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IBM Sample Problem Using Speed

Question 1
A policeman starts chasing a thief 30 minutes after the thief had run from a spot. With an average speed of 20km per hour, he takes 2 hours to catch the thief. What is the average speed of the thief?
a)16km/hr b)25km/hr c)24km/hr d)18km/hr
Answer : a)16km/hr
Solution:
As given, the average speed of the policeman = 20km/hr.
He takes 2 hours to catch the thief, so from formula, "distance = speed x time" we have
The total distance covered by the police to catch the thief = 20 x 2 = 40 km (This value is also equal to the distance run by thief before being caught by Police.)
Policeman had started late by 30 minutes and took 2 hours to catch the running thief.
Above means that the thief takes (30minutes + 2 hours =) 5/2 hours to reach 40km.
So the speed of the thief = 40/(5/2) = 40 x 2 / 5 = 16 km/hr.
Hence the answer is 16km/hr.
Question 2
From a particular spot, Tom started to chase Jerry which had left the spot before 30 minutes. Tom ran across a highway and…

Infosys Sample Distance Calculation Questions

Question 1
A train starts from station P towards Q with certain speed. Due to a problem, after crossing 50km, the train slows down to 2/3 rd of its actual speed and it reaches Q 50 minutes later than the planned time. Suppose the technical problem had happened after crossing 60 km and the train would had reached 40 minutes late. What is the actual(original) speed of the train and what is the distance between P and Q ?
a)30km/hr, 100km b)20km/hr,80km c)40km/hr,150km d)50km/hr,150km
Answer : a)30km/hr, 100km
Solution:
Let the distance between P and Q be X and the speed initially be V.
Note that the train travels 50km with speed V and the remaining distance(X-50)km with speed 2/3 of V.
According to the above condition, with the formula " distance/speed = time", we can have
Time Taken For First 50 Km + Time Taken For Remaining (X-50) Km = Planned Time + Extra Time Due To Problem
[50/V]+[(X-50)/(2V/3)] = [X/V]+5/6 (here 50 minutes = 5/6 hours)
[100+3X-150-2X]/2V = …

TCS 4 Verbal Synonym Sample Questions

Choose the appropriate meaning from among the options.
Question 1
Fluent
a. Voluble b. Loud c. Audible d. Shout
Answer : a. Voluble
Being voluble means to be very fluent in speaking.
Question 2
Volatile
a. Oscillate b. Transient c. Penetrate d. Light
Answer : b. Transient.
Transient can be used to refer something that is temporary and volatile.
Question 3
Cryptic
a. terse b. difficult c. complex d. Unknown
Answer : a. Terse
Cryptic, Terse etc can be used to describe things that are brief.
Question 4
Correlate
a. to join b. to match c. to refer d. to fix
Answer : b. to match

Accenture Sample Logical Equation Questions

Question 1
If 11 + 21 = 3, 34 + 45 = 32 and 42 + 53 = 23 then 64 + 75 = ?
a)59 b)43 c)69 d)53
Answer : a)59
Solution :
Let AB be the first number of the addition and CD be the second one. By closely observing, we can conclude that questions are in the general form AB + CD = (A x B) + (C x D).
i.e.,11 + 21 = 1 x 1 + 2 x 1 = 1 + 2 = 3
34 + 45 = 3 x 4 + 4 x 5 = 12 + 20 = 32
42 + 53 = 4 x 2 + 5 x 3 = 8 + 15 = 23.
Therefore, the result for 64 + 75 = 6 x 4 + 7 x 5 =24 + 35 = 59.
Question 2
If 15 + 22 = 32, 19 + 35 = 62 and 23 + 49 = 101 then 34 + 52 = ?
a)91 b)101 c)172 d)72
Answer : c)172
Solution:
By observing the question we can generalize the equations to, AB + CD = (AB x C) + D
Then,
15 + 22 = (15 x 2)+ 2 = 30 + 2 = 32
19 + 35 = (19 x 3)+ 5 = 57 + 5 = 62
23 + 49 = (23 x 4)+ 9 = 92 + 9 = 101
Similarly,34 + 52 = (34 x 5)+ 2 = 170 + 2 = 172
Hence 172 is the required answer.
Question 3
If 3 + 5 = 19, 5 + 9 = 61 and 9 + 12 = 117 then 13 + 14 = ?
a)128 b)152 c)124 d)183
Answer :

Capgemini Sample Dimensions Based Questions

Question 1
A manufacturer reduces the size of his machine of Material1 from 30" to 28" and that of Material2 from 24" to 23". He normally spends Rs.1440 for Material1 and Material2. 3/4th of the amount spent would be for Material1. How much will he save under the size of new size (Assuming that the costs incurred by the machines are directly proportional to their sizes)?
a)Rs.85 b)Rs.87 c)Rs.86 d)Rs.88
Answer : b)Rs.87
Solution:
The manufacturer totally spends Rs.1440 for both material 1 and 2.
Out of this 3/4th is entirely for Material1.
Amount spent on Material1 = 1440 x 3/4 = Rs.1080
i.e, he spent Rs.1080 for Material1 of 30" and he reduces 2".
Size Amount 30" Rs.1080 2" ? Amount saved on Material1 = 1080 x 2/30 = Rs.72
Amount spent on Material2 = total amount - amount for Material1 = 1440 - 1080 = Rs.360.
i.e., he spent Rs.360 for Material2 of 24" and he reduces 1".
Size Amount 24" Rs.360 1" ? Amoun…

Read this small story; Hope that makes a BIG change in YOU !!

Read this small story;
Hope that makes a BIG change in YOU !!
At The End Don't Forget To Share It If It Inspires You..

The Professor began his class by holding up a glass with some water in it.
He held it up for all to see & asked the students "How much do you think
this glass weighs?"

'50gms!'..... '100gms!' .....'125gms' ...the students answered.

"I really don't know unless I weigh it," said the professor, "but, my
question is:
What would happen if I held it up like this for a few minutes?"

'Nothing' ...the students said.

'Ok what would happen if I held it up like this for an hour?' the professor
asked.

'Your arm would begin to ache' said one student.

"You're right, now what would happen if I held it for a day?"

"Your arm could go numb; you might have severe muscle stress & paralysis &
have to go to hospital for sure!"

... Ventured another student & all the…

Explain Thrashing

Thrashing is said to occur when the system spends a large amount of time transferring shared data blocks from one node to another, compared to the time spent doing the useful work of executing application  processes. It is serious performance problem with DSM systems that allow data blocks to migrate from one node to another. Thrashing may occur in the following situations:

a.  When interleaved data accesses made by processes on two or more nodes causes a data block to move back and forth from one node to another in quick succession.

b.  When blocks with read-only permissions are invalidated soon after they are replicated. Following methods may be used to solve the thrashing problem in DSM systems:

a.  Providing  application-controlled  locks.  Locking  data  to  prevent  other  nodes  from accessing that data for a short period of time can reduce thrashing.

b.  Nailing a block to a node for a minimum amount of time. Another method to reduce thrashing is to disallow a block to be taken a…