1) A merchant had a diamond, cost of which varies as a
square of its weight. The
merchant broke the diamond into 3 pieces in the ratio (based
on weights) 4:5:6. When
the pieces were sold he incurred a loss of Rs.444000. What
could be the original price of
the diamond?
Options:
a)Rs.750000
b)Rs.665000
c)Rs.600000 d)Rs.675000
Answer: d) Rs.675000.
Solution:
As given, the weights of the broken pieces are in the ratio
4:5:6.
Let the actual weights of broken pieces be 4X,5X and 6X.
Then weight of the original diamond = 4X + 5X + 6X = 15X
Since the cost of a diamond varies as its square of the
weight, the original cost will be
(15X)
2
= 225(X
2
). ...(1)
The total cost of individual pieces will be (4X)
2
+ (5X)
2
+ (6X)
2
= (16 + 25 + 36)x(X
2
) =
77(X
2
). ...(2)
The loss value = (1) - (2) = 225(X
2
)-77(X
2
) = 148(X
2
).
But the above loss value is given to be Rs.444000
Therefore, 148(X
2
)=Rs.444000
X
2
=Rs.444000/148.
=Rs.3000
Original cost = 225(X
2
) = 225 x 3000
= 675000
Hence the answer is Rs.675000.
2) X and Y can
complete a work in 12 days and 10 days respectively. With the help of Z,
X and Y can together complete the work in 5 days for a total
wage of Rs.6000. Then
what wage should be paid to Z for his part of the work?
Options:
a) Rs.1000 b)
Rs.1500 c) Rs.500 d) Rs.2000 Questions/Articles Extracted
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Page 23
Answer : c) Rs.500
Solution:
From given data,
X's 1 day work = 1/12
Y's 1 day work = 1/10
If X,Y and Z would complete the work in 5 days, then Z's 1
day work = One day work of
X,Y and Z combined - (One day work of X + One day work of Y)
=
1/5 -[(1/12)+(1/10)].
= 1/5 - 11/60 = 1/60.
Now we have to find Z's share of the salary.
X's share : Y's share : Z's share = 1/12 : 1/10 : 1/60 = 5 :
6 : 1
Z's share of wage from the total wage of Rs.6000 = 1/(5 + 6
+ 1) x 6000 = Rs.500
3) Gautam is good in sculpturing. He makes a sculpture of
height 4 feet 8 inches and
places them on a sandal pedestal. If the total height of
sculpture and the pedestal put
together is 7 feet 3 inches, what is the height of the
pedestal?
Options:
a) 2 feet 4 inches
b) 5 feet 2 inches c) 2 feet
7 inches d) none of these
Answer : c) 2 feet 3 inches
Solutions:
Total Height = Height of the sculpture + Height of the
pedestal
7 feet 3 inches = 4 feet 8 inches + Height of the pedestal
Height of the pedestal = 7 feet 3 inches - 4 feet 8 inches
Writing all the measurements in units of inches we get,
Height of the pedestal = 7 x 12 + 3 inches - 4 x 12 + 8
inches = 87 inches - 56 inches =
31 inches = 2 x 12 inches and 7 inches = 2 feet 7
inches
4) John, Chief Executive Officer of Infacan , Mumbai drove
by his Benz car from Mumbai
to New Delhi. He started at 2 pm and drove at 60 kmph.
Around 2.30 pm his personal
assistant called him over his cell phone and told that John
had not taken an important
paper with him and sought instructions from him. John
advised his personal assistant to Questions/Articles Extracted From
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Page 24
send those papers through Don Lee. Don Lee started from
Mumbai at 2.45 pm and he
was driving his Hyundai car at 75 kmph. At what time Don Lee
will overtake John
assuming they travel at speeds indicated and they do not
stop in between?
Options:
a) 5.15 pm b)
5.45 pm c) 5.30pm d) 6.00 pm
Answer: b) 5.45 pm
Solution:
Distance travelled by John before Don Lee starting from
Mumbai is 45 km.
Relative speed of Don Lee with respect to John = 75 - 60 =
15 kmph.
Time taken by Don Lee to overtake John = Distance travelled
by John before Don Lee
starting / Relative speed = 45/15 = 3 hours
So Don Lee will overtake at 2.45 + 3 = 5.45 hours
5) Raghavan and Krishnan started from Chennai to Mumbai in
two different cars at 10
am one day. Raghavan was driving his car at 50 kmph and
Krishnan was driving his car
at 40 km per hour. Around 11 am their friend Gopalan started
in his car from Chennai
and drove his car at 70 kmph. At what time Gopalan will
overtake Raghavan and
Krishnan?
Options:
a) 1.00pm and 12.20 pm
b) 1.30 pm and 12.10
pm
c) 1.30 pm and 12.20 pm d) none of these.
Answer: c) 1.30 pm and 12.20 pm
Solution:
Raghavan and Krishnan would have travelled 50 km and 40 km
by the time Gopalan
starts at 11 am.
Gopalan is driving at 70 kmph.
Relative speed – Gopalan with respect to Raghavan = 70 - 50
= 20 kmph
Distance travelled by Raghavan when Gopalan started = 50
km.
To catch up with Raghavan, time taken by Gopalan = Distance
covered by Raghavan
when Gopalan starts / Relative speed of Gopalan with respect
to Raghavan = 50/20 = 2
½ hours.
Questions/Articles Extracted From www.CareersValley.com
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Therefore, he will overtake Raghavan at 11.00 + 2.30 = 1.30
pm.
Relative speed of Gopalan with respect to Krishnan = 30 kmph
Distance travelled by Krishnan when Gopalan started = 40 Km.
Time taken by Gopalan to cross Krishnan = Distance covered
by Krishnan when
Gopalan started / Relative speed of Gopalan with respect to
Krishnan = 40/30 = 1.33
hours = approximately 1 hour 20 minutes
Gopalan is starting at 11 am. So he will overtake Krishnan
at 11 + 1hr 20 min = 12.20
pm
6) A 2 mb PCM (Pulse Code Modulation) has
Options
a) 32 voice channels
b) 30 voice channels
& 1 signalling channel
c) 31 voice channels
& 1 signalling channel
d) 32 channels out of which 30 voice channels, 1 signalling
channel and 1
synchronization channel.
Answer: 31 voice channels & 1 signalling channel
7) Word alignment is
Options
a) aligning the address
to the next word
boundary of the machine
b) aligning to even
boundary
c) aligning to word boundary
d) none of the
above
Answer : aligning the address to the next word
boundary of the machine
8) To send a packet data using datagram, when a connection
will be established
Options
a) before data transmission
b) connection is not
established before data transmission
c) no connection required
Questions/Articles Extracted From www.CareersValley.com
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d) none of the
above
Answer: no connection required
9) The status of the kernel is?
Options
a) task
b) process
c) not defined
d) none of the
above
Answer : Process
(Next 8 questions are of interview type)
10) Briefly can you explain a 'database trigger'?
Answer:
A database trigger is a predefined code that gets executed
based on certain events on
tables. For example, a trigger can be written that executes
automatically whenever a
particular column gets updated in a table. Though they are
memory intensive, they can
be used for critical background tasks to maintain data
integrity.
11) Tell a common error encountered by programmers when
using stack data structures
without proper precautions?
Answer:
Stack Overflow is a common error when dealing with stack
data structures. This occurs
when data is pushed onto stack till a point when there is no
further memory is available.
Good programming and understanding of the resource
limitations of the underlying
machine can prevent these errors.
12) When a program is
under execution, what does a Program Counter (PC) hold?
Answer:
When a program is in execution, the program counter holds
the address of the next
instruction to be executed.Questions/Articles Extracted From
www.CareersValley.com
Page 27
13) (Generally) In an application development life cycle,
give the proper order of the
following stages: Testing, Test Plan Preparation,
Development, and Requirement
Gathering.
Answer:
Correct order would be: Requirements Gathering, Test Plan Preparation,
Development
and Testing.
14) Can you guess why IPv6 (Internet Protocol version 6)
addresses came into
existence while still IPv4 is being used by many machines?
Answer:
The only reason could be the enormous growth in the number
of machines using v4
addresses. This forces concerned agencies to raise the
address pool by some means.
One of such ways was the introduction of IPv6 so that more
machines can be
accommodated.
15) Consider two network layer devices, one operating at
network layer and the other
operating at data link layer. In very generic terms, which
one is more intelligent? Explain
with example.
Answer:
Generally a device that operates at a higher layer in OSI
model is intelligent. For
example, there are switches that operate at data link layer
and some others that operate
at network layer. Simply due to the additional capabilities
of the switches that operate in
network layer they can be considered more intelligent.
16) What is the main drawback when using Hubs in networks?
Answer:
Hubs are least intelligent devices transmitting whatever
data they receive into all the
output ports. Though this behaviour can be favourable in
some cases, most of the times
this behaviour results in unnecessary traffic in the
network. More network consumption
leads to problems like collision and congestion.
Questions/Articles Extracted From www.CareersValley.com
Page 28
17) In socket communication happening between a client and a
server, how can you tell
which one is a client and which is a server?
Answer:
Consider two network devices communicating through network
sockets. In many cases a
client initiates a transaction. For example, your computer
could request a web page from
a server. This implies that your computer is initiating the
connection and is a client. The
responding machine will be the server. Questions/Articles
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