Apptitude For IT Job

Apptitude For IT Job – The placement session for any company generally has 3 – 4 rounds. The first round is the written test, consisting of quantitative aptitude, reasoning (logical ability), English (verbal ability) and computer science subjects (C, C++, Data Structures, Algorithms, DBMS, OS, Networks, etc.). For some companies, you might also have a coding round, which is sometimes taken as a separate round and sometimes is clubbed with the written test.

Apptitude For IT Job

Q 1.2/3rd of the balls in a bag are blue, the rest are pink. if 5/9th of the blue balls and 7/8th of the pink balls are defective, find the total number of balls in the bag given that the number of non defective balls is 146?

a) 216

b) 649

c) 432

d) 578

Solution- let total no of balls =x



total no of defective balls = 10x/27 +7x/24


non defective balls=x-143x/216=146


Q 2. Find no of ways in which 4 particular persons a,b,c,d and 6 more persons can stand in a queue so that A always stand before B. B always stand before C, And C always stand before D.





Solution – a,b,c,d are grouped ie consider them as one and remaining as 6. total 6+1 = 7! Ways

Q 3. 100 students appeared for two different examinations 60 passed the first,50 the second and 30 both the examinations.Find the probability that a student selected at random failed in both the examination








so B is the answer.

Q 4. There are 10 points on a straight line AB and 8 on another straight line AC none of them being point A. how many triangles can be formed with these points as vertices?

a. 680

b. 720

c. 816

d. 640

Solution-To form a triangle we need 3 points select 2 points from the 10 points of line AB & 1 from the 8 on AC

= (10C2)*(8C1)

select 2 points from the 8 points of line AC & 1 from the 10 on AB=


total no. of triangles = (10C2)*(8C1)+ (8C2)*(10C1) = 640

Apptitude For IT Job

Q 5. From a bag containing 8 green and 5 red balls,three are drawn one after the other .the probability of all three balls beings green if the balls drawn are replace before the next ball pick and the balls drawn are not replaced , are respectively.


b) 512/2197,336/1716

c) 336/2197,512/2197

d) 336/1716,512/1716





Q 6. find the greatest number that will divide 148 246 and 623 leaving remainders 4 6 and 11 respectively.

a) 20 b) 12 c)6 d)48

Solution-Hcf ( (148-4), (246-6), (623-11))=12

Q 7. a mother her little daughter and her just born infant boy together stood on a weighing machine which shows much does the daughter weigh if the mother weighs 46kg more than the combined weight of daughter and the infant and the infant weighs 60% less than the daughter.



c)cannot be determined


Solution-daughter weight is x

infant weight is 60% less than daughter i.e., 0.4x

mother weight is (x+0.4x+46)

total weight = x + 0.4x + (x+0.4X+46) = 74

solving the eq. x = 10

option d is correct

Q 8. find the number of ways a batsman can score a double century only in terms of 4’s & 6’s?





Solution-4’s and 6’s

50 0

47 2

44 4

41 638 8

35 10

32 12

29 14

26 16

23 18

20 20

17 22

14 24

11 26

8 28

5 30

2 32

So total 17 ways but here it is 4’s & 6’s both so don’t consider 1st one

Final ans : 16 ways

Q 9. 98. Thomas takes 7 days to paint a house completely whereas Raj would require 9 days to paint the same house completely. How many days will take to paint the house if both them work together. (Give answers to the nearest integer)?

A. 4

B. 2

C. 5

D. 3

Solution-work done by thomas in day=1/7

work done by other in a day=1/9work done by both in a day=1/7 +1/9 =16/63

days required by both = 63/16


Apptitude For IT Job

Q 10. how many positive integers less than 4300 of digits 0-4.

a) 560 b)565 c)575 d)625

Solution-Ans is 575

one digit no =4 (0 is not a positive integer)

two digit no=4*5=20

three digit no=4*5*5=100

four digit no=3*5*5*5=375(the possibility for 1,2,3 will come in the first

position)four digit no=1*3*5*5(the possibility of 4is fixed in the first position and

then 0,1,2 is comes in second position)and the last digit is 4300 we include this

number also

Ans is 4+20+100+375+75+1=575

Q 11. A person travels from Chennai to Pondicherry in cycle at 7.5 Kmph. Another person travels the same distance in train at a speed of 30 Kmph and reached 30 mins earlier. Find the distance.

A)5 Km

B)10 Km



Solution-Let, time taken by 🚲 =t

//ly, time taken by train 🚉 =t-30

We know that…..speed=distance /time

Speed of bicycle 🚲, 7.5=d/t….(1)Speed of 🚉, 30=d/(t-30/60)….(2)

Sol 1&2,we get t=0.666

By sub and value in equal(1)

We d=4.999~5km

Q 12. A bag contains 8 white balls, and 3 blue balls. Another bag contains 7 white, and 4 blue balls. What is the probability of getting blue ball?

a)3/7 b)7/22 c)7/25 d)7/15

Solution-First we have to select a bag and then we will draw a ball.

Probability of selection of both bags is equal =1/2

Now probability of blue ball taken from first bag = ( 1/2) x (3/11)

and probability of blue ball taken from second bag = (1/2) x (4/11)

So probability of blue ball = ( 1/2) x (3/11) + (1/2) x (4/11) = 7/22

Q 13. In a 3*3 square grid comprising 9 tiles each tile can be painted in red or blue color. When the tile is rotated by 180 degree, there is no difference which can be spotted. How many such possibilities are there?

a) 16 b)32 c)64 d)256

Solution-ans is 32 .

ans grid has to be rotated at 180 degrees.









cell 11-33 can be altered in 2 ways (as thr are 2 colours)

cell 12-32 can be altered in 2 ways

cell 13-31 can be altered in 2 ways

cell 21-23 can be altered in 2 ways

and , cell 22 can be altered in 2 ways

, so

2X2X2X2X2 = 32.

Q 14. Jake can dig a well in 16 days. Paul can dig the same well in 24 days. Jake, Paul and Hari together dig the well in 8 days. Hari alone can dig the well in





Solution-1/x = 1/8 – (1/16 + 1/24)

so x=48

ans = 48 days

Q 15. For any two numbers we define an operation $ yielding another number, X $ Y such that following condition holds:

• X $ X = 0 for all X

• X $ (Y $ Z) = X $ Y + Z

Find the Value of 2012 $ 0 + 2012 $ 1912

a) 2112

b) 100

c) 5936

d) Can not be determined

Solution-here $ represent the – operator so X-X=0 first condition


it can be written as x$y+z

2012-0+2012-1912=2112 so ans is 2112

Apptitude For IT Job

Q 16. On a toss of two dice, A throws a total of 5. Then the probability that he will throw another 5 before he throws 7 is

a)40% b) 45% c)50% d)60%

Solution-ans: 40%


total probabilities for getting 5 = 4/36

total probabilities for getting 7 = 6/36

Total Probability = 10/36

We need only 5, hence prob of getting only 5 is (4/36)/(10/36)


Q 17. 1,2,2,3,3,3,4,4,4,4,1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4,………………………………….. Then what is the 2320 position of the number in the sequence?

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

Solution-answer is b)1

1,2,3,4(1-1time 2- 2times 3-3 4-4)=10 terms /completes cycle and starts from 1

1,2,3,4(1-2 2-4 3-6 4-8)= 20 terms /completes cycle and starts from 1

(1,2,3,4)each digit 3 time to its value =30 terms/completes cycle and starts from 1


x is nearer value to 2320 solving n(n+1)/2

10.(21.22)/2= 2310

analysing it 2310 completes cycle and starts from 1 again

now it 22 times

(1-22 times 2-44 times ……)

2320 position will occupied by 1

Q 18. In 2003, there are 28 days in February and there are 365 days in the year. In 2004, there are 29 days in February and there are 366 days in the year. If the date March 11, 2003 is a Tuesday, then which one of the following would be the date March 11, 2004 be?

A. Wednesday

B. Tuesday

C. Thursday

D. Monday

Solution-Every year day is increased by 1 odd day. Or in leap year it is increased by 2 odd days.

so 11 march 2003 is tuesday,

11 march 2004 is thursday

C. Thursday

Q 19. How many 6 digit even numbers can be formed from digits 1 2 3 4 5 6 7 so that the digit should not repeat and the second last digit is even?

a)6480 b)320 c)2160 d)720

solution-given 6th digit even number , so last digit 2 or 4 or 6-> 3 ways

” 5th digit should be even…so there will be 2 ways(rep. not allowed)

so,therefore we get 5*4*3*2*2*3=720 ways

Q 20. There are 5 letters and 5 addressed envelopes. If the letters are put at random in the envelops, the probability that all the letters may be placed in wrongly addressed envelopes is.

a)119 b)44 c)53 d)40

Solution-If there is one letter and one envelope then no way you can put it


If there are 2 letters and 2 envelopes then you can put them wrong in 1 way(S2).

If there are 3 letters and 3 envelopes then you can put them wrong in 2 ways(S3).

If there are 4 then you can put them wrong in 9 ways(S4).

If there are 5 then you can put them wrong in 44 ways(S5).

If you observe you can find a pattern.




In general, Sn=(Sn-2 + Sn-1)*(n-1)

So, if there are 5 letters then S5=(S3+S4)*4=(2+9)*4=44

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