## 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
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.
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 three streets. After travelling 1 hour Tom met Jerry at a distance if 120 meters. Find the ratio of the speed of Tom to that of Jerry.
a)X=2,Y=1 b)X=3,Y=2 c)X=5,Y=2 d)X=4,Y=3
Solution :
Part :1 To find the speed of Tom
As given in the question, the total time taken by Tom = 1 hour and
The total distance = 120 meters = 0.12 km
Now, the speed of Tom = distance / time = 0.12 / 1 = 0.12km/hr
Part :2 To find the speed of Jerry
The total time taken by Jerry = 30 minutes + 1 hour = 3/2 hour
Distance = 0.12km
Then, the speed of Jerry = 0.12/(3/2) = 0.08km/hr.
Part :3 To find the ratio of the speed of Tom to that of Jerry
Based on part 1 and 2, ratio of the speed of Tom and Jerry = 0.12:0.08 = 3:2
Question 3
A call taxi started from certain place A at 7.00am. The taxi reached its destination, waited there for 30 minutes and returned back to A. The destination is 100 meters from A. The speed of the taxi is 50 meters per hour. In return journey, taxi travels with 20% greater speed (as compared to onward journey speed). At what time it will return to A?
a)11.00am b)11.10am c)11.20am d)12.10pm
Solution:
From the question, the distance between A and the destination is 100 meters and speed of the taxi is 50mph.
Then the time taken to reach the destination = Time = distance / speed = 100/50 hr = 2 hours.
Since the taxi started at 7.00 am, it should had reached the destination at 9.00am.
It is given that the taxi stayed at destination for 30 minutes. Therefore, it should have started from destination at 9.30am.
On return journey its speed had increased by 20%. This means, its new speed = Original speed + 20% of Original speed = 50 + (20/100)50 = 60mph.
Then the time taken to reach A = 100/60 = 1 + 2/3 hours = 1 hour and 40 minutes.
So it should had reached A at 11.10am.

## 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
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 = 5/6
3X-5V = 150 ...eqn1
Suppose the technical problem had happened after crossing 60 km and the train would had reached 40 minutes late.
Using the same logic we used to arrive at eqn1, we get,
[60/V]+[(X-60)/(2V/3)] = [X/V]+2/3 (here 2/3 is for 40 minutes in hour.)
[120+3X-180-2X]/2V = 2/3
3X-4V = 180 ....eqn2
Solving eqn1 & eqn2, we get
X=100 & V=30
Hence the distance between P and Q is 100km and the actual speed of the train is 30km/hour.
Question 2
A train starts from A towards B with some velocity. Due to an engine problem, after travelling 3/8 of its journey, it slows to 3/5 of its actual velocity. The train reaches B 1 hour later than the actual planned time. If the engine had failed after travelling 80km and if it would had slowed down to 4/5th of its initial velocity for another 80km and covered remaining distance with 1/2 of its initial velocity, the train would had reached the destination one and half hours late. What is the distance between A and B in meters?
a)10000 b)48000 c)24000 d)52000
Solution:
Let the distance between A and B be X and the speed initially be V.
The train travels 3X/8km with speed V and the remaining distance(X - 3X/8)km with speed 3/5 of V. Ultimately the train was late by 1 hour.
According to the above condition with the formula " distance/speed = time", we can have
[(3X/8)/V]+[(X-(3X/8))/(3V/5)] = [X/V]+1
[3X/8V] + 5(8X-3X)/24V = [X/V]+1
9X+25X-24X / 24V = 1
10X-24V = 0 ..........eqn1
According to the question, if the train travelled 80km with speed V, another 80km with 4/5 th of V and the remaining distance(X-160)km with speed 1/2 of V then
[80/V]+[80/(4V/5)]+[(X-160)/(1V/2)] = [X/V]+3/2
80/V + 100/V + (2X-360)/V = X/V + 3/2
X-180 / V = 3/2
2X-3V = 360 .........eqn2
solving eqn1 and eqn2
we have, X=480 and V=200
Thus the distance between A and B is 480km and the speed of the train is 200km/hour.
Hence 480km = 480000meters is the answer.
Question 3
A man rides a bike with speed V for a distance X km. After completing 1/2 of his journey he slows down to 1/2 of his initial speed and completes his ride half an hour late than planned time. What will be the ratio of the distance to the speed?
a)1:2 b)2:1 c)1:1 d)2:3
Solution:
Let the distance be X and the speed initially be V.
Using same logic as we used for previous two questions, we get,
X/2V + [X-(X/2)]/(V/2) = X/V + 1/2
X/2V + [4X-2X]/2V = X/V + 1/2
X/2V = 1/2
X = V or X/V = 1
Thus 1:1 is the required ratio.

## 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
Being voluble means to be very fluent in speaking.
Question 2
Volatile
a. Oscillate b. Transient c. Penetrate d. Light
Transient can be used to refer something that is temporary and volatile.
Question 3
Cryptic
a. terse b. difficult c. complex d. Unknown
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

## 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
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
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
Solution:
The above question seems to be in the general form A + B = (A x B)+(A - B)2
3 + 5 = (3 x 5)+(3 - 5)2 = 15 + 4 = 19
5 + 9 = (5 x 9)+(5 - 9)2 = 45 + 16 = 61
9 + 12 = (9 x 12)+(9 - 12)2 = 108 + 9 = 117
Similarly, 13 + 14 = (13 x 14)+(13 - 14)2 = 182 + 1 = 183.

## 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
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"    ?```
Amount saved on Material2 = 360 x 1/24 = Rs.15
Therefore total amount saved after size reduction = 72 + 15 = Rs.87
Question 2
The dimensions of a certain machine are X" x Y" x Z". If the average of its dimensions equals 680" and none of the dimensions is less than 640", what is the greatest possible length of one of the dimensions?
a)760" b)800" c)720" d)745"
Solution:
Given (X+Y+Z)/3 = 680".
Therefore the total length of the dimensions is X + Y + Z = 3 x 680 = 2040.
Let us assume Z is the dimension with greatest possible length.
From above equation, Z = 2040 - X - Y ...(1)
Based on eq (1), for Z to be of maximum value, X and Y should be as low as possible. Since it is given that none of the dimensions is lesser than 640", the lowermost value that X and Y can get is 640".
Substituting X = Y = 640" in eq (1), we get,
Z = 2040 - 640 -640 = 760".
Question 3
What will be the ratio of the shortest side of new dimensions to the greatest side of the old dimensions if the size of a certain machine is increased proportionally until the sum of its dimensions equals 800" and the dimensions of old size are 130" x 60" x 360" ?.
a)1:2 b)3:5 c)5:18 d)18:3
Solution:
From the given data, we have
The dimensions of the machine before size increment = 130" x 70" x 360".
Sum of the dimensions = 130+70+360 = 560"
Note that the greatest side = 360" & smallest side = 70". ...(A)
It is given that the sum of the dimensions of new size = 800"
Since dimensions are increased proportionately, the smallest side of the old dimensions before increment will be the smallest side of new dimension after size increment.
``` Sum  Side
560"   70"
800"    ?
Smallest side will be increased to 800 x 70 / 560 = 100"
i.e., the smallest side of the new dimensions is 100".
We have already found that the the greatest side of old dimensions is 360".
Required Ratio = Smallest side of the New Dimensions : Greatest side of the Old
Dimensions
= 100 : 360 = 5 : 18
Hence the answer is 5 : 18.```

## 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 away from a node until a minimum amount of time t elapses after its allocation to that node. The time can either be fixed statically or be tuned dynamically on the basis of access patterns.

c.  Tailoring the coherence algorithm to the shared-data usage patterns. Thrashing can be minimized  by  using  different  coherence  protocols  for  shared  data  having  different characteristics.