After watching this video you will be able to crack such long questions in mind definitely
1.A person travels 300 km to his home , partly by train and partly by car . He takes 4 hours if he travels 60 km by train and the rest by car .He takes 4 hours 10 minutes if he travels 100 km by train and the rest by car .
Find the speed of train and the train ?
2.Rohan travels 600 km from his home partly by train and partly by car. He takes eight hours if he travels 120 km by train and rest by car. he takes 20 minutes more if he travels 200 km by train and rest by car. find the speed of train and car.
A liquid P is 1(3/7) times heavy as water and water is 1(2/5) times at heavy as another liquid `Q'. The amount of liquid 'P' that must be added' to 7 litres of the liquid "Q' so that the mixture may weigh as much as an equal volume of water, will be
Seventy percent of the employees in a multinational corporation have VCD
players, 75 percent have microwave ovens ,80 percents have Acs and 85 percent
have washing machines .At least what percent of employees has all four gadgets
?
solution
The least percentage of people with all 4 gadgets would happen if all the
employees who are not having any one of the four objects is mutually exclusive
thus 100-30-25-20-15=10
Let us call the people P1, P2, P3... P100
P = 30 people
don't have VCD. Let them be P1, P2.. P30
Q = 25 people
don't have ovens. Let them be P31, P32, P33... P55
R = 20 people
don't have ACs. Let them be P56, P57, P58... P75.
S = 15 people
don't have Washing Machines. Let them be P76, P77... P90.
Now, P91, P92...
P100 have everything. So, 10 people have everything.
If you make some
people common in P, Q, R, and S; then the number of people who have everything
will be more that 10.
=> The minimum number of people
who have everything is 10.
This is again a very important question asked in
many aptitude exams
Two men
undertake to do a piece of work for Rs 1400. First man alone can do this work
in 7 days while the second man alone can do this work in 8 days. If they
working together complete this work in 3 days with the help of a boy, how
should money be divided?
So funda is money is divided according to work
performed so first man in 3 days will be
able to do 3/7 of work in 3 days o 3/7 of 1400 is Rs 600
Second
man can do 3/8 of work in 3 days 3/8 of 1400 is Rs 525
and rest money should be given to boy 1400-600 -525= Rs 275
Today i want to discuss a very important question asked in many exams SSC CGL/BANK PO/ AFACAT /MBA solution of this question is given by very long method in many books
Two
typists of varying skills can do a typing job in 6 minutes if they work
together. If the first typist typed alone 4 minutes and then the second typist
typed alone for 6 minutes, they would be left with 1/5 of the whole work. How
many minutes would it take the slower typist to complete the typing job working
alone?
now think logically they both can complete the work in 6 minutes now the first typist typed alone 4 minutes and then the second typist typed alone for 6 minutes, it means if first typist work 2 more minutes after 4 minutes the work will be completed it means he can do 1/5th of work in 2 minutes so whole work in 5 * 2 = 10 minutes it means first can do the work alone in 10 minutes and together it is given in the question they complete the work in 6 minutes so you can easily calculate by basics of time & work that he will complete a work alone in 15 minutes. i.e. our answer that slower typist can complete the work in 15 minutes.
This is very important algebra question i am posting with my own unique trick if you master these tricks you can solve SSC CGL ALGEBRA QUESTIONS with this trick in few seconds
Three men A, B and C working together can do a job
in 6 hours less time than A alone, in 1 hour less time than B alone and in one
half the time needed by C when working alone. Then A and B together can do the
job in
This is my second
post today I am going to discuss 1000 doors problem it is world famous puzzle
asked in many aptitude exams like CAT/E LITMUS/GRE/GMAT and interviews the
puzzle is
There are 1000 doors in a high school with 1000 students.
The problem begins with the first student opening all 1000 doors next the
second student closes doors 2,4,6,8,10 and so on to door 1000 the third student
changes the state (opens doors closed, closes doors s open) on doors
3,6,9,12,15 and so on the fourth student Changes the state of doors 4, 8, 12,
16 and so on. This goes on until every student has had a turn. How many doors
will be open at the end?
before discussing this puzzle I want to clear your
basic concepts to understand this puzzle
if you list the factors of 24, you get (1, 24), (2, 12), (3, 8),
(4, 6). thing to notice here are Since
the factors are listed in pairs, there are an even number of them.
if you list the
factors of 36, you get (1, 36), (2, 18), (3, 12), (4, 9), and(6, 6). The last pair
of factors is just the square root of 36 listed twice.
Since 6 only
counts once as a factor of 36, 36 has nine factors, an odd number.
so core funda to
solve this puzzle is perfect square hace odd number of factors
initially all
the doors are closed
person 1 came
opens all of the doors, they're all open. Now person 2 goes, and all of
the even numbered doors means multiples of 2 (see in below table) are shut. Now it's 3's turn: door 3 is shut,
door 6 is open again, and 9 is shut. but
if we continue to solve in this type of manner it will take lot of time to work
on 1000 doors and 1000 persons
doors
Door 1(close)
Door 2(close)
Door 3 (close)
…………………
Door1000(close)
Person
1
Open
Open
Open
……………….
open
Person
2
Open
Close
Open
……………….
Close
Person
3
Open
Open
Close
……………………
close
…………
Person
1000
………………………
………………………..
……………………………
now think in
this manner suppose think door number 36 who can change the state of door
number 36 yes those persons number that are factor of 36 so see in table person
1 came he will open the closed door person 2 will close person 3 will open
person 4 close person 6 open and in last…….person 36 open .as door 36 is a perfect
square having odd number of factors the last condition is open so every perfect
square door will have open state in the
end .up to 1000 there are 31 perfect squares doors number last one is
312 =
961
DOOR NUMBER
36
FACTORS
1
OPEN
2
CLOSE
3
OPEN
4
CLOSE
6
OPEN
9
CLOSE
12
OPEN
18
CLOSE
36
OPEN
now think about door
number 24 which is not a perfect square number it have even number of factors
so will have closed position in the end
DOOR NUMBER 24 FACTORS
1
OPEN
2
CLOSE
3
OPEN
4
CLOSE
6
OPEN
8
CLOSE
12
OPEN
24
CLOSE
so answer of this puzzle is 31 doors will be open in the end
In this post I would like to discuss
some of the fundamental ideas that can be used to solve number of
trailing zeros in a product important for CAT,E LITMUS,PLACEMENTS,GMAT,GRE/SSC
CGL tier 2. If you have just started your preparation you might find this
article helpful.first of all we should have idea about number of trailing zerostrailing
zeros are a sequence of 0s after which no other digits follow.
think number 210 have one trailing
zeros but have you ever thought why in this number only one trailing zero
because 210 can be written as in form of prime factors
2 × 5× 3 × 7 = 210 only one pair of
2 and 5 so there is only there is only one zero at the end of the product
so number of trailing zeros depends upon number of pairs of 2’s and 5’s
or We can say that for n number of zero’s at the end of the product we need exactly
n combinations of “5 × 2 “. For example
