Infosys Placement Papers, Aptitude Online Test, Infosys Careers

Infosys Placement Papers, Aptitude Online Test, Infosys Careers

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J2EE Questions:

Question 1

JSP files can be found in

a) application client module

b) enterprise JavaBeans module

c) web module

d) resource adapter module

Answer : c) web module

Question 2

File holding generic libraries of java classes, resources, auxiliary files etc. is called as

a) Jar files

b) Ear files

c) War files

d) Rar files

Answer : a) Jar files

Question 3

JSF stands for

a) Java Servlet Faces

b) Java Server Faces

c) Java Servlet Float

d) Java Server Float

Answer : b) Java Server Faces

Question 4

Expansion for HQL is

a) High Query Language

b) Hiber Query Language

c) Higher Query Language

d) Hibernate Query Language

Answer : d) Hibernate Query Language.

Problems on Boats and Streams:

Formulas to remember:

Downstream/Upstream:

1. In water, the direction along the stream is called downstream. And, the direction against the stream is called upstream.

2. If the speed of a boat in still water is u km/hr and the speed of the stream is v km/hr, then:

Speed downstream = (u + v) km/hr.

Speed upstream = (u – v) km/hr.

3. If the speed in downstream is a km/hr and the speed in upstream is b km/hr, then:

Speed in still water = (a + b)/2 km/hr.

Rate of stream =(a – b)/2 km/hr

Question 1

A motorboat can cover 10 1/3 km in 1 hour in still water. And it takes twice as much as time to cover up than as to cover down the same distance in running water. The speed of the current is:

a)3 4/9 km/hr b) 2 1/3 km/hr c) 4 km/hr d) none of these

Answer : a) 3 4/9 km/hr

Solution :

Let the speed of upstream be X km/hr.

Then, speed in downstream = 2X km/hr (since boat takes twice as much as time to cover up than as to cover down the same distance in running water).

Speed in still water = (2X+X)/2 km/hr. (formula 3)
= 3X/2 km/hr.

Given that, boat covers 10 1/3 km in 1 hour in still water.

Therefore, 3X/2 = 10 1/3
X = 62/9

So, speed in upstream = 62/9 km/hr.
And, speed in downstream = 2 x 62/9 = 124/9 km/hr

Hence, speed of the current = [(124/9 – 62/9)]/2 km/hr
= 62/9×2 = 34/9 = 3 4/9 km/hr.

Question 2

A man can row a certain distance downstream in 2 hours while he takes 3 hours to come back. If the speed of the stream be 6 km/hr then the speed of the man in still water is:

a) 15km/hr b) 30km/hr c) 25km/hr d) 29km/hr

Answer : b) 30km/hr

Solution :

Let the speed of the man in still water be X km/hr.

Given that, speed of the stream = 6 km/hr.
Therefore, speed in downstream = (X+6) km/hr (by using formula 2)
And, speed in upstream = (X-6) km/hr

Distance covered in downstream in 2 hours = (X+6)2 km

Distance covered in upstream in 3 hours = (X-6)3 km

Therefore, (X+6)2 = (X-6)3
2X+12 = 3X-18
X = 30km/hr.

Question 3

A man can take the same time to row 13 km downstream and 7 km upstream. His speed in still water 5 km/hr. The speed of the stream is:

a) 5/2 km/hr b) 3/2 km/hr c) 7/2 km/hr d) 2 km/hr

Answer : b) 3/2 km/hr

Solution :

Given that, the speed in still water = 5 km/hr
Let the speed of the stream be X km/hr.
Then speed in downstream = (5+X) km/hr
And, speed in upstream = (5-X) km/hr

The time taken to cover 13 km downstream = 13/(5+X)
The time taken to cover 7 km upstream = 7/(5-X)

Therefore, 13/(5+X) = 7/(5-X)
13(5-X) = 7(5-X)
65 – 13X = 35+7X
30 = 20X
X = 30/20 = 3/2

Hence the required answer is 3/2 km/hr.

Question 4

A boat takes 7 hours to cover 24 km distance and comes back. And, it can cover 2 km with the stream in the same time as 1.5 km against the stream. The speed of the stream is:

a) 1 km/hr b) 2 km/hr c) 3 km/hr d) 4 km/hr

Answer : a) 1 km/hr

Solution :

Let the boat takes X hours to cover 2 km in downstream.
Then, speed in downstream = (2/X) km/hr

and, speed in upstream = (1.5/X)km/hr

Given that, the boat takes 7 hours to cover 24 km distance and comes back.

That is, 24/(2/X) + 24/(1.5/X) = 7
24X/2 + 48X/3 = 7
168X/6 = 7
X = 42/168 = 1/4

So, speed in downstream = 2/X = 2 /(1/4) = 8 km/hr
Speed in upstream = 1.5/X = 1.5 /(1/4) = 6 km/hr.

Speed of the stream = (8-6)/2 km/hr (by using the formula 3)
= 1 km/hr.

Simplification Problems:

Question 1

In an entrance exam of 200 objective questions, a student can score 1 point for every correct answer, loss ¼ points for every wrong answer and loss 1/2 point for every unanswered question. If he attempts only 160 questions and he scores 100 points then the number of questions answered by him correctly is:

a) 132 b) 126 c) 139 d) 128

Answer : d) 128.

Solution :

Let C be the number of questions answered correctly.
Let W be the number of questions answered wrongly.
Let U be the number of unanswered questions.
Total number of questions = 200 = C + W + U

Given that, he attempts only 160 questions.
i.e., U = 200 – 160 = 40.
And then, C + W = 160 ……(1)

Total score obtained by him = 100 points
Therefore, C – ¼ W – ½ U = 100
Putting U value in above, we get
C – ¼ W = 120 ….(2)
Solving (1) and (2), we get C = 128.
Hence the number of correct answer is 128.

Question 2

An exam contains 120 objective questions and the examiner calculate the score by using the formula S = 60 + 2C – ½ W where C is the number of correct answers and W is the number of wrong answers. If a candidate attempts 80 questions of 120 questions and his total score becomes 150 then the number of questions answered by him correctly is:

a) 42 b) 58 c) 39 d) 52

Answer : d) 52

Solution :

Total number of questions = 120.
Number of questions answered by a student = 80
Given that, C = number of correct answers and W = number of wrong answers.
Therefore, C + W = 80 ……(1)

His score = 150 points.
Then, S = 150 = 60 + 2C – ½ W
300 = 120 + 4C – W
4C – W = 180 ……(2).

Solving (1) and (2), we get C = 52.
Hence, the required number of correct answers is 52.

Question 3

In a game of Chess, a boy lost 3 rounds less than he won. He can score 4 points for a win and loss 2 points for a loss. How many rounds, in all, have he played if his score is 46?

a) 37 b) 17 c) 46 d) none of these.

Answer : a) 37.

Solution :

Let the number of rounds lost by him = X
Then, number of rounds won by him = X+3.
Total rounds played by him = X + X + 3 = 2X+3.

He score 4 points for a win and loss 2 points for a loss and his total score is 46.
Therefore, 4 ( X+3 ) – 2X = 46.
4X + 12 – 2X = 46
2X = 34
X =17
Required total number of rounds = 2X + 3 = 37.

Algebra Problems:

Question 1

If 26, 25, 24,…,3, 2, 1 are the position of the alphabets Z, Y, X,…, C, B, A, respectively and each alphabet has a value which equals the position of that alphabet raised to the value of succeeding alphabet, then which of the following is the exact value of the product (X-A)(X-B)(X-C)….(X-Y)(X-Z)?

a) cannot be determined b) 0 c) 2426 d) 1

Answer : b) 0

Solution :

We have to find,

(X-A)(X-B)(X-C)….(X-Y)(X-Z)

= (X-A)(X-B)(X-C)(X-D)….(X-W)(X-X)(X-Y)(X-Z)

= (X-A)(X-B)(X-C)(X-D)….(X-W)(0)(X-Y)(X-Z) (since X-X = 0)

= 0 [Product of any term multiplied with zero always results in zero].

Hence the answer is zero.

Question 2

Each alphabet a, b, c,… is a constant and the value of them are a=1,b=2,c=32,d=49,….. That is, b = 21, c = 3b, d = 4c,…, y = 25x, z = 26y; Find how many sum of products in the expression (X-a)(X-b)(X-c)….(X-y)(X-z);where X is a variable?

a) 27 b) 2626 c) 26 d) 2726

Answer : a) 27.

Solution :

We have to find the sums of the products in (X-a)(X-b)(X-c)….(X-y)(X-z);where X is a variable and a, b, c, d,..z are constants.

Consider;
(X-a)(X-b) = X2 – aX – bX + ab = X2 -(a+b)X + ab

Two binomials (or 2 factors) give 3 sum of product terms.

(X-a)(X-b)(X-c) = X3 – aX2 – bX2 -cX2 + abX + acX + bcX – abc
= X3 – (a+b+c)X2 + (ab+ac+bc)X – abc

Three binomials (or 3 factors) give 4 sum of product terms.

Here, 26 factors in (X-a)(X-b)(X-c)….(X-y)(X-z).

(X-a)(X-b)(X-c)…(X-y)(X-z) =

X26
– (sum of all products of constant terms taken 1 at a time)X25
+ (sum of all products of constant terms taken 2 at a time)X24
– (sum of all products of constant terms taken 3 at a time)X23
+ (sum of all products of constant terms taken 4 at a time)X22 … + or – …..
+ (sum of all products of constant terms taken 24 at a time)X2
– (sum of all products of constant terms taken 25 at a time)X
+ (product of constant terms taken 26 at a time)

Hence, the required number of terms is 27.

Question 3

Each alphabet a, b, c,… is a constant and the value of them are a=1,b=2,c=32,d=49,….. That is, b = 21, c = 3b, d = 4c,…, y = 25x, z = 26y; How many terms are in the sum of co-efficient of X in the expression (X-a)(X-b)(X-c)….(X-y)(X-z);where X is a variable?

a) 27 b) 2626 c) 26 d) 2726

Answer : c) 26.

Solution :

We have to find the number of terms in the sum of co-efficient of X in (X-a)(X-b)(X-c)….(X-y)(X-z);where X is a variable and a, b, c, d,..z are constants.

Consider,
(X-a)(X-b) = X2 – aX – bX + ab = X2 -(a+b)X + ab
Here, co-efficient of X is -(a+b).
Two binomials (or 2 factors) give 2 terms in the sum of co-efficient of X.

(X-a)(X-b)(X-c) = X3 – aX2 – bX2 -cX2 +abX +acX + bcX – abc
= X3 – (a+b+c)X2 +(ab+ac+bc)X – abc

Here, co-efficient of X is (ab+ac+bc) .
Three binomials (or 3 factors) give 3 terms in the sum of co-efficient of X.

Here, 26 factors in (X-a)(X-b)(X-c)….(X-y)(X-z).

(X-a)(X-b)(X-c)…(X-y)(X-z) =

X26
– (sum of all products of constant terms taken 1 at a time)X25
+ (sum of all products of constant terms taken 2 at a time)X24
– (sum of all products of constant terms taken 3 at a time)X23
+ (sum of all products of constant terms taken 4 at a time)X22 … + or – …..
+ (sum of all products of constant terms taken 24 at a time)X2
– (sum of all products of constant terms taken 25 at a time)X
+ (product of constant terms taken 26 at a time)

Here,co-efficient of X = -(sum of all products of constant terms taken 25 at a time).

i.e., 26 binomials (or 26 factors) give 26 terms in the sum of co-efficient of X.

Hence, the required number of terms is 26.

Probability Problems:

Question 1

In a bus stand, there are two services namely A and B. Every 10 minutes buses will leave from A and this service works from 6.10 am to 2 pm. The service at B starts at 2.20 pm and for every 20 minutes buses will leave from the bus stand. Find the probability of getting bus from service B between 2.20 pm to 2.50 pm, if service A is late by 1 hour.

a) 1/2 b) 1/3 c) 1/4 d) 1/5

Answer : b) 1/3

Solution :

Usually service A starts at 6.10 am.
Since the service is late by 1 hour, the first and last bus will leave by 7.10 am and 3 pm respectively.
Note that, there is a bus for every 10 minutes.
Number of buses leaving between 2.20 pm to 2.50 pm is 4 and the timings are 2.20 pm, 2.30pm, 2.40pm and 2.50pm.

There is a bus for every 20 minutes from service B.
Number of buses leaving between 2.20 pm to 2.50 pm is 2 and they start at 2.20 pm and 2.40 pm.
Therefore, 4 buses from A and 2 buses from B are available.
The probability of getting bus from B = buses from B / total number of buses from A and B.
= 2/6 = 1/3.

Question 2

From a railway station, trains leave for every 15 minutes and 25 minutes to city A and city B respectively. First train to city A and city B start at 9 am and 10.15 am respectively. If a man arrives to the station in between 11.25 am and 12.25 pm then the probability of getting train for city A is:

a) 1/4 b) 4/7 c) 3/5 d) 2/5

Answer : b) 4/7.

Solution :

The man wants to go to city A and he arrives station in between 11.25 am and 12.25 pm.
First train to city A is at 9 am and there is a train for every 15 minutes.
Trains for city A will leave at the following times : 9 am, 9.15 am, 9.30 am,…,11.30 am, 11.45 am, 12 pm, 12.15pm, and so on.
Number of trains for city A between 11.25 am and 12.25 pm is 4.

First train to city B is at 10.15 am and there is a train for every 25 minutes.
Trains for city B will leave at the following times: 10.15 am, 10.40 am, 11.05 am, 11.30 am, 11.55 am, 12.20 pm, and so on.
Number of trains for city B between 11.25 am and 12.25 pm is 3.
Probability of getting train for city A between 11.25 am and 12.25 pm = Number trains for city A from 11.25 am to 12.25 pm / Total number of trains for city A and B from 11.25 am to 12.25 pm
= 4/7.

Question 3

There are two bus stands, namely X and Y. Buses leave from X for every 30 minutes and its first bus starts at 8.05 am. Every hour number of buses leaving from Y increases by 1 and its first bus starts at 7 am. From Y there is only 1 bus for the 1st hour. Any bus from either of the bus stations takes 15 minutes to reach a nearby bus stop. Suppose a person reaches the stop in between 12.15 pm and 1.15 pm. The probability that the person will get a bus from Y is:

a) 3/4 b) 1/3 c) 1 d) 1/4

Answer : a) 3/4

Solution :

From bus stand X :
The first bus will leave by 8.05 am and reach the bus stop in 15 minutes, i.e. at 8.20 am
Second bus will leave after 30 minutes i.e. at 8.35 am and will reach the stop at 8.50 am
Therefore, buses will reach the stop at the following times: 8.20am, 8.50am, 9.20 am,…,12.20 pm, 12.50 pm, 1.20 pm and so on.
Between 12.15 pm and 1.15 pm, two buses will reach the stop at 12.20 pm and 1.20 pm.
Therefore, the person will get 2 buses from X.

From bus stand Y :
The first bus will leave by 7 am and reach the bus stop in 15 minutes, i.e. at 7.15 am.
There is only one bus for first 1 hour.
i.e., the second bus will leave after 8 am.
Note that, the number of buses leaving from Y is increased by 1 per hour.

From 8 am to 9 am, two buses will leave from Y and reach the stop between 8.15 am to 9.15 am.
And from 9 am to 10 am, 3 buses will leave from Y and reach the stop between 9.15 am to 10.15 am.
Proceeding like this, we have,
From 12 pm to 1 pm, 6 buses will leave from Y and reach the stop between 12.15 pm to 1.15 pm.

Therefore, the person will get 6 buses from Y between 12.15 pm to 1.15 pm.
Probability of getting bus from Y between 12.15 pm to 1.15 pm = Number buses from Y in between 12.15 pm to 1.15 pm / Total number of buses from X and Y in between 12.15 pm to 1.15pm = 6/(2+6) = 6/8 = 3/4.

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