Saturday 29 September 2012

TCS New Pattern Placement paper 2013 (OpenSeesame)


As you already aware of the fact that TCS has changed the database of questions for its aptitude test.  The questions below give you an overview of the models to be prepared.  But don't depend on these models only.  We solved these questions only as an indicative purpose.  You are requested to go through all the arithmetic topics given in this site so that you become confident of sitting for TCS or any other written test.  All the best... The question below have been taken from https://campus.tcs.com/OpenSeesame/

1.  If 3y + x > 2 and x + 2y$ \le $3, What can be said about the value of y?
A. y = -1
B. y >-1
C. y <-1
D. y = 1
Answer: B

Multiply the second equation with -1 then it will become - x - 2y$ \ge $ - 3.  Add the equations.  You will get y > -1.

2. If the price of an item is decreased by 10% and then increased by 10%, the net effect on the price of the item is
A. A decrease of 99%
B. No change
C. A decrease of 1%
D. An increase of 1%
Answer: C

If a certain number is increased by x% then decreased by x% or vice versa, the net change is always decrease.  This change is given by a simple formula $ - {\left( {\displaystyle\frac{x}{{10}}} \right)^2} =  - {\left( {\displaystyle\frac{{10}}{{10}}} \right)^2} =  - 1\% $.  Negitive sign indicates decrease.

3. If m is an odd integer and n an even integer, which of the following is definitely odd?
A. (2m+n)(m-n)
B. $(m + {n^2}) + (m - {n^2})$
C. ${m^2} + mn + {n^2}$
D. m +n
Answer: C and D (Original Answer given as D)

You just remember the following odd $ \pm $ odd = even; even $ \pm $ even = even; even $ \pm $ odd = odd
Also odd x odd = odd; even x even = even; even x odd = even.

4.  What is the sum of all even integers between 99 and 301?
A. 40000
B. 20000
C. 40400
D. 20200
Answer: D

The first even number after 99 is 100 and last even number below 301 is 300.  We have to find the sum of even numbers from 100 to 300.  i.e., 100 + 102 + 104 + ............... 300.
Take 2 Common.  2 x ( 50 + 51 + ...........150)
There are total 101 terms in this series.  So formula for the sum of n terms when first term and last term is known is $\displaystyle\frac{{\rm{n}}}{{\rm{2}}}\left( {{\rm{a + l}}} \right)$
So 50 + 51 + ...........150 = $\displaystyle\frac{{101}}{{\rm{2}}}\left( {{\rm{50 + 150}}} \right)$
So 2 x $\frac{{101}}{{\rm{2}}}\left( {{\rm{50 + 150}}} \right)$ = 20200

5. There are 20 balls which are red, blue or green.  If 7 balls are green and the sum of red balls and green balls is less than 13, at most how many red balls are there?
A. 4
B. 5
C. 6
D. 7
Answer: B

Given R + B + G = 17; G = 7; and R + G < 13.  Substituting G = 7 in the last equation, We get R < 6.  So maximum value of R = 6

6.  If n is the sum of two consecutive odd integers and less than 100, what is greatest possibility of n?
A. 98
B. 94
C. 96
D. 99
Answer : C

We take two odd numbers as (2n + 1) and (2n - 1).
Their sum should be less than 100. So (2n + 1) + (2n - 1) < 100 $ \Rightarrow $ 4n < 100.
The largest 4 multiple which is less than 100 is 96

7. ${x^2}$ < 1/100, and x < 0 what is the highest range in which x can lie?
A. -1/10 < x < 0
B. -1 < x < 0
C. -1/10 < x < 1/10
D. -1/10 < x
Answer: A

Remember:
(x - a)(x - b) < 0 then value of x lies in between a and b.
(x - a)(x - b) > 0 then value of x does not lie inbetween a and b. or ( $ - \infty $, a) and (b, $ - \infty $) if a < b
${x^2}$ < 1/100 $ \Rightarrow $
$({x^2} - 1/100) < 0 \Rightarrow ({x^2} - {(1/10)^2}) < 0 \Rightarrow (x - 1/10)(x + 1/10) < 0$
So x should lie inbetween - 1/10 and 1/10.  But it was given that x is -ve. So x lies in -1/10 to 0

8.  There are 4 boxes colored red, yellow, green and blue.  If 2 boxes are selected, how many combinations are there for at least one green box or one red box to be selected?
A. 1
B . 6
C. 9
D. 5
Answer: 5

Total ways of selecting two boxes out of 4 is $4{C_2}$ = 6.  Now, the number of ways of selecting two boxes where none of the green or red box included is only 1 way.  (we select yellow and blue in only one way).  If we substract this number from total ways we get 5 ways.

9. All faces of a cube with an eight - meter edge are painted red.  If the cube is cut into smaller cubes with a two - meter edge, how many of the two meter cubes have paint on exactly one face?
A. 24
B. 36
C. 60
D. 48
Answer : A

If there are n cubes lie on an edge, then total number of cubes with one side painting is given by $6 \times {\left( {n - 2} \right)^2}$.  Here side of the bigger cube is 8, and small cube is 2.  So there are 4 cubes lie on an edge. Hence answer = 24

10. Two cyclists begin training on an oval racecourse at the same time.  The professional cyclist completes each lap in 4 minutes; the novice takes 6 minutes to complete each lap.  How many minutes after the start will both cyclists pass at exactly the same spot where they began to cycle?
A. 10
B. 8
C. 14
D. 12
Answer: D

The faster cyclyst comes to the starting point for every 4 min so his times are 4, 8, 12, .........  The slower cyclist comes to the starting point for every 6 min so his times are 6, 12, 18, .........  So both comes at the end of the 12th min.

11. M, N, O and P are all different individuals; M is the daughter of N; N is the son of O; O is the father of P; Among the following statements, which one is true?
A. M is the daughter of P
B. If B is the daughter of N, then M and B are sisters
C. If C is the granddaughter of O, then C and M are sisters
D. P and N are bothers. 
Answer: B

From the diagram it is clear that If B is the daughter of N, then M and B are sisters.  Rectangle indicates Male, and Oval indicates Female.



12. In the adjoining diagram, ABCD and EFGH are squres of side 1 unit such that they intersect in a square of diagonal length (CE) = 1/2.  The total area covered by the squares is

A. Cannot be found from the information
B. 1 1/2
C. 1 7/8
D. None of these
Answer:  C

Let CG = x then using pythogerous theorem ${\rm{C}}{{\rm{G}}^{\rm{2}}}{\rm{ + G}}{{\rm{E}}^{\rm{2}}}{\rm{ = C}}{{\rm{E}}^{\rm{2}}}$
$ \Rightarrow $  ${x^2} + {x^2} = {(1/2)^2} \Rightarrow 2{x^2} = 1/4 \Rightarrow {x^2} = 1/8$
Total area covered by two bigger squares = ABCD + EFGE - Area of small square = 2 - 1/8 = 15/8


13. There are 10 stepping stones numbered 1 to 10 as shown at the side.  A fly jumps from the first stone as follows; Every minute it jumps to the 4th stone from where it started - that is from 1st it would go to 5th and from 5th it would go to 9th and from 9th it would go to 3rd etc.  Where would the fly be at the 60th minute if it starts at 1?
A. 1
B. 5
C. 4
D. 9

Answer : A

Assume these steps are in circular fashion. 
Then the fly jumps are denoted in the diagram.  It is clear that fly came to the 1st position after 5th minute.  So again it will be at 1st position after 10th 15th .....60th. min.

So the fly will be at 1st stone after 60th min.

 

14. What is the remainder when ${{\rm{6}}^{{\rm{17}}}}{\rm{ + 1}}{{\rm{17}}^{\rm{6}}}$  is divided by 7?
A. 1
B. 6
C. 0
D. 3
Answer: C

${{\rm{6}}^{{\rm{17}}}}$ = ${{\rm{(7 - 1)}}^{{\rm{17}}}}$ =
${\rm{17}}{{\rm{C}}_{\rm{0}}}{\rm{.}}{{\rm{7}}^{{\rm{17}}}}{\rm{ - 17}}{{\rm{C}}_{\rm{1}}}{\rm{.}}{{\rm{7}}^{{\rm{16}}}}{\rm{.}}{{\rm{1}}^{\rm{1}}}.....{\rm{ + 17}}{{\rm{C}}_{{\rm{16}}}}{\rm{.}}{{\rm{7}}^{\rm{1}}}{\rm{.}}{{\rm{1}}^{{\rm{16}}}}{\rm{ - 17}}{{\rm{C}}_{{\rm{17}}}}{\rm{.}}{{\rm{1}}^{{\rm{17}}}}$

If we divide this expansion except the last term each term gives a remainder 0.  Last term gives a remainder of - 1.

Now From Fermat little theorem, ${\left[ {\displaystyle\frac{{{{\rm{a}}^{{\rm{p - 1}}}}}}{{\rm{p}}}} \right]_{{\rm{Rem}}}}{\rm{ = 1}}$
So ${\left[ {\displaystyle\frac{{{{17}^6}}}{7}} \right]_{{\rm{Rem}}}}{\rm{ = 1}}$

Adding these two remainders we get the final remainder = 0

15. In base 7, a number is written only using the digits 0, 1, 2, .....6.  The number 135 in base 7 is 1 x ${{7^2}}$ + 3 x 7 + 5 = 75 in base 10.  What is the sum of the base 7 numbers 1234 and 6543 in base 7.
A. 11101
B. 11110
C. 10111
D. 11011
Answer: B

In base 7 there is no 7.  So to write 7 we use 10.  for 8 we use 11...... for 13 we use 16, for 14 we use 20 and so on.
So from the column d, 4 + 3 = 7 = 10, we write 0 and 1 carried over.  now 1 + 3 + 4 = 8 = 11, then we write 1 and 1 carried over.  again 1 + 2 + 5 = 8 = 11 and so on



16. The sequence ${\rm{\{ }}{{\rm{A}}_{\rm{n}}}{\rm{\} }}$ is defined by ${{\rm{A}}_1}$ = 2 and ${{\rm{A}}_{{\rm{n + 1}}}}{\rm{ = }}{{\rm{A}}_{\rm{n}}}{\rm{ + 2n}}$ what is the value of ${{\rm{A}}_{100}}$
A. 9902
B. 9900
C. 10100
D. 9904
Answer: A
We know that ${{\rm{A}}_1}$ = 2 so ${{\rm{A}}_{\rm{2}}}{\rm{ = }}{{\rm{A}}_{{\rm{1 + 1}}}}{\rm{ = }}{{\rm{A}}_{\rm{1}}}{\rm{ + 2(1) = 4}}$
${{\rm{A}}_3}{\rm{ = }}{{\rm{A}}_{{\rm{2 + 1}}}}{\rm{ = }}{{\rm{A}}_2}{\rm{ + 2(2) = 8}}$
${{\rm{A}}_4}{\rm{ = }}{{\rm{A}}_{{\rm{3 + 1}}}}{\rm{ = }}{{\rm{A}}_3}{\rm{ + 2(3) = 14}}$
So the first few terms are 2, 4, 8, 14, 22, ......
The differences of the above terms are 2, 4, 6, 8, 10...
and the differences of differences are 2, 2, 2, 2.  all are equal.  so this series represents a quadratic equation.
Assume  ${{\rm{A}}_{\rm{n}}}$ = $a{n^2} + bn + c$
Now ${{\rm{A}}_1}$ = a + b + c = 2
${{\rm{A}}_2}$ = 4a + 2b + c = 4
${{\rm{A}}_3}$ = 9a + 3b + c = 8
Solving above equations we get a = 1, b = - 1 and C = 2
So substituting in ${{\rm{A}}_{\rm{n}}}$ = ${n^2} + bn + c$ = ${n^2} - n + 2$
Substitute 100 in the above equation we get 9902.

17.Find the number of rectangles from the adjoining figure (A square is also considered a rectangle)
A. 864
B. 3276
C. 1638
D. None

Answer: C

To form a rectangle we need two horizontal lines and two vertical lines.  Here there are 13 vertical lines and 7 horizontal lines.  The number of ways of selecting 2 lines from 13 vertical lines is $13{C_2}$ and the number of ways of selecting 2 lines from 7 horizontals is $7{C_2}$. So total rectangles = $7{C_2}x13{C_2}$

18. A, B, C and D go for a picnic.  When A stands on a weighing machine, B also climbs on, and the weight shown was 132 kg.  When B stands, C also climbs on, and the machine shows 130 kg.  Similarly the weight of C and D is found as 102 kg and that of B and D is 116 kg.  What is D's weight
A. 58kg
B. 78 kg
C. 44 kg
D. None
Answer : C

Given A + B = 132; B + C = 130; C + D = 102, B + D = 116
Eliminate B from 2nd and 4th equation and solving this equation and 3rd we get D value as 44.

19.  Roy is now 4 years older than Erik and half of that amount older than Iris.  If in 2 years, roy will be twice as old as Erik, then in 2 years what would be Roy's age multiplied by Iris's age?
A. 28
B. 48
C. 50
D. 52
Answer: 48

20. X, Y, X and W are integers.  The expression X - Y - Z is even and the expression Y - Z - W is odd.  If X is even what must be true?
A. W must be odd
B. Y - Z must be odd
C. W must be odd
D. Z must be odd
Answer: A or C (But go for C)

21.  Mr and Mrs Smith have invited 9 of their friends and their spouses for a party at the Waikiki Beach resort.  They stand for a group photograph.  If Mr Smith never stands next to Mrs Smith (as he says they are always together otherwise). How many ways the group can be arranged in a row for the photograph?
A. 20!
B. 19! + 18!
C. 18 x 19!
D. 2 x 19!
Answer: C

22.  In a rectanglular coordinate system, what is the area of a triangle whose vertices whose vertices have the coordinates (4,0), (6, 3) adn (6 , -3)
A. 6
B. 7
C. 7.5
D. 6.5
Answer: A

23. A drawer holds 4 red hats and 4 blue hats.  What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer and immediately returning every hat to the drawer before taking out the next?
A. 1/2
B. 1/8
C. 1/4
D. 3/8
Answer: B

24. In how many ways can we distribute 10 identical looking pencils to 4 students so that each student gets at least one pencil?
A. 5040
B. 210
C. 84
D. None of these
Answer: C

25. The prime factorization of intezer N is A x A x B x C, where A, B and C are all distinct prine intezers.  How many factors does N have?
A. 12
B. 24
C. 4
D. 6
Answer: A

26. Tim and Elan are 90 km from each other.they start to move each other simultanously tim at speed 10 and elan 5 kmph. If every hour they double their speed what is the distance that Tim will pass until he meet Elan 
A. 45
B. 60
C. 20
D. 80
Answer: B

27. A father purchases dress for his three daughter. The dresses are of same color but of different size .the dress is kept in dark room .What is the probability that all the three will not choose their own dress.
A.  2/3
B.  1/3
C.  1/6
D.  1/9
Answer: B

28. N is an integer and N>2, at most how many integers among N + 2, N + 3, N + 4, N + 5, N + 6,  and N + 7 are prime integers?
A. 1
B. 3
C. 2
D. 4
Answer: C

29. A turtle is crossing a field.  What is the total distance (in meters) passed by turtle? Consider the following two statements
(X) The average speed of the turtle is 2 meters per minute
(Y) Had the turtle walked 1 meter per minute faster than his average speed it would have finished 40 minutes earlier

A. Statement X alone is enough to get the answer
B. Both statements X and Y are needed to get the answer
C. Statement Y alone is enough to get the answer
D. Data inadequate
Answer: B

30. Given the following information, who is youngest?
C is younger than A; A is talled than B
C is older than B; C is younger than D
B is taller than C; A is older than D

A. D
B. B
C. C
D. A
Answer: B

31. If P(x) = ${\rm{a}}{{\rm{x}}^{\rm{4}}}{\rm{ + b}}{{\rm{x}}^{\rm{3}}}{\rm{ + c}}{{\rm{x}}^{\rm{2}}}{\rm{ + dx + e}}$ has roots at x = 1, 2, 3, 4 and P(0) = 48, what is P(5)
A. 48
B. 24
C. 0
D. 50
Answer: A

Thursday 27 September 2012

TCS-drive in Jadavpur Univeristy,Kolkata(September 19, 2012 )


TCS recruitment process consist of 4 rounds for those who does not have 75% through-out his/her carrier Tata Consultancy Services,TCS fresher job recruitment process,TCS Aptitude Test,TCS Technical Round,TCS Hr round TCS aptitude test consists 30 lengthy questions for 80 minutes TCS aptitude test topics are Time and work,Profit,Probabality,Permutation combination Volume etc.....questions.......
1Aptitude Test
2Technical Interview(1st round)
3Technical Interview(2nd round)
4HR Interview
And only 2 rounds for those who have above 75% through-out his/her carrier.
NOTE :: In case of CGPA system % will be calculated as=(CGPA-0.75)*100%.....
TCS On line Aptitude Test
1Aptitude Test ::
You can find so many placement paper on various site on internet. But you can be remain sure that a single one you will not get in Current TCS Aptitude Test. Aptitude test consist of 30 question for 80minutes.
There are 5-6 question from “Time & Work(lengthy calculation)”. And the rest of the question you can’t be guessed. In my question They were from volume & surface area, average, profit & loss, partnership, probability. But may be different in your test. Many question were too much lengthy to read. They consumes 1-1.2min just to read in.
So my suggestion is that do practice as much as possible from “Dr.R.S.Aggarwal Quantitative Aptitude”.
Now If you can clear the Aptitude Test, you are eligible for the next round..
2Technical Interview(1st round) ::
First 30 min there was only 1 interviewer in my panel. After 30min one another joined with him..
 
Interviewer: How was your day today?
Me: Not good sir. I have not took anything from morning till now(6pm).
Interviewer: So you can take your meal. After that we can continue.
Me: No thanks sir. Its ok.
Interviewer: You are doing M.tech. Why are you want to join industry?(common question for M.tech candidate)
Me: told.(But it takes  couple of minutes to convince him)
Interviewer: Suppose you are told to do project which programming language you will choose c/c++?
Me:  c++.
Interviewer: Why? What is the difference between c & c++ other than syntax?
Me: told.
Interviewer: What is class?
Me: told the definition of class from a renowned book(but he was not satisfied  ).
After that I told another definition from another book(but he was not satisfied ). Then I took pen & paper and wrote a sample class. And then elaborate the class.(now I little bit convinced)
Interviewer: What is object?
Me: told.
Interviewer: Give a real life example of class?
Me: told .But it takes a while.
Interviewer: Do you know the syntax of C function?
Me: wrote.
Interviewer: Do you know recursion?
Me: yaa.
Interviewer: Ok. Write a code using recursion?
Me: wrote factorial program.
Now he began to ask the pros & cons of the program.  Wheather  it will run? How it will run? And many more.
Interviewer: Told me some name of the subject you read.
Me: told.
Interviewer: What is difference between microprocessor & microcontroller?
Me: told.
Interviewer: where they are used?
Me: told.
After that another interviewer came.
Interviewer: told me something about your B.tech final year project.
Me: told.
Interviewer: You have used JSP. How we can connect mysql database through JSP.
Me: wrote the code(but not 100% right. A little bit syntax error  but algo was right).
Interviewer: What are the different types  of driver in Mysql?
Me: told.
Interviewer: Which one you have used?
Me: told.
Interviewer: Which one is the best?
Me: told.
Interviewer: Why?
Me: told.
Interviewer: How can you identify which version of java is being installed in a machine without opening program files folder?
Me: told for the linux based operating system(as I have done all the programming in linux). But can’t able to tell for windows operating environment.
Interviewer: Ok. Thank you. You may go.
 
 
 
This was the 1st round technical interview of mine so far I can remember and it was about 50-60min. This was an Elimination round. If you survive in this round you are qualified for the 2nd round technical.
TCS Technical Round
3Technical Interview(2nd round) ::
In this round  there were 2 interviewer from the beginning.
In this round  they have asked from c, c++, java, dbms.
Interviewer: told me the entities & attributes(they have told me that I have to computerized the event management system like “fresher welcome” or “collage annual function” )
Me: told.
Then They have asked various question from the above mentioned subject. I have answered 70-80%  to their satisfaction. This round was going near about 30min.
TCS HR Interview
4 HR Interview ::
This round was short than the other two & it was about 10-15min.
Hr: Why TCS?(common question. Be prepared)
Me: told.
Hr: If you got in both CTS & TCS. Which one do you prefer?(CTS & TCS both came in 0 slot)
Me: TCS.
Hr: told.
Hr: why?
Me: told.                                                                                    
Hr: Do you have any year gap?
Me: told(If you have don’t lie).
Hr: do u have any backlog ?
Me: told(If you have don’t lie).
Hr: Do you have any relocation problem?
Me: told.
And this the overall my interview experience so far I can remember…..

Wednesday 26 September 2012

TCS Pattern : 08.09.2012 Drive


Round 1 ; Day 1 ; Aptitude
Type of College: Thiagarajar College of Engineering ,Madurai
Test; Online
Pattern: Totaly 30 questions only Quants and Reasoning . Majority of the questions kept varying student to student.
Time Limit;80 minutes
Negative Marking: 1/3
Expected interview call cutoff: 15

Simple quant questions - around 10

1.Areas : Average, Age, Time Speed Distance, Time and Work( 2 people working at a time independently), Percentage etc.
Most importantly all the answers were interim say if we get an answer in time and work question, that answer had to be used in an equation. The answer to the time and work question was not the final answer.

2.Geometrv - Circle properties. Line properties

3.Prime factors based question

4.Tenali Raman Flower type question. There is a garden with n flowers. A person picks 2 flowers on Monday, it is replaced by 5 flowers on Tuesday.On Tuesday he picks 3 flowers, it is replaced by 7 flowers, .._ Find the final number of flowers in the garden on Sunday.

5. 3 people in a family, all the 3 stand on the weighing machine then the weight  is = some number *variable, then 2 on the weighing machine then the weight is = some number + variable and then 1 person on the weighing machine then the weight is = some number + variable. Find each person’s Individual weight. All the measurements were given In a comparative mode.

6.Functions based question. F[n] = ..._ then F[l00]=?

7.Data Arrangement question with overall directions and 3 to 4 questions with directions and Inputs changing for each question.

8.Cubes with a.b.c what Is on the opposite side, diagram based

9. Cubes painted is with 2 colors. Cut and then painted and not painted faces.

10. Series question AP and GP based. Say a frog jumps x distance and In 2000 jumps It will cover-- - distance and then the AP/GP question

11. A$A=0
(A$B)$C = (A$B)+C
2012$0+2012$(10$1)=?

12. 1 person said one of us is lying
      2 person said two of us are lying
      3 person said three of us are lying
       5 person said all of us are lying
Who is saying the truth or who is lying?

13. Puzzles

14. Sudoku wIth 3 or 4 numbers given. The numbers to be found were marked
as x, y. Then there was an equation which had to be solved. The answer was
a 3 digIt number

Tuesday 25 September 2012

TCS Placement paper (Conducted at JNTUK on 13.9.12)

 
If f(x) = (1+x+x2+x3+.......x2012)2x2012
g(x) = 1+x+x2+x3+.......x2011 
Then what is the remainder when f(x) is divided by g(x)
Let us multiply g(x) with x on the both sides
x.g(x) = x+x2+x3+.......x2012
add 1 on both sides
x. g(x) + 1 = 1+x+x2+x3+.......x2012
Substitute this value in f(x)
then f(x) = (x.g(x)+1)2x2012
f(x) = x2.g(x)2+2.g(x)+1x2012
Now f(x) is divisible by g(x) first two terms are exactly divisible by g(x) and we get 1 - x2012
But 1 - x2012 = (1 - x)(1+x+x2+x3+.......x2011)
if this expression is divisible by g(x) we get (1 -x) as remainder.

A number has exactly 3 prime factors, 125 factors of this number are perfect squares and 27 factors of this number are perfect cubes. overall how many factors does the number have?
We know that the total factors of a number N = ap.bq.cr ....
Now the total factors which are perfect squares of a number N = ([p2]+1).([q2]+1).([r2]+1)....
where [x] is greatest intezer less than that of x. 
Given ([p2]+1).([q2]+1).([r2]+1).... = 125
So [p2]+1 = 5;  [q2]+1 = 5; [p2]+1 = 5
[p2] = 4  p = 8 or 9, similarly q = 8 or 9, r = 8 or 9
Given that 27 factors of this number are perfect cubes
so ([p3]+1).([q3]+1).([r3]+1).... = 27
So [p3]+1 = 3  [p3] = 2
 p = 6, 7, 8
By combining we know that p = q = r = 8
So the given number should be in the format = a8.b8.c8 ....
Number of factors of this number = (8+1).(8+1).(8+1) = 729


In a class there are 60% of girls of which 25% poor.  What is the probability that a poor girl is selected is leader?
Assume total students in the class = 100
Then Girls = 60% (100) = 60
Poor girls = 25% (60) = 15
So probability that a poor girls is selected leader = Poor girls / Total students = 15/100 = 15%

A and B are running around a circular track of length 120 meters with speeds 12 m/s and 6 m/s in the same direction.  When will they meet for the first time?
A meets B when A covers one round more than B.
A's relative speed = (12 - 6) m/s.  So he takes 120 / 6 seconds to gain one extra round. 
So after 20 seconds A meets B.

A completes a work in 20 days B in 60 days C in 45 days.  All three persons working together on a project got a profit of Rs.26000 what is the profit of B?
We know that profits must be shared as the ratio of their efficiencies.  But efficiencies are inversely proportional to the days.  So efficiencies of A : B : C = 1/20 : 1/60 : 1/45 = 9 : 3 : 4
So B share in the total profit = 3 / 13 X 26000 = Rs.6000

A completes a piece of work in 3/4 of the time in B does, B takes 4/5 of the time in C does.  They got a profit of Rs. 40000  how much B gets?
Assume C takes 20 Days.  Now B takes 4/5 (20) = 16 days.  A takes 3/4(16) = 12
Now their efficiencies ratio = 1/20 : 1/16 : 1/12 = 12 : 15 : 20
B's share in the profit of Rs.40000 = 15/47 (40000) = Rs.12765


An empty tankbe filled with aninlet pipe ‘A’ in 42 minutes. after 12 minutes an outlet pipe ‘B’ is opened which can empty the tank in 30 minutes. After 6 minutes another inlet pipe ‘C’ opened into the same tank, which can fill the tank in 35 minutes and the tank is filled find the time taken to fill the tank?
Assume total tank capacity = 210 Liters
Now pipe A fill this tank in 210/42 = 5 min
B empties this tank in 210 / 30 = 7 min
C fills this tank in 210 / 35 = 6 min
Assume tank gets filled in x min.
So x × 5 - (x-12)×7 + (x - 18) × 6 = 210  x = 61.25

Mother, daughter and an infant combined age is 74, and mother's age is 46 more than daughter and infant.  If infant age is 0.4 times of daughter age, then find daughters age.
Assume M + D + I = 74; .................(1)
Also given M - D - I = 46  M = D + I + 46
Also I = 0.4 D  I = 2/5 D
Substituting M and I values in the first equation we get D - 25D - 46 + D + 25D = 74
Solving D = 10


A Grocer bought 24 kg coffee beans at price X per kg. After a while one third of stock got spoiled so he sold the rest for $200 per kg and made a total profit of twice the cost. What must be the price of X?
Total Cost price = 24×X
As 1/3rd of the beans spoiled, remaining beans are 2/3 (24) = 16 kgs
Selling price = 200 × 16 = 3200
Profit = Selling price - Cost price = 3200 - 24×X
Given Profit = 2 × Cost price

3200 - 24×X = 2 × (24×X)
Solving X = 44.44

Bhanu spends 30% of his income on petrol on scooter 20% of the remaining on house rent and the balance on food. If he spends Rs.300 on petrol then what is the expenditure on house rent? 
Given 30% (Income ) = 300  Income = 1000
After having spent Rs.300 on petrol, he left with Rs.700.
His spending on house rent = 20% (700) = Rs.140

Let exp(m,n) = m to the power n. If exp(10, m) = n exp(2, 2) where to and n are integers then n = 
Given 10m=n.22
 2m×5m=n.222m2×5m=n
For m = 2 we get least value of n = 25, and for m > 2 we get infinite values are possible for n.


How many kgs. of wheat costing Rs. 5 per kg must be mixed with 45 kg of rice costing Rs. 6.40 per kg so that 20% gain may be obtained by  selling the mixture at Rs. 7.20 per kg ?
If the selling price of the mixture is Rs.7.2 when sold at 20% profit then
CP ×120100 = 7.2  CP = Rs.6
Now by applying weighted average formula = K×5+45×6.4K+45=6
 K = 18 kgs

The diagonal of a square is twice the side of equilateral triangle then the ratio of Area of the Triangle to the Area of Square is?
Let the side of equilateral triangle = 1 unit. 
We know that area of an equilateral triangle = 34a2
As side = 1 unit area of the equilateral triangle = 34
Now Diagonal of the square = side of the equilateral triangle = 1
We know that area of the square = 12D2 where D = diagonal
So area of the square = 12
Ratio of the areas of equilateral triangle and square = 34 12  3:2

Raj tossed 3 dices and there results are noted down then what is the probability that raj gets 10?
Always remember when 3 dice are rolled the number of ways of getting n ( where n is the sum of faces on dice)
(n1)C2 where n = 3 to 8
= 25 where n = 9, 12
= 27 where n = 10, 11
(20n)C2 where n = 13 to 18
The required probability = 2563 25216
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