Accenture Solved Placement Papers

Below are three probability based questions to practice for Accenture papers.

Question 1

A box contains 4 blue and 5 white balls and another box contains 5 blue and 4 white balls. One ball is to be drawn from either of the two boxes. What is the probability of drawing a blue ball?
a) 2/9 b) 1/9 c) 4/9 d) 1/2

Answer : d) 1/2

In Accenture papers, you can expect few questions from probability. These would generally of conventional nature and hence you can answer these with good practice.

Question 1

Accenture generally gives due importance to aptitude section.However, a good number of questions from aptitude section can be expected to be moderate and from sections like distances, averages, ratios etc. If you can answer these questions in quick time, you get enough time to answer more difficult ones in the same section as well as other sections like verbal .

Generally recruitment process of accenture would include a group discussion stage. Below are few points on the nature of topics you can expect. Also you can find tips to stay prepared for GDs. These tips not only apply for Accenture but can help your preparation to tackle GDs of other companies as well.

Below are four sample accenture type technical interview questions with answers. To be successful in accenture interviews, you need to stay well prepared in basic programming concepts including C. OOPS, OS related concepts etc.

Interview Question 1

What is the difference between strcmp and strncmp functions in C ?

Answer:

Here are some solved questions from Accenture  Placement Papers

1) 3 blocks are chosen randomly on a chessboard. What is the probability that they are in the same diagonal?

Answer
There are total of 64 blocks on a chessboard. So 3 blocks can be chosen 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 require probability is
= 112 / 41664
= 1 / 372
= 0.002688

2) 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 colinear. Hence, it is a straight line.
Hence area of triangle is 0.

3) Silu and Meenu were walking on the road.Silu said, "I weigh 51 Kgs. How much do you weigh?"Meenu replied that she wouldn't reveal her weight directly as she is overweight.
But she said, "I weigh 29 Kgs plus half of my weight."How much does Meenu weigh?

Answer
Meenu weighs 58 Kgs.
It is given that Meenu weighs 29 Kgs plus half of her own weight. It means that
29 Kgs is the other half. So she weighs 58 Kgs.
Solving mathematically, let's assume that her weight is X Kgs.
X = 29 + X/2
2*X = 58 + X
X = 58 Kgs

4) 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.

Here are some solved questions from Accenture Placement Papers

1)  Using two 2's and two 3's and using a maxm of three mathematical signs, symbols, can you have a result in between 14 and 15 ? concatanation ( clubbing of digits ) allowed.

Solution:
( 23 + 3! ) / 2 = 14.5

2) what is the 28383rd term in the series 1234567891011121314............
 
a) 3
b) 4
c) 7
d) 9

Solution:
there are 9 no. of single digit
there are 180 no. of double digit
there are 2700 no. of three digit
now total 2889 no. till 999
remaining no. are 25494 that is devided by 4 and the q is 6373 with reminder of 2 so 28381 is
6373+999=737(2)
n next no is 7(3)73
so ans is 3

3)  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
2,3,9,18
2,3,9,27,54
2,3,9,27,81,162

4) 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 = (729 * (7/9))=567ml
Water = (729-567)= 162ml
Let water to be added be x ml 567/(162+x) = 7/3 1701 = 1134 + 7x x = 81ml

Here are some solved questions from Accenture Placement Papers

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

Solution:
Let the number be x. Then X/7 - x/11 =100 11x-7x = 7700 x=1925.

2) 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.

3) The smallest number which when diminished by 3 is divisible by 21,28,36 and 45 is...
(i) 869
(ii) 859
(iii) 4320
(iv) 1263

Solution:
The required number = l.c.m. of (21,28,36 ,45)+3=1263

4) 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:
Since 653xy is divisible by 2 as well as by 5, so y = 0
Now 653x0 is divisible by 8 so 3x0 is also divisible by 8.
By hit and trial x=6 and x+y = 6

Here are some solved questions from Accenture  Placement Papers

1) What are identifiers and what is naming convention?

Ans : 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.

2) What is the return type of program’s main( ) method?

Ans : void

3)  What is the use of bin and lib in JDK?

Ans : Bin contains all tools such as javac, applet viewer, awt tool etc., whereas Lib contains all packages and variables.

4) The Java source code can be created in a Notepad editor.

a) True
b) False

Ans: True

Here are some solved questions from Accenture Placement Papers

1) The Java  interpreter is used for the execution of the source code.

True
False

Ans: True

2) What declarations are required for every Java application?

Ans: A class and the main( ) method declarations.

3) What are the two parts in executing a Java program and their purposes?

Ans: Two parts in executing a Java program are:
Java Compiler and Java Interpreter.
The Java Compiler is used for compilation and the Java Interpreter is used for execution of the application.

4) What are the three OOPs principles and define them?

Ans : 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.