Infosys Aptitude Test Previous Papers

Hy I am searching for the Infosys Aptitude Test Previous Papers so please can you provide me the papers?

[Arun Vats]

As you required Infosys Aptitude Test Previous Papers so here I Am providing you the same


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.

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.

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.

Question 1

In a dance school, there are 14 girls and 15 boys. In how many ways, a group consisting of 3 girls and 7 boys can be formed for a competition?

a) 2342340 b) 2232340 c) 2132340 d) 2132840

Answer : a) 2342340

Solution :

Now, use the combination method, nCr = n! / r! (n-r)!
3 are selected from 14 girls in 14C3 ways = 14! / 3! (14 - 3)! = 14! / 3!11! = 364 ways.
And 7 are selected from 15 boys in 15C7 ways = 15! / 7! (15 - 7)! = 15! / 7!8! = 6435 ways
Hence the required number of possible ways = 364 x 6435 = 2342340 ways.

Question 2

The players for district football team are selected from three different colleges and each college has 5 top players. Except captain and goalkeeper, each 3 are selected from each 3 colleges. Find the number of possible ways of such selection.

a) 1000 b) 1500 c) 4500 d) none of these

Answer : a) 1000

Solution :

Total number of players = 11
2 players(captain & goalkeeper)selected in 1 way.
Now, use the combination method, nCr = n! / r! (n-r)!
3 out of 5 can be selected in 5C3 ways = 5! / 3! (5 - 3)! = 10
Similarly, other 3 are selected from other two colleges.
Hence total number of possible ways = 1 x 10 x 10 x 10 = 1000

Question 3

In a college cricket club of 22 players, captain and wicketkeeper are selected from seniors team. For remaining players they have to select from 8 seniors and 12 juniors. Now, for a new team of 11 players except captain and wicketkeeper, out of 9 they select 4 from juniors and 5 from seniors. Find the number of methods to select players of new team.

a) 12220 b) 27720 c) 15870 d) none of these

Answer : b) 27720

Solution :

To form a new team of 11 players, they select 5 seniors out of 8 and 4 juniors out of 12.
Now, use the combination method, nCr = n! / r!(n-r)!
2 players (captain & wicketkeeper) are already selected = 1 possible way.
5 from seniors can be selected in 8C5 ways = 8! / 5! (8 - 5)! = 8! / 5!3! = 56 ways
4 from juniors can be selected in 12C4 ways = 12! / 4! (12 - 4)! = 495 ways
Hence the total number of possible ways = 56 x 495 x 1 = 27720