Infosys puzzles
Technical Training
Infosys Placement Puzzles - III (Ten Puzzles completely solved )

1)Two trains are approaching each other at 60mph each. The length
of each train is 1/6 miles. When the two locomotives(engines) meet,
how many seconds will it take for them to pass each other completely?
Time taken t in hr = sum of lengths / relative speed in miles per hour
t = (1/6 + 1/6) / 120
that gives t = 2/720 hour
multiply by 3600 to get seconds
So in 10 seconds both the train will cross each other

2)If i multiplied by i is me. me multiplied by multiplied by me is
she. What is the value of she?
i is single digit
me two digit
she three digit
only possible values for single digit which gives three digit in fourth power is 4 or 5
but 5*5 gives 25 again 5 in unit digit that is not the case here
So 4*4 gives 16 and 16*16 gives 256
that gives i = 4, me = 16 and she = 256

3) A man jumps into a river from a overbridge ,at the same time a hat drops in the river which flows with the stream of the water. That man travels 10 minute upstream and returns back where he was asked to catch the hat. He catches hat at a distance of 1000 yard down at a second bridge . Find the speed of the river.
Let us take x as the distance travelled by the man in ten min upstream
x/(v-u) = 10  ( Note : v-u is the relative speed where v is mans speed and u river speed)

total time swim by the man upstream and down stream is equal to the time taken by the hat to travel 1000 yards with the speed of river

1000/u = 10 + (x+1000)/(u+v)
now substitute x = 10(v-u) and simplify
you will get
100v = 2uv
that gives 2u = 100 and u = 50 yards per minute

4) There is a 1 km long wire placed on some number of poles which are in equal distance. If the number of poles is reduced by 1 then the distance of wire between each poles
increases 1(2/3).How many poles are there initially.
Let there are N poles intially
Then the distance between two adjacent poles is =1000/(N-1) mts
After removing one pole, distance between two adjacent poles is= 1000/(N-2) mts
Now  1000/(N-1) +  (5/3)  = 1000/(N-2)
(N-1) * (N-2) =600 
By Solving ans will be 26

5). For every one hour two trains will start at the same time,one from Bangalore to Madras another from Madras to Bangalore On parallel tracks. All trains travel with same speed and each train will take 5 hours to reach the other station.How many trains, one train will meet during the journey?
The train that started 5 hour before from another station just arrived to the station when a train about to start.
But the train will meet Train that started 4 hours before while on journey simillarly all the train that started from 4 hours before to the time it reaches there at another station

So Totally in Nine hours 9 trains, the last train not on journey but in station
Hence the train will meet 8 trains on journey from one station to other

6)A and B write a test A  says " i got a third of the ques. wrong" b  says " i got 5 wrong"   together they got three quarters of the questions many did A get correct.

Let us take there are x questions then  (2/3)*x + (x–5)  = (3/4)* 2x  which gives x= 30

7)In how many ways you can arrange the numbers on the die .There is only condition that  1,6 and 2,5 and 3,4 will always be on  the opposite sides.
For n positions r objects can be arangened in nPr ways here 6 positions 3 numbers because remaining 3 positions are occupied by corresponding pairs
Hence the answer is 6P3 which equal to 120

8) There are 3 societies A, B, C. A Lent Tractors to B and C as many as they had
already. After some time B gave as many tractors to A and C as many as they have. After sometime C did the same thing.At the end of this transaction each one of them had 24.
Find the cars each originally had.This question was bit differently asked with the same thing.
At last All A, B, C has 24 tractors
So Before C did the transaction A and B has only half of that i.e A = 12 and B = 12 also it means C = 48 (24+12+12)
Before B did the transaction A and C has half of that, i.e A = 6 and C = 24 and also B = 42  ( 12+24+6)
Before A did the transaction B and C has half of that, i.e B = 21 and C = 12 and also A = 39  (6+21+12)
So intially A has 39 tractors, B has 21 and C has 12

9)Quarter of the time from midnight plus half of the time to the next midnight equals that time. What is the time?
Let time be x
Time From mid night is x-0
Time to next mid night is 24-x
So Given that
x/4 + 12-x/2 = x
x = 48/5 which equal to 9(3/5)hr or 9 hr 36 minutes

10)A publishing company advertisd the job opening.The response was good. One hundred people applied for the post. The company wanted someone expert in both physics and  chemistry. Out of which 10 did not had any specalization.70 of them had got physics training and 82 got training in many applicants had got training in both
physics and chemistry?
A -- Physics
B -- Chemistry

Out of 100, 10 has no Specialization
So the remaining 90 has specialization in Physics or Chemistry or both
n(A) = 70
n(B) = 82
n(AorB) = n(A) + n(B) - n(AandB)
90 = 70 + 82 - x
So x = 62