Reasoning Preparation

CLASSIFICATION TYPE

Read the following information carefully and answer the questions
that follow:

1)There are six cities A,B,C,D,E and F
A is not a hill station.
B and E are not historical places.
D is not an industrial city.
A and D are not historical
A and B are not alike.

1.Which two cities are industrial centers?
a)A,B b)E,F* c)C,D d)B,F e)A,D

2.Which two cities are historical places?
a)A,C b)B,F c)C,F* d)B,E e)A,D

3.Which two cities are hill stations?
a)A,B b)C,A c)B,D* d)A,F e)none

4.Which city is a hill station and an industrial center but not
historical place?
a)E* b)F c)A d)B e)C

5.Which two cities are neither historical places nor industrial
centers?
a)A,B b)D,E c)F,C d)B,D* e)none

Solution:
A B C D E F


Historical x x y x x y

Industrial y x y x y y

Hillstation x y y y y y

2) Five friends Indu,Pinki,Sai,Srujan,Pavan  travelled to five different
cities of Chennai,Calcutta,Delhi,Bangalore and Hyderabad by five
differnet modes of transport of Bus,Train,Aeroplane,Car and Boat from
Mumbai. The person who travelled to Delhi did not travel  by Boat.
Sai went to Banalore by Car and Pinki went to Calcutta by Aeroplane.
Srujan travelled by Boat whereas Pavan travelled by Train.
Mumbai is not connected by bBus to Delhi and Chennai.

1.Which of the following combinations of person and mode is not correct?
a)Indu-Bus b)Pinki-Aeroplane c)Sai-Car  d)Srujan-Boat e)Pavan-Aeroplane*

2.Which of the following combinations is true for Srujan?
a)Delhi-Bus b)Chennai-Bus c)Chennai-Boat* d)Data inadequate

3.Which of the following combinations of place and mode is not correct?
a)Delhi-Bus* b)Calcutta-Aeroplane c)Bangalore-Car d)Chennai-Boat

4.Person travelling to Delhi went by which of the following modes?
a)Bus b)Train* c)Aeroplane d)Car e)Boat

5.Who among the following traj_konkepudiravelled to Delhi?
a)Sai b)Srujan c)Pavan* d)Data inadequate e)none

Solution:
Sai travels by Car.
Pinki travels by Aeropraj_konkepudilane.
Srujan travels by Boat.
Pavan travels by Train.
Indu travels by Bus.
Sai goes to Bangalore.
Pinki goes to Calcutta.
Bus facility is not there for Delhi or Chennai.
Indu goes to Hyderabad by Bus.
From given information it is clear that Srujan goes by Boat but not
to Delhi.
Hence Srujan goes to Chennai.
Pavan goes to Delhi.

PLACE MODE

Indu Hyderabad Bus

Pinki Calcutta Aeroplane

Sai Bangalore Car

Srujan Chennai Boat

Pavan Delhi Train



3)Four youngmen Thirbhuvan,Thrishanth,trinath,Trived are friendly
with four girls Indira,Madhuri,Swetha and Dimple.Indira and Swetha
are friends.Trinath's girlfriend does not like Indira and Swetha.
Madhuri does not care for Trinath.Trishanth's girlfriend is
friendly with Indira.Indira does not like Thribhu.

1.Who is Thribhu's girlfriend?
a)Indira b)Madhuri* c)Swetha d)Dimple

2.With whom is Indira friendly?
a)Thribhuvan b)Thrishanth c)Trinath d)Trived*

3.who is Dimple's boyfriend?
a)Trived b)Trinath* c)Thrishanth d)Thribhuvan

4.Who does not like Indira and Swetha?
a)Dimple* b)Thribhuvan   c)Trived d)Trinath

Solution:
Given Indira and Swetha are friends.
Thrishanth's girlfriend is friendly with Indira.
Hence,Thrishanth's girlfriend is Swetha.
Given,Trinath's girlfriend does not like Indira and Swetha.
=>She might be Madhuri or Dimple.
But Madhuri does not care for Trinath.
=>Trinath's girlfriend is Dimple.
Given,Indira does not like Thrinbhuvan.
=>Thribhuvan's girlfriend is Madhuri.
Clearly,Trived's girlfriend is Indira.


BOY GIRL
Thribhuvan Madhuri
Thrishanth Swetha
Trinath Dimple
Trived Indira
                                                   
COMPARISION TYPE QUESTIONS

1)
Clues will be given regarding comparisions among a ssset of persons
or things with respest to one or more qualities.After analysing a
proper ascending ,descending sequence is formed and then are
supposed to answer.
There are five friends-Sachin,Sourav,Rahul,Zaheer and Yuv

Race concept

RACES AND GAMES OF SKILL

Races :- A contest of speed in running ,riding,driving,sailing or rowing is called a Race.

Race Course :- The ground or path on which contests are made is called
a race course

STARTING POINT :- The point from which a race begins is called starting point.

Winning point or goal:- The point set to bound a race is called a winning point.

Dead Heat Race:- If all the persons contesting a race reach the goal exactly
at the same time,then the race is called a dead heat race.

Start :- suppose A and B are two contestants in a race .If before the start of

 the race, A is at the starting point and B is ahead of A by 12 meters.

 Then we say that "A gives B a start 12 meters.

► To cover a race of 100 meters in this case,A will have to cover 100 m while B

  will have to cover 88 m=(100-12)

► In a 100 m race 'A can give B 12 m' or 'A can give B a start of 12 m or A beats

  B by 12 m means that while A runs 100 m B runs 88 m.

GAMES:- A game of 100 m, means that the person among the contestants who
scores 100 points first is the winner.

If A scores 100 points while B scores only 80 points then we say that 'A can give
B 20 points'.

Puzzles

Introduction:

Puzzles are dealt in a detailed manner with certain solutions.

Different puzzles are gathered from Shakuntala Devi puzzle

books. Keeping in mind certain puzzles for Infosys some

reasoning problems are also dealt. Puzzle name at the top of

each problem will give a brief idea regarding the mode of

application.

SELECTING A CANDIDATE

For an advertisement of six local posts,twelve persons applied

for the job.Can you tell in how  many different ways the

selection can be made?

Solution:

 6^12

SET OF BAT AND BALL

When I wanted to buy a bat and ball, the shopkeeper said they

would together cost Rs.3.75.But I did not want to buy a ball.

The shopkeeper said that bat would cost 75 paise more than the

ball.What was the cost of bat and the ball?

Solution:

 Given that bat and ball together cost Rs.3.75 = 75 paise

 Let cost of the ball alone be x.

 Given cost of the bat is 75p greater than cost of the ball.

 So cost of the bat = x+75

 x+x+75 = 375

 2x = 375 - 75

 2x = 300

 x = 150p

 Hence cost of the ball = Rs.1.50

 =>Cost of the bat = 1.50 + 75 = Rs.2.25

PLAYING CHILDREN

A group of boys and girls are playing.15 boys leave.There remain

2 girls for each boy.Then 45 girls leave.There remain 5 boys for

each girl.How many boys were in the original group?

Solution:

Let B and G represent no.of boys and girls in the original group

respectively.

 G ---------> 2

 B-15 ----------> 1

 G/B-15 = 2/1

        i.e., 2 girls are left for 15 boys who are alone.

 G-45 -------------------->1

 B-15 ------------------------->5

 5 boys are left out when 15 girls are alone.

        =>G/B-15=2/1    --------------------------------(1)

 =>G-45/B-15 =1/5   ----------------------------(2)

 (1) & (2) =>

 G = 2B-30

 5G - 225 = B - 15

 5 ( 2B - 30 ) = B - 15 + 225

 10B = B - 15 + 225 + 150

 9B = 360

 B = 40

 (1)=> G/40-15 = 2

       G=50 girls.

                                               

Problems

1.Reshma appeared for a maths exam.She was given 100 problems to

solve.She tried to solve all of them correctly but some went

wrong.But she scored 85. Her score was calculated by subtracting

two times th no.of wrong answers from the no.of correct answers.

How many problems did Reshma do correctly?

Solution:

 Assume  W as wrong answers and R as correct answers

 Given total no.of questions as 100

 R+W=100  ---------------------------(1)

    Score is calculated by subtracting 2 times wrong answers(2W)

 from right answers(R) and given as 85

 R-2W=85 -------------------------------(2)

 (2)-(1)

               R-2W=85

   R+W=100

         ---------------------

   W=5

 Hence,100-5=95 is the no.of correct answers  of Reshma.

2.A RUNNING RACE

Sneha, Shilpa, Sushma join a running race.The distance is 1500 Meter.

Sneha beats Shilpa by 30 Meter and Sushma by 100 Meter. By how much could Shilpa beat Sushma over the full distance if they both ran as before?

Solution:

 Total distance covered by Sneha=1500m

      Shilpa=1500-30=1470

      Sneha =1500-100=1400

        Distance covered by Shilpa=1500*1400/1470=1428.6

Distance to be covered by Shilpa to beat Sushma over full distance

1500-1428.6=71.4m

3.FILLING A CISTERN

Pipe S1 can fill a cistern in 2 hours and pipe S2 in 3 hours.Pipe S3

can empty it in 5 hours.Supposing all the pipes are turned on when

the cistern is completely empty,how long will it take to fill?

Solution:

 S1 fills cistern in 1/2 hours

 S2 fills cistern in 1/3 hours

 S3 empties it in 1/5 hours

 A the pipes S1,S2,S3 working i.e.,filling the cistern

                         1/2+1/3-1/5=15+10-6/30=25-6/30=19/30

 No.of hours to fill=30/19=1 11\19 hours.

4.SEQUENCE PROBLEMS

What are the next two terms in the sequence?1,1,5,17,61,217.........

Solution:

 The order in this cases is

 Tn=3*Tn-1 +2*Tn-2

     = 3(217)+2(61)

     = 773

 Tn+1=3(773)+2(271)

                     =2319+542

                     =2753

5.SEQUENCES

What are the next two terms in the sequence?

1,1,5,17,61,217.................

Solution:

 Tn=3Tn-1+2Tn-2

Profit & Loss

Important Facts:

Cost Price: The price at which an article is purchased,
is called its cost price,abbreviated as C.P.

Selling Price: The price at which an article is sold,
is called its selling price,abbreviated as S.P.

Profit or Gain: If S.P. Is greater than C.P. The seller
is said to have a profit or gain.

Loss: if S.P. Is less than C.P., the seller is said to
have incurred a loss.

Formula
                         
1.Gain=(S.P-C.P)
2.Loss=(C.P-S.P)
3.Loss or Gain is always reckoned on C.P.
4.Gain%=(gain*100)/C.P
5.Loss%=(loss*100)/C.P
6.S.P=[(100+gain%)/100]*C.P
7.S.P=[(100-loss%)/100]*C.P
8.C.P=(100*S.P)/(100+gain%)
9.C.P=(100*S.P)/(100-loss%)
10.If an article is sold at a gain of say,35%,then S.P=135% of C.P.
11.If an article is sold at a loss of say,35%,then S.P=65% of C.P.
12.When a person sells two similar items, one at a gain of say,
x%,and the other at a loss of x%,then the seller always incurs a
loss given by Loss%=[common loss and gain %/10]2=(x/10)2
13.If a trader professes to sell his goods at cost price,but uses
false weight,then Gain%=[(error/(true value-error))*100]%
14.Net selling price=Marked price-Discount

                                               

Simple Problems

1.A man buys an article for Rs.27.50 and sells it for Rs.28.60
Find the gain percent.

Sol:     C.P=Rs 27.50        S.P=Rs 28.60
         then Gain=S.P-C.P=28.60-27.50=Rs 1.10
        Gain%=(gain*100)/C.P%
             =(1.10*100)/27.50%=4%

2.If a radio is purchased for Rs 490 and sold for Rs 465.50
Find the loss%?

Sol:  C.P=Rs 490      S.P=Rs 465.50
        Loss=C.P-S.P=490-465.50=Rs 24.50
         Loss%=(loss*100)/C.P%
              =(24.50*100)/490%=5%

3.Find S.P when C.P=Rs 56.25 and Gain=20%

Sol:   S.P=[(100+gain%)/100]*C.P
      S.P=[(100+20)/100]56.25=Rs 67.50

4.Find S.P when C.P=Rs 80.40,loss=5%

Sol:    S.P=[(100-loss%)/100]*C.P
        S.P=[(100-5)/100]*80.40=Rs 68.34

5.Find C.P when S.P=Rs 40.60,gain=16%?

Sol:     C.P=(100*S.P)/(100+gain%)
         C.P=(100*40.60)/(100+16)=Rs 35

6.Find C.P when S.P=Rs 51.70 ,loss=12%?

Sol:     C.P=(100*S.P)/(100-loss%)
         C.P=(100*51.70)/(100-12)=Rs 58.75
 
7.A person incurs 5% loss by selling a watch for Rs 1140 . At
what price should the watch be sold to earn 5% profit?

Sol:  Let the new S.P be  Rs x then,
        (100-loss%):(1st S.P)=(100+gain%):(2nd S.P)
             (100-5)/1140=(100+5)/x
       x=(105*1140)/95=Rs 1260

8.If the cost price is 96% of the selling price,then what is
the profit percent?

Sol:       let S.P=Rs 100 then C.P=Rs 96
             profit=S.P-C.P=100-96=Rs 4
             profit%=(profit*C.P)/100%
                    =(4*96)/100=4.17%

9.A discount dealer professes to sell his goods at cost price
but uses a weight of 960 gms for a Kg weight .Find his gain %?

Sol:             Gain%=[(error*100)/(true value-error)]%
                      =[(40*100)/1000-40)]%=25/6%

10.A man sold two flats for Rs 675,958 each .On one he gains
16% while on the other he losses 16%.How much does he gain or
lose in the whole transaction?

Sol:         loss%=[common loss or gain%/10]2=(16/10)2=2.56%

11.A man sold two cows at Rs 1995 each. On one he lost 10% and
on the other he gained 10%.what his gain or loss percent?

Sol:      If loss% and gain% is equal to 10
             then there is no loss or no gain.

12.The price of an article is reduced by 25% in order to
restore the must be increased by ?

Sol:      [x/(100-x)]*100 =[25/(100-25)]*100
                          =(25/75)*100=100/3%

13.Two discounts of 40% and 20% equal to a single discount of?

Sol:    {[(100-40)/100]*[(100-20)/100]}%=(60*80)/(100*100)%
                                        =48%
        single discount is equal to (100-48)%=52%
                                               

Problems On Numbers

Simple problems:

1.What least number must be added to 3000 to obtain a number
exactly divisible by 19?
Solution:
On dividing 3000 by 19 we get 17 as remainder
Therefore number to be added = 19-17=2.

2.Find the unit's digit n the product 2467 153 * 34172?
Solution:
Unit's digit in the given product=Unit's digit in 7 153 * 172
Now 7 4 gives unit digit 1
7 152 gives unit digit 1
7 153 gives 1*7=7.Also 172 gives 1
Hence unit's digit in the product =7*1=7.

3.Find the total number of prime factors in 411 *7 5 *112 ?
Solution:
411 7 5 112= (2*2) 11 *7 5 *112
= 222 *7 5 *112
Total number of prime factors=22+5+2=29

4.The least umber of five digits which is exactly
divisible by 12,15 and 18 is?
a.10010   b.10015    c.10020    d.10080
Solution:
Least number of five digits is 10000
L.C.Mof 12,15,18 s 180.
On dividing 10000 by 180,the remainder is 100.
Therefore required number=10000+(180-100)
=10080.
Ans (d).

5.The least number which is perfect square and is divisible
by each of the numbers 16,20 and 24 is?
a.1600 b.3600 c.6400 d.14400
Solution:
The least number divisible by 16,20,24 = L.C.M of 16,20,24=240
=2*2*2*2*3*5
To make it a perfect square it must be multiplied by 3*5.
Therefore required number =240*3*5=3600.
Ans (b).

6.A positive number which when added to 1000 gives a sum ,
which is greater than when it is multiplied by 1000.
The positive integer is?
a.1 b.3 c.5 d.7
Solution:
1000+N>1000N
clearly N=1.

7.How many numbers between 11 and 90 are divisible by 7?
Solution:
The required numbers are 14,21,28,...........,84.
This is an A.P with a=14,d=7.
Let it contain n terms
then T =84=a+(n-1)d
=14+(n-1)7
=7+7n
7n=77 =>n=11.

8.Find the sum of all odd numbers up to 100?
Solution:
The given numbers are 1,3,5.........99.
This is an A.P with a=1,d=2.
Let it contain n terms 1+(n-1)2=99
=>n=50
Then required sum =n/2(first term +last term)
=50/2(1+99)=2500.

9.How many terms are there in 2,4,6,8..........,1024?
Solution:
Clearly 2,4,6........1024 form a G.P with a=2,r=2
Let the number of terms be n
then 2*2 n-1=1024
2n-1 =512=29
n-1=9
n=10.

10. 2+22+23+24+25..........+28=?
Solution:
Given series is a G.P with a=2,r=2 and n=8.
Sum Sn=a(1-r n)/1-r=Sn=2(1-28)/1-2.
=2*255=510.

11.Find the number of zeros in 27!?
Solution:
Short cut method :
number of zeros in 27!=27/5 + 27/25
=5+1=6zeros.
                                             
Medium Problems:

12.The difference between two numbers 1365.When the larger
number is divided by the smaller one the quotient is 6 and
the remainder is 15.The smaller number is?
a.240   b.270    c.295    d.360
Solution:
Let the smaller number be x, then larger number =1365+x
Therefore 1365+x=6x+15
5x=1350 => x=270
Required number is 270.

13.Find the remainder when 231 is divided by 5?
Solution:
210 =1024.
unit digit of 210 * 210 * 210 is 4
as 4*4*4 gives unit digit 4
unit digit of 231 is 8.
Now 8 when divided by 5 gives 3 as remainder.
231 when divided by 5 gives 3 as remainder.

14.The largest four digit number which when divided by 4,7
or 13 leaves a remainder of 3 in each case is?
a.8739 b.9831 c.9834 d.9893. Solution:
solution:
Greatest number of four digits is 9999
L.C.M of 4,7, and 13=364.
On dividing 9999 by 364 remainder obtained is 171.
Therefore greatest number of four digits divisible by 4,7,13
=9999-171=9828.
Hence required number=9828+3=9831.
Ans (b).

15.What least value must be assigned to * so that th number
197*5462 is divisible by 9?
Solution:
Let the missing digit be x
Sum of digits = (1+9+7+x+5+4+6+2)=34+x
For 34+x to be divisible by 9 , x must be replaced by 2
The digit in place of x must be 2.

16.Find the smallest number of 6 digits which is exactly
divisible by 111?
Solution:
Smallest number of 6 digits is 100000
On dividing 10000 by 111 we get 100 as remainder
Number to be added =111-100=11.
Hence,required number =10011.

Problems On Ages

Simple problems:

1.The present age of a father is 3 years more than three times
  the age of his son.Three years hence,father's age will be 10
  years more than twice the age of the son.Find the present age
  of the father.

  Solution: Let the present age be 'x' years.
            Then father's present age  is 3x+3 years.
            Three years hence
            (3x+3)+3=2(x+3)+10
                        x=10
            Hence father's present age = 3x+3 = 33 years.


2. One year ago the ratio of Ramu & Somu age was 6:7 respectively.
   Four years hence their ratio would become 7:8. How old is Somu.

   Solution: Let us assume Ramu &Somu ages are x &y respectively.
             One year ago their ratio was 6:7
             i.e  x-1 / y-1 = 7x-6y=1
             Four years hence their ratios,would become 7:8
             i.e   x-4  / y-4 = 7 / 8
                          8x-7y= -4
              From the above two equations we get y= 36 years.
                  i.e Somu present age is 36 years.


3. The total age of A &B is 12 years more than the total age of
   B&C. C is how many year younger than A.

   Solution:   From the given data
                            A+B = 12+(B+C)
                  A+B-(B+C) = 12
                             A-C = 12 years.
               C is 12 years younger than A


4. The ratio of the present age of P & Q is 6:7. If  Q is 4 years
   old than P. what will be the ratio of the ages of P & Q after
   4 years.

   Solution:     The present age of P & Q is 6:7 i.e
                                                 P / Q  = 6 / 7
                 Q is 4 years old than P i.e Q = P+4.
                                               P/ P+4 = 6/7
                                                 7P-6P = 24,
                     P = 24 , Q = P+4 =24+4 = 28
                 After 4 years the ratio of P &Q is
                     P+4 : Q+4
                   24+4 : 28+4 = 28:32 = 7:8      

5. The ratio of the age of a man & his wife is 4:3.After 4 years this
   ratio will be 9:7. If the time of marriage  the ratio was 5:3,
   then how many years ago were they married.

   Solution:     The age of a man  is 4x .
                 The age of his wife is 3x.
                 After 4 years their ratio's will be 9:7 i.e
                       4x+4 / 3x+4  =  9 / 7
                       28x-27x=36-28
                            x = 8.
                  Age of a  man is  4x = 4*8 = 32 years.
                  Age of his wife is 3x = 3*8 = 24 years.
                  Let us assume  'y' years ago they were married ,
                  the ratio was 5:3 ,i.e
                       32-y /  24-y = 5/ 3
                                       y = 12 years
                  i.e 12 years ago they were  married
                                                     
6. Sneh's age is  1/6th of her father's age.Sneh's father's age will
   be twice the age of Vimal's age after 10 years. If Vimal’s eight
   birthday was celebrated two years before,then what is Sneh's
   present age.

    a)  6 2/3 years  b) 24 years   c) 30 years   d) None of the above
   
   Solution:  Assume Sneh’s age  is 'x' years.
              Assume her fathers age is 'y' years.
              Sneh’s age is 1/6 of her fathers age i.e x = y /6.
              Father’s age will be twice of Vimal's age  after 10
              years.
              i.e  y+10 = 2( V+10)( where 'V' is the Vimal's age)
              Vimal's eight  birthday  was celebrated two years before,
              Then  the Vimal's present age is 10 years.
                       Y+10 = 2(10+10)
                          Y=30 years.
              Sneh's present age x = y/6
                    x = 30/6 = 5 years.
              Sneh's present age is  5 years.

7.The sum of the ages of  the 5 children's born at the intervals of
   3 years each is 50 years what is the age of the youngest child.

   a) 4 years  b) 8 years  c) 10 years  d)None of the above

   Solution:   Let the age of the children's be  
               x ,x+3, x+6, x+9, x+12.
               x+(x+3)+(x+6)+(x+9)+(x+12)  = 50
                                          5x+30  = 50
                                           5x = 20
                                           x = 4.
                    Age of the youngest child is x = 4 years.

8. If 6 years  are subtracted from the present age of Gagan and
   the remainder is divided by 18,then the present age of his
   grandson Anup is obtained. If Anup is 2 years younger to Madan
   whose age is 5 years,then  what is Gagan's present age.

     a) 48 years  b)60 years    c)84 years  d)65 years
    Solution:  Let us assume Gagan present age is 'x' years.
                   Anup age  =  5-2 = 3 years.
                              (x-6) / 18 = 3
                                       x-6 = 54
                                  x = 60 years

9.My brother is 3 years elder to me. My father was 28 years of
  age when my sister was born while my father was 26 years of age
  when i was born. If my sister was 4 years of age when my brother
  was born,then what was the age my father and mother respectively
  when my brother was born.
  a) 32 yrs, 23yrs  b)32 yrs, 29yrs c)35 yrs,29yrs d)35yrs,33 yrs

   Solution: My brother was born 3 years before I was born & 4
             years after my sister was born.
             Father's age when brother was born  = 28+4 = 32 years.
             Mother's age when brother was born = 26-3 = 23 years.