Infosys Placement Paper 2012

1) A boy asks his father, " what is the age of grand father?". Father replied " He is x years old in x^2 years", and also said, "we are talking about 20th century". what is the year of birth of grand father?

solution: 1892

explanation :

20th century means 1900 to 2000
Find out square root of 2000. 
Sqrt(2000)=44.72 . So 44^2 1936.
so Grandfather is 44 years old in 1936
Year of birth of grandfather is =1936-44 =1892

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2) A boy travels in a scooter after covering 2/3rd of the distance the wheel got punctured he covered the remaining distance by walk. Walking time is twice that of the time the boys riding time. How many times the riding speed as that of the walking speed?

solution: riding speed is 4 times of walking speed.

explanation :

let the total distance =x

distance travelled by scooter=2/3 * x
time taken by scooter(t1) = (2/3 * x)/ speed of the scooter

distance travelled by walk = 1/3 * x
time taken by walk(t2) = (1/3 * x)/speed of the person

Walking time is twice that of the time the boys riding time

t2=2*t1
(1/3 * x)/speed of the person = 2* [(2/3 * x)/ speed of the scooter]
1/speed of the person = 4/ speed of the scooter
speed of the scooter/ speed of the person = 4:1

riding speed is 4 times of walking speed.

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3)In a Knockout tournament 51 teams are participated, every team thrown out of the tournament if they lost twice. How many matches to be held to choose the winner?

solution: 101

explanation :

After every two matches one team is eliminated.
so to eliminate 50 team there will be require 100 matches + 1 for deciding winner.
therefore total 101 matches to be held to choose the winner 

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4)A man sold 2 pens. Initial cost of each pen was Rs. 12. If he sell it together one at 25% profit and another 20% loss. Find the amount of loss or gain, if he sells them separately?

solution:  60 Paise

explanation :

Initial cost of each pen was Rs. 12
CP =12 Rs 

Given 1 pen is sold at 25% profit 
therefore SP = 12+(25/100*CP) = 12+3 = 15 Rs

Given other at loss of 20%
SP = 12-(20/100*CP) = 12-2.4 = 9.6 Rs

Total SP= 15+9.6 = 24.6 Rs
Total CP=12+12=24 Rs

Total gains = SP - CP = 24.6-24 = 0.6 Rs = 60 Paise

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5)Find the 3 digits no. whose last digit is the sqaure root of the first root of the first digits and second digit is the sum of the other two digits. And the number is divisble by 2,3,6,7. what is the number?

solution: 462

explanation :

let x,y,z be the digits of no.
then no = 100x+10y+z

last digit is the square root of the first root of the first digits
z=sqrt(x)

second digit is the sum of the other two digits
y=x+z
y=x+sqrt(x)

so no. is =100x+10y+z
              =100x+10( x+sqrt(x) )+sqrt(x)
              =110x+11*sqrt(x)

if x=1, no. is =121
if x=4, no. is =462
if x=9, no. is =990+33=1023 (not 3 digit)

two no's 121 & 462

now no. is divisible by 2,3,6,7 
so divisible by LCM OF 2,3,6,7 = 42
only 462 is divisible by 42
hence that number is 462.

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6)Meera was playing with her brother using 55 blocks.She gets bored playing and starts arranging the blocks such that the no. of blocks in each row is one less than that in the lower row. Find how many were there in the bottom most row?

solution: 10

explanation :

total no. of blocks=55 

let the no. of blocks at the bottom most row be x. 

If she starts arranging the blocks in descending order then no. of blocks in the next row will be (x-1),                                     

then in next row it will be (x-2). Similarly the top most row will have only 1 block.

As we know the sum of first x natural numbers = [x(x+1)]/2
So, here x(x+1)/2= 55
x^2+x=110
(x-10)(x+11)=0
hence x=10

therefore no. of blocks in the bottom most row= 10.

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7)Two people are playing with a pair of dies. Instead of numbers, the dies have different colors on theirsides. The first person wins if the same color appears on both the dies and the second person wins if the colors are different. The odds of their winning are equal. If the first dice has 5 red sides and 1 blue side, find the color(s) on the second one.

solution: 3 red, 3 blue

explanation :

Solution 1: 

As question says "The odds (chance)of their winning are equal"

As there 5 red and 1 blue if suppose there are again 5 red and 1 blue on the other side
 
so no.of favourable cases will 5red*5red+1blue*1blue = 25+1 = 26
and no.of unfavouarable cases will 1blue*5red+5red*1blue = 5+5 = 10
but he gave odds equal 

so now let's take 4 red and 2 blue 
so no.of favourable cases 5red*4red+2blue*1blue=20+2=22 
and no.of unfavouarable cases will 5red*2blue+1blue*4red=20+4=24 
odds not equal again.

so now let's take 3red 3 blue 
no.of favourable cases 5red*3blue+1blue*3blue=15+3=18 
and no.of unfavouarable cases will 5red*3blue+1blue*3red=15+3=18 
so 3 red 3 blue

Solution 2: 

p(red)=5/6 
p(blue)=1/6

Let x sides are red then (6-x) are blue.

Probability To get both colors are same 
[(5/6)(x/6)+(1/6)(6-x)/6] =1/2
5x+6-x/36=1/2
4x+6/36=1/2
4x+6=36/2=18
4x=18-6
4x=12
x=3
side with red color= x =3
side with blue color= 6-x =6-3=3

logical answer:

As question says "The odds (chance)of their winning are equal"
case 1: first dice = blue
Without writing out all of the possible combinations, it can be seen that if the first die comes up "blue",
the second die will match it half of the time (B-B) and not match half of the time (B-R). 
This will give each player the same odds of winning.
case 2: first dice = red
Likewise, if the first die comes up "red", the second die will still match it half of the time (R-R) and half not (R-B).

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8)A person travels in a car with uniform speed. He observes the milestone,which has 2 digits. After one hour he observes another milestone with same digits reversed. After another hour he observes another milestone with same 2 digits separated by 0. Find the speed of the car?

solution: 45 Km/Hr

explanation :

First milestone :  10x +y
Second milestone : 10y +x
Third milestone:   100x+y

Since the speed is uniform so 

Distance covered in first Hr = Distance covered in Second Hr 
(10y+x)-(10x+y) = (100x+y)-(10y+x)
9y-9x=99x-9y
18y=108x
y=6x

Therefore
First milestone :  10x +y=10x+6x=16x
Second milestone : 10y +x=60x+x =61x
Speed of the car = (distance between two milestone)/time
                           = (61x - 16x)/1
                           = 45x km/hr

by just putting x = 1,2,3,4,....................
only possible combination is x=1
 
So average speed = 45 Km/Hr

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9)Three persons A, B & C went for a robbery in different directions and they theft one horse, one mule and one camel. They were caught by the police and when interrogated gave the following statements
A: B has stolen the horse
B: I didn't rob anything.
C: both A & B are false and B has stolen the mule.

The person who has stolen the horse always tell the truth and
The person who has stolen the camel always tell the lie.

Find who has stolen which animal?

solution: 

A- camel
B- mule
C- horse

explanation :

B lies so A also lies. 
so C is correct B has stolen the Mule. 
C has stolen the horse and A has stolen the Camel.

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10)One quarter of the time till now from midnight and half of the time remaining from now up to midnight adds to the present time. What is the present time?

solution: Time = 09:36 am 

explanation :

Let the current time be T
one quarter of current time from midnight[00:00hrs]   = T/4      ....(A)
Half of the time remaining from now to midnight       = (24-T)/2 ....(B)
Current time (T)= A + B = (T/4)+(24-T)/2

Therefore 
T  = (T+48-2T)/4
4T = 48 - T 
5T = 48
T  = 48/5 
T  = 9.6 
T  = 9.0 + 0.6

i.e. 9 minutes + 0.6 minutes
i.e. 9 minutes + 36 seconds

Time = 09:36 am 

Cross Check:

1] 1/4 of 09.36 HRS is 02.24 hrs
2] Remaining Hours till midnight from 09:36 am is 14:24 HRS (24:00-09:36)
Half of 14:24 is 07:12

Current time 02.24+07:12 = 09:36

Therefore Current Time is 9Hrs 36Min

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11)After world war II three departments did as follows First department gave some tanks to 2nd &3rd departments equal to the number they are having. Then 2nd department gave some tanks to 1st & 3rd departments equal to the number they are having. Then 3rd department gave some tanks to 2nd &1st departments equal to the number they are having. Then each department has 24 tanks. Find the initial number of tanks of each department?

solution:  Initially 1st had 39 ,2nd had 21 and 3rd had 12 Tanks. 
explanation :

Method 1:

Here we are solving problem from the end.


formula:

      Result of after  transaction nth transaction= 1         2     3       4        n 

formula for before transaction nth transaction=  1/2     2/2   3/2    4/2     n+1/2+2/2+3/2...



let's apply,

Now each of them got 24



Before 3rd  transaction:

here n is 3.

value for 1st  dept =24/2=12

value for 2nd dept =24/2=12

value for 3rd dept = value of 3rd dept+ 1/2 previous value of 1st  dept+ 1/2 previous value of 2nd dept

                             = 24+24/2+24/2=48

Before 2nd  transaction:

here n is 2.

value for 1st  dept =12/2=6

value for 2nd dept = value of 2nd  dept+ 1/2 previous value of 1st  dept+ 1/2 previous value of 3rd dept

                                     = 12+12/2+48/2=42

value for 3rd dept =48/2=24



Before 1st  transaction:

here n is 2.

value for 1st  dept = value of 1st  dept+ 1/2 previous value of 2nd dept+ 1/2 previous value of 3rd dept

                              = 6+42/2+24/2=39

value for 2nd dept =42/2=21

value for 3rd dept =24/2=12


look at table



Transaction

1st  dept

2nd dept

3rd dept


 final

24

24

 24


Before 3rd  transaction

24/2=12 

24/2=12

 24+24/2+24/2=48


Before 2nd  transaction

12/2=6

12+12/2+48/2=42

48/2=24


Before 1st  transaction

12+42/2+24/2=39 

42/2=21

24/2=12




so initially 1st had 39 ,2nd had 21 and 3rd had 12 Tanks.


Method 2:
let's assume 1st had x tanks,2nd had y tanks,3rd had z tanks

1st lends y tanks to 2nd and z tanks to 3rd.

So now no of tanks:

1st: x-(y+z)

2nd: 2y

3rd: 2z

2nd lends x-(y+z) tanks to 1st and 2z tanks to 3rd.

So now no of tanks:

1st: 2x-2y-2z

2nd: 2y-(x-y+z)

3rd: 4z

3rd lends 2x-2y-2z tanks to 1st and 2y-(x-y+z) tanks to 2nd.

So now no of Tanks:

1st: 4x-4y-4z

2nd: 6y-2x-2z

3rd: 4z -(x+y-3z)

now equating,

4x-4y-4z = 24

x-y-z = 6 ..........(i)

6y-2x-2z = 24

-x+3y-z = 12 .......(ii)

-x-y+7z = 24 .......(iii)

Solving (i) (ii) and (iii) we get,

x = 39, y = 21, z = 12

Hence 39, 21 and 12 tanks each in order for 1st, 2nd and 3rd depts

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12)A, B, C, D & E are having their birthdays on consecutive days of the week not ecessarily in the same order. A 's birthday comes before G's as many days as B's birthday comes after E's. D is older than E by 2 days. This time G's birthday came on wednesday. Then find the day of each of their birthdays?

solution: 

D _ Sunday
A _ Monday
E _ Tuesday
G _ Wednesday

B _ Thursday

explanation :

A 's birthday comes before G's as many days as B's birthday comes after E's
from this, A _ G & D _ E and D is older than E by 2 days means in between D and E one person is there,
from these, D A E G B.. will be the correct sequence.

D _ Sunday
A _ Monday
E _ Tuesday
G _ Wednesday
B _ Thursday

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13)A girl 'A' told to her friend about the size and color of a snake she has seen in the beach. It is one of the colors brown/black/green and one of the sizes 35/45/55.

If it were not green or if it were not of length 35 it is 55.
If it were not black or if it were not of length 45 it is 55.
If it were not black or if it were not of length 35 it is 55.

a) What is the color of the snake?
b) What is the length of the snake?

solution:  (a) brown;  (b) 55;

explanation :

it is clearly mentioned that it is not of length 35,45 .so its obvious that length is 55.
 
After we came to know that it is of length 55 then by using the three statements.
we comes know that only color left is brown color.

Therefore
a) brown
b) 55

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14)There are 2 persons each having same amount of marbles in the beginning. after that 1 person gain 20 more from second person and he eventually lose two third of it during the play and the second person now have 4 times marble of what 1st person is having now. find out how much marble did each had in the beginning?

solution:  100 marbles .

explanation :


Let two A & B initially have X marbles.



After 1st play

A gain 20 more from B

A=X+20

B=X-20



After 2nd play

A eventually lose two third of it during the play and the B now have 4 times marble of what A is having

after losing 2/3(X+20)

A=1/3(X+20)

B=(X-20)+2/3(X+20)



and the B now have 4 times marble of what A is having

B=4*A

(X-20)+2/3(X+20)=4*1/3(X+20)

X=100



Thus initially each have 100 marbles in the beginning.

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15)A lady was out for shopping. she spent half of her money in buying A and gave 1 doller to bagger. futher she spent half of her remaining money and gave 2 doller to charity. futher she spent half of remaining money n gave 3 dollor to some childrans. now she has left with 1 doller. how much she had in the beginning?

solution: Total money = 42

explanation :

lets assume total money she had in beginning was x

She spent half of her money in buying
money = x/2 
 
gave 1 dollar to bagger
money = (x/2)-1 = (x-2)/2

Further she spent half ofremaining money
money = (x-2)/4

gave 2 dollar to charity
money = (x-2)/4 - 2=(x-10)/4

Further she spent half of remaining money  
money = (x-10)/8

gave 3 dollars to some children
money = (x-10)/8 - 3= (x-34)/8 

Now she is been left with 1 dollar
(x-34)/8  = 1
x-34=8
x=42

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16)There are certain diamonds in a shop.
1 thief stole half of diamonds and 2 more.
2 thief stole half of remaining and 2 more
3. same as above
4 same as above.
5 came nothing was left for that.

how many diamonds was there?

solution: 60

explanation :

let the total diamonds x.

first thief took  half of diamonds and 2 more
remaining diamonds = (x/2)-2

second thief took  half of diamonds and 2 more
remaining diamonds = ( (x/2)-2 )/2-2

third thief took  half of diamonds and 2 more
remaining diamonds = ( ((x/2)-2)/2-2 )/2-2

fourth thief took  half of diamonds and 2 more
remaining diamonds = ( (((x/2)-2)/2-2)/2-2 )/2-2

now shop is left with 0 diamonds
((((x/2)-2)/2-2)/2-2)/2-2 =0 
((((x/2)-2)/2-2)/2-2)/2 =2
((((x/2)-2)/2-2)/2-2) =4
((x/2)-2)/2-2)/2   =6
(((x/2)-2)/2-2) =12
(x/2)-2)/2 =14
(x/2)-2) =28
(x/2) =30
x=60

so, first person took 60/2 +2 =32
2nd person took 28/2 +2 = 16
3rd person took 12/2 +2 =8
4th person took 4/2 +2 =4
so we are left with 0 diamonds in the shop.

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17)There are three friends A B C.
1. Either A or B is oldest
2. Either C is oldest or A is youngest.

Who is Youngest and who is Oldest?

solution: A is youngest n B is oldest.

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18) Father says my son is five times older than my daughter. my wife is 5 times older that my son. I am twice old from my wife and altogether (sum of our ages) is equal to my mother 's age and she is celebrating her 81 birthday. so what is my son's age?

solution: 5 years

explanation :

let daughter's age be x
son's age= 5x
wife's age=25x
his age= 50x
sum = x+5x+25x+50x = 81
81x=81
x=1

now 
daughter's age: x=1
sons' age: 5x=5

son is 5 years old

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19)In Mulund, the shoe store is closed every Monday, the boutique is closed every Tuesday, the grocery store is closed every Thursday and the bank is open only on Monday, Wednesday and Friday. Everything is closed on Sunday.

One day A, B, C and D went shopping together, each with a different place to go. They made the following statements:
A:- D and I wanted to go earlier in the week but there wasn?t day when we could both take care of our errands.
B:- I did not want to come today but tomorrow I will not be able to do what I want to do.
C:- I could have gone yesterday or the day before just as well as today.
D:- Either yesterday or tomorrow would have suited me.

Which place did each person visit ?

solution: 

A - BOUTIQUE
B - BANK
C - GROCERY STORE
D - SHOE STORE 

explanation :

They made the following statements: 
A :
- D and I wanted to go earlier in the week but there wasnt day when we could both take care of our errands. 

B :
- I did not want to come today but tomorrow I will not be able to do what I want to do. 

C :
- I could have gone yesterday or the day before just as well as today. 

D :
- Either yesterday or tomorrow would have suited me. 

Which place did each person visit? 

      Shoe  Butq  Grcy   Bank 
Mon    X     O       O        O
Tue     O     X       O        X
Wed    O     O       O       O
Thu     O     O        X       X
Fri       O     O        O      O
Sat      O     O        O      X
Sun     X      X        X      X

1st Clue:
- One day A, B, C and D went shopping together, 

2nd Clue
- each with a different place to go.

From this we know they they must have gone shopping either, Wednesday or Friday, 
because these are the only two days on which all the different Shops are open.

As statement:
"D and I wanted to go earlier in the week but there wasnt day when we could both take care of our errands."

Thus we know, A and D must have tried for either Monday or Tuesday shopping since ealier in the week is 
either Monday or Tuesday, and that A or D must have wanted to go to either the Shoe Store, the Boutique, 
or the Bank. Thus either A or D must have wanted to go to the Shoe Store, and the other (A or D) must 
have then wanted to go to either the Boutique or the Bank.

Bs statement:
"I did not want to come today but tomorrow I will not be able to do what I want to do."

Thus we know that, given the only 2 days in which A, B, C, and D could shop together, that B must have 
wanted to go to either the Grocery or the Bank

Cs statement:
"I could have gone yesterday or the day before just as well as today."

If today is Wednesday, then
C could have only gone to the Grocery

If today is Friday, then
C could have gone to either the Shoe Store or the Boutique.

Ds statement: 
"Either yesterday or tomorrow would have suited me."

If today is Wednesday, then
D must have wanted to go to the Shoe Store and
A must want to go to either the Boutique or the Bank

If today is Friday, then 
D must have wanted to go to either the Shoe Store or the Boutique

Now that is all spelled out, first find what day "today" is:

If today was Friday, then 
Case #1:
- If D goes to the Shoe Store then
- C goes to the Boutique
- A goes to the Bank
- B goes to the Grocery
- Case #1 is false since Bs statement would be false

Case #2
- If D goes to the Boutique Then
- C goes to the Shoe Store
- A goes to the Bank
- B goes to the Grocery
- Case #2 is false since Bs statement would be false

Thus, given that neither Case#1 or Case#2 can be True, that means D must have gone on a Wednesday

Since today must be Wednesday, then:

D - The Shoe Store
C - The Grocery
B - The Bank (given Bs statement, B could only have wanted go to either the Grocery or the Bank)
A - The Boutique (only shop left)

Thus the only correct answer is: 

A - BOUTIQUE
B - BANK
C - GROCERY STORE
D - SHOE STORE 





Note: only for your understanding
[ Also, thus this answer could not possible be right:

D - Boutique, C - Shoe Store, B - Bank, A - grocery store

Why? If D went to the Boutique and A went to the Grocery, then As statement: "D and I wanted to go earlier 
in the week but there wasnt day when we could both take care of our errands." 
would be false since D and A could both have gone to the Boutique and the Grocery on Monday!

Just a few breaks in logic lead to the mistake of:

D - Boutique, C - Shoe Store, B - Bank, A - grocery store

Breaking down the steps that lead to that incorrect answer, its possible to see where things went wrong:

The day of the week is:-
1. Cant be a sunday
(True, since everything is closed Sunday)

2. Cant be a saturday ( since D says yesterday and tomorrow) 
(True since everything is closed Sunday)

3. Cant be a monday ( since D says yesterday and tomorrow)
(True since everything is closed Sunday)

4. Cant be a tuesday ( since C says he could have gone day before yesterday also)
(True since everything is closed Sunday)

5. Cant be a wednesday ( since C says he could have gone day before yesterday also)
(FALSE, C could have gone to the Grocery on Wednesday, and could have gone "yesterday" 
which would have been Tuesday and could have gone "or the day before" which would have 
been Monday. This is the mistake that cascades into other logic mistakes)

6. Has to be a thursday or friday
(FALSE, it can not be Thursday since "One day A, B, C and D went shopping together", and it can not be Friday 
since there would be no why for all the statements of A, B, C and D to be true if today was Friday)

7. its clearly evident that B goes to the Bank.
(True)

8. So, it has to be a friday
(FALSE, it can not be Friday since there would be no why for all the statements of A, B, C and D to be true 
if today was Friday)

9. C goes to the shoe store.
(FALSE, If C goes to the shoe store and B goes to the Bank, then D and A are going to the Boutique 
or the Grocery Store, but they could have done that earlier in the week on Monday making As statement false)

10. Since, A and D cant go together on M,Tue,Wed and Thu its only possible if A goes to the grocery store. (ie. Thu is ruled out).
(FALSE, A and D can go together on Wednesday. Only Monday and Tuesday are earlier in the Week)

11. D goes to the Boutique. 
(FALSE, If D goes to the Boutique and A goes to the Grocery Store, then As statement is false 
since they could have gone to the Boutique and the Grocery on Monday)
]

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20)5 hunters Doe, Deer, Hare, Boar and Row kill 5 animals. Each hunter kills an animal that does not correspond to his name. Also each hunter misses a different animal which again does not correspond to his name.
a) The Deer is killed by the hunter, known by the name of the animal killed by Boar.
b) Doe is killed by the hunter, known by name of animal missed by Hare.
c) The Deer was disappointed to kill only a Hare and missed the Roe.

solution: 

Deer killed Hare and missed Roe. 
Doe killed Deer and missed Hare.
Hare killed Roe and missed Boar.
Roe killed Boar and missed Doe.
Boar killed Doe and missed Deer.

explanation :

Start with a matrix of the Hunters versus animals killed and missed. 


 


  

             Animal Killed

  

               Animal Missed

 

  



   


    

Hunter

    

doe

    

deer

    

hare

    

boar

    

roe

   

    

doe

    



    



    



    



    



   

    

deer

    



    



    



    



    



   

    

hare

    



    



    



    



    



   

    

boar

    



    



    



    



    



   

    

roe

    



    



    



    



    



   




  



   


    

Hunter

    

doe

    

deer

    

hare

    

boar

    

roe

   

    

doe

    



    



    



    



    



   

    

deer

    



    



    



    



    



   

    

hare

    



    



    



    



    



   

    

boar

    



    



    



    



    



   

    

roe

    



    



    



    



    



   




 

Fill in what we know -- the diagonals are definitely not possible 
since each hunter did not kill or miss the animal with his/her own 
name and Deer killed a hare and missed a roe. Indicate "definite" 
items with Y and "definite not" items with X. Whenever we know 
something (Y), all other entries in that row and column are X. 
This gives us: 


 


  

            Animal Killed

  

                Animal Missed

 

  



   


    

Hunter

    

doe

    

deer

    

hare

    

boar

    

roe

   

    

doe

    

X

    



    

X

    



    



   

    

deer

    

X

    

X

    

Y

    

X

    

X

   

    

hare

    



    



    

X

    



    



   

    

boar

    



    



    

X

    

X

    



   

    

roe

    



    



    

X

    



    

X

   




  



   


    

Hunter

    

doe

    

deer

    

hare

    

boar

    

roe

   

    

doe

    

X

    



    



    



    

X

   

    

deer

    

X

    

X

    

X

    

X

    

Y

   

    

hare

    



    



    

X

    



    

X

   

    

boar

    



    



    



    

X

    

X

   

    

roe

    



    



    



    



    

X

   




 

Using statement 2, the animal missed by Hare is either doe, deer, 
or boar. The doe was killed by Hare, Boar, or Roe. Thus it must be 
Boar who killed the doe. And it must be Hare that missed the boar. 
Add in that any animal killed (or missed) cannot be missed (or 
killed) by that same hunter. Also put Y in any row or column that 
has only one cell that is not X. This gives us: 

 


  

           Animal Killed

  

                Animal Missed

 

  



   


    

Hunter

    

doe

    

deer

    

hare

    

boar

    

roe

   

    

doe

    

X

    



    

X

    



    



   

    

deer

    

X

    

X

    

Y

    

X

    

X

   

    

hare

    

X

    



    

X

    

X

    



   

    

boar

    

Y

    

X

    

X

    

X

    

X

   

    

roe

    

X

    



    

X

    



    

X

   




  



   


    

Hunter

    

doe

    

deer

    

hare

    

boar

    

roe

   

    

doe

    

X

    



    



    

X

    

X

   

    

deer

    

X

    

X

    

X

    

X

    

Y

   

    

hare

    

X

    

X

    

X

    

Y

    

X

   

    

boar

    

X

    



    



    

X

    

X

   

    

roe

    

Y

    

X

    

X

    

X

    

X

   




 


Using statement 1, the animal missed by Roe is the doe. So, 
the deer was killed by Doe. Filling in Xs horizontally and 
vertically allows us to complete the killed side. Add in 
that any animal killed (or missed) cannot be missed 
(or killed) by that same hunter. Also put Y in any row or 
column that has only one cell that is not X. This gives us 
the completed solution matrix:

 


  

             Animal Killed

  

                 Animal Missed

 

  



   


    

Hunter

    

doe

    

deer

    

hare

    

boar

    

roe

   

    

doe

    

X

    

Y

    

X

    

X

    

X

   

    

deer

    

X

    

X

    

Y 

    

X

    

X

   

    

hare

    

X

    

X

    

X

    

X

    

Y

   

    

boar

    

Y

    

X

    

X

    

X

    

X

   

    

roe

    

X

    

X

    

X

    

Y

    

X

   




  



   


    

Hunter

    

doe

    

deer

    

hare

    

boar

    

roe

   

    

doe

    

X

    

X

    

Y

    

X

    

X

   

    

deer

    

X

    

X

    

X

    

X

    

Y

   

    

hare

    

X

    

X

    

X

    

Y

    

X

   

    

boar

    

X

    

Y

    

X

    

X

    

X

   

    

roe

    

Y

    

X

    

X

    

X

    

X

   

Therefore
Deer killed Hare and missed Roe. 
Doe killed Deer and missed Hare.
Hare killed Roe and missed Boar.
Roe killed Boar and missed Doe.
Boar killed Doe and missed Deer.

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21)The local weatherman says No Rain, and his record is 2/3 accuracy of prediction. But the Federal Meteorological Service predicts rain, and their record is 3/4.
With no other data available, what is the chance of rain? 

solution: 3/5

explanation :

Assuming the two sources are independent:
P(rain) = P(weatherman is wrong and FMS is right) = 1/3 * 3/4 = 1/4
P(no rain) = P(weatherman is right and FMS is wrong) = 2/3 * 1/4 = 1/6
Actual probability of rain = P(rain) / (P(rain) + P(no rain)) = (1/4) / (1/4 + 1/6) = 3/5.

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22)10 1 9 2 8 3 7 4 6 5 5 6 4 7 3 8 2 __
Write the next elements in the series.

solution: 9,1

explanation :

in the series there are two sequences of numbers. 
1]descending to ascending(10....2). [ 10 __ 9 __ 8 __ 7 __ 6 __ 5 __ 4 __ 3 __ 2 __ ? ] 
2]ascending to descending(1....8).  [      1 __ 2 __ 3 __ 4 __ 5 __ 6 __ 7 __ 8 __ ? ]
so the next elements are according to the sequence are. 9 and 1 respectively. 

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23)A Man is sitting in the last coach of train could not find a seat, so he starts walking to the front coach ,he walks for 5 min and reaches front coach. Not finding a seat he walks back to last coach and when he reaches there, train had completed 5 miles. What is the speed of the train.

solution: 30 mph 

explanation :

he walks for 5 min and reaches front coach. Not finding a seat he walks back to last coach and when he reaches there.
total time = 10 min= 10/60 hr=1/6 hr

train had completed 5 miles.
so distance is 5 miles.

speed of train = distance /time =5 / (1/6) = 5*6 = 30 mph 

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24)The Old car of Mary requires tyres to be changed after each 24000km. If she wants to go for 42000 km journey then how many minimum number of tyres she will need?

solution:  7 tyres

explanation :

Total distance travelled by 4 wheels = 42000km * 4 =168000
so, minimum no of tyres she requires = 168000/24000 = 7 

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25)A coin is so unbalanced that it may come both heads in 2 tosses as it may come tails in a single toss. What is the probability of getting a head in a single toss.

solution: 1/2

explanation :

2c1*(1/2)*(1/2)=1/2

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26)The virgo club members used to meet every week to play cards. Each time they used to seat around a round table and for their memory they used all the possiblecombinations of postions each for a single time only. Can you tell for how many times they met?

solution: (n-1)! Days

explanation :

If there are n members in the club then the no. of different possible combinations in a circle is ( n – 1 ) !
Thus they meet for (n-1)! Days

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27)A person needs 6 steps to cover a distance of one slab. If he increases his foot length (step length) by 3 inches he needs only 5 steps to cover the slabs length. What is the length of the each slab?

solution: 90 inches or 7.5 feet

explanation :

let his foot length is x inches.
Then it takes him 6 steps to cover the distance of 1 slab. 
So the length of the slab = 6 * x----------------(1)
Now when he increases his foot length by 3 inches then, it takes him only 5 steps 
now the length of the slab = 5 * ( x+3) ----------(2)
from (1) & (2)

6x = 5x + 15
x=15 inches 
thus the slab length is 15 *6 = 90 inches
distance of each step is 90 inches or 90/12 i.e 7.5 feet

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28)There are 19 red balls and one black ball. Ten balls are put in one jar and the remaining 10 are put in another jar. What is the possibility that the black is in the right jar?

solution:  1/2

explanation :

The black ball can go either in the right jar or left jar. So the probability that it goes in a right jar is 1/2. 
same for the left jar.

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29)There is one lily in the pond on 1st june. There are two in the pond on 2nd june . There are four on 3rd june and so on. The pond is full with lilies by the end of the june.
(i) On which date the pond is half full?
(ii) If we start with 2 lilies on 1st june when will be the pond be full with lilies.

solution: (i) 29th june; (ii) 29th june;

explanation :

after every day lilies get doubled.

june has 30 days.so at 29th june no of lilies are half of 30 june.

if we start with 2 than pond be full with lilies at 29th june

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30)A Lorry Starts From Banglore To Mysore At 6.00 Am, 7.00 Am, 8.00 Am.....10 Pm. Similarly on another side starts from Mysore to Banglore at 6.00 Am, 7.00 Am, 8.00 Am.....10.00 Pm. A Lorry Takes 9 Hours To Travel From Banglore To Mysore And Vice Versa.

(I) A Lorry Which Has Started At 6.00 Am Will Cross How Many Lorries?
(II) A Lorry Which Had Started At 6.00 Pm Will Cross How Many Lorries?

solution: 

(i)10

(ii)14

explanation : 

Case I :
It will arrive at its destination at 3 pm so it will cross paths with a lorries that leave before 3 pm ie the ones that leave the destination at 6 am,7 am, 8 am. 9 am 10 am 11 am ,12 noon, 1 pm, 2 pm  and 3 pm.

Thus the no of lorries =10

Case II :
A lorry which has started it journey at 6 p.m from bangalore will come across the following lorries:-
The concerned lorry starting from b'lore at 6 pm will reach mysore at 3am.
a lorry starting at 9 am from mysore will reach b'lore at 6 pm

thus the lorries starting from mysore at 9 am,10 am,11 am,12 pm,1 pm,2 pm,3 pm ,4 pm,5 pm,6 pm, 7 pm,8 pm,9 pm,10 pm  will all cross the concerned lorry.

Thus the no of lorries =14

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31)Every day a cyclist meets a train at a particular crossing. The road is straight before the crossing and both are travelling in the same direction. Cyclist travels with a speed of 10 Kmph. One day the cyclist comes late by 25 min. and meets the train 5km before the crossing. What is the speed of the train.

solution:  60 km/hr.

 
explanation :


first part:

when the cyclist meets the train, he has 5 km to go

so for the cyclist Time require to travel 5km is

Distance = Speed * Time

Cyclist:

Speed = 10 km/h

Distance = 5 km

Time = 5/10 h = 1/2 h = 30 min

The cyclist takes 30 min to travel 5 km.

second part:

The cyclist is already 25 minutes late, so it takes the train t=30-25= 5 minutes to travel 5 km.

So the speed should have been 60 km/h



Hence, the speed of the train is 60 km/hr.

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32) A theif steals half the total no of loaves of bread plus 1/2 loaf from a backery. A second theif steals half the remaing no of loaves plus 1/2 loaf and so on. After the 5th theif has stolen there are no more loaves left in the backery. What was the total no of loaves did the backery have at the beginning?

solution: 

total number of loaves =31
after 1st thief remaining loaves= a  = 15
after 2nd thief remaining loaves= b  = 7
after 3rd thief remaining loaves= c  = 3
after 4th thief remaining loaves= d  = 1
after 5th thief remaining loaves= e  = 0

explanation :

let y be the total number of loaves
a be the remaining loaves after 1st thief
b be the remaining loaves after 2nd thief
c be the remaining loaves after 3rd thief
d be the remaining loaves after 4th thief
e be the remaining loaves after 5th thief

a = y - (y/2 + 1/2) = y/2 - 1/2    ...(I)
b = a - (a/2 + 1/2) = a/2 - 1/2    ...(II)
c = b - (b/2 + 1/2) = b/2 - 1/2    ...(III)
d = c - (c/2 + 1/2) = c/2 - 1/2    ...(IV)
e = d - (d/2 + 1/2) = d/2 - 1/2    ...(V)

After the 5th theif has stolen there are no more loaves left in the backery.
therefore,
e = 0
d/2 - 1/2 = 0
d/2 = 1/2
d=1

substituting in eq (IV)
d = c/2 - 1/2 
1 = c/2 - 1/2
3/2 =c/2
c=3 

substituting in eq (III)
c = b/2 - 1/2
3 = b/2 - 1/2
7/2 =b/2
b = 7

substituting in eq (II)
b = a/2 - 1/2
7 = a/2 - 1/2
15/2 = a/2
a = 15


substituting in eq (I)
a  = y/2 - 1/2
15 = y/2 - 1/2
31/2 = y/2
y=31

so total number of loaves =31
after 1st thief remaining loaves= a  = 15
after 2nd thief remaining loaves= b  = 7
after 3rd thief remaining loaves= c  = 3
after 4th thief remaining loaves= d  = 1
after 5th thief remaining loaves= e  = 0

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33)A gardener plants 100 meters towards east, next 100 meters towards north,next 100 meters towards west. 98 meters towards east, 96 meters towards north and 96 meters towards west, 94 meters towards south. and 94 meters towards east and so on. If a person walks between the trees what is the total distance travelled by him before he reaches the center?

solution: 

|---------------|
| |
| |
| | |
--------|- |
---------------------| -

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34) From a vessel, 1/3rd of the liquid evaporates on the first day. On the second day 3/4th of the remaining liquid evaporates. What fraction of the volume is present at the end of the second day?

solution: 1/6

explanation :

Let the vessel contain xL of the liquid.

1st day --> 
liquid evaporate = x/3
liquid left = 2x/3

2nd day --> 
liquid evaporate = 2x/3 * 3/4 = x/2, 
liquid left = 2x/3 - x/2= x/6

Fraction of the volume present at the end of second day =(x/6)/x = 1/6

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35)There is a 4 inch cube painted on all sides. This is cut down into of 1 inch cubes. What is the no of cubes which have no pointed sides?

solution: 8

explanation :

click here for tricks to solve cube problem

0 sides painted = (n-2)^3  = (4-2)^3 = 2^3 = 8

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