Series Completion – Sequences, Types of Series, Concepts & Simple Tricks

Series Completion

Are you facing the problem with Series Completion? Dont worry we have provided Concept, Simple Tricks, Practice Questions on Series Completion to help you out. Learning with different examples of Series Completion helps to remember the concept. We have gathered Series Completion Questions from different levels that are useful for the candidates who are preparing for the competitive & placement exams.

Series Completion – Verbal Reasoning Questions with Answers

1. Choose the correct alternative that will the same pattern and fill in the blank spaces.: 1, 4, 9, 16, 25, x

  • A. 35
  • B. 36
  • C. 48
  • D. 49

Answer: B (36)

2. Find the missing terms: 2 3 B _ 6 _ F G _ 5 D _ 8 _ H I

  • A. C, 7, 4, E, 9
  • B. D, 8, 6, C, 7
  • C. E, 8, 7, D, 9
  • D. W, 8, 7, I, 9

Answer: A (C, 7, 4, E, 9)

3. In series 2, 6, 18, 54, …… what will be the 8th term?

  • A. 4370
  • B. 4374
  • C. 7443
  • D. 7434

Answer: B (4374)

4. One term in the number series is wrong. Find out the wrong term 121, 143, 165, 186, 209

  • A. 143
  • B. 165
  • C. 186
  • D. 209

Answer: C (186)

5. Choose the missing terms out of the given alternatives WFB, TGD, QHG, ?

  • A. NIJ
  • B. NIK
  • C. NJK
  • D. OIK

Answer: B (NIK)

6. Choose the correct alternative gfe _ ig _ eii _ fei _ gf_ ii

  • A. eifgi
  • B. figie
  • C. ifgie
  • D. ifige

Answer: C (ifgie)

7. In the series 357, 363, 369, …. what will be the 10th term?

  • A. 405
  • B. 411
  • C. 413
  • D. 417

Answer: B (411)

8. How many terms are there in the series 201, 208, 215, ….., 369?

  • A. 23
  • B. 24
  • C. 25
  • D. 26

Answer: C (25)

9. A random alphanumeric sequence is given. In this series, which of the following is third to the left of fifth from the right end?

R * @ 2 D F % ^ E 3 G # 1 Y 9 & R

  • A. 3
  • B. G
  • C. Y
  • D. 9

Answer: A (3)

10. Odd man out: 196, 169, 144, 121, 80

  • A. 80
  • B. 121
  • C. 169
  • D. 196

Answer: A (80)

Classification – Verbal Reasoning Questions and Answers | Concept & Tricks on Classification

Classification based on Series

Alphabet classification

  • In this type of classification, letters of the alphabet are classified in a group using a particular method or rule.

Words classification

  • In this type of classification, different objects are classified on the basis of common features and properties.

Miscellaneous classification

  • In this type of classification, any rule other than alphabet or words can be implemented for grouping.

Classification Questions

Classification Test a student’s ability to identify the common features among the given set of words. The different types of classification questions are as follows.

  • Identify the odd word: Words signify some things/objects classified on the basis of common properties and features.

Example: artificial, man-made, natural, synthetic. As obvious, ‘natural’ is the odd word here.

  • Identify the odd pair of words: A pair of words are related to each other in some way.

Example: Apple: Jam, Lemon: Citrus, Orange: Squash, Tomato: Pury.

  • Identify the group of odd letters: Groups of random letters are given which follow a common pattern

Example: ACE, GIK, BEG, PRT

Classification – Verbal Reasoning Questions and Answers

1. Choose the word which is different from the rest.

  • A. Chicken
  • B. Snake
  • C. Swan
  • D. Crocodile

Answer: A (Chicken)

2. Three of the following four are the same in a certain way and hence form a group. Find out the one which does not belong to that group.

  • A. CPU
  • B. Computer
  • C. Monitor
  • D. Keyboard

Answer: B (Computer)

3. In the following question, select the odd word from the given alternatives.

  • Pen
  • Paper
  • Marker
  • Highlighter

Answer: D (Highlighter)

4. Choose the word which is different from the rest.

  • A. Gallon
  • B. Ton
  • C. Quintal
  • D. Kilogram

Answer: A (Gallon)

5. Choose the pair in which the words are differently related.

  • A. Car: Road
  • B. Ship: Sea
  • C. Rocket: Space
  • D. Airplane: Pilot

Answer: D (Aeroplane: Pilot)

6. Choose the word which is different from the rest.

  • A. Copper
  • B. Silver
  • C. Gold
  • D. Platinum

Answer: A (Copper)

7. Choose the pair in which the words are differently related.

  • A. Apple: Jam
  • B. Lemon: Citrus
  • C. Orange: Squash
  • D. Tomato: Pury

Answer: B (Lemon: Citrus)

8. Choose the odd one out

  • A. YDWB
  • B. TKRI
  • C. QNOM
  • D. HLFJ

Answer: C (QNOM)

9. Choose the pair in which the words are differently related to the rest.

  • A. Shirt : Dress
  • B. Boy : Girl
  • C. Mango: Fruit
  • D. Table: Furniture

Answer: B (Boy : Girl)

10. Choose the word which is different from the rest.

  • A. Wool
  • B. Honey
  • C. Silk
  • D. Wax

Answer: A (Wool)

Letter Series | Logical Reasoning Questions & Answers on Alphabet Series!!

Letter Series

Basically, Letter Series is also one part of the Reasoning. The letters in the series are arranged in a particular order. The letter series is easy to learn. Letter series or sequence consists of questions where you are supposed to understand the logic behind the series or sequence of letters. Letter series classified into three types:

  • One Lettered Series: In One lettered series each term contains one letter and follows a certain pattern.

Example: What is the next letter in the series B, F, J, N, _______?

  • Two lettered series: In Two lettered series each term contains two-letter and follows a certain pattern.

Example: What is the next letter in the series AC, DB, FG, HL,…….

  • Three lettered series: In Three lettered series each term contains three-letter and follows a certain pattern.

Example: What is the next letter in the series CNL, BLI, AJF _______?

Letter Series – Logical Reasoning Questions and Answers

1. Fill the blank in the middle of the series or end of the series SCD, TEF, UGH, ____, WKL

  • A. CMN
  • B. UJI
  • C. VIJ
  • D. IJT

Answer: C (VIJ)

2. RQP, ONM, _, IHG, FED, find the missing letters.

  • A. CDE
  • B. LKI
  • C. LKJ
  • D. BAC

Answer: C (LKJ)

3. Find the missing letters in the series, E3FG, _, E5FG, E6FG, E7FG.

  • A. EF4G
  • B. E3F4G
  • C. E4FG
  • D. EF3G4

Answer: C (E4FG)

4. What is the missing letter in the series, U, O, I, _, A?

  • A. E
  • B. K
  • C. F
  • D. E

Answer: D (E)

5. Which letter should come next in the series, Q, J, E, B, _?

  • A. A
  • B. Z
  • C. C
  • D. X

Answer: B (Z)

6. P5QR, P4QS, P3QT, _____, P1QV Find middle series

  • A. PQW
  • B. PQV2
  • C. P2QU
  • D. PQ3U

Answer: C (P2QU)

7. Which letter should come next in the series F,G,H,J,K,L,M,N,P,_?

  • A. Q
  • B. R
  • C. T
  • D. O

Answer: A (Q)

8. Find the missing pair in the series, EQ, FS, HW, KA, __, TL.

  • A. OF
  • B. OB
  • C. AL
  • D. AM

Answer: A (OF)

9. CKDL, EKFL, GKHL, _, KKLL, find the missing letters.

  • IJKL
  • IKJL
  • MNOP
  • MNPQ

Answer: B (IKJL)

10. ELFA, GLHA, ILJA, _____, MLNA Find missing Letter Series

  • A. OLPA
  • B. KLMA
  • C. LLMA
  • D. KLLA

Answer: D (KLLA)

Number Series Logical Reasoning Questions and Answers with Tips & Tricks

Number Series

Number Series will present you with mathematical sequences that follow a logical rule, based on elementary arithmetic. Number Series will classify into the following categories.

  • Series with a constant difference: In this kind of series, any 2 consecutive numbers have the same difference between them.

Example: 1 , 5 , 9 , 13 , ?

  • Series with an increasing difference: The difference between two consecutive terms keeps on increasing as we move forward in a series.

Example: 1,2,4,7,11,16,?

  • Series with a decreasing difference: The difference between two consecutive terms keeps on decreasing as we move forward in a series.

Example: 16,11,7,4,2, ?

  • Squares/ Cubes series: We can have a series where the terms are related to the squares/ cubes of numbers. We can have a lot of variations here.

Example: 1, 9, 25, 49 , ? & 1 , 1 , 2 , 4 , 3 , 9 , 4 , ?

  • Combination of different operations: This kind of series has more than 1 type of arithmetic operations which have been performed or it can also have 2 different series which have been combined to form a single series.

Example: 1, 3 , 6 , 2 , 6 , 9 , 3 , 9 , ?

  • Miscellaneous: Some series do not come under any of the above-mentioned categories but are very important and also asked in many examinations.

Example: 9, 25 , 49 , 121 , ?

Number Series Logical Reasoning Questions and Answers

1. Look at this series: 7, 10, 8, 11, 9, 12, … What number should come next?

  • A. 7
  • B. 10
  • C. 12
  • D. 13

Answer: B (10)

2. Find the missing number in the series? 4, 18, ?, 100, 180, 294, 448

  • A. 48
  • B. 50
  • C. 58
  • D. 60

Answer: A (48)

3. Look carefully for the pattern, and then choose which pair of numbers comes next 28 25 5 21 18 5 14

  • A. 11 5
  • B. 10 7
  • C. 11 8
  • D. 5 10

Answer: A (11 5)

4. Look at this series: 664, 332, 340, 170, ____, 89, … What number should fill the blank?

  • A. 85
  • B. 97
  • C. 109
  • D. 178

Answer: D (178)

5. Find the next term 496, 492, 483, 467,?

  • A. 442
  • B. 431
  • C. 462
  • D. 437

Answer: A (442)

6. What should come in place of question mark (?) in the following number series?  132 156? 210 240 272

  • A. 196
  • B. 182
  • C. 199
  • D. 204

Answer: B (182)

7. Find out the wrong term in series 2, 3, 4, 4, 6, 8, 9, 12, 16

  • A. 9
  • B. 12
  • C. 16
  • D. 8

Answer: A (9)

8. Complete the number series given below?  13   19   37   ?   141   243

  • A. 58
  • B. 67
  • C. 75
  • D. 96

Answer: C (75)

9. Which fraction comes next in the sequence 2/3, 1/3, 5/27, 7/81, ?

  • A. 1/27
  • B. 11/278
  • C. 9/48
  • D. 7/123

Answer: A (1/27)

10. Choose the correct alternative that will continue the same pattern and replace ‘x’ in series 4,9,25,x,121,169,289,361

  • A. 81
  • B. 36
  • C. 64
  • D. 49

Answer: D(49)

BSPHCL Syllabus 2020 | Get Bihar SPHCL Assistant Operator, Switch Board Operator & GR-IV Exam Pattern

BSPHCL Assistant Operator Syllabus 2020

Hello Folks!! ABSPHCL Assistant Operator Syllabus 2020 is available here. Bihar State Power Holding Company Limited will conduct the written test for Assistant Operator, Switch Board Operator, Jr Lineman & Technician GR-IV Posts. If you are looking for a better job oppurtunity then you can utilize this chance to achieve your goal. Refer to the Bihar BSPHCL Switch Board Operator Syllabus 2020 and Exam Pattern to begin your preparation. Without knowing the bsphcl.bih.nic.in Jr Lineman & Technician Syllabus 2020 candidates can’t start their preparation. Go through the below sections of this page to get detailed BSPHC Syllabus 2020 Subject Wise. You can also download the Bihar SPHCL Switch Board Operator Syllabus 2020 by clicking on the free pdf link shared at the end of the page.

By checking the Bihar State Power Holding Company Syllabus 2020 you will get to know the required topics which will be asked in the written exam. Go through this full article to know more information about the BSPHCL Syllabus 2020. And by making the perfect schedule with the use of given Bihar SPHCL Syllabus 2020 shared here on this All India Jobs page onwards start your preparation. Proper planning and preparation only candidates can score qualifying marks in the written exam. The official website of BSHPCL is www.bsphcl.bih.nic.in. To make yourself perfect read each and every topic shared in the following sections.

Bihar BSPHCL Switch Board Operator Syllabus 2020

Organization Name Bihar State Power Holding Company Limited
Name of the Post  Assistant Operator, Switch Board Operator, Jr Lineman & Technician GR-IV Posts
Category  BSPHCL Syllabus
Job Location Bihar
Selection Process Written Exam and Interview
Official Website www.bsphcl.bih.nic.in

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BSPHCL Exam Pattern 2020 for Various Posts

S.No. Exam Type  Name of the Subject  No of Marks(%)
1. Objective Type  Questions General Knowledge 10
2. Logical Reasoning 10
3. General Hindi 10
4. Basic knowledge of Computer 10
5. General English & Comprehension 5
6. Technical paper as per final Year syllabus of ITI in Electrician Trade 60
Total 105

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bsphcl.bih.nic.in Jr Lineman & Technician Syllabus 2020

BSPHCL Assistant Operator Syllabus 2020 is very helpful to the candidates who are preparing for the Written Exam. Check out thebsphcl.bih.nic.in Jr Lineman & Technician Syllabus 2020 along with APSSB Exam Pattern to easily know the exam strategy. Bihar BSPHCL Switch Board Operator Syllabus 2020 will consist of important topics for the written exam. This article is updated for the searchers who are looking for the BSPHCL Syllabus 2020. The subjects that are going to ask in the written exam are General Knowledge, Logical Reasoning, General Hindi, Basic knowledge of Computer, General English & Comprehension, and Technical Paper. Moreover, you can also take the assistance of Bihar SPHCL Assistant Operator Previous Papers with Answers to score good marks in the written exam. By solving the bsphcl.bih.nic.in Junior Lineman Old Question Papers & Model Papers candidates can improve subject knowledge and timing skills.

Bihar BSPHCL Switch Board Operator Syllabus 2020 – General Knowledge

  • Famous Books & Authors.
  • Biology.
  • Tourism.
  • Countries and Capitals.
  • Geography.
  • Indian History.
  • Environmental Issues.
  • Heritage.
  • Rivers, Lakes and, Seas.
  • Indian Parliament.
  • Indian Politics.
  • Famous Places in India.
  • Civics.
  • Famous Days & Dates.
  • Current Affairs.
  • Literature.
  • Indian Economy.
  • Sports.
  • General Science.
  • Artists.
  • Inventions and Discoveries.

Bihar State Power Holding Company Syllabus 2020 – Logical Reasoning

  • Analogy.
  • Coding-Decoding.
  • Cubes and Dice.
  • Syllogism.
  • Alphabet Series.
  • Number Ranking.
  • Decision Making.
  • Clocks & Calendars.
  • Mirror Images.
  • Arithmetical Reasoning.
  • Directions.
  • Data Interpretation.
  • Non-Verbal Series.
  • Blood Relations.
  • Statements & Conclusions.
  • Embedded Figures.
  • Number Series.
  • Statements & Arguments.

Latest BSPHCL Syllabus 2020 – General Hindi

  • Parts of Speech
  • Antonyms
  • Grammar
  • Vocabulary
  • Synonyms
  • Active & Passive Voice

Syllabus for BSPHCL Assistant Operator Exam – General English & Comprehension

  • Prepositions.
  • Sentence Completion.
  • Error Correction (Underlined Part).
  • Para Completion.
  • Substitution.
  • Error Correction (Phrase in Bold).
  • Sentence Arrangement.
  • Antonyms.
  • Spotting Errors.
  • Active Voice and Passive Voice.
  • Idioms and Phrases.
  • Joining Sentences.
  • Synonyms.
  • Passage Completion.
  • Sentence Improvement.
  • Fill in the blanks.

Bihar SPHCL Switch Board Operator Syllabus 2020 – Basic knowledge of Computer

  • MS Office
  • Net-Surfing
  • Downloading from Web Page
  • Uploading in Web Page
  • Internet use
  • PowerPoint Presentation
  • MS Excel
  • MS Word

Download BSPHCL Syllabus 2020 PDF for Various Posts 

Probability – Basic Concepts, Formulas, Tricks, Solved Problems

Probability

Probability is a measure of the likelihood of an event to occur. Probability is a type of ratio where we compare how many times an outcome can occur compared to all possible outcomes.

Probability P (E) = Number of Wanted Outcomes / Number of Possible Outcomes

probability line

Probability is always between 0 and 1

Probability Important Formulas

Experiment – An operation that can produce some well-defined outcomes is called an experiment.

Random Experiment – An experiment in which all possible outcomes are known and the exact output cannot be predicted in advance, is called a random experiment.

Examples:

  • Rolling an unbiased dice.
  • Tossing a fair coin.
  • Drawing a card from a pack of well-shuffled cards.
  • Picking up a ball of a certain color from a bag containing balls of different colors.

Sample Space – When we perform an experiment, then the set S of all possible outcomes is called the sample space.

Probability of Occurrence of an Event

Let S be the sample space and E be an event Then, E<= S.

  • P(E) = n(E) / n(S)

Results on Probability

  • P(S) = 1
  • 0<= P(E) <= 1
  • P(Ø) = 0
  • For any events A and B, we have P(A U B) = P(A) + P(B) – P(A ∩ B)

Practice Probability Aptitude Questions & Answers

1. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

  • A. 1 / 2
  • B. 2 /5
  • C. 8 / 15
  • D. 9 /20

Answer: Option D (9 /20)

2. Three unbiased coins are tossed. What is the probability of getting at most two heads?

  • A. 3 / 4
  • B. 1 / 4
  • C. 3 / 8
  • D. 7 / 8

Answer: Option D (7 / 8)

3. In class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

  • A. 21 / 46
  • B. 25 / 117
  • C.  1 / 50
  • D. 3 / 25

Answer: Option A (21 / 46)

4. From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?

  • A. 1 / 15
  • B. 25 / 57
  • C. 35 / 256
  • D. 1 / 221

Answer: Option D (1 / 221)

5. One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen, and King only)?

  • A. 1 / 13
  • B. 3 / 13
  • C. 1 / 4
  • D. 9 / 52

Answer: Option B (3 / 13)

6.  Sam and Joan are playing a tennis match. If the probability of Sam’s win is 0.59, then find the probability of Joan’s win.

  • A. 0.47
  • B. 0.36
  • C. 0.41
  • D. 0.25

Answer: Option C (0.41)

7. Let A and B be events on the same sample space, with P (A) = 0.6 and P (B) = 0.7. Can these two events be disjoint?

  • A. Yes
  • B. No
  • C. None of the above
  • D. All the above

Answer: Option B (No)

8. Which of these numbers cannot be a probability?

  • A. -0.00001
  • B. 0.5
  • C. 1.001
  • D. 1

Answer: Option D (1)

9. The probability that Soumya will get married within 365 days is ‘a’ and the probability that her colleague Alma gets marry within 365 days is ‘b’. Find the probability that only one of the two gets marries at the end of 365 days.

  • A. a-b-2ab
  • B. a+b-2ab
  • C. a-b+2ab
  • D. ab-a-b

Answer: Option B (a+b-2ab)

10. A card is drawn from a pack of 52 cards. The probability of getting a queen of the club or a king of heart is

  • A. 1/13
  • B. 2/13
  • C. 1/26
  • D. 1/52

Answer: Option C (1/26)

True Discount Quantitative Aptitude (MCQ) Questions & Answers – Shortcuts & Tricks for Competitive Exams

Discount

  • Present Value: The money to be paid before the due date to clear off debt is called present worth.
  • True Discount: The difference between the amount due and the present value or worth of the amount is called the true discount. In other words, it is an interest in the present worth.
  • Banker’s Discount: It is the simple interest on the face value or amount due for the period from the date on which the bill was discounted till the legally due date or for the unexpired time.

Aptitude Discount Formulas

Let rate = R% per annum and Time = T years. Then,

  • P.W. = 100 x Amount / 100 + (R x T) = 100 x T.D / R x T
  • T.D. = (P.W.) x R x T / 100 = Amount x R x T / 100 + (R x T)
  • Sum = (S.I.) x (T.D.) / (S.I.) – (T.D.)
  • (S.I.) – (T.D.) = S.I. on T.D.
  • When the sum is put at compound interest, then P.W. = Amount / (1 + R /100) T

Discount Aptitude Questions and Answers

1. A trader owes a merchant Rs. 10,028 due 1 year hence. The trader wants to settle the account after 3 months. If the rate of interest 12% per annum, how much cash should he pay?

  • A. Rs. 9025.20
  • B. Rs. 9200
  • C. Rs. 9600
  • D. Rs. 9560

Answer: Option B (Rs. 9200)

2. A man buys a watch for Rs. 1950 in cash and sells it for Rs. 2200 at a credit of 1 year. If the rate of interest is 10% per annum, the man:

  • A. gains Rs. 55
  • B. gains Rs. 50
  • C. loses Rs. 30
  • D. gains Rs. 30

Answer: Option B (gains Rs. 50)

3. The true discount on Rs. 1760 due after a certain time at 12% per annum is Rs. 160. The time after which it is due is:

  • A. 6 months
  • B. 8 months
  • C. 9 months
  • D. 10 months

Answer: Option D (10 months)

4. A shopkeeper professes to sell all things at a discount of 20% but increases the selling price of each article by 30%. His gain on each article is

  • A. 3%
  • B. 4%
  • C. 5%
  • D. 6%

Answer: Option B (4%)

5. A merchant decided to mark up his goods by 20% and then offered a 10% discount. What will be a percentage of profit or loss?

  • A. 8% profit
  • B. 8% loss
  • C. 4% loss
  • D. 4% profit

Answer: Option A (8% profit )

6. If the true discount on s sum due 2 years hence at 14% per annum be Rs. 168, the sum due is:

  • A. Rs. 768
  • B. Rs. 968
  • C. Rs. 1960
  • D. Rs. 2400

Answer: Option A (Rs. 768)

7. A merchant sold goods at a 10% discount and earned a profit of 20%. What would be the percentage of profit if no discount was offered?

  • A. 32.32 %
  • B. 33.33 %
  • C. 34. 5 %
  • D. 35. 6 %

Answer: Option B (33.33 %)

8. The true discount on a bill due 9 months hence at 16% per annum is Rs. 189. The amount of the bill is:

  • A. Rs. 1386
  • B. Rs. 1764
  • C. Rs. 1575
  • D. Rs. 2268

Answer: Option B (Rs. 1764)

9. Which of the following will yield the maximum discount on Rs 6896?

1) 2 successive discounts of 5% and 5%
2) The single discount of 10%
3) 2 successive discounts of 8% and 2%

  • A. 3
  • B. 2
  • C. 1
  • D. All will yield the same discount

Answer: Option B (2)

10. If Rs. 10 be allowed as true discount on a bill of Rs. 110 due at the end of a certain time, then the discount allowed on the same sum due at the end of double the time is:

  • A. Rs. 20
  • B. Rs. 21.81
  • C. Rs. 22
  • D. Rs. 18.33

Answer: Option D (Rs. 18.33)

Simple and Compound Interest – Practice Questions, Shortcuts, Important Formulas

Simple and Compound Interest

  • Principal: The money borrowed or lent out for a certain period is called the principal or the sum.
  • Interest: Extra money paid for using other’s money is called interest.
  • Simple Interest (SI): If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called simple interest.
  • Compound Interest: Interest added to the principal of a loan or deposit, so that the added interest also earns interest. This process is called compounding.

Important Formulas & Equations

If P = Principal, A = Amount, R = Rate percent per year, T = T years, S.I = Simple interest, C.I = Compound interest, Then,

Simple Interest

  • S.I = (P × R × T)/100
  • A = P + S.I

Compound Interest

  • When interest is compounded yearly,

  • When interest is compounded half-yearly,

  • When interest is compounded quarterly,

  • When time is in a fraction of a year, say 2⅕

  • If the rate of interest in 1st year, 2nd year …………..… nth year is R1%, R2% …………. Rn% respectively, then,

Simple Interest and Compound Interest Aptitude Questions

1.Mr. Thomas invested an amount of Rs. 13,900 divided into two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?

  • A. Rs. 6400
  • B. Rs. 6500
  • C. Rs. 7200
  • D. Rs. 7500

Answer: Option A (Rs. 6400)

2. How much time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest?

  • A. 3.5 years
  • B. 4 years
  • C. 4.5 years
  • D. 5 years

Answer: Option B (4 years)

3. A sum of Rs. 12,500 amounts to Rs. 15,500 in 4 years at the rate of simple interest. What is the rate of interest?

  • A. 3%
  • B. 4%
  • C. 5%
  • D. 6%

Answer: Option D (6%)

4.  A man took a loan from a bank at a rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was

  • A. Rs. 2000
  • B. Rs. 10,000
  • C. Rs. 15,000
  • D. Rs. 20,000

Answer: Option C (Rs. 15,000)

5. What will be the ratio of simple interest earned by a certain amount at the same rate of interest for 6 years and that for 9 years?

  • A. 1 : 3
  • B. 1: 4
  • C. 2 : 3
  • D. Data inadequate

Answer: Option C (2:3)

6. A certain amount earns simple interest of Rs. 1750 after 7 years. Had the interest been 2% more, how much more interest would it have earned?

  • A. Rs. 35
  • B. Rs. 245
  • C. Rs. 350
  • D. Cannot be determined

Answer: Option D (Cannot be determined)

7. Find out the C.I on Rs.5000 at 4% p.a. compound half-yearly for 1 1/2 years.

  • A. Rs.420.20
  • B. Rs.319.06
  • C. Rs.306.04
  • D. Rs.294.75

Answer: Option C (Rs.306.04)

8. Rs.8000 becomes Rs.9261 in a certain interval of time at the rate of 5% per annum of C.I. Find the time?

  • A. 4 years
  • B. 6 years
  • C. 2 years
  • D. 3 years

Answer: Option D (3 years)

9. The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 1. The sum (in Rs.) is:

  • A. 625
  • B. 630
  • C. 640
  • D. 650

Answer: Option A (625)

10. There is a 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate?

  • A. Rs. 2160
  • B. Rs. 3120
  • C. Rs. 3972
  • D. Rs. 6240

Answer: Option C (Rs. 3972)

Sambalpur DCCB Previous Papers | sambalpurdccb.com Support Staff (Peon) Solved Question Papers with Answers PDF

Sambalpur DCCB Peon Previous Papers

Well, its time to get ready for the preparation for the written exam which is going to be conducted for Support Staff (Peon) Posts in the forthcoming days. Applied candidates for Sambalpur DCCB Recruitment 2020 can hurry up and begin the preparation using the required study material shared here on this page. Sambalpur DCCB Peon Previous Papers is one of the best sources to clear the test with excepted marks. In order to support the candidates, we have uploaded the Sambalpur District Cooperative Central Bank Model Papers for each subject in pdf file at the finishing point of this page. Sambalpur District Cooperative Central Bank officials have not yet announced the exam dates. So candidates will have enough time to prepare well for the written exam. Make a proper schedule and prepare for the written exam according to it without wasting your time.  You can get sambalpurdccb.com Support Staff Previous Year Question Papers with Answers for better preparation.

Details of Sambalpur District Cooperative Central Bank Model Papers

Organization Name  Sambalpur District Cooperative Central Bank
Name Of The Post Support Staff (Peon) Posts
Total Posts 30 Vacancies
Category Sambalpur DCCB Previous Papers
Type of Job Bank Jobs
Work Location Sambalpur District, Odisha State
Starting Date to Apply  6th May 2020
Application Last Date 25th May 2020
Date of Written Exam Update Soon
Official Website www.sambalpurdccb.com

Sambalpur DCCB Support Staff (Peon) Sample Papers

The perfect way of preparation for the written exam is to refer to Sambalpur DCCB Peon Previous Papers. If you would like to obtain qualifying marks in the exam then we advised those candidates to take the assistance of Sambalpur DCCB Support Staff (Peon) Sample Papers. Download and practice sambalpurdccb.com Support Staff Old Question Papers with Answers available here on this All India Jobs page. One-click on the below-enclosed links you can get Sambalpur District Cooperative Central Bank Model Papers and Old Papers. By referring to the Sambalpur DCCB Previous Papers candidates can easily know the difficulty level of the written exam. Moreover, exam participants should get an idea about the pattern of the question paper before appearing. Go ahead and read this complete article to know the detailed information about Sambalpur DCCB Bank Sub Staff Model Papers.

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Sambalpur DCCB Support Staff (Peon) Exam Pattern 2020

Let’s have a look at the Sambalpur DCCB Support Staff (Peon) Exam Pattern 2020 tabulated below which helps to give you a better understanding of the scheme of the test. The Subjects that are going to be asked in the upcoming exam are General Knowledge, Arithmetic, Reasoning, etc. By knowing the Exam Pattern candidates will get to know details such as Asked Subjects, Questions asked from each subject, Marks to be secured for each subject, Duartion of the test. Before appearing for the written exam candidates should know the specified Sambalpur DCCB Syllabus 2020 and Exam Pattern for Peon Vacancy.

Sl.No Subject Name  No Of Questions No Of Marks Time Duration
1 General Knowledge 35 35 2 Hours
2 Arithmetic 35 35
3 Reasoning 30 30
4 Total 100 100

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sambalpurdccb.com Support Staff Old Question Papers with Answers

Download Sambalpur DCCB Peon Previous Papers and Old Papers with the help of free pdf links available below. Increase your timing skills by solving as possible as Sambalpur District Cooperative Central Bank Sample Papers for each subject. This article is updated to help out the candidates who are eagerly searching for the sambalpurdccb.com Support Staff Old Question Papers. Hurry up and go through this full article until the end of the page to get more information regarding the Previous Year Papers for Sambalpur DCCB Online Test. Keep visiting our website for more updates related to the written exam.

Sambalpur District Cooperative Central Bank Sample Papers

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Aptitude Boats and Streams Concepts, Important Formulas, Solved Questions & Answers,

Boats and Streams

  • Still water: The water of a river or any other water body which is not flowing is known as still water.
  • Stream: It is the flowing water of a river that is moving at a certain speed.
  • Upstream: The boat or a swimmer moving against the stream is known as moving upstream i.e. against the flow of water.
  • Downstream: The boat or a swimmer moving along the stream is known as moving downstream i.e. along the flow of water.

Important Formulas

  • Speed of this boat downstream = (u + v) km/hr
  • Speed of this boat upstream = (u -v) km/hr.
  • Also, if the downstream speed is ‘a’ km/hr (suppose) and the speed upstream is b km/hr, then:
  • Speed in still water will be = [1/2](a + b) km/hr
  • Rate of stream is = [1/2](a – b) km/hr
  • When the distance covered downstream and upstream are equal, we can write: (u + v)t1 = (u – s)t2 where t1 and t2 are a different time taken.

Aptitude Boats and Streams Questions with Answers

1.A boat can travel with a speed of 13 km/hr in still water. If the speed of the stream is 4 km/hr, find the time taken by the boat to go 68 km downstream.

  • A. 2 hours
  • B. 3 hours
  • C. 4 hours
  • D. 5 hours

Answer: Option C (4 hours)

2. A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and the speed of the water current respectively?

  • A. 2: 1
  • B. 3: 2
  • C. 8 : 3
  • D. Cannot be determined

Answer: Option C (8 : 3)

3. A man takes 20 minutes to row 12 km upstream which is a third more than the time he takes on his way downstream. What is his speed in still water?

  • A. 41 km/hr
  • B. 36 km/hr
  • C. 42 km/hr
  • D. 45 km/hr

Answer: Option C (42 km/hr)

4. How long will it take to row 20 km upstream if one can row 10 km in 10 minutes in still water and the same distance in 8 minutes with the stream?

  • A. 12 min
  • B. 13.33 min
  • C. 24 min
  • D. 26.67 min

Answer: Option D (26.67 min)

5. The speed of a boat in still water is 15 km/hr and the rate of current is 3 km/hr. The distance traveled downstream in 12 minutes is:

  • A. 1.2 km
  • B. 1.8 km
  • C. 2.4 km
  • D. 3.6 km

Answer: Option D (3.6 km)

6. A boat takes 90 minutes less to travel 36 miles downstream than to travel the same distance upstream. If the speed of the boat in still water is 10 mph, the speed of the stream is:

  • A. 2 mph
  • B. 2.5 mph
  • C. 3 mph
  • D. 4 mph

Answer: Option A (2 mph)

7. A man rows ‘k’ km upstream and back again downstream to the same point in H hours. The speed of rowing in still water is s km/hr and the rate of the stream is r km/hr. Then

  • A. (s2-r2) =2sk /H
  • B. (r + s) = kH / (r -s)
  • C. rs = kH
  • D. None of the above

Answer: Option A ((s2-r2) =2sk /H)

8. A man rows 24 km upstream in 6 hours and a distance of 35 km downstream in 7 hours. Then the speed of the man in still water is

  • A. 4.5 km/hr
  • B. 4 km/hr
  • C. 5 km/hr
  • D. 5.5 km/hr

Answer: Option A (4.5 km/hr)

9. A man can row 30 km upstream in 6 hours. If the speed of the man in still water is 6 km/hr, find how much he can row downstream in 10 hours.

  • A. 70 km
  • B. 140 km
  • C. 200 km
  • D. 250 km

Answer: Option A (70 km)

10. A boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along with the current in 10 minutes. How long will it take to go 5 km in stationary water?

  • A. 40 minutes
  • B. 1 hour
  • C. 1 hr 15 min
  • D. 1 hr 30 min

Answer: Option C (1 hr 15 min)

Pipes and Cisterns Solved Problems & Solutions with Explanation

Pipes and Cisterns

A pipe is connected to a tank or cistern which will be used to fill or empty the tank. Accordingly, it is known as an inlet or an outlet. Problems on pipes and cisterns are similar to problems on time and work.

  • Inlet: A pipe connected to fill a tank is known as an inlet.
  • Outlet: A pipe connected to empty a tank is known as an outlet.

Formulas of Pipes and Cisterns

  •  If an inlet connected to a tank fills it in X hours, part of the tank filled in one hour is = 1/X
  • If an outlet connected to a tank empties it in Y hours, part of the tank emptied in one hour is = 1/Y
  • An inlet can fill a tank in X hours and an outlet can empty the same tank in Y hours. If both the pipes are opened at the same time and Y > X, the net part of the tank filled in one hour is given by = (1 / X) – (1 / Y).
  • If a pipe can fill a tank in X hours and another pipe can empty the full tank in Y hours, then on opening both the pipes, the remaining part emptied in 1 hour = (1 / Y) – (1 / X).

Pipes and Cisterns Problems With Solutions

1. A pump can fill a tank with water in 2 hours. Because of a leak, it took 2 1/3 hours to fill the tank. The leak can drain all the water of the tank in:

  • A.  4 1/3 hours
  • B. 7 hours
  • C. 8 hours
  • D. 14 hours

Answer: Option D (14 hours)

2. A tank is filled with three pipes with uniform flow. The first two pipes operating simultaneously fill the tank at the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe to fill the tank is

  • A. 6 hours
  • B. 15 hours
  • C. 10 hours
  • D. 30 hours

Answer: Option B (15 hours)

3. Two pipes A and B can separately fill a tank in 2 minutes and 15 minutes respectively. Both the pipes are opened together but 4 minutes after the start the pipe A is turned off. How much time will it take to fill the tank?

  • A. 9 min
  • B. 10 min
  • C. 11 min
  • D. 12 min

Answer: Option B (10 min)

4. A cistern has a leak that would empty the cistern in 20 minutes. A tap is turned on which admits 4 liters a minute into the cistern, and it is emptied in 24 minutes. How many liters does the cistern hold?

  • A. 480 liters
  • B. 600 liters
  • C. 720 liters
  • D. 800 liters

Answer: Option A (480 liters)

5. Two pipes A and B can fill a cistern in 12 and 15 minutes respectively. Both are opened together but after 3 minutes A is turned off. After how much more time will the cistern be filled?

  • A. 3 1/4 min
  • B. 5 1/4 min
  • C. 8 1/4 min
  • D. 9 1/4 min

Answer: Option C (8 1/4 min)

6. Two pipes A and B can fill a tank in 15 minutes and 40 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?

  • A. 20 min 10 sec
  • B. 25 min 20 sec
  • C. 10 min 10 sec
  • D. 29 min 20 sec

Answer: Option D (29 min 20 sec)

7. Two pipes can fill a tank in 25 and 30 minutes respectively and a waste pipe can empty 3 gallons per minute. All three pipes working together can fill the tank in 15 minutes. The capacity of the tank is:

  • A. 250 gallons
  • B. 450 gallons
  • C. 150 gallons
  • D. 120 gallons

Answer: Option B (450 gallons)

8. Two pipes can fill a tank in 18 minutes and 15 minutes. An outlet pipe can empty the tank in 45 minutes. If all the pipes are opened when the tank is empty, then how many minutes will it take to fill the tank?

  • A. 12
  • B. 13
  • C. 11
  • D. 10

Answer: Option D (10)

9. A cistern can be filled by a tap in 3 hours while it can be emptied by another tap in 8 hours. If both the taps are opened simultaneously, then after how much time will the cistern get filled?

  • A. 1.8 hr
  • B. 3.6 hr
  • C. 4.8 hr
  • D. 2.4 hr

Answer: Option C (4.8 hr)

10. Bucket P has thrice the capacity as bucket Q. It takes 80 turns for bucket P to fill the empty drum. How many turns it will take for both the buckets P and Q, having each turn together to fill the empty drum?

  • A. 60
  • B. 30
  • C. 45
  • D. 80

Answer: Option A (60)

Problems on Ages – Basic Concepts, Shortcuts & Aptitude Questions

Ages

Learn and practice Problems on Ages with the help of Solved Aptitude Questions & Answers given here. Each given question of the Ages topic is accompanied by a clear and easy explanation with diagrams, formulae, shortcuts, and tricks. So let’s check out them in an easy understanding of the concept.

Formulas and Quick Tricks on Age Problems

  • If the current age is x, then n times the age is nx.
  • If the current age is x, then age n years later/hence = x + n.
  • If the current age is x, then age n years ago = x – n.
  • The ages in a ratio a: b will be ax and bx.
  • If the current age is x, then 1 / n of the age is x / n

Age Aptitude Questions & Answers with Explanation

1. The sum of ages of 5 children born at intervals of 3 years each is 50 years. What is the age of the youngest child?

  • A. 4 years
  • B. 8 years
  • C. 10 years
  • D. None of these

Answer: Option A (4 Years)

2. The age of father 10 years ago was thrice the age of his son. Ten years hence, the father’s age will be twice that of his son. the ratio of their present ages is:  

  • A. 5: 2
  • B. 7 : 3
  • C. 9: 2
  • D. 13: 4

Answer: Option B (7:3)

3. The sum of the present ages of two persons A and B is 60. If the age of A is twice that of B, find the sum of their ages 5 years hence?

  • A. 50
  • B. 60
  • C. 70
  • D. 80

Answer: Option C (70)

4. A man said to his son, “I was two-third of your present age when you were born”. If the present age of the man is 48 years, find the present age of the son?

  • A. 25.7 years
  • B. 28 years
  • C. 29.3 years
  • D. 28.8 years

Answer: Option D (28.8 years)

5. Present ages of Sameer and Anand are in the ratio of 5: 4 respectively. Three years hence, the ratio of their ages will become 11: 9 respectively. What is Anand’s present age in years?

  • A. 24
  • B. 27
  • C. 40
  • D. Cannot be determined

Answer: Option A (24)

6. A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of his son is:

  • A. 14 years
  • B. 18 years
  • C. 20 years
  • D. 22 years

Answer: Option D (22 Years)

7. The ratio of the present age of two brothers is 1:2 and 5 years back, the ratio was 1:3. What will be the ratio of their ages after 5 years?

  • A. 1:4
  • B. 2:3
  • C. 3:5
  • D. 5:6

Answer: Option C (3:5)

8. The sum of the present ages of a father and his son is 60 years. Six years ago, the father’s age was five times the age of the son. After 6 years, the son’s age will be:

  • A. 12 years
  • B. 14 years
  • C. 18 years
  • D. 20 years

Answer: Option A (12 years)

9. The present ages of three persons in proportions 4: 7: 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years).

  • A. 8, 20, 28
  • B. 16, 28, 36
  • C. 20, 35, 45
  • D. None of these

Answer: Option B (16, 28, 36)

10. Q is as much younger than R as he is older than T. If the sum of the ages of R and T is 50 years, what is definitely the difference between R and Q’s age?

  • A. 1 year
  • B. 2 years
  • C. 25 years
  • D. Data inadequate
  • E. None of these

Answer: Option D (Data inadequate)

Volume and Surface Area – Solid Figures Aptitude Questions & Answers, Important Topics!!

Solid Figures

Anything that occupies space is called a Solid. With Area, a solid figure has Volume also. Length, Breadth, and Height are three dimensions.

CUBOID

Let length = l, breadth = b and height = h units. Then

  1. Volume = (l x b x h) cubic units.
  2. Surface area = 2(lb + bh + lh) sq. units.
  3. Diagonal = Square root of l2 + b2 + h2 units.

CUBE

Let each edge of a cube be of length a. Then,

  1. Volume = a3 cubic units.
  2. Surface area = 6a2 sq. units.
  3. Diagonal = square root of 3a units.

CYLINDER

Let radius of base = r and Height (or length) = h. Then,

  1. Volume = (pi r2h) cubic units.
  2. Curved surface area = (2 pi rh) sq. units.
  3. Total surface area = 2 pi r(h + r) sq. units.

CONE

Let radius of base = r and Height = h. Then,

  1. Slant height, l = h2 + r2 units.
  2. Volume = [1/3 pi r2h] cubic units.
  3. Curved surface area = (pirl) sq. units.
  4. Total surface area = (pirl + pir2) sq. units

SPHERE

Let the radius of the sphere be r. Then,

  1. Volume = 4/3 pi r3 cubic units.
  2. Surface area = (4pir2) sq. units.

HEMISPHERE

Let the radius of a hemisphere be r. Then,

  1. Volume = 2/3 pi r3 cubic units.
  2. Curved surface area = (2 pi r2) sq. units.
  3. Total surface area = (3 pi r2) sq. units.Note: 1 litre = 1000 cm3.

Volume and Surface Area of Solid Figures Aptitude Moderate Questions and Answers

1. The surface area of a cube is 726 m2. Its volume is?

  • A. 1300 m3
  • B. 1331 m3
  • C. 1452 m3
  • D. 1542 m3

Answer: Option B (1331 m3)

2. Three metal cubes of sides 5 cm, 4 cm, and 3 cm are melted and recast into a new cube. The length of the edge of this cube, is?

  • A. 6 cm
  • B. 8 cm
  • C. 10 cm
  • D. None of these

Answer: Option A (6 cm)

3. What is the maximum number of pieces of 5cm x 5cm x 10 cm of cake that can be cut from a big cake of 5 cm x 30 cm x 30 cm size?

  • A. 10
  • B. 15
  • C. 18
  • D. 30

Answer: Option C (18)

4. In a shower, 5 cm of rainfalls. The volume of water that falls on 1.5 hectares of the ground is:

  • A. 75 cu. m
  • B. 750 cu. m
  • C. 7500 cu. m
  • D. 75000 cu. m

Answer: Option B (750 cu. m)

5. A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:

  • A. 720
  • B. 900
  • C. 1200
  • D. 1800

Answer: Option C (1200)

6. A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/cm3, then the weight of the pipe is:

  • A. 3.6 kg
  • B. 3.696 kg
  • C. 36 kg
  • D. 36.9 kg

Answer: Option B (3.696 kg)

7. A boat having a length of 3 m and breadth 2 m is floating on a lake. The boat sinks by 1 cm when a man gets on it. The mass of the man is:

  • A. 12 kg
  • B. 60 kg
  • C. 72 kg
  • D. 96 kg

Answer: Option B (60 kg)

8.  A cistern 6m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is

  • A. 49 m2
  • B. 50 m2
  • C. 53.5 m2
  • D. 55 m2

Answer: Option A (49 m2)

9. A cistern of capacity 8000 liters measures externally 3.3 m by 2.6 m by 1.1 m and its walls are 5 cm thick. The thickness of the bottom is:

  • A. 90 cm
  • B. 1 dm
  • C. 1 m
  • D. 1.1 cm

Answer: Option B ( 1 dm)

10. A large cube is formed from the material obtained by melting three smaller cubes of 3, 4, and 5 cm sides. What is the ratio of the total surface areas of the smaller cubes and the large cube?

  • A. 2: 1
  • B. 3: 2
  • C. 25: 18
  • D. 27: 20

Answer: Option C (25: 18)

How to Solve Dice Problems | Dice Aptitude Question & Answers for Competitive Exams!!

Dice Problems

Dice is a cube. And cube having 6 faces. Dice is a part of the logical reasoning section of aptitude. And check below figure to know complete about Problems on Dice.

  • 6 Faces – ABCG, GCDE, DEFH, BCDH, AGEF, and ABHF.
  • Always four faces are adjacent to one face.
  • Opposite of ABCG is DEFH and so on.
  • CDEG is the upper face of the cube.
  • ABHF is the bottom of the cube.

Dice Reasoning Questions and Answers for Competitive Exams

1.A dice with six faces are marked with six numbers 1, 2, 3, 4, 5, and 6 respectively. This dice is rolled three times and three positions are shown as:

Find the number opposite to 1.

  • A. 2
  • B. 6
  • C. 5
  • D. 4

Answer: Option C (5)

2. Which of the following patterns can be formed from the piece of cardboard (X) as shown below?

  • A. 4
  • B. 2
  • C. 1
  • D. 3

Answer: Option C (1)

3. Refer to the following four positions of the dice and find out the color which is opposite the face grey?

  • A. Golden
  • B. Purple
  • C. Brown
  • D. Green

Answer: Option A (Golden)

4. Which number is in the opposite plane of 3? 

  • A. 1
  • B. 4
  • C. 6
  • D. 5

Answer: Option B (4)

5. Here 4 positions of a cube are shown. Which sign will be opposite to ‘*’?

  • A. @
  • B.!
  • C. ^
  • D.?

Answer: Option C (^)

6. Two positions of dice are shown below. How many points will appear on the opposite to the face containing 5 points?

  • A. 3
  • B. 1
  • C. 2
  • D. 4

Answer: Option D (4)

7. Two positions of a dice are shown below. Which number will appear on the face opposite to the face with the number 5?

  • A. 2/6
  • B. 2
  • C. 6
  • D. 4

Answer: Option C (6)

8. 

  • A. 2 and 3 only
  • B. 1, 3 and 4 only
  • C. 2 and 4 only
  • D. 1 and 4 only

Answer: Option B (1, 3 and 4 only)

9. From the four positions of a dice given below, find the symbol which is opposite to Y?

  • A. O
  • B. B
  • C. V
  • D. Ro

Answer: Option D (Ro)

10. Four usual dice are thrown on the ground. The total of numbers on the top faces of these four dice is 13 as the top faces showed 4, 3, 1, and 5 respectively. What is the total of the faces touching the ground?

  • A. 12
  • B. 13
  • C. 15
  • D. Cannot be determined

Answer: Option C (15)

Sambalpur DCCB Syllabus 2020 | Download sambalpurdccb.com Support Staff (Peon) Exam Pattern

Sambalpur DCCB Peon Syllabus 2020

Dear Candidates, Sambalpur DCCB Peon Syllabus 2020 is available here. Sambalpur District Cooperative Central Bank is going to conduct the written exam in the coming days. So candidates who are interested in participating in the written exam are advised to check out the sambalpurdccb.com Support Staff (Peon) Syllabus 2020 and Exam Pattern provided on this page. Go ahead and know the topics that are involved in Sambalpur DCCB Syllabus 2020 to begin the preparation as soon as possible. Therefore, Sambalpur District Cooperative Central Bank Peon Syllabus 2020 plays an important role for every candidate who would like to prepare well for the written exam.

In the following sections of this page, we have shared the complete information regarding the Sambalpur DCCB Sub Staff (Peon) Syllabus 2020. Make yourself perfect to give the best performance in the test conducted for Support Staff (Peon) Posts. This All India Jobs article will be more helpful for you to have better preparation. The official website of Sambalpur DCCB – www.sambalpurdccb.com.

Details of Sambalpur District Cooperative Central Bank Syllabus 2020

Organization Name  Sambalpur District Cooperative Central Bank
Name Of The Post Support Staff (Peon) Posts
Total Posts 30 Vacancies
Category Sambalpur DCCB Syllabus
Type of Job Bank Jobs
Work Location Sambalpur District, Odisha State
Starting Date to Apply  6th May 2020
Application Last Date 25th May 2020
Date of Written Exam Update Soon
Official Website www.sambalpurdccb.com

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Sambalpur DCCB Exam Pattern 2020 for Support Staff (Peon) Vacancy

Sl.No Subject Name  No Of Questions No Of Marks Time Duration
1 General Knowledge 35 35 2 Hours
2 Arithmetic 35 35
3 Reasoning 30 30
4 Total 100 100

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sambalpurdccb.com Support Staff (Peon) Syllabus 2020

Sambalpur District Cooperative Central Bank has given the required syllabus and exam pattern for Support Staff (Peon) Vacancy on its webportal for candidates reference. Detailed sambalpurdccb.com Support Staff (Peon) Syllabus 2020 are clearly provided below. Sambalpur DCCB Peon Syllabus 2020 will give you an idea about what to prepare for the test and which topics are important for the written exam. The subject wise Sambalpur DCCB Syllabus 2020 will be more helpful for the candidates to prepare well. Moreover, you can also solve the Sambalpur District Cooperative Central Bank Peon Previous Papers and Old Papers shared here.

At the end of the page, you can have the Sambalpur DCCB Peon Syllabus 2020 downloading link in the form of a pdf file. So interested candidates can download the Sambalpur DCCB Syllabus 2020 for free of cost. Know more details about sambalpurdccb.com Support Staff (Peon) Syllabus 2020 by reading this full article till the end. Keep visiting our blog to get the latest updates on the written exam.

Sambalpur DCCB Syllabus 2020 for General Knowledge

  • Sports & Games
  • Important Days
  • About SDCCB
  • Function of DCCBs
  • Banking & Indian Financial System
  • Awards & Honours
  • History
  • Indian Polity
  • General policy and scientific research
  • Geography
  • Economic Scene
  • International organizations like IMF, BBB, BIS, UNO, SWIFT, IBA
  • RBI and its functions
  • Culture
  • General Awareness of Environment
  • Knowledge of Current Events
  • Latest banking & economical news
  • Indian Banking History
  • Indian Constitution
  • Budget & Banking Schemes/ Policies
  • Capital Marking & Money Market
  • India and its neighboring countries
  • Reforms in banking
  • Sectors in the Banking Industry
  • Banking terms and abbreviation
  • Fiscal and Monetary Policies
  • Currencies
  • Banking Awards
  • Locations/ headquarters of Banks
  • Important banking sector acts, functions/ regulations
  • Alternate names of banks
  • Slogans of Banks

Sambalpur District Cooperative Central Bank Syllabus 2020 for Arithmetic

  • Profit and Loss
  • Simplification
  • Percentage
  • Averages
  • Time & Work
  • Ratio & Proportion
  • Square Roots
  • Mixture and Alligation
  • Time and distance
  • Algebra
  • Decimals, Fractions
  • Cube Roots
  • Simple Interest
  • Compound Interest
  • LCM, HCF
  • Trigonometric Ratio
  • Heights and Distances
  • Discount
  • Partnership Business

sambalpurdccb.com Peoon Syllabus for Reasoning

  • Semantic Series
  • Problem Solving
  • Emotional Intelligence
  • Symbolic/ Number Analogy
  • Trends, Figural Analogy
  • Symbolic/ Number Classification
  • Drawing inferences
  • Space Orientation
  • Figural Classification
  • Punched Hole/Pattern-Folding & Unfolding
  • Semantic Analogy
  • Symbolic operations
  • Figural Pattern-Folding and Completion
  • Number Series
  • Semantic Classification
  • Venn Diagrams
  • Embedded Figures
  • Word Building
  • Figural Series
  • Problems on Ages
  • Coding and De-coding
  • Critical Thinking
  • Social Intelligence

Download Sambalpur DCCB Support Staff (Peon) Syllabus 2020 PDF 

Mixture and Alligation Practice Questions & Answers with Explanation

Mixture and Alligation

Allegation is a rule allowing one to find the ratio in which two or more ingredients must be combined at the specified price to create a mixture of the desired amount. The segment on Mixture and Alligation Practice Questions has questions about Alligation, Mean Price, and the Allegation Law.

  • Alligation: It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of desired price.
  • Mean Price: The cost of a unit quantity of the mixture is called the mean price.

Rule of Alligation

Ratio =(M – B)=(B – M)(A – M)(M – A)

If a vessel contains A liters of milk and if B litres of milk is withdrawn and replaced by water, and again if B litres of mixture is withdrawn and replaced by water and this operation is replaced n times in all, then

  • (Quantity of milk left after nth operation)=[(A – B)]n(Initial quantity of milk)A
  • Quantity of milk left after nth operation = A x[A(1 – (B/A))n]A
  • Simplified formula to calculate quantity of milk left after nth operation = [A(1 – (B/A))n]

Practice Alligations and Mixtures Questions

1.8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the water is 16: 65. How much wine the cask hold originally?

  • A) 18 litres
  • B) 24 litres
  • C) 32 litres
  • D) 42 litres

Answer: Option B (24 litres)

2. A merchant has 1000 kg of sugar part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The Quantity sold at 18% profit is

  • A) 400 kg
  • B) 560 kg
  • C) 600 kg
  • D) 640 kg

Answer: Option C (600 kg)

3. The cost of Type 1 rice is Rs. 15 per kg and Type 2 rice are Rs.20 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2 : 3, then the price per kg of the mixed variety of rice is

  • A) Rs. 19.50
  • B) Rs. 19
  • C) Rs. 18
  • D) Rs. 18.50

Answer: Option C (Rs. 18)

4. A mixture of 150 liters of wine and water contains 20% water. How much more water should be added so that water becomes 25% of the new mixture?

  • A) 10 liters
  • B) 20 liters
  • C) 30 liters
  • D) 40 liters

Answer: Option A (10 liters)

5. The ratio of expenditure and savings is 3 : 2 . If the income increases by 15% and the savings increases by 6% , then by how much percent should his expenditure increases?

  • A) 25
  • B) 21
  • C) 12
  • D) 24

Answer: Option B (21)

6. Find the ratio in which rice at Rs. 7.20 a kg be mixed with rice at Rs. 5.70 a kg to produce a mixture worth Rs. 6.30 a kg.

  • A) 1:3
  • B) 2:3
  • C) 3:4
  • D) 4:5

Answer: Option B (2:3)

7. The ratio of petrol and kerosene in the container is 3:2 when 10 liters of the mixture is taken out and is replaced by the kerosene, the ratio become 2:3. Then the total quantity of the mixture in the container is:

  • A) 25
  • B) 30
  • C) 45
  • D) cannot be determined

Answer: Option B (30)

8. A container contains 50 liters of milk. From that 8 liters of milk were taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?

  • A) 24.52 litres
  • B) 29.63 litres
  • C) 28.21 litres
  • D) 25.14 litres

Answer: Option B (29.63 liters)

9. A vessel is filled with liquid, 3 parts of which are water and 5 parts of syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

  • A) 1/3
  • B) 1/4
  • C) 1/5
  • D) 1/7

Answer: Option C (1/5)

10. A milkman sells the milk at the cost price but he mixes the water in it and thus he gains 9.09%. The quantity of water in the mixture of 1 liter is

  • A) 83.33 ml
  • B) 90.90 ml
  • C) 99.09 ml
  • D) can’t be determined

Answer: Option A (83.33 ml)

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