# Management Trainee Recruitment COAL INDIA 2020

https://www.coalindia.in/career/en-us/managementtrainee201920.aspx

posted Dec 19, 2019
18
Like
3
Love
0
Haha
0
Wow
0
Angry
0

@Shivateja MST No it is correct becoz atmost O(log n) comparisons are performed in BST

@s_dr_13 Then what about left or right skewed case?In that height becomes same as no of nodes.If they have mentioned as balanced binary tree then O(logn) would have been right.

What needs to be selected as Option ID while objecting incorrect answer keys?( whether the answer key given by CIL or the answer we think is correct), I tried asking CIL support but they said refer to advertisement but nothing is mentioned in it.

@Shivateja MST  yes bro.

In the question they havn't mentioned anything about worst case/bestcase/avg case

so it could be O(log n) or O(n). only one of the option matches with best case. so option will be C.

if they have given O(log n) and O(n) as two of the options. then it will have ambuiguity.

@rjking7403 Yaa that's also a point.Thank you.

I think this question has incorrect answer key as none of the option matches
@Shivat

I think seeing options u should have guessed it ( they are implicitly assuming the tree to be balanced,otherwise one of the options would have been O(n)  !!

anyhow,comparisons are atmost O(height of the tree)

It was a tricky question ....LOL

@s_dr_13 Yaa lets see what they gonna do.

anyway the answer O(logn) is wrong rest all three can be accepted since they are not asking the tightest upper bound ,so any upper bound will do.but i don't think they will correct it beacuse its not gate
option D is correct, because if f(n) = O(log  n) it means, $$f(n) \leq c. log(n)$$

also, $$f(n) \leq c_1. n^2, \\ f(n) \leq c_2. n. log(n),\\ f(n) \leq c_3. n^2. log(n)$$

we need to consider the highest upper bound, which is D.

source : wiki

so it should be O(logn) or O(n)

Generally, it is O(n) but we can also represent O(n) as O(n^2) or O(n^2 log n).

The given question can be converted to the following question.

The first positive number is less than or equal to ____________

a) 1

b) 2

c) 3

d) 4

@ all options are correct.
Searching an element in tree can't be converted  like above question. As we have  only n elements in tree it can not exceed O(n) to search an element in the tree how can you give time complexity as O(n^2logn).

According to your explanation , answer for the below one is  D as you choose highest upper bound.

 Q.90 The time complexity of searching an element in linked list of length n will be: Ans A. O(n log n) B. O(log n) C. O(n) D. O(n2)
Actually the problem is with the big O notation, if the options are given in theta notation then your answers will be correct.

what are the questions you guys are going to challenge?

my list:

1. The multiplicand register & multiplier register of a hardware circuit
implementing booth’s algorithm have (11101) & (1100). The result shall
be
A. (812)
B.  (-12)
C. (12)
D. (-812)

ans should be 12.

11101 = -3

1100 = -4

so -3 * -4 = 12

2.

 A ______ is the combination of one or more column values in a table that make a row of data unique within the table.

A.primary key

B.foreign key

C.candidate key

D.natural key

should be both a & d

3.

 In hashing, collision resolution is carried out by close addressing. Which of the following is close addressing technique – I. Buckets (for contiguous storage) II. Chains (for linked storage)

A. Only I
B. Only II
 C. I and II D. None

Buckets for continuous storage is like arrays so it will use linear probing (open addressing)

should be only II.

4.

Q.24 Generation of intermediate code based on an abstract machine model is useful in compilers because:
Ans A. Syntax translations are easier for intermediate code generation
B. It enhances the portability of the compiler system program
C. It is difficult to generate executable code from high level language program.
D. It makes implementation of lexical and syntax analysis easier

it should be option A. (previous gate question)

5.

A. 4 seconds
B. 5 seconds
 C. 6 seconds D. 7 seconds

it should be 6 seconds if Scheduling is done only at arrival or completion of process. so how can you pre empt the process while it is executing?

6.

 Q.50 Write true ( T ) / false ( F )for each of the following statements: A. Symbol table is used only in first phase of a compiler. B. A pretty printer analyses a program and prints it in such a way that the structure of the program becomes early visible. C. YACC build up LALR parsing table. D. If a grammar is LALR( 1 ), it is not necessarily SLR( 1) Ans A. T T F T B. T T T F C. F T T T D. F T T F

it should be 'c'.

7.

 Q.59 Solve the following recurrence relation T(n) = 4T(n/2) + n Ans A. O(n /2) B. Ο(n) C. O(log n) D. O(v2)

i think D option should be given in terms of 'n' not in terms of V. it is not there in the recurrence equation and will become constant.

8.

 Q.76 The following grammar is: S → Aa | b A c | dc | bda A → a Ans A. LALR(1) and SLR(1) B. Neither LALR(1) nor SLR(1) C. LALR(1) but not SLR(1) D. Not LALR(1) but SLR(1)

option should be A.

please tell me if i missed any question which should be challenged.

@rjking7403 There is one question which has options containing LALR(1) and SLR(1).For that the answer should be both LALR(1) and SLR(1).

And is there any additional fee for raising objection?

thre is  no additional fee for raising objections.

This one??

 Q.76 The following grammar is: S → Aa | b A c | dc | bda A → a Ans A. LALR(1) and SLR(1) B. Neither LALR(1) nor SLR(1) C. LALR(1) but not SLR(1) D. Not LALR(1) but SLR(1)

this is correct only.

refer this document..

http://people.cs.vt.edu/~ryder/515/f05/homework/hw1ans.pdf

@rjking7403 Bro just now i drew the states and checked it is SLR(1) and therefore LALR(1).

In the document they proved the Grammar is LALR(1) and not SLR(1) by taking A->d instead of A->a.

But in the question (in CIL as well as in document) they mentioned it as A->a.That makes the difference.

yes bro. i missed this. i choose A as option in exam then we can challenge this also.

It should be challenged.

any question apart from these 8?

Tq @ .

I think even CIL people might have just seen the example and framed the question without completely looking into it.

@rjking7403 can you just list those 8 in short so it will be easy to see at a place.

ok [email protected] MST

for Q24 option (a) is correct for intermediate code generation.

https://gateoverflow.in/1664/gate1998-1-27

the following is the sequence of insertion in a binary tree 45 65 35 40  33 70  60 75  691

1. LST=3 , RST =5

2. LST=6, RST =2

3 LST =5, RST =3

4 LST =2, RST =6

Answer given is option 1 , but answer should be option 3 , because option 1 is the answer of binary search tree but in question binary tree is given.

@Shatakshi_mishra
If it is not BST or heap we cant  tell exactly how many nodes in RST and LST.
it can be anything because normal Binary tree does not have any criteria like BST or heap while inserting nodes in tree so you can can insert  all noeds in either in LST or RST or distribute over LST and RST.
So It can be heap or BST. As per the given options it is tree is not heap so it must be unbalanced BST.
BST is a special type of tree-data structure  in Binary Tree.
So I think information hidden intentionally (didn't mention it as Binary Search Tree) same as GATE 2020 Aptitude Question P,R,T given asked Q+S.

So either LST=3,RST=5 should be the correct option or wrongly framed question.

let me know if i am wrong.

While submitting any objection in CIL MT objection link , when the given answer key is incorrect , what we have to fill in option id ?

1. Answer key given by CIL.

2. Answer key we think is correct.

selection nature of Objection:

Nature of Objection: Incorrect Answer Key

In Remarks:

Ex.

@rjking7403

But there is one more choice of option id , what to fill in that?
I think we need to fill it with the CIL Answer Key.
@rjking7403

okky 🙂

Bro. How many question you are going to challenge ?

This is my list.

Coal INDIA MT2020 Objections

am i provided enough explanation?

let me know if i miss any question to be challenged.

A person marks his goods 40% above the cost price. He sold 30% of goods at marked
price, 60% of the remaining at a discount of 20% and remaining at 40% discount. What
is his overall gain percent?

In this question, all options are wrong. I was getting 12.56% gain in the exam...
(100-(1.4*30+0.8*1.4*60+0.6*1.4*10))/100=17.6%
They said "60% of the remaining"

not 60% of original items, as u did.

the wording of the question are very vague
yes.. I missed that.. Have you challenged it?
How is the selection exactly done? They said interview is going to be of 10 marks only. So is the final merit list going to be generated out of 210 marks(200 for written and 10 for interview) or the written marks is going to get scaled down?
Coal India results were announced.
Can we check our marks of CBT, just like ISRO ? Or there is any other provision ?
They have just provided the list of shortlisted candidates.
Can we check our individual marks ?

How we can review our response sheet again ?
did the interview and final selection happen after that  for Systems?