X 15 Multiplication Tables (Blank) X 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 2 3 4: Fill & Download for Free

The syllabus for IBPS PO is huge and you will do well to take note of not just the topics in each section but also what weightage they have and what type of questions come into those topics. This information will be helpful for your preparations in the Preliminary examinations. Here you go:Quantitative Aptitude Syllabus(1) Simplification (5 questions)Types of Qs: BODMAS Rule, Approximation, Decimals, Fractions, Surds & Indices, Percentages etc.Easy to Solve, Less Time Consuming, High Accuracy Possible.(2) Average (1 question)Types of Qs: Average Weight/Height/Age/Marks, Average Money Expenditure, Average Temperature etc. But the concept remains simple. We have to use a single formula i.e. Average = Sum of Quantities/Number of QuantitiesEasy to Solve, Less Time Consuming, High Accuracy Possible.(3) Percentage (1–2 questions)Types of Qs: Calculation-oriented basic percentage.Easy to Solve, Less Time Consuming, High Accuracy Possible.(4) Ratio and Proportion (1–2 questions)Types of Qs: Simple Ratios, Compound Ratios, Componendo and Dividendo, Direct & Indirect Proportions.Easy to Solve, Less Time Consuming, High Accuracy Possible.(5) Data Interpretation (1–3 sets)Types of Qs: Line Graphs, Bar Graphs, Pie Charts, Tables, Miscellaneous Infographics, Missing Data Type Caselets.Easy to Moderate in Difficulty, Moderately Time Consuming, High Accuracy Possible(6) Mensuration (1–3 questions)Types of Qs: Area/Volume of Square, Trapezium, Parallelogram, Semicircle, Rectangle, Triangle, Cone, Rhombus, Circle, Cylinder, Cube, Cuboid. Also, Moulding from one shape to another etc.Easy to Moderate in Difficulty, Moderately Time Consuming, High Accuracy Possible.(7) Quadratic Equations (0–5 questions)Types of Qs: Comparing values of x and y when two equations given.Scoring, Easy if Proper Steps Followed, Less Time Consuming.(8) Interest (1–2 questions)Types of Qs: Simple Interest, Mixed Interest, Installments.Scoring, Easy if Formulas & Proper Steps Followed, Moderately Time Consuming.(9) Problems on Ages (0–1 questions)Types of Qs: Ages of family using algebra, ratio of ages, average of ages etc.Scoring, Easy if Proper Steps Followed, Less Time Consuming.(10) Profit and Loss (1–3 questions)Types of Qs: Successive Selling type questions, Dishonest Dealings, Partnerships, Discount & MP.Scoring, Easy if Proper Steps Followed, Moderately Time Consuming.(11) Number Series (5 questions)Types of Qs: Complete the Series, Find the Missing Term, Find the Wrong Term etc.Scoring, Moderate in Difficulty, Moderately Time Consuming(12) Speed, Time and Distance (1–3 questions)Types of Qs: Relative Speed, Average Speed, Problems on Train, Boat & Stream.Moderate in Difficulty, Moderately Time Consuming(13) Time and Work (1–2 questions)Types of Qs: Work Efficiency, Pipes & Cisterns, Work & Wages, etc.Moderate in Difficulty, Moderately Time Consuming(14) Number System (0–1 question)Types of Qs: Divisibility & Remainder, Integers, LCM & HCF, Multiples & Factors.Could be tricky, Moderately time Consuming(15) Data Sufficiency (0–5 questions)Types of Qs: Percentages, Ratio & Proportion, Profit & Loss, Simple & Compound Interest, Time Speed & Distance, Time & Work, DS questions based on Number System, Averages, Problems on Age, Mensuration, Mixtures & Alligations.Moderate in Difficulty, Moderately Time Consuming, Tricky(16) Linear Equations (0–1 question)Types of Qs: Basic linear equations with 1 or 2 variables.Could be tricky, Easy if Proper Steps Followed, Moderately Time Consuming(17) Permutation & Combination / Probability (0–1 question)Types of Qs: Basic Selection & Arrangement Problems, Very basic problems on Probability.Less Time Consuming, Scoring, Easy if all cases considered(18) Mixtures & Alligations (1 question)Types of Qs: Mixture of two or more entities/mixtures, dishonest dealings.Tricky, Time ConsumingReasoning Ability Syllabus(1) Coded Inequalities/ Mathematical Inequalities (5 questions)Types of Qs: Coded relations between letters to be decoded and solved.Scoring, Easy if Steps Followed properly, Moderately time consuming, High Accuracy possible.(2) Syllogisms (5 questions)Types of Qs: Direct questions with 2-3 statements & 2-3 conclusions, possibility based questions, negative information based questions.Easy to Moderate in difficulty, Less time consuming, Moderate Accuracy possible, Very likely to appear.(3) Coding & Decoding (5 questions)Types of Qs: Coding by Letter Shifting, Coding in Fictitious Language, Coding Letters of a Word, Coding by Analogy.Moderate in Difficulty, Moderate to Highly Time Consuming, Moderate Accuracy possible.(4) Circular Seating Arrangement (5 questions)Types of Qs: Uni- & Bi-directional problems on circular, square, rectangular, hexagonal tables. May be combined with other information to make it a Blood Relations or Double Lineup type problem as well.Moderate to Difficult, Moderate to Highly Time Consuming, Moderate to High Accuracy possible, Requires clean charting with proper technique and step-by-step solving.(5) Linear Seating Arrangement (5 questions)Types of Qs: Single or Double rows facing each other/ away from each other/ in the same direction. May be combined with other information to make it a Blood Relations or Double Lineup type problem as well.Moderate to Difficult, Could be tricky, Moderate to Highly Time Consuming, Moderate to High Accuracy possible, Requires clean charting with proper technique and step-by-step solving.(6) Blood Relations (2–4 questions)Types of Qs: Coded Blood Relations, Family Tree Problems.Moderate in difficulty, Low to Moderately time consuming, Moderate to High Accuracy possible, Requires clean charting with proper technique and step-by-step solving.(7) Directions & Distances (2–3 questions)Types of Qs: Arranging by age, height, weight, rank, order in row etc.Easy but could be tricky, Less time consuming, High Accuracy possible, Requires clean charting with proper technique and step-by-step solving.(8) Ordering & Ranking (2–5 questions)Types of Qs: Floor Based Puzzles, Arranging by age, height, weight, rank, order in row etc.Easy to Moderate in difficulty, Less time consuming, High Accuracy possible.(9) Arrangement & Pattern (0–5 questions)Types of Qs: Random Sequence of Alphabets (based on group of 5 terms each of 3 alphabets OR series of alphabets), Number Arrangements (based on group of 5 terms each of 3 digits OR series of numbers), Mixed Series Dictionary or Alphabet based Arrangements etc.Easy to Moderate in difficulty, Less time consuming, High Accuracy possible.(10) Double Lineup (0–5 questions)Moderate to Difficult, Could be tricky, Moderate to Highly Time Consuming, Moderate Accuracy possible, Requires clean charting with proper technique and step-by-step solving.(11) Scheduling (0–5 questions)Moderate to Difficult, Could be tricky, Moderate to Highly Time Consuming, Moderate Accuracy possible, Requires clean charting with proper technique and step-by-step solving.(12) Analogy (0–2 questions)Types of Qs: Meaning based, Letter based, Number based.Easy to Moderate, Quick Solve, Moderate to High Accuracy possible.(13) Classification (0–3 questions)Types of Qs: Letter based, Meaningful Words based, GK based, Number based.Easy to Moderate, Quick Solve, High Accuracy possible.(14) Data Sufficiency (0–3 questions)Types of Qs: Ordering & Ranking, Blood Relations, Coding in Fictitious Language, Circular Arrangements, Directions & Distances, Ages & Birthdates.Moderate Difficulty, Moderately Time Consuming, Moderate Accuracy possible.(15) Series (0–2 questions)Types of Qs: Complete the Alphabet Series, Number Series and Mixed Series.Easy to Moderate, Less Time Consuming, Moderate to High AccuracyEnglish Language Syllabus(1) Reading Comprehension (7–10 questions)Types of Qs: Fact and Inference based questions – easy to moderate, Vocab & Theme based questions – moderate to tough. The passage will be focussed on topics like Banking & Economy, Business, Social Issues etc.Moderate to More Time Consuming, Moderate Accuracy Possible.(2) Grammar (5–10 questions)Types of Qs: Error Spotting, Phrase Replacement/ Sentence Correction, Fill in the Blanks.Sub-Topics: Tenses, Articles, Prepositions, Subject Verb Agreement, Parallelism, Pronouns, Adverbs, Non-Finite Verbs, Common Errors Degrees of Comparison, Sentence Construction, Phrasal Verbs etc.Easy to Moderate in Difficulty, Less to Moderately Time Consuming, Moderate Accuracy Possible.(3) Vocabulary (7–10 questions)Types of Qs: Error Spotting, Phrase Replacement, Cloze Test (single passage with 5-10 questions), Fill in the Blanks (Double Blanks).Easy to Moderate Difficulty, Quick Solve based on Contextual usage, Moderate to High Accuracy possible.(4) Verbal Ability (0–5 questions)Types of Qs: Para JumblesEasy, Low to Moderately Time Consuming, High Accuracy possible.You can find important tips on each of these topics and take quizzes and tests for all the questions on Testbook. Just click on this detailed syllabus article and you will be provided a huge amount of online resource guide.IBPS PO - Detailed SyllabusAll the best for your preparations!

A2A: I found this to be a very interesting question. It admits more than one interpretation. However, for any interpretation I have come up with, my answer differs from the two which had been posted when I posted this answer.I prefer to think of such matters in terms of count rate. So, in this situation, we are talking about rational count rates [math]R = \frac{10}{11x}[/math] per second, for integers [math]x[/math][math][/math] in [math][1,9].[/math] Let us index our count observations, [math]c_t,[/math] by integer times [math]t[/math]. The values we can observe are given by [math]c_t = \lfloor p [/math][math][/math][math]+ tR \rfloor,[/math] where [math]p[/math] is a real number which relates to the phase relationship between the ticker clock (rate [math]R[/math]) and the observation clock (rate [math]1[/math][math][/math]). Without loss of generality, we can simplify this a bit. Let us assume that we make our first observation a [math]t=0.[/math] Adding a constant to all observed counts does not change the information they provide, so we can assume that [math]c_0=0.[/math] Now we can be assured that [math]p\in [0,1).[/math]For a given [math]x[/math][math],[/math] if [math]10t[/math] is a multiple of [math]11x,[/math] then [math]tR[/math] is an integer and [math]c_t[/math] can only be [math]tR.[/math] Otherwise, [math]c_t[/math] can be either [math]\lfloor tR \rfloor[/math] or [math]1[/math][math][/math] more than that.Now, as for the question of how many times one would need to check the value of [math]c_t,[/math] that is just twice! We have already assumed that we have checked [math]c_0[/math] (to make the adjustment so that it is [math]0[/math]). We only have to check one other value. E.g. for [math]t=2772,[/math] which is the LCM of all the integers from [math]1[/math][math][/math] through [math]11[/math][math][/math] with the [math]10[/math][math][/math] which is already in the numerator divided out, then [math]tR[/math] is guaranteed to be an integer. Thus [math]c_t = tR[/math] and the average rate [math]c_t/t[/math] over [math]t[/math] seconds is exactly [math]\frac{10}{11x}[/math] from which it follows that [math]x[/math][math] =\frac{10t}{11c_t}.[/math]Maybe you don’t want to wait 46 minutes. So we can ask how long we would have to wait so that [math]x[/math][math][/math] is determined unambiguously with just two looks. Since distinguishing between [math]x=8[/math] and [math]x=9[/math] is the hardest problem, it suffices for [math]t[/math] to be large enough that the larger possible value of [math]c_t[/math] for [math]x=8[/math] is guaranteed to be smaller than the smaller possible value of [math]c_t[/math] for [math]x=9.[/math] That happens when [math]t=141.[/math] With that [math]t,[/math] for [math]x=8 [/math][math][/math][math][/math], [math]c_t[/math] can only be [math]16[/math] or [math]17;[/math] and, for [math]x= [/math][math]9[/math][math], c_t[/math] can only be [math]14[/math][math][/math] or [math]15[/math][math].[/math] (If you check it for [math]t=140,[/math] you will see that you can get [math]c_t=15[/math] for either [math]x[/math][math].)[/math] So with [math]t =141, x=9[/math] and [math]x=8[/math] are distinguishable. So we got it down to only [math]141[/math] seconds after the first count to get [math]x[/math][math][/math] unambiguously with only [math]2[/math][math][/math] looks.Another interpretation of the question would seek the smallest [math]t[/math] for which the rate can be determined from [math]\{c_i\ [/math][math][/math][math]|\ i \in [0,t]\}.[/math] For this purpose, I like to go to a run-length representation of the sequence [math]c_0, c_1, … [/math][math][/math][math], c_t.[/math] [math](l_0, l_1, …, l_m)[/math] represents a sequence with [math]l_i[/math] consecutive count values of [math]i[/math] for [math]i \in [0,m].[/math] E.g., the first value of [math]j[/math] for which [math]c_j=k[/math] is [math]\sum_{i<k}l_i.[/math] For any run length [math]l[/math] you might observe, [math]l-1[/math] is a lower bound on [math]x[/math][math];[/math] so a length of [math]10[/math][math][/math] guarantees [math]x=9.[/math] The first and last runs do not generally provide as much information since we don’t necessarily know what could have preceded or what might follow. I will refer to the other run lengths as being delimited. I.e., we know exactly where in the sequence each such run starts and ends. A delimited run length must be either [math]x[/math][math][/math] or [math]x+1.[/math] Indeed, there will be [math]10[/math][math][/math] consecutive runs of length [math]x[/math][math][/math] separating any pair of length [math]x+1.[/math] Thus a delimited run length is an upper bound on [math]x+1.[/math] Once we have seen at least two consecutive delimited run lengths, we know [math]x[/math][math].[/math] If they are the same length, that length is [math]x[/math][math].[/math] If they have different lengths, [math]x[/math][math][/math] is the smaller of those lengths.What we seek now is the maximum length of a count sequence for which it is not possible to infer [math]x[/math][math].[/math] There is only one such sequence and that is the [math]25[/math] count sequence [math](8,9,8).[/math] [math]x[/math][math][/math] could be either [math]8[/math][math][/math] or [math]9[/math][math].[/math] If [math]c_{25} = [/math][math]2[/math][math],[/math] we go to [math]([/math][math]8[/math][math], [/math][math]9[/math][math], [/math][math]9[/math][math]),[/math] which can only be satisfied by [math]x=9.[/math] (If we were to continue, [math]l_2[/math] could only be [math]9[/math][math][/math] or [math]10[/math][math][/math] since [math]x[/math][math][/math] cannot exceed [math]9[/math][math].)[/math] If [math]c_{25}=3,[/math] we go to [math](8,9,8,1)[/math] with two delimited run lengths confirming [math]x=8.[/math] So we need to be able to look at (some of) at least [math]26[/math] consecutive counts in order to be sure of determining [math]x[/math][math].[/math] (In case you might be wondering why you cannot get [math](9,9)[/math] if [math]x=8:[/math] The time between the first and last of [math]3[/math][math][/math] consecutive ticks would be [math]2\cdot11\cdot8/10=17.6[/math] seconds. So there would have to be at least [math]2[/math][math][/math] ticks in any [math]18[/math] second interval, not just [math]1[/math][math].)[/math]Now we can ask, “Given [math]26[/math] consecutive counts of the sequence which are available for examination, how many of the values do we actually have to look at in order to determine [math]x[/math][math][/math]?” You clearly don’t have to look at all of them since they tend to be redundant. E.g., if [math]c_{i+k} = c_i+k,[/math] then we know without looking that [math]c_{i+j} = c_i+j[/math] for all [math]j \in [1,k).[/math] Similarly, if [math]c_{i+k} = c_i,[/math] we know [math]c_{i+j} = c_i[/math] for [math]j \in [1,k).[/math]I can’t give a definitive answer on this last question because I can imagine a number of algorithms to go about it and I have not gone far enough to find a guaranteed best one. Indeed, this looks like a hard problem to me. One obvious approach is that we’d like to try to determine [math]2[/math][math][/math] delimited run lengths, so we need to find the indices of run-starts. I.e., determine [math]l_0, l_1,[/math] and [math]l_2.[/math] My approach would be sort of a heuristic algorithm which I will not describe in great detail. However, I’ll offer my thinking after what follows.Given time indices [math]i \lt j,[/math] let [math]D=c_j-c_i[/math] and [math]d = j - i.[/math] I refer to such a pair of [math]D[/math] and [math]d[/math] values as a “Dd pair”. Given such a Dd pair there is a restriction on the values of [math]x[/math][math][/math] which can produce such a pair. I wrote a little Python script to enumerate those restrictions:def can_get(x, d, D): # Return True iff can get c[d]==D with x.  tRi, r = divmod(10*d, 11*x) # integer part of tR and remainder from div   return D==tRi if r==0 else (D==tRi or D==(tRi+1))  column_heads = '\n\n d D->' + ''.join([str(j).rjust(3) for j in range(24)]) print column_heads  for d in range(1, 26):  print '\n' + str(d).rjust(2) + 6*' ',  for D in range(d+1):   xs = [x for x in range(1, 10) if can_get(x, d, D)]  print ' ' if len(xs)==0 else str(min(xs))+str(max(xs)),  print column_heads I call its output the Dd Chart: d D-> 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23   1 19 19   2 29 19 11   3 39 29 12 11   4 49 29 23 11 11   5 59 39 24 22 11 11   6 69 39 25 22 11 11   7 79 49 36 23 22 11 11   8 89 49 37 23 22 11 11   9 99 59 38 34 22 22 11 11  10 59 49 34 23 22 11 11  11 69 49 34 33 22 11  12 69 49 35 33 22 22 11 11  13 69 49 35 33 22 22 11 11  14 79 59 46 34 33 22 22 11 11  15 79 59 46 34 33 22 22 11 11  16 89 59 47 34 33 22 22 11 11  17 89 69 47 45 33 33 22 22 11 11  18 99 69 58 45 34 33 22 22 11 11  19 99 69 58 45 34 33 22 22 11 11  20 79 59 46 44 33 33 22 22 11 11  21 79 59 46 44 33 33 22 22 11 11  22 79 69 56 44 33 33 22 11  23 79 69 56 45 34 33 22 22 11 11  24 89 69 57 45 44 33 33 22 22 11 11  25 89 69 57 45 44 33 33 22 22 11 11    d D-> 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 The 2 digit pair in column [math]D[/math] for row [math]d[/math] is the min/max pair for the values of [math]x[/math][math][/math] which the Dd pair admits. Impossible pairs result in blank fields.When you have seen [math]n[/math] counts, you have [math]n(n-1)/2[/math] possible index pairs for [math]d,[/math] many of which will lead to different values of [math]D[/math] and a new restriction on [math]x[/math][math].[/math] I thought it was promising to think that there would be a good strategy for picking indices to look at the count so that intersecting the intervals from all the resulting Dd pairs would uniquely determine [math]x[/math][math].[/math] I was not very successful in finding such an approach. The toughest problem lies in distinguishing [math]x=8[/math] and [math]x=9.[/math] But you may see that there are very few of the [math]x[/math][math][/math] intervals which include one and exclude the other. Furthermore, even if you use a [math]d[/math] for which there is such a [math]D,[/math] there is no guarantee that you will get that [math]D.[/math] The more I thought about it, it seemed to come down to finding the run-starts to delimit runs.Nevertheless, that table is helpful for getting started with a more heuristic sort of algorithmic approach. In particular, after the first look at [math]c_0,[/math] make the second look at [math]c_{25}.[/math] Looking at row [math]25[/math] the Dd Chart, you can see that any [math]D[/math] (aka [math]c_{25}[/math] here) bigger than 5 actually resolves [math]x[/math][math][/math] immediately. In fact, it will do so for any of [math]1[/math][math], [/math][math]2[/math][math],[/math] or [math]3[/math][math][/math] for [math]x[/math][math][/math] and may do so for [math]4[/math][math].[/math] Furthermore, if you get [math]45[/math] (for [math]D = [/math][math]5[/math][math][/math]), checking [math]c_{22}[/math] will resolve that in one more look. So the interesting cases after [math]2[/math][math][/math] looks are those for [math]D[/math] being [math]2[/math][math], [/math][math]3[/math][math],[/math] or [math]4[/math][math].[/math] We are armed with a range of possible values for [math]x[/math][math][/math] and we seek run-starts. We want to find the index of the first [math]1[/math][math].[/math] I’d start by using an index which is one more than about half the mid-point of the [math]x[/math][math][/math] range. You can proceed in a binary search style. Each of these looks at small indices, relative to [math]c_{25},[/math] will generate new large [math]d[/math] restrictions which may useful in further limiting the possible range for [math]x[/math][math].[/math] I expect [math]3[/math][math][/math] more looks will suffice to determine [math]l_1.[/math] We're up to [math]5[/math][math][/math] looks. By now, we should have a fairly tightly controlled range on [math]x[/math][math][/math] for how far forward to look to find the next run-start. I’d expect [math]3[/math][math][/math] more looks to suffice. We’re up to [math]8[/math][math].[/math] Now no more than [math]2[/math][math][/math] more looks are needed to find the 3rd run-start. We know a delimited run length. If it has length [math]k[/math] and starts at [math]j,[/math] we examine [math]c_{j+2k}.[/math] If that is [math]c_{j+k}+1[/math] then [math]x=k-1,[/math] otherwise [math]x[/math][math][/math] is [math]k[/math] or [math]k+1[/math] and we need one more look unless the previous established length is already known to be the max for [math]x[/math][math][/math] based on Dd look-ups (not unusual). So I don’t know it for sure, but I think you can determine [math]x[/math][math][/math] with no more than [math]10[/math][math][/math] looks. Perhaps someone else might wish take up the challenge of designing a complete algorithm which they can prove to be optimal.I decided to actually play the game myself. So I wrote a script to choose a random [math]p[/math] and a random [math]x[/math][math][/math] and generate a 26-count sequence. I did it for 8 sequences. Playing along the lines suggested above, the number of looks it took for each such sequence were as follows: 8 7 2 10 2 7 10 9. This suggests that my guesses are not too far off. The two 2-look solutions are not surprising. What was unusual in my brief experience was that, on two occasions, [math]l_1[/math] was [math]x+1,[/math] saving one look. (If anyone is curious, I can post the Python script which generated the sequences and the commented console dialogues for my play.)

CCA175 : Cloudera Hadoop and Spark Developer certifications tips :-1. Preparation: CCA175 Questions link and practice the code provided by http://www.HadoopExam.com . ( Gone through all the Spark Professional training module as well)2. No. Of Questions: Generally you will get 10 questions in real exam: Topic will be coverings are Sqoop, Hive, Pyspark and Scala and avro-tools to extract schema (All questions are covered in CCA175 Certification Simulator).3. Code Snippets: will be provided for Pyspark and Scala. You have to edit the snippets accordingly as per the problem statement.4. Real Exam Environment: Gateway node will be accessible for execution of the problems during the exam. Keep in mind there will not be any on-screen timer available during the exam. You have to keep asking for the time left. There are three sections for each problem i.e.InstructionsData SetOutput Requirements.Please go through all the three sections carefully before start developing the code.Note: If you started developing code right after looking at the Instruction part of the question, then later you will realized the exact details of the table like name of the table and HDFS directory are also mentioned. This can waste your time if have to redo the code or might as well cost you a question.5. Editor: nano, gedit are not available. So if you have to edit any code snippets, you have to use vi alone. Please make yourself familiar with vi editor if you are not.6. Fill in blanks: You dont have to write entire code for Python and Scala for Apache Spark, generally they will ask you to do fill in the blanks.7. Flume: Very few questions on flume.8. Difficulty Level: If you have enough knowledge, you will feel exam is quite easy. The questions were logically easy and can be answered in the first attempt if you read the question carefully (all three sections).9. Common mistake in Sqoop: People use connector as localhost which is wrong, you have to use full name instead of localhost (Avoid wasting your time). Use given hostname10. Hive: Have initial knowledge of hive as well.11. Spark: Using basic transform functions to get desired output. For instance filter according particular scenario, sorting and ranking etc.12. Avro-tool : avro-tool to get schema of avro file. (Very nicely covered in CCA175 HadoopExam.com Simulator)13. Big Mistake: Avoid accidently deleting your data: good practice is necessary to avoid such mistakes. (Once you delete or drop hive table, you have to create it entirely once again.) Same is instructed by www.HadoopExam.com during their videos session provided at http://cca175cloudera.training4exam.com/ (Please go through sample sessions)14. Spark-sql: They will not ask questions based on Spark Sql learn importantly aggregate, reduce, sort.15. Time management: It is very important, (That’s the reason you need too much practice, use CCA175 simulator to practice all the questions at least a week or two before your real exam).16. Data sets in real exam is quite larger, hence it will take 2 to 5 mins for execution.17. Attempts: try to attempt all questions at least 9/10, hence you must be able to score 70%.18. File format: In most of questions there was tab delimited file to process.19. Python or Scala: You will get a preloaded python or scala file to work with, so you don't have(Now you can choose) a choice whether you want to attempt a question via scala or pyspark. (I have gone through all the Video sessions provided by www.HadoopExam.com here20. Connection Issue: If you got disconnected during exam, you may need to contact the proctor immediately. If he/she is not available log back into Schedule an Exam and use their online help.21. Shell scripts: Have good experience to use shell scripts.22. Question types as mentioned in syllabus : Questions were from Sqoop(import and export), Hive(table creation and dynamic partitioning), Pyspark and Scala(Joining, sorting and filtering data), avro-tools. Snippets of code will be provided for Pyspark and Scala. You have to edit the snippets accordingly as per the problem statement and can the script file(which is another file apart from snippet) to get the results.23. Overall exam is easy, but require lot of practice to complete on time and for accurate solutions of the problem. Hence go through the all below material for CCA175 (It will not take more than a month, if you are new and already know the Spark and Hadoop then 2-3 weeks are good enough.· CCA175 : Hadoop and Spark Developer Certification practice questions· Hadoop professional training· Spark professional training.Features of the CCA 175 (Cloudera Hadoop and Spark Developer certification)1. Entire syllabus will be covered.2. All questions are scenario based and step by step solutions will be given3. Same will be executed by our exper technical team and complimentary selected recorded videos will be shared here.4. Almost all scenarios will be covered for real exams.5. Any future updates will be free for lifetime on single machine.6. Solutions are already executed on Cloudera CDH, hence same can be used for real exam.7. Our expert regularly update the simulator.8. It will help you gain confidence and reduce study time.9. Always updated and correct/incorrect way of solutions explanationImportant Notes:-The pattern, you should follow it.1. Every question in hands-on, no objective2. Every problem we need write the code & execute on their cdh5 cluster must.3. Result must & should be their expected form, if not 0 scoreexamples:expected output x,y => you got output x, y or x , ythey will not consider as a correct answer also.such a minor points also you have to remember.In code also very carefully we need to write it. other wise they will make it as a wrong answer only.simple example for that code need to written in single line in spark scala or python code but readability purpose you might return in multiple lines then the answer will treat as a wrongval wordcount = file.flatMap(line => line.split(" ").map( word => (word, 1))We can above statement like this also.val words = file.flatMap(line => line.split(" ")val wordcount = words.map( word => (word, 1))4. Please read the questions clearly .. becausesome questions they will give partial solution file in specific location on cdh5 vm , we need to give the correct solution on that file onlysome questions we need to read data from hdfssome questions we need to read data from local file systemsome questions we need to read data from RDBMSsome questions we need to read data from HDFS in different file formats also using spark with scala or python or hive or sqoopLike avro, parquet, jsonfew questions they given partial information only.I got 1 question like that.I almost waste my time 10 mins there only. finally solved it by deeper observation in data.5. Might face some issues in writing examfont is very low, difficult to read the questions & giving the answers alsocluster is very slow, don't do multiple tasks it hangssame issue i faced, almost it is taken for me 2-3 mins to come up again. like 2 -3 times happen also.it may kill your time around 5-10 mins also***Remember locations of normal cdh5 software's and this exam clustercdh5 will be differ as i observed.Because of this 1 Question is very difficult to answer from my side. taken more time to solve.I used my previous experience on Big Data knowledge to solve that problem. Other wise i may lose the 1 Question answer.Note:- This answer has taken from google just to help people.