1. Introduction
2. TCS Sample Solved Questions
3. Infosys Sample Solved Questions
4. Wipro Sample Solved Questions
5. CTS Sample Solved Questions
6. HCL Sample Solved Questions
7. Accenture Sample Solved Questions
8. IBM Sample Solved Questions
9. Syntel Sample Solved Questions
10. HP Sample Solved Questions
11. Honeywell Sample Aptitude Questions
12. General Aptitude & Verbal Questions
(For All IT Companies like L&T, Amazon, 3i Infotech, Alcatel
etc.,)
What to Expect and What Not to Expect in this
eBook?
Don't expect a huge number of questions which obviously is not
possible in a simple e-book.
But expect a limited number of the most important types of
questions commonly being asked in
placement tests and interviews.
Don't expect this e-book to be the only resource to prepare
completely.
But expect this e-book to give you a right head start. This e-book
is NOT A replacement for our
very comprehensive Placement Success Book or careersvalley.com's
solved placement papers
section.
So, What’s inside this Free eBook?
This e-book contains solved sample questions in the first section
and career related tips/articles
in the second part. The first part will not have too many
questions but most important question
types that get repeated very often with minor differences. Most of
the questions have been
extracted from those posted on our website.
Disclaimer:
The placement papers/questions discussed are
SAMPLE and Unofficial and not from the
respective companies. They have been framed
based on the inputs and discussions from
students based on their memory, to help the
student community stay prepared. Contact Us using
the contact form on the website
careersvalley.com for any discrepancies and policy related
questions.
Page 4
Section I – Placement Papers Section
TCS Sample Questions
1) There is a rectangular shaped park and a circular garden inside
the park. If the length
and
breadth of the park are 18 m and 14 m respectively then what will be the
maximum
possible
area of the circular garden?
a) 154
sq.cm b) 308 sq.cm c) 218 sq.cm d) 174 sq.cm
Answer :
a) 154 sq.cm
Solution
:
Given
that a circular garden which is inside a rectangular park.
We have
to find the maximum possible area of the garden.
i.e., we
would find the area of the inscribed garden in rectangular park.
Also
given that length of the park is 18 m and breadth of the park is 14 m.
Then the
largest possible diameter of the inscribed garden is 14 m.
And the
largest possible radius of the inscribed garden is 7 m.
Therefore,
the required area = pi(r2) sq.m =
(22/7) x (72) = 22 x
7 = 154 sq.m
Hence
the answer is 154 sq.m.
Page 5
2) There
are 6 bottles of drinks out of which one is poisoned. If a man tastes a drop of
that, he
will die within 10 hours (Note : This part is different from that of the first
question.
In
previous question it was "exactly at the 10th hour" but here it is
"within 10 hours"). A
doctor
decides to check it out by using some number of mice within 16 hours. Then the
minimum
number of mice needed to find out the poisoned bottle is:
a)2 b)4
c)6 d)3
Answer :
d)3
Solution
:
If mouse
would die exactly at 10 hours then definitely 1 will be the answer but here the
given
condition says that mouse will die within 10 hours.
so this
question can be solved by set theory.
There is
a Formula to use for these types of problems :
Consider
there are b bottles to be tested. Then the minimum value of m such that 2m > =
b gives
the minimum number of mice required.
In our
case b = 6.
The
minimum value of m for 2m > 6
is 3. (21=2, 22 = 4, 23 = 8. Only when m = 3, 2m
exceeds
6.)
3) In
the adjoining diagram, ABCD and EFGH are squares of side 10 units such that
they
intersect
in a rectangle of 4 units long with diagonal length (CE) = 5 units . Find the
total
area
covered by the squares is:
a) 100 b)
2 00 c) 188 d) 150
Answer :
c) 188
Solution
:
Page 6
Since
the side of the squares are of 10 units then the area of each square is 100
unit2.
The
diagonal of any rectangle of l units long and b units width is sqrt [ l2 + b2].
It is
given that sqrt [ l2 + b2] = 5 and l = 4
Then, b2 = 25 - 16 = 9
b = 3
units.
Therefore
the area of the rectangle lb = 12 units2.
Then,
the total area covered by ABCD and EFGH in adjoining diagram = area of ABCD
+ area
of EFGH - intersecting area between ABCD and EFGH(area of the rectangle)
= 100 +
100 - 12 = 188 unit2 = 188
Hence
the answer is 188.
4) Find
the approximate distance between the points of X-intercept and Y-intercept
respectively
of the plane determined by the equation 11x - y = 17.
a) 17 b)
11 c) 1 d) 17/11
Answer :
a)17
Solution
:
Given
that the equation, 11x - y = 17.
To find
x-intercept, put y = 0 on the given equation:
i.e.,
11x - 0 = 17 x = 17/11
This
gives x = 17/11, i.e., x-intercept point is (17/11,0)
Page 7
To find
y-intercept, put x = 0.
i.e.,
11(0) - y = 17
This
gives y = -17, i.e., y-intercept point is (0,-17).
Now, we
have to find the distance between (17/11,0) and (0,-17).
We know
that," the distance between the points (x1,y1) and (x2,y2) is sqrt [(x2 -
x1)2 +
(y2 -
y1)2]
Then,
the required distance = sqrt [(0 - 17/11)2
+ (-17-0)2]
sqrt [172 ( 1/112 + 1)]
17sqrt [
1/121 + 1]
17sqrt
[122/121]
17/11
sqrt(122)
17/11 x
11.04
17.06
Hence
the answer is 17(approximately)
5) On
dividing (18377)10 by 11
the remainder will be
a)1 b)10
c)9 d)2
Answer :
a)1
Solution
:
To
Remember :
Fermat's Little Theorem states : "If a is an integer and not divisible by p then a(p-1) - 1
is an integer multiple of p".
i.e., we get the remainder 1 when we divide a(p-1) by p.
For example, consider a = 7 and p = 3. Here a is not divisible
by p. a(p-1) - 1 = 7(3 - 1)
- 1 =
7(2) - 1 = 49 - 1 = 48 is a multiple of 3
Here we
have to divide (18377)10 by a
prime 11.
Now,
take a = 18377 and p = 11
Then , a(p-1) = (18377)10
Page 8
Now,
according to Fermat's Little Theorem, a(p-1)
- 1 i.e (18377)10
- 1 should be an integer
which is
a multiple of 11
(18377)10 - 1 when divided by 11 gives no
remainder
This
implies (18377)10 when
divided by 11 gives 1 as remainder
Hence
the answer is 1.
6) A
circle has 4 equal-length chords such that the points of intersection form a
square
and the
diagonals of the square are the diameter of the circle, the number of points
equidistant
from all the 4 chords is
a)1 b)2 c)
0 d) 4
Answer :
a)1
Solution
:
The
center of the circle is the only point.
First we
select the 4 chords as follows:
Take two
diameters AB, CD of the circle which are perpendicular to each other.
Now
consider 4 chords with same length - AC, AD, BC and BD.
(Now the
diagonals of the square are AB, CD and sides are AC, AD, BC, BD).
Obviously,
the center of the circle is the only point which is equidistant from all the 4
chords.
Hence
the answer is 1.
Page 9
7) At a
distance of 2 meters from Arun, there is a circular dartboard with radius 50cm.
What is
the probability that Arun hit the board at a point P on the circle such that P
is
closer
to the center of the circle than the periphery?
a)1/2 b)1
c)1/4 c)2/3
Answer :
c)1/4
Solution:
(These
types of problems can be solved using an interesting method as described
below.)
According
to the requirement, we have to locate P closer to center in comparison to the
circumference.
For doing this we shall make a circle on that board whose radius will be
half of
the radius of the dartboard.
Now we
will compare the area of the inner circle to the total area of the dartboard.
Area of
whole circle = pi x (50 cm)2 = 2500pi sq.cm. (this is according to formula area =
pi x
(radius)2)
Similarly
the area of inner circle = pi x (25cm) 2
=625pi sq.cm
Now the
probability that Arun hit center = Area of inner circle/ Area of dart board =
625pi
/ 2500pi
= 1/4
8) 4
equal aged boys and 6 equal aged girls were regularly
attending a guitar class. 22
is the
average age of all and 56 is the age of two boys and three girls. After 3 years
a
boy is
replaced by a new boy aged 15. Now, what will be the average age of boys?
Options
a) 13.5
years b) 15.4years c) 16.3 years d) 14 years
Answer :
a) 13.5 years.
Solution:
Let X
and Y be the age of a boy and a girl respectively.
Given
that 22 is the average age of all.
Then
4X/4 + 6Y/6 = 22
i.e., X
+ Y = 22 ...eqn1
Page 10
Also
given that, 56 is the age of two boys and three girls.
Then, 2X
+ 3Y = 56 ...eqn2
Solving
the above equations, we get X=10, Y=12.
That is,
the age of each boy is 10 and age of each girl is 12.
After 3
years, 1 among 4 boys leaves the group but replaced by a new 15 year old boy.
Sum of
the ages of 3 left over boys + new boy = Ages of original 3 boys after 3 years
+
Age of
new boy = 3(X+3) + 15 = 3x13 + 15 = 54 years.
Their
average age = Total Age / 4 = 54/4 = 13.5 years.
13.5
Years is our answer.
9) When
8 is subtracted out from two third of X, the result is 25 more than one third
of X.
Find the
largest prime number which is less than X.
Options :
a) 113 b)
97 c) 89 d) 131
Answer :
b) 97
Solution:
Part-1 :
To find X.
Below
equation can be formed based on the data in question.
(2/3)x X
- 8 = 25 + 1/3 x X
(2/3)x X
- (1/3)x X = 25+8
X/3 = 33
X=99.
Part-2 :
To find the required prime
97 is the
largest prime number which is lesser than 99. Hence 97 is the answer.
10) 6
reduced from 1/5 of a number is 7 reduced from half of the same number. Find
the
number.
Options
a)10/3
b)22/3 c)7/3 d)none of these
Page 11
Answer:
a) 10/3
Solution:
X/5 - 6
= X/2 - 7
(X-30)/5
= (X-14)/2
2X-60 =
5X-70
5X-2X =
70-60
3X = 10
X = 10/3
11) Four
friends namely Rahul, Ravi, Rajesh and Rohan contested for a dairy milk
chocolate.
To decide which friend will get the chocolate they decided to throw two dice.
Every
friend was asked to choose a number and if the sum of the numbers on two dice
equals
that number, the concerned person will get the chocolate. Rahul's choice was7,
Ravi's
choice was 9, Rajesh's choice was 10 and Rohan's choice was 11. Who has the
maximum
probability of winning the amount?
Options :
a) Rahul
b) Ravi c) Rajesh d) Rohan
Answer :
a) Rahul
Solution
:
Number 7
will appear more often –(1,6), (2,5), (3,4), (4,3), (6,1), (5,2) --- 6 cases
Number 9
-- (3,6),(6,3), (4,5) (5,4) ---4 cases
For
number 10 -- (4,6) ,(6,4) (5,5) ---3 cases
For
number 11 -- (5,6),(6,5)...2 cases.
Since
number 7 has the maximum chance of appearing, it will have the maximum
probability
as well. Hence, Rahul will most probably be the winner.
( Next 4 questions are of interview type )
12)
Assume you are one of the developer of a new Operating System. Please tell any
one
logical approach you would employ to get the OS out of deadlock scenarios?
Answer :
Page 12
Many
approaches are possible. One possible and widely used approach is to roll back
and
restart in a different manner so that processes don’t indefinitely wait for
each other.
For
example if two processes are responsible for deadlocks, post rollback, I would
schedule
the processes so that one of them starts only after the other completes.
13) What
is the difference between a = ++b and a = b++ ?
Answer :
The
first statement a = ++b can be broken down into two statements b = b + 1
followed
by a =
b. Hence b will be incremented first and will be assigned to a. Hence values of
a
and b
will be same at the end.
The
second statement a = b++ can be broken down into two statements a = b followed
by b = b
+ 1. Hence b will be assigned to a. Only after this assignment b will be
incremented.
Hence values of a and b will be different at the end.
14) Is
it necessary to use register modifier on all compilers irrespective of whether
they
are old
or modern ones ?
Answer :
No,
compilers nowadays are intelligent enough to decide which variables need to be
kept in
CPU registers depending upon the estimated usage.
15) Can
a program be invoked from another program ? If yes, how it can be done ?
Answer :
Yes, a
program can invoke another program using system calls.
16)
Which of the following program structure/component/statement is not an example
for
the
implementation of modularization ?
Options
a) DLL b)
Functions c) type casting
Answer :
Option
c) type casting. DLL and Functions help in modularization of a program while
typecasting
just converts from one data type to another.
Page 13
Useful Tip :: Prepare
For Placement Papers By Exploring Common
Sections
Almost
all of the companies have a placement test in their recruitment process. It is
a
common
tendency to prepare specifically for every company's paper at the
announcement
of a recruitment drive either on campus or off campus. But most
youngsters
fail to explore the common sections that exist across placement papers of
different
companies so that they can complete their preparation in one shot.
What are
the Common Sections?
With few
exceptions, all most all companies give due importance to quantitative aptitude.
From
chapters like time and speed, age, linear equations, averages etc.
Verbal
sections like synonyms and antonyms (GRE type), filling with right tenses etc.
are
again common across different placement papers.
Software
placement papers also have questions from basic programming languages
like C.
Preparing
on these common sections can help you to save much time.
Page 14
How
should you Plan your Preparation?
My first
advice is that you should not wait till the last minute. Keep preparing on the
common
sections mentioned above as much as you can.
CareersValley.com’s
Placement Success Book priced at Rs. 400 will help you to a
great
extent. In that book, we have explored and presented common sections and
repeated
question types. When you get that book, you never have to prepare for every
company
separately but can prepare simultaneously for all companies. For details visit
Infosys Sample Questions
1) Is
the integer k is divisible by 40?
1) 8 is
a factor of k
2) k is
a factor of 10
Options :
(a)
statement 1 alone is sufficient, but statement 2 alone is not sufficient to
answer the
question
(b)
statement 2 alone is sufficient, but statement 1 alone is not sufficient to
answer the
question
(c) both
statements taken together are sufficient to answer the question, but neither
statement
alone is sufficient
(d) each
statement alone is sufficient
(e)
statements 1 and 2 together are not sufficient, and additional data is needed
to
answer the
question.
Answer :
(b) statement 2 alone is sufficient, but statement 1 alone is not sufficient to
answer
the question.
Solution
:
From
statement 1,
suppose
8 is a factor of k then k can be 8, 16, 24 and so on.
Only
some multiples of 8 are only divisible by 40
Therefore
the statement 1 alone is not sufficient to derive the answer.
From
statement 2,
The
factors of 10 are 1,2,5 and 10.
If k is
a factor of 10 then k belongs to {1,2,5,10}
No
element in the above set is divisible by 40.
Hence k
is certainly not divisible by 40. Hence using statement 2, we can give a
conclusive
answer that k is not divisible by 40.
Thus
statement2 alone is sufficient to answer the question.
Page 16
Hence
the answer is option b.
2) From
a bus stand, buses leave for every 15 minutes to both temple and Railway
junction.
First bus towards junction is at 7.00 am and towards temple is at 7.10 a.m.
Suppose
you have to visit both temple and junction and decide to go by the first bus
you
encounter.
Then the probability to get a bus towards temple is ___ .
a)0.67
b)0.5 c)0.75 d)0.33
Answer :
a)0.67
Solution
:
Let us
discuss an interesting method to solve such problems :
Let's
simulate a 60 minute time period:
A period
of 60 Minutes broken into intervals of 5 minutes each:
0 - 5 -
10 - 15 - 20 - 25 - 30 - 35 - 40 - 45 - 50 - 55 - 60(or "0" again)
Arrival
of BUS : ( t = temple, j = junction, X = no bus arrives at that particular
time)
j - X -
t - j - X - t - j - X - t - j - X - t - j
TO MAKE
IT EASY TO UNDERSTAND WE HAVE WRITTEN THE SAME THING
AGAIN.
0 - 5 -
10 - 15 - 20 - 25 - 30 - 35 - 40 - 45 - 50 - 55 - 60
j - X -
t - j - X - t - j - x - t - j - x - t - j
0 10 5
10 5 10 5 10 5
The line
written immediately above (that is the 3rd line of data) has numbers
corresponding
to "t" and "j" which shows how many minutes a person has to
wait till he
gets the
"First bus".
Therefore,
t= 10 +
10 + 10 + 10 = 40 Minutes
j= 0 + 5
+ 5 + 5 + 5 = 20 Minutes
Therefore
in 40 minutes of standing time for the bus you will get t bus and in 20 minutes
of
standing time for the bus you will get j bus.
You will
also see that this is the schedule for every hour after 7.00 A.M.
Page 17
This
means that the probability to get t bus is = 40 Minutes / 60 Minutes = 2/3 =
0.67
Also the
probability to get j bus is = 20 Minutes / 60 Minutes = 1/3 = 0.33
Hence
the answer is 0.67
3) In a
family there are several brothers and sisters. Every 2 boys have brothers as
many as
sisters and each girl has 2 brothers less than twice as many brothers as
sisters.
Now find
the number of boys and girls.
a)6,8
b)8,6 c)12,10 d)6,4
Answer :
b)8,6
Solution:
Let b be
the number of brothers and s be the number of sisters in the family.
Consider
any two boys. They would be having b - 2 brothers (excluding the two). But this
number
is equal to the number of sisters they have.
Therefore,
b - 2 =
s
or b - s
= 2 ----- (1)
Each
girl will have s - 1 sisters. Twice the number of sisters = 2(s - 1).
Since
each girl has twice as many brothers as sisters, we have, 2(s-1)-2 = b
2s - 4 =
b ---- (2).
Sub b =
2s - 4 in eq 1, we get
2s - 4 -
s = 2
s = 6
Sub s in
eq 1 we get,
b - 6 =
2
b = 8
4) Find
the largest five digit number whose 4th digit is 2/3 of the 5th digit and
thrice of
the 4th
digit is 3rd digit and there are exactly 2 pairs of digits where difference
between
the
numbers in each pair is 6.
a) 99623
b) 99485 c) 99678 d) 99523
Page 18
Answer :
a) 99623
Solution
:
4th
digit is 2/3 of the 5th digit, so the 5th digit should be a multiple of 3.
The
possibilities of 4th and 5th digits are
1) 2, 3
2) 6, 9
3) 4, 6
The 2nd
condition states that thrice of 4th digit is 3rd digit.
The 2nd
and 3rd possibilities are ruled out since 18, 12 cannot be the digit of that
number.
And the
thrice of 2 is 6 which is the 3rd digit
So 3rd
digit is 6, 4th digit is 2 and 5th is 3.
Since,
We have to find the greatest 5 digit number, the 1st and 2nd digit may be 9.
And so,
the number may be 99623.
By
checking the 3rd condition that there is exactly 2 pairs of digits whose
difference is 6.
i.e.,
1st digit - 5th digit = 6 and 2nd digit - 5th digit = 6 and we have no other
choices.
Hence
the answer is 99623.
5) The
Calendar for the year 2013 is the same as for the year?
a) 2020 b)
2018 c) 2017 d) 2019
Answer :
d) 2019
Solution:
For
calendar to repeat exactly, the dates and days have to match perfectly.
Consider 2014 :
Any date
on 2014 will correspond to same date on 2013 advanced by one day. (same
logic
used in first question.) For example if Jan 1 is Tuesday on 2013, then Jan 1
will be
Wednesday
on 2014.
Year
2014 2015 2016 (leap year) 2017 2018 2019
Advanced
days 1 1 2 1 1 1
Page 19
In 2019
the total number of advancements will be 1 + 1 + 1 + 2 + 1 + 1 = 7 . Any week
has
seven days. Hence advancement of 7 days also means the days are going to be the
same for
any dates. That is if 1st Jan on
2013 is Tuesday, then 1st Jan on
2019 will also
be
Tuesday.
Hence
the calendar for the year 2013 and 2019 is the same.
6) Two
persons - Srikanth and Kashyap start at --6 am on 19th February 2012 from two
places –
Coimbatore and Hyderabad ( 900 km away at ) — each going towards the other
end in
two cars. Rupashri starts from Coimbatore at 7 am on the same day from
Coimbatore
towards Hyderabad. Srikanth, Rupashri and Kashyap travel at 40 km/hour,
60
km/hour and 50 km/hour respectively. Which pair will meet on the way – Srikanth
–
Kashyap
or Kashyap-Rupashri or Srikanth-Rupashri.
a)
Srikanth-Kashyap b) Srikanth-Rupashri c) Kashyap-Rupashri d) none of these.
Answer :
b) Srikanth –Rupashri.
Solution
:
Case I : Consider Srikanth and Kashyap
Distance
between Coimbatore and Hyderabad - 900 km
Srikanth
and Kashyap start at 6 am and their speeds are respectively 40 km/hour and 50
km/hour.
So relative speed is 40 + 50 = 90 km. Hence they will meet after [distance
between
them/relative speed] = 900/90 = 10 hours from 6 am i.e 4 pm
Case II: Consider Rupashri and Kashyap
Rupashri
starts at 7 am from Coimbatore – speed 60 km/hour. But since Kashyap had
started
at 6 am (1 hour earlier at 50Km/hr) he would have travelled 50 km by 7am.
Distance
between them = 900 - 50 =850 km. Relative speed = 60 + 50 = 110.
Rupashri
and Kashyap will meet after [distance between them/relative speed] = 850/110
= 7 and
8/11 hours.
Case III: Consider Rupashri and Srikanth
On the
other hand When Rupashri starts – 7 AM Srikanth would have travelled 40 km
Page 20
only (as
Srikanth would had started at 6 am travelling at 40 Km/hr).
The
speed of Rupashri and Srikanth = 60 and 40 km/hour. Relative speed of Rupashri
with
respect to Srikanth = 60 - 40 = 20 km per hour (we are subtracting the speeds
in
relative
speed calculation as they are both travelling in same direction) and hence
Rupashri
will cross/meet Srikanth after [distance between them/relative speed] = 40/20 =
2 hours
i.e. at 9 am.
Inference : Considering all discussed cases above, Srikanth-Rupashri will
meet first at 9
am.
7) 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 have slowed down to 4/5th of its initial velocity for another 80km and
covered
the remaining
distance with 1/2 of its initial velocity, the train would have reached the
destination
one and half hours late. What is the distance between A and B in meters?
a)10000
b)480000 c)240000 d)520000
Answer :
b)480000
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
Page 21
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.
8) Agil
is twice as fast as Mugil. Mugil is thrice as fast as Annie. The distance
covered
by Annie
in 54 minutes will be covered by Mugil in ___ minutes.
a)27
minutes b)9minutes c)38 minutes d)18 minutes
Answer :
d)18 minutes
Solution:
Let
Annie's speed be X km/hr.
Its
given, Mugil is thrice as fast as Annie. Therefore Mugil's speed is a 3X km /
hr.
Also it
is said Agil is twice as fast as Mugil. Therefore, Agil's speed = Mugil's speed
x 2 =
6X
km/hr.
Therefore
ratio of their speeds = 6X : 3X : X = 6 : 3 : 1
For a
given distance, the time taken will be inversely proportional to speed.
Therefore,
Ratio of times taken by the three friends Agil, Mugil and Annie = 1/6 : 1/3 : 1
=
1 : 2 :
6
If Annie
takes 6 minutes then Mugil takes 2 minutes.
If Annie
takes 54 minutes then Mugil takes [2/6 x 54]= 18 minutes.
Hence
the answer is 18 minutes
9) At
the time of his marriage Arunachalam was 29 years old. During marriage, his
wife
was
younger by 3 ½ years.20 years afterwards the total ages of Arunachalam, his
wife
and
their only son Kamalraj is 110 years. What will be the age of Kamalraj forty
years
from
now?
a) 35 1/2
years b) 45 1/2 years c) 25 years d) 30 years
Page 22
Answer :
a) 35 1/2 years
Solution
:
I. During
Marriage
Arunachalam's
age = 29 years
Wife's
age = 29 - 3.5 = 25.5 years
II. 20
years afterwards
Arunachalam's
age = 29 + 20 = 49 years
Wife's
age = 25.5 + 20 = 45.5 years
Let
Son's age (20 years after marriage) = S
It is
given that, 20 years after the wedding, Arunachalam's age + Wife's age + S =
110
Or 49 +
45.5 + S = 110
S = 15.5
years
III. 40
years afterwards.
Since we
have found that the son was 15.5 years old 20 years after marriage, adding 20
to 15.5
will give age of Kamalraj after 40 years.
Therefore
answer = 15.5 + 20 = 35.5 years =35 1/2 years.
10)
Eight friends Harsha, Fakis, Balaji, Eswar, Dhinesh, Chandra, Geetha, and Ahmed
are
sitting in a circle facing the center.
Balaji
is sitting between Geetha and Dhinesh. Harsha is third to the left of Balaji
and
second
to the right of Ahmed. Chandra is sitting between Ahmed and Geetha and
Balaji
and Eshwar are not sitting opposite to each other. Who is third to the left of
Dhinesh?
Answer: Fakis
Explanation:
Arranging the friends as per the question statement we can arrive at the
following
diagram
Page 23
Ahmed
Fakis
Chandra
Harsha
Geetha
Eswar
Balaji
Dhinesh
Hence
correct answer is Fakis.
11)
Today is 4.11.09. Keeping that figure 41109 in mind, i have arrived at the
following
sequence: 2, 1, 9, 5, _. Which of the following four numbers can fill the
dash?
Options
a) 7 b) 65
c) 4563 d) 262144
Answer : d) 262144
14 + 1 = 2
11 + 0 = 1
01 + 9 = 9
90 + 4 = 5
So next is 49 + 1 = 262144
12)
There is a unique number of which the square and the cube together use all
ciphers
from 0 up to 9 exactly once. Which number is this?
Answer: The number is 69.
Explanation:
69^2=4761
and 69^3=328509
Page 24
13) In a
certain code RELATED to written as EFUBKDQ. How is RETAINS written in that
code?
a) SDQBTOJ
b) JOTBQDS
C) JOTBSDQ
d) TOJBSDQ
Answer :
d) TOJBSDQ
Reason :
Inspecting
how RELATED was written as EFUBKDQ, we can easily find that,
RELATED –
is first written as DETALER
DETALER
becomes EFUBKDQ (+1,+1,+1, +1,-1,-1,-1 – alphabetically)
Applying
the same logic to RETAINS we get,
RETAINS
is first written as SNIATER
SNIATER
becomes TOJBSDQ (applying the above rule)
14) Six
people – A,B,C,D, E and F are standing in a straight line facing north not
necessarily
in the same order. B is standing to the right of D. A is standing fourth to the
left of
F and F is not standing on the extreme end of the line. D is standing second to
the
left of
B.
In the
above arrangement, which of the following pair represents the people standing
at
the
extreme ends of the line?
a) DF b)
AC c) BC d) AF
Answer :
b) AC
Reason :
Based on
the information given seating arrangement is as follows:
north
A D E B
F C
south
Page 25
(Next
5 Questions are of interview type)
15) When the number of updates is
very less or none, how an array can be
advantageous
over a linked list?
Answer :
Though
arrays require more operations when it comes to updates, they require lesser
memory
than linked list. Hence, when there are none or less number of updates, an
array
can be advantageous.
16) Tell anyone important advantage
array has over linked list when it comes to element
access?
You can explain your answer considering an example operation of reading an
element
at some fixed location from start?
Answer :
An array
offers a kind of random access to its elements. This is possible only because
of
the
reason that the elements are contiguous in memory. For example, consider an
array
that has
5 elements. These 5 elements will be at contiguous memory locations. If a
program
wants to read the 4th element of the array, it knows very well that
incrementing
the
address pointer 4 times relative to the first element will land it at the exact
location of
the 4th
element. However, in case of a linked list, the program has to traverse element
by
element. After an element is encountered, based on the address pointer the
program
will
move on to the next element and so on.
17) Tell
any two advantages offered by a DLL (Dynamic Link Libraries) to programmers.
Answer :
DLLs are
shared across many processes. This sharing significantly reduces the memory
requirement
which in turn provides better memory management.
DLLs can
provide great functionality enhancements without much rework on applications
(like
recompilation, relinking etc.).
18) Tell
any two important features/characteristics of Telnet protocol?
Answer :
a.
Telnet is bidirectional. That is participants can both send and receive
information.
Page 26
b.
Telnet is a text oriented protocol.
19) Tell
any two features of 'Sessions' used in communications over the internet.
Answer :
a.
Generally sessions exist only for a predefined period of time. For example, if
you are
accessing
an eCommerce site, leaving the browser idle for an hour or so could
automatically
log you out.
b. By
default HTML communications are stateless. But sessions can make the
communication
'stateful'. For example, when you are accessing a shopping site, the
shopping
server will know that the subsequent requests are from the same buyer i.e.
you.
This state is maintained till session expiration.
Page 27
Wipro Sample Questions
1) Find
the value of P when the line through the points (2,2) and (5,7) is
perpendicular to
the line
3x+Py-9=0.
a)3/5 b)5
c)5/3 d)3
Answer :
b)5
Solution:
A(2,2)
and B(5,7) are the given points.
We know
that the slope of the line through the points (x1,x2)and(y1,y2) is y2-y1 /
x2-x1
Then the
slope of the line AB is m1 = 7-2 / 5-2 = 5/3.
The
given eqn of the other line is: 3x + Py - 9 = 0 ---(1)
We know
that the slope of the line ax+by+c=0 is "-co-efficient of x / co-efficient
of y".
then the
slope of the line 3x + Py - 9 = 0 is m2 = -3/P.
For two
lines to be perpendicular, the product of their slopes should be -1.
i.e.,
m1.m2 = -1
5/3 x
-3/P = -1.
-5/P =
-1
P = 5
Hence
the answer is 5.
2) The
product of the distinct roots of the equation (3x)(3x+2)(3x-4)(3x-6)= 64 is:
a)-32/27
b)-61/5 c)63/16 d)69/12
Answer :
a)-32/27
Solution:
Given
that, (3x)(3x+2)(3x-4)(3x-6)= 64
Let 3x -
2 = p
Then the
given eqn becomes
(p + 2)
(p + 4) (p - 2) (p - 4) = 64
(p2 - 4) (p2 - 16) = 64
Page 28
p4 - 20p2 + 64 = 64
p4 - 20p2 = 0
p2(p2 - 20) = 0
p2 = 0 or p2-20 = 0
p = 0 or
p = sqrt(20) or p = - sqrt(20)
then 3x
- 2 = 0, 3x - 2 = sqrt(20) or 3x - 2 = - sqrt(20)
and
x=2/3, x=[2 + sqrt(20)] / 3 or x = [2 - sqrt(20)] / 3
Now the
distinct roots of the given eqn are 2/3, [2 + sqrt(20)] / 3 and [2 - sqrt(20)]
/ 3
The
product of the distinct roots = 2/3 x [2 + sqrt(20)] / 3 x [2 - sqrt(20)] / 3 =
2[(22)-
(sqrt(20))2] / 27 = -32/27
Hence
the answer is -32/27.
3) 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.
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(X2). ...(1)
The
total cost of individual pieces will be (4X)2
+ (5X)2 + (6X)2 = (16 +
25 + 36)x(X2) =
77(X2). ...(2)
The loss
value = (1) - (2) = 225(X2)-77(X2) = 148(X2).
But the
above loss value is given to be Rs.444000
Therefore,
148(X2)=Rs.444000
X2 =Rs.444000/148.
Page 29
=Rs.3000
Original
cost = 225(X2) = 225
x 3000
= 675000
Hence
the answer is Rs.675000.
4) 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?
a) Rs.1000
b) Rs.1500 c) Rs.500 d) Rs.2000
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
5)
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?
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
:
Page 30
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
6) If a
varies inversely as 1/(b+3) and a = 10 when b = 2 then express a in terms of b.
a)
a=2(b+3) b) a=1/(b+3) c) a=(b+3) d)a=10b
Answer :
a) a=2(b+3)
Solution
:
Given
that, a = k/(1/b+3) where k is a constant
a(1/b+3)
= k .....(1)
Put a =
10 & b = 2, we have
10(1/5)
= k
k = 2
Substitute
k value in eqn(1),
a / b+3
= 2
a =
2(b+3)
Hence,
the answer is a = 2(b+3)
7)
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?
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
Page 31
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.
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
(Next
8 questions are of interview type)
8) What
is the basic difference between Stacks and Queues in data structures?
Answer :
A stack
is a data structure where the last element inserted is processed first i.e if
an
element
E4 is inserted after E3 into a stack, the first element to be retrieved will be
E4
followed
by E3. In Queue processing order is exactly opposite to that of Stack. The
first
element
inserted will be the first one to be processed/served.
9) Any
OS plays a major role in a very prominent task called "Scheduling".
This is
basically
an implementation of scheduling algorithms developed over years. Which of
the
following resources are affected by this implementation?
a.
Throughput time of processes
b. Waiting
time of processes
Page 32
c.
Turnaround time of processes
d. All of
the above
Answer :
d. All of the above.
Explanation
: In simple terms, Scheduling algorithms determine when and what
resources
need to be allocated to which processes/threads. Hence all the performance
related
parameters associated with processes are affected in totality.
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.
13)
(Generally) In an application development life cycle, give the proper order of
the
following
stages : Testing, Test Plan Preparation, Development, Requirement Gathering.
Answer :
Page 33
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.
17) In
socket communication happening between a client and a server, how can you tell
which
one is a client and which is a sever ?
Answer :
Page 34
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.
Page 35
CTS Sample Questions
1) There is a group of 5 boys and 2
girls. The two groups working together can do four
times as
much work as a boy and a girl. Ratio of working capacities of a boy and a girl
is:
a)1:3
b)1:2 c)2:1 d)2:3
Answer :
c)2:1
Solution
:
Let 1
boy's 1 day's work = x
And 1
girl's 1 day's work = y
Now, (5
boys + 2 girls)'s work = 5x + 2y
Given
that 5x + 2y is equal to 4 times work done by a boy and a girl.
i.e., 5x
+ 2y = 4(x+y)
5x + 2y
= 4x + 4y
x = 2y
x/y =
2/1
Hence,
the required ratio is 2:1
2) A
group of teachers from Venkat Engineering College went to a hotel for
celebrating a
happy
event with an understanding all of them will share the expenses. The bill
amount
was
Rs.2560. Since four teachers did not bring their purse other teachers had to
contribute
Rs.32 more. How many teachers contributed towards the bill settlement?
a) 24 b)
16 c) 20 d) none of these.
Answer :
b) 16
Solution
:
Let the
number of teachers be x.
Total
bill amount = Rs. 2560.
Share of
each teacher = 2560/x
But, 4
teachers failed to bring their purse.
Number
of teachers who actually paid = x - 4
Each one
had paid Rs. 32 more than the planned share of 2560/x so as to settle the bill
Page 36
amount
of Rs. 2560
Therefore,
(x- 4)(2560/x + 32) = 2560
2560 +
32x - 10240/x - 128 = 2560
Multiply
both sides by x.
2560x +
32x2 - 10240
- 128x=2560x
32x2 - 128x - 10240=0
x2 - 4x - 320=0
Factorizing
the above equation, we get
(x -
20)(x + 16) = 0
x = +20
or -16
Since x
cannot be negative, x = 20 = total number of teachers.
Since, 4
teachers did not carry their purses, number present at the party = 20 - 4 = 16
Therefore,
our answer = 16.
3) What
number should come at the place of the question mark ?
16, 136,
1096, ?
a) 4998 b)
6884 c) 8776 d) none of these
Answer :
c) 8776
Solution
:
(16 x 8)
+ 8 = 136
( 136 x
8) + 8 = 1096
(1096 x
8) + 8 = 8776
4) Find
the number that should replace the question mark.
4200,
1680, 672, ?
a) 268.8
b) 324.8 c) 242.8 d) 122.8
Answer :
a) 268.8
Solution
:
Page 37
4200/2.5
= 1680
1680/2.5
= 672
672/2.5
= 268.8
5)
Kashinath, can row 5 kmph in still water.When the rate of flow of the river is
2.0 kmph
, it
takes Kashinath 1 hour to row to a ghat and return. How far is Kashinath’s
starting
place
from the ghat ?
(a) 2 km
(b) 2.1 km (c)2.5km (d) 2.8 km
Answer :
b) 2.1 km
Solution:
Let d be
the distance of ghat from the point where Kashinath starts to swim.
Let Sk
be the speed of Kashinath in still water
Let Sr
be the speed of the stream (river)
Then,
Sdown = Sk + Sr ...(1)
And Sup
= Sk - Sr ...(2)
Time
taken to travel downstream to ghat = d / Sdown
Time
taken to travel upstream from ghat = d / Sup
Total
time taken, T = d / Sdown + d / Sup
Substituting
values from eq1 and eq2 in the above equation we get,
Or, T =
d/(Sk + Sr) + d/(Sk - Sr)
T = d(Sk
- Sr) + d(Sk + Sr) / (Sk + Sr)(Sk - Sr)
T = 2dSk
/ (Sk2-Sr2)
Or d = T
(Sk2-Sr2) / 2Sk ...(3)
Note :
Remember the above formula as this can prove to be a shortcut.
Substitute
T = 1 hour, Sk = 5 Kmph and Sr = 2 Kmph in equation 3 we get
d = 1(52 - 22) / 2x5 = 21 / 10=2.1 km
6)
Janaki gets onto the elevator at the 8th floor of a building and rides up at
the rate of
60
floors per minute. At the same time, Ajay gets on another elevator at the 71st
floor of
the same
building and rides down at the rate of 66 floors per minute. If they continue
travelling
at these rates, then in which floor will their paths cross?
a.31st
floor b. will never cross c.38th floor d.12thfloor
Answer :
c. 38th floor
Solution:
Speed of
Janaki upwards = 60 floors/min
Speed of
Ajay downwards = 66 floors/min
No of
floors in between 8th and 71st floors = 71 - 8 = 63 floors
Let t be
the time after which they cross each other. In other words, after t minutes the
number
of floors covered by Ajay downwards added to the number of floors Janaki
covers
upwards should be equal to 63 floors. Putting this in the form of an equation
we
get :
60t +
66t = 63
Or 126t
= 63
Or t =
63/126 = 1/2 minutes
Therefore,
in half a minute both will cross each other. Floors travelled by Janaki upwards
in 1/2
min can be found as below :
Time
Floors
1 60
1/2 ?
Floors
travelled by Janaki upwards in 1/2 min = 60/2 = 30 floors
Since
Janaki is already starting from 8th floor, the floor of crossing counted from
ground
floor
will be 8 + 30 = 38th floor.
(
Next 4 questions are of interview type )
7) Which
of the following is not true regarding Normalization of databases ?
a.
Redundancy of data is increased.
b. Data
integrity is efficiently maintained
c.
Concurrent processing becomes easier for heavily transactional databases
d. Data
dependencies are minimized
Page 39
Answer :
a. Redundancy of data is increased
Explanation
:
One of
the major goals of normalization is to reduce redundancy (i.e duplication of
data
when not
really needed.)
8) Tell
at least two uses of NORMALIZATION in designing tables and other structures in
a
database ?
Answer :
Normalization
ensures that a) there is no redundant data storage and b) data
dependencies
across tables (foreign keys) are suitable for the level of concurrency.
9) How
Databases handle concurrent SQL operations on the same row of a table?
Answer :
Concurrent
operations are handled by the use of "Locks". To be more clear, if
one sql
query is
updating (INSERT/UPDATE) a particular row of a table, no other sql query
would be
able to update the same row at the same point in time. However other sql
queries
can read (SELECT) columns from the row under updation.
10) In
SDLC, what is black box testing (which comes under testing phase) ?
Answer :
Black
box testing is nothing but testing the actual functionality of a module/program
at a
high
level. Here one may test different outputs for different sets of inputs. Also
the one
may not
worry about the module/program internals under testing.
Page 40
HCL Sample Aptitude Questions
1) One day when Ram was walking on
the street, one boy requested him to donate for
cancer
patients welfare fund. He gave him a rupee more than half the money he had. He
walked a
few more steps. Then came a girl who requested him to donate for poor
people's
fund for which he gave two rupees more than half the money he had then. After
that,
again a boy approached him for an orphanage fund. He gave three rupees more
than half
of what he had. At last he had just one rupee remaining in his hand.
How much
amount did Ram have in his pocket when he started?
a)Rs.72
b)Rs.58 c)Rs.35 d)Rs.42
Answer :
d)Rs.42
Solution:
Let X be
the rupees Ram initially had.
he gave
a rupee more than half the money he had for cancer's fund.
i.e., he
gave Rs.1 + X/2.
so the
remaining money = X-(1 + X/2)= X/2 - 1.
then he
offered 2 rupees more than half the money he had for poor people's fund.
i.e., he
gave 2 + 1/2 x(X/2 - 1)= 2+ (X-2)/4 = (6+X)/4
so the remaining
money = X/2 - 1 - (6+X)/4 = (X-10)/4.
again he
gave 3 rupees more than half of what he had for orphanage
i.e., he
gave 3+ 1/2x(X-10)/4 = 3+ (X-10)/8 =(14+X)/8
now the
remaining money(X-10)/4 - (14+X)/8 = (2X-X-20-14)/8 = (X-34)/8
As
given, finally he had one rupee remaining so (X-34)/8 = 1
i.e.,
X-34 = 8
X= 8+34
= 42
Hence
Ram had Rs.42 initially in his pocket
2) A man buys 50 mangoes from a shop
for Rs 5 per fruit. In the next shop he buys 100
mangoes
at Rs 4 per fruit. He mixes mangoes purchased from both shops. Now at what
price he
must sell each mango to get a profit of 25% on each fruit he sells.
Page 41
a)Rs.5.4
b)Rs.5.2 c)Rs.5.3 d)Rs.5.5
Answer :
a)Rs.5.4
Solution
:
Total
money spent on first 50 mangoes = Rs. 50 x 5 = Rs. 250
Total
money spent on next 100 mangoes = Rs. 100 x 4 = Rs. 400
Net
amount spent on all 150 mangoes - Rs. 250 + Rs. 400 = Rs. 650
Effective
cost price of each mango = Rs. 650/150 = Rs. 13/3
Selling
price of each mango so as to gain 25% profit = cost price x 125/100 = 13/3 x
125/100
= Rs 5.4 (approximately)
3) A group of students were
proceeding on an education excursion. One of the students
told:The
arithmetic mean of 2 numbers is 32.5 and their geometric mean is 30. One of
the
numbers will be
a) 20 b)
42 c) 12 d) none of these.
Answer :
a) 20
Solution
:
Let the
numbers be x and y
Arithmetic
mean of the two numbers 32.5
Then,
(x+y) / 2 = 32.5
So, x +
y = 65 or y = 65 - x ...(1)
Geometric
mean of x and y = _ xy = 30
Hence xy
= (30)2 = 900
Substituting
the value of y from equation 1 in the above equation we get,
x (65-x)
= 900
or x2 -65x = 900
x2 – 65x -900 = 0
(x-45)(x-20)
= 0
Hence x
= 45 or 20
Or y =
20 or 45
Page 42
4) Rajarajan retired after serving
in Indian army as Lt. Colonel. Rajarajan's age is 20
times
that of number of daughters he has. Each of his daughters has as many daughters
as they
have sisters. If total number of grand daughters of Rajarajan is 1/3rd of the
number
of daughters, find the age of Rajaran.
a) 72 b)
90 c) 80 d) 70
Answer :
c) 80
Solution I
:
Let the
number of daughters of Rajarajan be - x
No. of
sisters each daughter has - (x-1)
No. of
daughters for each daughter - (x-1)
Then
total number of granddaughters - Number of daughters x Number of
granddaughter
per daughter = x (x - 1)
It is
given that total number of granddaughters of Rajarajan is 1/3rd of the number
of
daughters.
Therefore
x = 1/3 (x (x - 1))
Or x - 1
= 3 or x = 4.
It is
given that Rajarajan's age is 20 times that of number of daughters.
Therefore,
his age = 20 X 4 = 80.
Solution
II : (short cut)
Actually
there is a simpler short cut to this problem. Since it is given that
Rajarajan's age
is 20
times that of number of daughters , his age should be divisible by 20. Among
the
options
given only 80 is divisible by 20. Hence it is the answer.
5) Murari can complete a task in 14
days. Murari and Karthikeyan together can do it in
40 days.
In Project titled 'ABC' What part of the work was carried out by Murari?
I. During
project 'ABC' Murari completed the job alone after Murari and Karthikeyan
worked
together for 5 days.
II. Part
of the work done by Murari during project 'ABC' will be lesser than that of
Alamelu's
work when Karthikeyan and Alamelu work together for 6 days.
Page 43
Options :
a) if the
data in Statement I alone is/are sufficient to answer the question, while the
data in
Statement II alone are not sufficient to answer the question.
b) if the
data in Statement II alone is/are sufficient to answer the question, while the
data in
Statement I alone are not sufficient to answer the question.
c) if the
data either in Statement I or in Statement II alone is/are sufficient to answer
the
question.
d) if the
data even in both Statements I and II together are not sufficient to answer
the
question.
e) if the
data in both Statements I and II together are necessary to answer the
question.
Answer :
a) data in Statement I alone is/are sufficient to answer the question, while
the
data in
Statement II alone are not sufficient to answer the question.
Solution
:
To prove
this, let us try to solve the question with the data from statement I in hand :
Murari
can do the job in 14 days. One day of Murari's work = 1/14. -> eq 1
Murari
and Karthikeyan can do the job in 40 days. So in one day Karthikeyan can do
1/14 -
1/40 = 13/280 of the work
Statement
I says both of them work together for 5 days during project 'ABC'. Therefore
work
completed by both during the first 5 days of the project
= (one
day's work of Murari + one day's work of Karthikeyan)x 5
= (1/14
+ 13/280)x5 = 33/280 x 5 = 33/56
Work
that remains in project ABC = whole unit of work - part of work completed
during
first 5
days = 1 - 33/56 = 23/56 -> eq 2
After 5
days, the remaining work i.e 23/56 of work will be taken care by Murari alone.
Therefore
work done by Murari alone in the project 'ABC' = five days of Murari's work +
remaining
work of Murari after 5 days
= (one
day work of Murari) x 5 + remaining work of Murari after 5 days
Substituting
values from eq 1 and eq 2 in the above equation we get
Page 44
work
done by Murari alone in the project 'ABC' = (1/14)5 + 23/56
(23 +
20) divided by 56 = 43/56
Statement
II doesn’t help us in any way because it just says Murari's contribution would
be
lesser than that of Alamelu if Karthikeyan and Murari work together. With this
we can't
calculate
the work done by Murari alone.
(Next
12 questions are of interview type)
6) Who is the creator of 8086
microprocessor?
a.
Microsoft
b. Intel
c. Apple
d. Mac
Answer :
b. Intel.
7) The
phase that identifies an efficient execution plan for evaluating a query that
has
the
least estimated cost is referred to as___.
Options
a) Query
optimization
b) Query
String
Answer :
a) Query optimization
8)
Can a non maskable interrupt in 8086 be turned off by
a programmer?
a. Yes b.
No
Answer :
No, a non maskable interrupt cannot be turned off.
9) A
collection of conceptual tools for describing data, data relationships data
semantics
and Constraints is called as_____.
Options
a) Data
base
b) Table
Page 45
c) Data
model
Answer:
Data model
5)
______ is copying the three sets of files (database files, redo logs, and
control file)
when the
instance is shut down. This is a straight file copy, usually from the disk
directly
to tape.
You must shut down the instance to guarantee a consistent copy.
Options
a) cold
backup
b) hot
backup
c)
Armstrong Rules
Answer :
cold backup
11)
_______ is a program module, which ensures that database remains in a
consistent
state
despite system failures and concurrent transaction execution proceeds without
conflicting.
Options
a)
Transaction manager
b) File
manager
c) None of
these
Answer :
Transaction manager
12)
_______ is a program module that provides the interface between the low-level
data
stored
in database, application programs and queries submitted to the system.
Options
a) Buffer
manager
b) Storage
manager
C) None of
these
Answer :
Storage manager
13) A
_____ with respect to DBMS relates to user commands that are used to interact
with a
database.
Page 46
Options
a)
Connection string
b) Query
String
c) Query
Answer :
Query
14) In Very
Simple Terms, Can you describe what is actually a Database Server
instance?
Answer :
Every
Database Server Instance is a complete server by itself with own set of
databases,
login credentials etc. A single machine that has many server instances can
be
logically compared to multiple machines with their own Database Server
installations.
15) Will
executing/operating more than one Database Server instances on a single
machine
affect performance of applications adversely?
Answer :
More
than one instance would require more memory and processor related resources
than
single instance on the same machine. But, more than one instance can speed up
applications
by allowing multiple applications to operate on multiple instances of
Database
Server simultaneously.
16) What
could be the role of a server administrator to keep up the performance of
machine
with multiple instances?
Answer :
The
server administrator is responsible for striking a balance between number of
allowed
SQL
instances and system's hardware resources.
17) Can
databases on multiple instances have same names?
Answer :
Yes. Any
database on an instance is completely local to that database.
Page 47
Accenture Solved Questions
Note: some questions are directly given with answers and solutions
without having
options.
(Next
12 questions are of interview type)
1)
(State True or False). The Java interpreter is used for the execution of the
source
code.
Options
True
False
Answer :
True
2) What
declarations are required for every Java application?
Answer :
A class and the main( ) method declarations.
3) What
are the two primary components involved in executing a Java program and their
purposes?
Answer :
Two parts in executing a Java program are:
Java
Compiler and Java Interpreter.
4) What
are the three basic OOPs principles and define them?
Answer :
Encapsulation, Inheritance and Polymorphism are the three OOPs Principles.
Encapsulation
:
Is the
Mechanism that binds together code and the data. It manipulates, and keeps
both
safe from outside interference and misuse.
Inheritance
:
Is the
process by which one object acquires the properties of another object.
Polymorphism
:
Is a
feature that allows one interface to be used for a general class of actions.
5) What
are identifiers and what are their naming conventions in C?
Page 48
Answer :
Identifiers are used for class names, method names and variable names. An
identifier
may be any descriptive sequence of upper case & lower case letters, numbers
or
underscore or dollar sign and must not begin with numbers.
6) What
is the return type of program’s main( ) method?
Answer :
void
7) What
is the use of bin and lib in the JDK?
Answer :
Bin contains all tools such as javac, applet viewer, awt tool etc.,
whereas
Lib contains all packages and variables.
8) The
Java source code can be created in a Notepad editor.
Options
a) True
b) False
Answer :
True
9) In C,
arrays can be passed by reference. State True or False.
Answer:
Yes, it
is true, arrays are passed by reference (Though individual or group of elements
of
the
array can be passed by value/values as well.)
10) Name
the function which takes two strings as arguments and copies the second
string
into the character array of the first string. After this the function returns
the value of
the
first string.
Answer:
strcpy is the answer.
11)
Blocks are chosen randomly on a chessboard. What is the probability that they
are
on the
same diagonal?
Answer :
There
are a total of 64 blocks on a chessboard. So 3 blocks can be chosen
Page 49
out of
64 in 64C3 ways.
So the
sample space is = 41664
There
are 2 diagonal on chessboard each one having 8 blocks. Consider one of
them. 3
blocks out of 8 blocks in diagonal can be chosen in 8C3 ways.
But
there are 2 such diagonals, hence favourables = 2 * 8C3 = 2 * 56 =
112 The
required probability is
= 112 /
1664
= 1 /
372
=
0.002688
12) What
is the area of the triangle ABC with A(e,p) B(2e,3p) and C(3e,5p)?
where p
= PI (3.141592654)
Answer:
A tricky
ONE.
Given 3
points are collinear. Hence, it is a straight line.
Hence
the area of a triangle is 0.
13) Three water tanks A,B and C where
A is twice that of B. Tank A and B can be filled
at the
steady rate of 20 litres per hour and the excess water is filled with tank C.
The
pipe is
opened for 2 days and 1/5th of tank A is filled after 8 hours. Find the amount
of
water(in
litres) in tank C after 2 days.
a)720
b)580 c)160 d)560
Answer :
a)720
Solution
:
1/5 th
of tank A is filled in 8 hours.
Time
taken to fill the tank A is 8 / (1/5) = 40 hours.
Water is
opened for 2 days i.e., 48 hours.
Since,
tank A is filled, the excess water poured into C for (48 - 40) 8 hrs = 8 x 20 =
160
litres
Page 50
Given
that A is two times of B.
i.e.,
the Volume of tank A = 2 (volume of tank B).
Time
taken to fill tank B is 40 / 2 = 20 hours.
Now, the
excess water poured into C from B for (48 - 20) = 28 hrs = 28 x 20 = 560
litres.
Therefore,
the total amount of water in tank C = 160 + 560 = 720 litres.
Hence
the answer is 720.
14)
Consider the sum: ABC + DEF + GHI = JJJ .If different letters represent
different
digits,
and there are no leading zeros, what does J represent?
Answer
The
value of J must be 9.Since there are no leading zeros, J must be 7, 8, or 9.
(JJJ =
ABC + DEF + GHI= 14? + 25? + 36? = 7??)Now, the remainder left
after dividing any
number by 9 is the same as the remainder left after dividing the
sum of the digits of that
number by 9. Also, note that 0 + 1 + ... + 9 has a remainder of
0 after dividing by 9 and
JJJ has a remainder of 0, 3, or 6. The number 9 is the only
number from 7, 8 and 9 that
leaves a remainder of 0, 3, or 6 if you remove it from the sum 0
+ 1 + ... + 9. Hence, it
follows that J must be 9.
15)
Using two 2's and two 3's and using a maxim of three mathematical signs,
symbols,
can you have a result in between 14 and 15? Concatenation (clubbing of
digits)
allowed.
Solution:
( 23 +
3! ) / 2 = 14.5
16)
a*b*c*d*e + b*c*d*e*f + a*c*d*e*f + a*b*d*e*f + a*b*c*e*f + a*b*c*d*f =
a*b*c*d*e*f
and a,b,c,d,e
and f are all positive nonrepeating integers then solve a,b,c,d,e, and f.
Solution
:
Start
with 1/2 + 1/2, then progressively split the last part x into 2x/3 + x/3. This
gives the
following
progression:
2,2
2,3,6
Page 51
2,3,9,18
2,3,9,27,54
2,3,9,27,81,162
17) 729
ml of a mixture contains milk and water in ratio 7:2. How much of the water is
to
be added
to get a new mixture containing half milk and half water?
(i) 79 ml
(ii) 81 ml
(iii) 72
ml
(iv) 91 ml
Solution:
Milk
Quantity = (729 * (7/9))=567ml
Water
Quantity = (729-567)= 162ml
Let
water to be added be x ml 567/(162+x) = 7/3 1701 = 1134 + 7x x = 81ml
18) If
one-seventh of a number exceeds its eleventh part by 100 then the number is…
(i) 770
(ii) 1100
(iii) 1825
(iv) 1925
Solution:
Let the
number be x. Then X/7 - x/11 =100 11x-7x = 7700 x=1925.
19) If
1.5x=0.04y then the value of (y-x)/(y+x) is
(i) 730/77
(ii) 73/77
(iii)
7.3/77
(iv) None
Solution:
x/y =
0.04/1.5 = 2/75
So
(y-x)/(y+x) = (1 - x/y)/(1 + x/y) = (1 - 2/75)/ (1 + 2/75) = 73/77.
Page 52
20) If x
and y are the two digits f the number 653xy such that this number is divisible
by
80, then
x+y is equal to:
(i) 2
(ii) 3
(iii) 4
(iv) 6
Solution:
80 = 2 x
5 x 10
Therefore,
for 653xy to be divisible by 80, it has to be divisible by 2,5 and 10 as well.
Since
653xy is divisible by 2 as well as by 5, so y = 0
Now
653x0 is divisible by 8 so 3x0 should also be divisible by 8.
Only
option which satisfies the above condition is x = 6.
IBM Solved Technical Questions
1) 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
Answer :
b)X=3,Y=2
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
2. Evaluate [(1.386 x 0.643 + (2.921-1.535)x 0.357)/(0.6 x 0.015 +
0.6 x 0.985)]x(3 x 2.7
+ 3 x
0.3)
a)20.00
b)18.19 c)20.79 d)19.19
Answer :
c)20.79
Solution:
The
given expression can be simplified as follows;
[(1.386
x 0.643 + (1.386) x 0.357)/(0.6 x 0.015 + 0.6 x 0.985)]x(3 x 2.7 + 3 x 0.3)
=
{[1.386 x (0.643 + 0.357)]/[0.6(0.015 + 0.985)]} x [3(2.7 + 0.3)]
=
{[1.386 x 1.000]/[0.6 x 1.000]} x [3x3]
= [1.386
/ 0.6] x [3 x 3] = 1.386 x 3/0.2 = 6.93 x 3 = 20.79
Page 54
3. Find the greatest number that exactly divides 7667, 4603, 12263
when each of these
is
reduced by 7.
a) 2298 b)
1572 c) 1532 d) 2546
Answer :
c) 1532
Solution:
(Actually
this is a simple problem dealing with finding just HCF of the numbers after
subtracting
7 from each of these numbers. Since it immediately follows the I problem,
some
readers could think it as complex as the first.)
the
required number = h.c.f of [(4603-7),(7667-7),(12263-7)] (given that the
numbers
leaves
7)
= h.c.f
of (4596, 7660, 12256)
= 1532.
(Next
3 questions are of interview type)
4) Can
you tell advantages of RDBMS over simple custom DBMS ?
Answer :
Any data
storage in any manageable structure can be called a DBMS. But, it is the
responsibility
of the programmer to maintain the data integrity and to take care of
constraints.
But on RDBMS, the system takes care of almost all of the data integrity
requirements.
Hence, the programmer could focus more on business logic rather than
worrying
about data integrity much.
5) Which
is the layer of an Operating System that takes care of the system resource
usage at
the lowest level ?
Answer :
Kernel Layer.
6) Can
you tell any two advantages of stored procedures ?
Answer :
Page 55
Stored
procedures can replace complex program segments requiring extensive SQL
statements
being executed from the code, hence they can make the application more
manageable.
Another advantage is that the usage of stored procedures can significantly
reduce
the network and bandwidth usage. This is because of the reduction in the
number
of requests and responses between business layer and database layer.
Page 56
Syntel Solved Aptitude Questions
1. Find
the missing number in the sequence 2,5,4,7,_
Options
a) 6 b) 4
c) 5 d) 7
Answer is : 6.
Just consider the sequence 3, 4, 5, 6, 7 and start subtracting
and adding 1 to
consecutive numbers which will get you the sequence in question.
2. In an
alien planet, the word "lion" is coded as "mhpm". Then how
the word "tiger" would
be coded
as?
Options
a) uhids
b) uhhds
c) uhhfs
d) uhhfq
Answer
is : uhhds.
Reason :
First letter in a word would be replaced by the next adjacent letter. Second
letter
would be replaced by the immediately preceding letter... and so on.
3. Find
the odd man out a) 123 b) 235 c) 135 d) 358
Answer
is: 135.
Adding
first two digits will give the third digit in all the other three options
except option
c).
4. Read
the following statements.
"Weather
is good in all northern cities of India.
Unlike
northern cities, sunny weather exists in most of eastern cities of India.
Warm
weather exists in all northern cities and some eastern cities of India."
Considering
above statements, which of the following statements is false
a) Warm
weather is considered good.
b) All
eastern cities experience bad weather
Page 57
c) Some
eastern cities experience good weather
Answer :
b) All eastern cities experience bad weather is false.
This is
because, the second statement clearly states that warm (good) weather prevails
in some
eastern cities.
5. A
bookseller sells a particular novel at 10% discount on the labelled price. Also
he is so
generous
that he gives a free book for every 15 books for wholesale buyers. In this
transaction
his gain is 35%. Then find the ratio of Ratio of Labelled Price to the actual
CP.
Answer :
Lets
assume the CP of each book be 100. Hence CP of 16 books would be
1600. SP
of 15 books = 1600 + (1600 * 35/100) = 2160.
SP of
each book would be 2160/15 = 144.
If SP of
each book is 90, labelled price would be 100 (since he gives at a 10%
discount).
Hence if SP is 144 marked price would be 144*(100/90) = 160.
Ratio of
Labelled Price to the actual CP = 160/100 = 8/5.
6. If a
pen is being sold at 4% profit instead of 4% loss the actual profit is Rs 16.
What is
the
actual cost price of the pen?
Answer :
Let x be
the CP. (104/100)x - (96/100 )x = 16.Solving we get x = Rs.200.
7. A
cake seller sells one cake at a profit of 10% and sells another at a loss of
5%. Let the
ratio of
the CPs of the cakes is 2:3 respectively. Find his net profit or loss
percentage.
Answer :
Let the
CPs of the cakes be 2x and 3x (so that they are in the ration 2:3 as per the
question.)
Hence
net CP = 5x.
SP of
first cake = (110/100)*2x = 220x/100 SP of second cake = (95/100)*3x = 285x/100
Net SP =
(220x/100) + (285x/100) = 505x/100 = 5.05x. SP is greater than CP
and his
profit is 5.05x - 5x = .05x. His profit percentage = (.05x/5x)% = .01%.
Page 58
HP Sample Questions
1) If one-seventh of a number exceeds its eleventh part by 100 then
the number is…
(i) 770
(ii) 1100
(iii) 1825
(iv) 1925
Answer: (iv) 1925
Solution
: Let the number be x. Then X/7 - x/11 =100 11x-7x = 7700 x=1925.
2) The
ratio of Rita's age to her mother's age is 3:8. The difference of their ages is
35
years. The ratio of their ages after 4 years will be:
(i) 7:12
(ii) 5:12
(iii)
38:43
(iv) 42:47
Answer: (ii) 5:12
Solution:
Let
their ages be 3x and 8x
8x - 3x
=35
x =7
Their
present ages are 21 and 56 years.
Ratio of
their ages after 4 years are 25:60 = 5:12
3) A tap
can fill the tank in 15 minutes and another can empty it in 8 minutes. If the
tank
is
already half full and both the taps are opened together, the tank will be:
(i) filled
in 12 min
(iii)
emptied in 12 min
(iv)
filled in 8 min
(v)
emptied in 8 min
Page 59
Answer: 8 minutes
Solution:
Rate of
waste pipe being more the tank will be emptied when both taps are
opened.
Net
emptying work done in 1min =(1/8 -1/16)= 1/16
So full
tank will be emptied in 16 min
Half
tank will be emptied in 8 minutes.
4) A man
can row 5 kmph in still water. If the river is running at 1kmph, it takes him
75
minutes to row to a place and back. How far is the place?
(i) 3km
(ii) 2.5
km
(iii) 4 km
(iv) 5 km
Answer: 3 Km
Solution:
Speed
downstream = (5+1)km/hr = 6 km/hr Speed upstream = (5-1)km/hr = 4 km/hr
Let the
required distance be x km x/6 + x/4 = 75/60 2x+3x = 15 x = 3km
5) If
log 0.317=0.3332 and log 0.318=0.3364 then find log 0.319?
(i)0.3396
(ii)0.3369
(iii)0.3368
(iv)0.3338
Answer:
0.3396
Solution:
log 0.317=0.3332 and log 0.318=0.3364,
then log
0.319=log0.318+(log(0.318-0.317)) = 0.3396
6) In a
hotel, rooms are numbered from 101 to 550. A room is chosen at random.
Page 60
What is
the probability that room number starts with 1, 2 or 3 and ends with 4, 5 or 6?
Answer
There
are total 450 rooms.
Out of
which 299 room number starts with either 1, 2 or 3. (as room number
100 is
not there) Now out of those 299 rooms only 90 room numbers end with
4, 5 or
6.So the probability is 90/450 i.e. 1/5 or 0.20.
7)
Difference between Bholu's and Molu's age is 2 years and the difference between
Molu's
and Kolu's age is 5 years. What is the maximum possible value of the sum of
the
difference in their ages, taken two at a time?
Answer
The
maximum possible value of the sum of the difference in their ages - taken two
at a
time -
is 14 years.
8) How
many even integers n, where, are divisible neither by seven nor by nine?
Solution:
There
are 101 integers in all, of which 51 are even. From 100 to 200, there are 14
multiples
of 7, of which 7 are even. There are 11 multiples of 9, of which 6 are even.
But
there is one integer (i.e. 126) that is a multiple of both 7 and 9 and also
even. Hence
the
answer is (51 – 7 – 6 + 1) = 39
9) Four
persons A, B, C and D are playing cards. Each person has one card, laid
down on
the table below him, which has two different colours on either side.
No card
has the same color on both sides. The colours visible on the table are Red,
Green,
Red and Blue respectively. They see the color on the reverse side and give
the
following comment.
A: Yellow
or Green
B: Neither
Blue nor Green
C: Blue or
Yellow
D: Blue or
Yellow
Given
that out of the 4 people 2 always lie find out the colours on the cards each
Page 61
person.
ANSWER:
Try all possible combinations. Keep in mind two things. THE
combination obtained
should satisfy the conditions
1.Two are lying and two are telling the truth
2.Neither two cards are similar nor are two sides of a card are
of same color
A YELLOW
B YELLOW
C GREEN
D RED
10)
Grass in lawn grows equally thick and in a uniform rate. It takes 40 days for
40
cows and
60 days for 30 cows to eat the whole of the grass. How many days does it
take for
20 cows to do the same?
ANSWER:
g -
grass at the beginning
r - rate
at which grass grows, per day
y - rate
at which one cow eats grass, per day
n - no
of cows to eat the grass in 96 days
g + 40*r
= 40 * 40 * y------- 1
g + 60*r
= 30 * 60 * y------- 2
g + n*r
= 20 * n * y-------- 3
from 1
and 2
r=10y
g=120r
from 3
nr=120r
Solving,
n = 120
Page 62
11)
Lucia is a wonderful grandmother. Her age is between 50 and 70.Each of her sons
has as
many sons as they have brothers. Their combined number gives Lucia’s age.
What is
the age?
ANSWER
Let the
no. of Lucia’s sons = n
No. of
brothers for each son = n-1
No. of
sons for each of Lucia’s son = n-1
Lucia’s
age = n-1 * n-1
= a
perfect square between 50 and 70
= 64
12) Gold
is 19 times as heavy as water and copper is 9 times as heavy as water. In what
ratio
should these be mixed to get an alloy 15 times as heavy as water?
(i) 1:1
(ii) 2:3
(iii) 1:2
(iv) 3:2
Solution:
Let 1gm
of gold be mixed with x gm of copper to give (1+x)gm of the alloy.
1G=19W,
1C = 9W and alloy = 15W 1gm gold + x
13) What
is not a part of OS?
a)swapper
b)compiler
c)device
driver
d)file
system
Answer is compiler
14)
Which is the protocol used by PING?
a)ICMP
Page 63
b)HTTP
c)SMTP
d)RTSP
Answer
is ICMP
Honeywell Sample Aptitude Questions
1) A
lamp post stands vertically on the top of a building. From a point 40m distance
from
the
bottom of the building on the ground, the angle of elevation of the bottom and
top of
the lamp
post are 45 degree and 60 degree respectively. Then the height of the vertical
lamp
post will be
a) 29.28m
b) 39.38m c) 49.48m d) 19.18m
Answer :
a) 29.28m
Solution
:
Let BC =
height of the building = h m
CD =
height of the lamp post = h1 m
And AB =
40 m
To find
the height of the lamp(CD), first we have to find the height of the
building(BC)
On
right-angled triangle ABC, tan (45degrees) = BC / AB = h / 40.
Since
tan (45 degrees) = 1 then h / 40 = 1 ==> h = 40 m.
Now, on
right-angled triangle ABD, tan (60 degrees) = BD / AB or sqrt(3) = BD / 40
BD = 40
x sqrt(3) = 40 x 1.732 = 69.28 m
Then, h1
= CD = BD - BC = 69.28 - 40 = 29.28 m
Hence
the height of the lamp post is 29.28 m.
2) Solve
the inequality 6(6x-2) < 1/36
a) x <
0 b) x > 0 c) x = 1/3 d) x = 6
Page 65
Answer :
a) x < 0
Solution
:
Given
inequality is 6(6x-2) < 1/36
Multiply
by 36 on both sides, we get
36[6(6x-2)] < 1
Make
each term as the power of 6
(62)[6(6x-2)] < 60
Now,
using the formula (ax)(ay)= a(x+y)
6(6x-2+2) < 60
6(6x) < 60
Cancelling
the base 6 on both sides,
6x <
0
x < 0
Hence
the answer is option a
3) A
railway half ticket( for kids below 5 years) costs Rs. 150 and full ticket(for
above 5
years)
costs Rs. 250. The daily report says that for a particular day, 5000 passengers
have
travelled and the total collection is Rs.10,50,000. How many kids(below 5
years)
have
travelled on that day?
a)1000
b)2000 c)2500 d)3500
Answer :
b)2000
Solution:
Let 's'
be the number of kids travelled on that day and 'b' be the number of
passengers(above
5 years) travelled on that day.
Therefore,
s + b = 5000 ... eqn (1)
Each
half ticket's cost is Rs.150 and
Each
full ticket's cost is Rs.250.
Total
collection of amount = 150s + 250b = 10,50,000
Or 150s
+ 250b = 10,50,000 ... eqn (2)
Page 66
Multiplying
equation (1) by 150, we get 150s + 150b = 7,50,000 ... eqn (3)
Subtracting
eqn (3) from eqn (2), we get 100b = 3,00,000
Or b =
3000
We know
that s + b = 5000
So, s =
5000 - b = 5000 - 3000 = 2000.
Hence,
2000 kids travelled on that particular day.
(Next
3 questions are of interview type)
4) The
code for creating stack in DOS is _____ .
a) STACKS
= (number),(size)
b) STACKS
= number
c) STACKS
= (size),(number)
d) STACKS
= size
Answer :
a) STACKS = (number),(size)
5) Unix
Operating System is an example of ______ .
a) Macro
or Monolithic Kernel
b) Micro
Kernel
c) Hybrid
Kernel
d) None of
these
Answer :
a) Macro or Monolithic Kernel
6) In
Linux, the data structure of file system is called as ______ .
a) inode
b) struct
inode
c) vnode
d) lnode
Answer :
b) struct inode
Page 67
General Aptitude & Verbal Questions
Can be used for all IT Companies
1) A starts a project of duration of 15 months with capital
Rs.60000. 3 months
after the start, B joins with A and invests one fourth of that
of A. After another 6
months, C joins with capital Rs. 90000. At the end of the year,
A withdrew the
partnership leaving the firm to be run by B and C. What will be
the shares of B
and C if total profit is Rs.50000?
a)Rs.25000, Rs.6250 b)Rs.6250,Rs.18750
c)Rs.18750, Rs.6250 d)Rs.25000,Rs.18750
Answer : b)Rs.6250,Rs.18750.
Solution:
A works since start of the project but withdrew 3 months before
completion.
Therefore, he works for 15 - 3 = 12 months
B works 3 months after start of the project and continues till
end. Therefore, his
duration of partnership = 15 - 3 = 12 months
C joins 9 months after A started business (6 months after B who
was already late
by 3 months) and works till completion. Therefore, his duration
= 15 - 9 = 6
The ratio of their profits =(60000 x 12):(1/4 x 60000 x
12):(90000 x 6)
=720000:180000:540000
=4:1:3
Now let us find the respective share of profits out of total
profit Rs.50,000
A's share= 4/8 x 50000 = 25000
B's share = 1/8 x 50000 = 6250
C's share = 3/8 x 50000 = 18750.
Based on above results, our answer is option b.
Page 68
2) Arun says that his weight is in the range of 58 kg to 64 kg.
His brother weight
is 62 kg and Arun's weight is not more than that of his
brother's. Also, as a matter
of fact, Arun's weight is in between 60 kg and 65 kg. If all the
above statements
are true, then find the average weight of Arun.
Answer : 60.5 kg
Solution:
Let the weight of Arun be x.
In Arun's view, 58 < x < 64 ...(1)
Based on his brother's estimation we have x < = 62 ...(2)
also given that 60 < x < 65 ...(3)
Combining, (1),(2) and (3) we have 58 < 60 < x < = 62
< 64 < 65
Simplifying the above inequality, we get, 58 < x < = 62
Probable values for x are 59,60,61 and 62
Average of the probables are(59+60+61+62)/4 = 242/4 = 60.5
3) Before 5 years, the sum of the ages of Ragu and Suresh was 80
years. The
sum will be 104 years after 7 years. What could be possible
combinations of
ages from among the options.
a) 50,40 b) 25,65 c) none of these d)both a & b
Answer : d) both a & b
Solution:
Let the age of Ragu be x and that of Suresh be y
Before 5 years, the sum of ages was 80, i.e x - 5 + y - 5 = 80
Or x + y = 90 ...(1)
After 7 years, the sum will be 104, i.e x + 7 + y + 7 = 104
Or
i.e. x + y = 90 ...(2)
Equations 1 and 2 indicate that the sum of the ages has to be
90.
Page 69
Both option a and option b have listed ages whose sum is 90.
Hence there are
two possible solutions from among options. Hence option d) is
correct answer.
4) During Summer holidays, Deepa and Ramya visited their uncle
who was
residing in a village. It was a different experience in the
village for these two girls.
Aunty engaged them teaching how to play games with marbles.In
one particular
game both Ramya and Deepa had the same number of marbles when
they
started playing.After sometime Deepa gained 50 marbles. After
some time
Deepa lost 3/5 th of what she had. At the same instance Ramya
had 3 times as
many marbles as Deepa had. Can you find out the number of
marbles the girls
had at the start ?
(a) 100 (b) 200 (c) 175 (d)140
Answer : b) 200
Solution:
It is stated that both Deepa and Ramya had the same number of
marbles to start
with. Let us assume both of them had x number of marbles when
the game
began. After sometime Deepa gained 50 marbles, which means she
had x+50
marbles and Ramya had x-50 marbles.
Later, Deepa lost 3/5th marbles. Therefore, Deepa had (x + 50) -
(3/5)( x + 50) =
2/5 (x+50)
These lost 3/5th of marbles by Deepa would had been gained by
Ramya
These 2/5 (x+50) marbles would have been gained by Ramya.
Therefore, she
would have had x - 50 + 3/5(x + 50). At that particular
instance, Ramya's marble
count was thrice that of Deepa.
i.e 3 * 2 / 5(x + 50)= x - 50 + 3 / 5(x + 50)
6 / 5 (x + 50)=8/5 x - 20 or
6x+300=8x-100 (By multiplying both sides by 5)
Page 70
2x= 400 or x=200
The number of marbles Deepa and Ramya had initially = 200.
5) Roushan was a B.E student pursuing studies in the city of
Vizag. During
summer holidays he visited his village.In the courtyard of his
house, there are
many hens and goats. Roushan asked his sister as to how many
hens are there
in the courtyard. His sister who is a village girl gave her
reply like this : If you
count all the heads you will get 300. But if you count the legs
you will get 700.
Can you help Roushan in calculating the number of hens.
(a) 200 (b) 250 (c) 600 (d) 400
Answer : b) 250
Solution:
Let us assume that x number of hens and y number of goats are
there in the
courtyard.
x+y = 300, -------(1) since the total head count is given as
300.
Its given, there are totally 700 legs when counted. We know the
hens have two
legs and the goats have 4 legs. Therefore
2x+4y= 700 -------(2)
(1)x2 will give, 2x+2y=600 -------(3)
(2)-(3) will give you, 2y=100 and y=50.
Substituting in equation (1) you get x value as 250
6) Green and blue coloured toys are 149 in number in a box. Two
green toys are
removed from the lot and now there are twice as many green ones
as blue ones.
How many green coloured toys are there now?
a) 76 b) 86 c) 90 d) 98
Page 71
Answer : d) 98
Solution:
Let x and y be the original number of green and blue toys
respectively. Originally
there were 149 toys. Hence,
x + y = 149 ----------(1)
After two green toys were removed, the number of green toys
becomes twice
that of the blue toys.
x-2 = 2 y -----------(2)
Solving eq 1 and eq 2 we get
3 y = 149-2=147
y= 49
The current number of green toys (after removal of two) is twice
as that of blue
ones.So the green toys are 98 in number.
7) Ganesh Ram starts in his luxurious Honda City for a week end
holiday trip to a
near by hill station. The road was in a poor condition and the
car tire got
punctured in the mid way. Before he could reach out for help he
wanted to
measure how long he can sustain the repair. The first puncture
by itself would
make the tire flat in 9 minutes. The second puncture by itself
would make the tire
flat in 18 minutes. How long will it take for both the punctures
together to make
the tire flat, if Ganesh Ram assumes that the air leaks at a
constant rate?
(a)3 3/5 minutes (b)13 1/2 minutes
(c) 6 minutes (d)27 minutes
Answer : c) 6 minutes
Solution:
This is a time and work problem.
In each minute, the first puncture will leak 1/9th of the air
and the second
Page 72
puncture in one minute will leak 1/18th of the air. When both
are punctured
together effective leak per minute will be 1/9 +1/18=3/18 or
1/6th of the air.
So, in 6 minutes time the entire tire will be flat. Hence,
correct answer is option
(c)
8) The district collectorate at Thanjavoor had a flag post with
the tri color flag
flying. The 18 m high flags post casts a shadow of length 42m.
Collectors
residential quarters casts a shadow of 28m under similar
conditions. Calculate
the height of the building?
(a)14m (b)15m (c)12m (d) 16.5m
Answer : c)12m
Solution:
Height of the flag post = 18m
Length of the shadow of flag post = 42m
Let the height of Collector's residential quarters be x m
Length of the shadow of Collector's residential quarters = 28m
Under similar conditions, we can safely assume that the shadows
are
proportional to the heights.
Therefore we can write, Length of the shadow of the flag post /
Height of the flag
post = Length of the shadow of Collector's residential quarters
/ Height of
Collector's residential quarters
Substituting length and height values in the above equation we
get
42/18 = 28/x
x = 28 x 18/42 = 4 x 18/6 = 4 x 3 = 12m
Therefore our answer is 12 m.
Page 73
9) Ranjit Kumar purchased a new watch in Burma Bazaar. Sooner he
found that
the uniformly gaining watch starts with a lag of 2 minutes at
noon on a particular
Monday and it is 4 minutes 48 seconds fast at 2PM on the
following Monday.
Please guide Ranjit Kumar when the clock would show the true
time ?
(a)2PM on Tuesday (b)2PM on Wednesday
(c)3PM on Thursday (d)1PM on Friday
Answer : b)2PM on Wednesday
Solution:
12 noon Monday to 2PM next Monday is 7 days and 2 hours. i.e. 7
x 24 +2 = 170
hours.
The new watch purchased by Ranjit Kumar was 2 minutes slow and
by 170
hours it was fast by 4 minutes 48 seconds. (4 minutes 48 seconds
can be written
as 4 48/60 or 4 4/5 minutes.)
Therefore the total gain in 170 hours = 2 minutes + 4 4/5
minutes = 34/5 minutes
= 408 seconds
Gain per hour = 408/170 = 2.4 seconds.
It is given that the watch was slow by 2 minutes (120 seconds)
initially. Therefore
the time at which the watch will make up for the lost 2 minutes
by gaining 2
minutes or 120 seconds will be our answer.
2.4 seconds are gained in 1 hour.
120 seconds will be gained in 1/2.4 x 120 =50 hours.
So the watch will show right time in 50 hours from Monday 12.00
noon ie at
Wednesday 2.00 PM.
10) Dilip is a book worm and his wife Rama Dilip is also a
voracious reader. Dilip
reads at an average rate of 40 pages per hour, while Rama Dilip
reads at an
Page 74
average rate of 50 pages per hour. Dilip starts reading a novel
at 4.30 PM and
Rama Dilip begins reading an identical copy of the same book at
5.20 PM. At
6.00 PM suddenly a family friend arrived and both husband and
wife had to
spend 30 minutes with the guest. Thereafter, they started
reading again. At what
time will both husband and wife will be reading the same page?
(a)10.00PM (b) 9.30 pm (c) 9.10 PM
(d) 8.20Pm (e) 7.30 PM
Answer : c) 9.10 PM
Solution:
We have to find out when Rama Dilip will catch up with Dilip.
Dilip reads at the
rate of 40 pages per hour and Rama Dilip reads at the rate of 50
pages per hour.
Dilip starts 50 minutes ahead of his wife. Since 50 minutes is
5/6 of an hour, by
the time Rama Dilip starts reading at 5.20 PM, Dilip has already
read 5/6
x40=200/6 pages. Please note that Rama Dilip is faster by 10
pages per hour as
compared to her husband.
Since Dilip started with 200/6 pages ahead at 5.20 PM to catch
up with Dilip it
should take Rama Dilip 200/6 pages / 10 pages per hour. ie
200/60 hours or 3
hours 20 minutes. And of course the common time spent by both
husband and
wife chatting with guest is 30 minutes. So to catch up with
Dilip, Rama Dilip
requires 3 hours 20 minutes + 30 minutes or 3 hours 50 minutes.
Rama Dilip and
Dilip will be reading the same page at 5.20 PM + 3 hours 50
minutes ie by 9.10
PM.
11) All Win Recreation Club is a famous club in the city of
Vijayawada. A general
body meeting of the club was convened to discuss the expenses
that was
incurred during an inter club tournament.The total expenses were
for Rs.5000/-.
Page 75
All the club members agreed in the first instance to share the
expenses among
themselves. However, five members of the Club chose to resign,
leaving the
remaining members to pay an extra Rs 50 each. What is the
original membership
of the club?
(a) 33 (b)35 (c)28 (d) 25
Answer : d) 25
Solution:
Let us assume there are originally T members in the club.
Total expense as per plan = Rs. 5000
This means everyone has to pay Rs. 5000/T
However 5 members resigned. This means the budget was short by 5
x 5000/T
...(1)
Number of remaining members = T - 5
This shortcoming was balanced by the remaining members as they
paid an extra
of Rs. 50 each.
Total extra amount paid = 50 x (T - 5) ...(2)
Since total shortcoming is balanced by extra amount paid, values
of 1 and 2
should be equal
Therefore, 5 x 5000/T = 50 x (T - 5)
25000/T = 50 (T-5)
Multiplying both sides by T, we get
25000= 50T2-250T or
50T2 - 250T-25000=0
Factorising you get,
(T-25) (T+20) = 0
T = 25 or T = -20
Since total members cannot be negative, T = 25
12) Find the odd option among : astronomy,science, telescope,
astrology
Answer : astrology
'Astronomy' refers to a 'Science' where celestial space and
bodies are studies
using devices such as 'Telescope'. 'Astrology' is an entirely
different concept
13) Find the odd option among:
a) Gaur b) blackbuck c) Sambar d) Swallow
Answer : d) Swallow
Reason :
First three options namely Gaur, Blackbuck and Sambar refer to
animals while
Swallow is a bird. Hence d is the answer.
14) ___ of Indian painters and musicians who lived during
medieval period is to
be wondered about and appreciated.
a. Artistry
b. Works
c. Sculptures
d. Innovation
Answer : a. Artistry
Artistry refers to work in the field of art. Though option b
Works fits the blank, it
may not as good choice as Artistry. Sculptures and Innovation do
not fit the blank
either. Hence option a is the right answer.
15)
Which of the following is an appropriate synonym for the word Debauch?
Options
a) Demoralize
b) Encourage
c) Cultivate
Answer :
a) Demoralize
16) Find
the synonym of Decreed?
Options
a) made up one's mind
b) disagree
c) decrease in quantity
Answer :
a) made up one's mind
17) What
is an appropriate synonym for Bifid?
Options
a) Divided
b) Divided
in two
c) Timid
Answer :
a) Divided
18) Find
the antonym for gaurish.
Options
a) Cheap
b) Flashy
c) Costly
Answer :
a) Cheap
19)
Choose an appropriate antonym for the word deliberate.
Options
a)
unintended
b)
targeted
c)
focussed
Answer :
a) Unintended
Page 78
20) Choose
the antonym for Sorrow.
Options
a) Joy
b) empathy
c)
sympathy
Answer :
a) Joy
Did you know that that you can earn money by locking selected areas of your blog or site?
ReplyDeleteSimply open an account on AdscendMedia and implement their Content Locking tool.
With RentalCars you can get the most affordable car hires from over 49000 locations worldwide.
ReplyDelete