

How will be your approach to this???

chk this out buddy!
http://www.pagalguy.com/discussions/the-shout-boxers-team-09-25043743


Hello,


I have grt amount of difficulty while i am solving DI section especially bcos of long divison. CAn u pls help me with different methods of divison you are using??

Lets take fr Eg:: 3218/5439

How will be your approach to this???
Hello everyone,


I have grt amount of difficulty :w00t: while i am solving DI section especially bcos of long divison. CAn u pls help me with different methods of divison you are using??

Lets take fr Eg:: 3218/5439

How will be your approach to this???

I've replied to your query here:

http://www.pagalguy.com/forum/quantitative-questions-and-answers/43728-number-system-ii-134.html#post1753091

1/(2!+3!) + 1/(3!+4!) + ...

How to find the summation of this series and any other similar series involving factorials ??

what is the last 2 digits of
a)(101*102*103*197*198*199)
b)(65*29*37*63*71*87)
plz do post the full method..!!!!!


Last two digit means remainder when devided by 100..

a) 101*102*103*197*198*199 / 100
=(01*99)*(02*9*(03*97) /100

now use negative remainder
=(- 1)*(-4)*(-6)/100
=(-24)/100
so remainder is 100 - 24 = 76..

Six positive numbers are taken at random and are multiplied together.Then what is the probability that the product ends in an odd digit ohter than 5?

Kindly solve and explain........

from a bag containing 100 balls, one ball weighs9 grams and all the other weigh 10 grams each.Using a simple balance where balls can be kept on either pan, what is the minimum weighs required to identify the defective ball?
Ans. A.)3 B.)4 C.)5 D.)7

from a bag containing 100 balls, one ball weighs9 grams and all the other weigh 10 grams each.Using a simple balance where balls can be kept on either pan, what is the minimum weighs required to identify the defective ball?
Ans. A.)3 B.)4 C.)5 D.)7