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# Josh magazine-ssc-higher-secondary-exam-2012-booklet-1

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• 1. SSC Higher Secondary Examwww.jagranjosh.com (Preparation Booklet)
• 2. TABLE OF CONTENTS Introduction ................................................................................................................................ 4 General Intelligence .................................................................................................................... 6 Analogy ................................................................................................................................... 6 Classification ........................................................................................................................... 6 Series ....................................................................................................................................... 6 Coding-Decoding ..................................................................................................................... 7 Sequence ................................................................................................................................. 7 Direction Test .......................................................................................................................... 7 Puzzles ..................................................................................................................................... 8 Mathematical Puzzles ............................................................................................................. 9 Syllogism ............................................................................................................................... 10 Non-Verbal ............................................................................................................................ 11 Quantitative Aptitude ............................................................................................................... 14 HCF & LCM ............................................................................................................................ 14 Ratio & Proportion ................................................................................................................ 15 Percentage ............................................................................................................................ 16 Average ................................................................................................................................. 17 Surds and Indices .................................................................................................................. 18 Profit & Loss .......................................................................................................................... 18 Interest Calculation ............................................................................................................... 19 Time, Speed & Distance ........................................................................................................ 22 Time & Work ......................................................................................................................... 23 Elementary Algebra .............................................................................................................. 24 Geometry .............................................................................................................................. 26
• 3. Mensuration.......................................................................................................................... 28 Trigonometry ........................................................................................................................ 30 Data Interpretation ............................................................................................................... 32English Language ....................................................................................................................... 35 Synonyms .............................................................................................................................. 35 Antonyms .............................................................................................................................. 35 Idioms and phrases ............................................................................................................... 36 One Word Substitution ......................................................................................................... 36 Spelling Test .......................................................................................................................... 38 Incorrect Sentences .............................................................................................................. 38 Sentence Reconstruction ...................................................................................................... 39 Spotting the Error ................................................................................................................. 39 Cloze Test .............................................................................................................................. 40
• 4. INTRODUCTIONSSC Combined Higher Secondary Level (10+2) Examination 2012 will be conducted by StaffSelection Commission (SSC) for the recruitment to the posts of Data Entry Operators and LowerDivision Clerk. SSC Combined Higher Secondary Level (10+2) Examination 2012 is going to beconducted on 21 October 2012 & 28 October 2012. To be eligible for SSC Combined HigherSecondary Level (10+2) Examination 2012, the candidate must have passed 12th Standard orequivalent or higher examination from a recognized Board or University. In addition, thecandidate’s minimum age should be 18 years and maximum age should be 27 years.The selection will be based on basis of 2 hours written examination carrying 200 marks, which isfollowed by Data Entry Skill Test/ Typing Test of shortlisted candidates. The writtenexamination consists of objective type paper including 4 parts that are as follows: General Intelligence (50 Questions) English Language (50 Questions) Quantitative Aptitude (50 Questions) General Awareness (50 Questions)The question will be set in both Hindi and English Language for Part I, II, and IV. There will benegative marking of 0.25 marks for each wrong answer.Syllabus for Objective PaperGeneral IntelligenceSemantic Analogy, Symbolic operation, Symbolic/Number Analogy, Trends, Figural Analogy,Space Orientation, Semantic Classification, Venn Diagrams, Symbolic/Number Classification,Drawing inferences, Figural Classification, Punched hole/pattern-folding & unfolding, SemanticSeries, Figural Pattern – folding and completion, Number Series, Embedded figures, FiguralSeries, Critical Thinking, Problem Solving, Emotional Intelligence, Word Building, SocialIntelligence, Coding and de-coding, Numerical operationsEnglish LanguageSpot the Error, Fill in the Blanks, Synonyms/Homonyms, Antonyms, Spellings/Detecting Mis-spelt words, Idioms & Phrases, One word substitution, Improvement of Sentences,Active/Passive Voice of Verbs, Conversion into Direct/Indirect narration, Shuffling of Sentenceparts, Shuffling of Sentences in a passage, Cloze Passage, Comprehension PassageQuantitative Aptitude1. ArithmeticNumber Systems: Computation of Whole Number, Decimal and Fractions, Relationshipbetween numbers
• 5. Fundamental Arithmetical Operations: Percentage, Ratio and Proportion, Square roots,Averages, Interest (Simple and Compound), Profit and Loss, Discount, Partnership Business,Mixture and Alligation, Time and distance, Time and work2. AlgebraBasic algebraic identities of School Algebra (and their simple applications) e.g. Formulas for(a + b) 2, (a - b) 2, (a + b)3, (a - b)3 , a3 - b3, a3 + b3, a2 - b2; if a + b + c=0, then a3 + b3 + c3 = 3abcetc. and Elementary surds (simple problems), and Graphs of Linear Equations3. GeometryFamiliarity with elementary geometric figures and facts:Triangle and its various kinds of centres viz. Centroid, In-centre, Orthocentre, CircumcentreCongruence and similarity of trianglesCircle and its chords, tangents, angles subtended by chords of a circle, common tangents to twoor more circles4. MensurationTriangle, Quadrilaterals, Regular Polygons (sum of the internal angles of a polygon), Circle, RightPrism, Right Circular Cone, Right Circular Cylinder, Sphere, Hemispheres, RectangularParallelepiped, Regular Right Pyramid with triangular or square Base5. TrigonometryTrigonometry (for acute angles 0 with 0≤0≤90): Trigonometric ratios, Degree and RadianMeasures, Standard Identities like sin20 + Cos20=1 etcComplementary Angles, Heights and Distances (simple problems only)6. Statistical ChartsUse of Tables and Graphs: Histogram, Frequency polygon, Bar-diagram, and Pie-chartGeneral AwarenessHistory, Culture Geography, Economic Scene, General policy and Scientific Research
• 6. GENERAL INTELLIGENCEANALOGYAnalogy: In this type of questions the students are required to identify the previous relationand find out the best suitable relation for the other term. These questions are asked in the formof proportion where two related terms are given and the next matching missing term can beasked. The questions of the analogy can be in the form of words, numbers, letters and codedletters etc. with any relation.For Example, Cricket : Pitch :: Tennis :: ? (a) Arena (b) Course (c) Court (d) RinkAnswer: - (c) Court. Here Cricket is played at Pitch and Tennis is played at Court.CLASSIFICATIONClassification: In this type of questions the students are required to identify the item whichdoes not belong to a group. It means the other items in the group are similar with respect tocertain logic but one is different. The students can get questions on word group, number group,letter group, coded letter group and numeral pair group etc.For Examples1. (a) 36 (b) 125 (c) 144 (d) 121Answers: - (b) All except 125 are squares.SERIESSeries: It can be understood as a sequence of numerical, alphabetical and alphanumerical termswhich follows a fixed pattern. The students are required to identify the missing term or wrongterm in the series.
• 7. For Example1. 6, 11, 18, 27, 38, ?. (a) 49 (c) 51 (b) 50 (d) 52Answer: - (c) The series is +5, +7, +9……CODING-DECODINGCoding: Coding is a process by which certain information is written into another form ofinformation on the basis of certain principles.Decoding: Decoding is used to infer the right information from the coded information.For Example, In a certain code language, BOOK is written as 2151511. In the same languagehow PEN be written?Answer:- In the above example the letters of BOOK is coded on the basis of their position in theEnglish alphabet. The letters B=2, O = 15, K = 11. In the same way we can encode PEN as 16514.SEQUENCEThe question on this topic can be of following types;1. Ascertaining meaningful word sequence.2. Ascertaining similar numerical sequence.3. Finding a position in a sequence.For Example,Find the meaningful order of the following words in ascending order.1. Line2. Angle3. Pentagon4. Rectangle5. TriangleAnswer: - The right order for these words will be Line – Angle – Triangle – Rectangle –Pentagon. Therefore the answer is 1, 2, 5, 4, and 3.DIRECTION TESTIn this unit, the problems are generally formulated with successive follow-up of the directionsand the candidates are required to find the final direction, the final position with respect to theinitial position or the distance between the final position and initial position.On paper, we can draw the directions as;
• 8. (Here, N = North, NE = North-East, E = East, SE = South-East, S = South, SW = South-West,W = West, NW = North-West)Example: If a person is going to the east of his home and turning to his left after that turning tohis right. His final positionWe can draw his path as; His HomeFor Example, A starts walking towards East, turns right, again turn right, turns left, again turnsleft, turns left, again turns left. In which direction is A walking now? (a) East (b) West (c) North (d) SouthAnswer:- (b)PUZZLESThe Puzzle questions can be in the form of following types;1. Blood Relation Problems2. Height and Position Related Problems3. Cube and Blocks Related Problems4. Embedded Words & Unscrambled Words Related Problem.Examples;1. P and Q are brothers. R and S are sisters. P’s son is S’s brother. How is Q related to R? (SSC 2011)Answer: In the question P’s son is S’s brother means P is the father of S and similarly P is thefather of R because S and R are sisters. Therefore Q is the uncle of R. because Q is the father’sbrother of R.2. There are five friends - Satish, Kishore, Mohan, Anil and Rajesh. Mohan is the tallest. Satish is shorter than Kishore but taller than Rajesh. Anil is little shorter than Kishore but little taller than Satish. Who is taller than Rajesh but shorter than Anil? (SSC 2011)
• 9. Answer: It is given that Kishore>Satish>Rajesh and Kishore>Anil>Satish. It means the sequenceof height is Mohan>Kishore>Anil>Satish> Rajesh. Therefore, the person who is taller thanRajesh but shorter than Anil is Satish.MATHEMATICAL PUZZLESIntroduction: In this unit we will study various types of questions based on the basicmathematical and logical concepts. These types of questions are regularly asked in the SSCexams. After analyzing previous papers, it is clear that in the exam you can get 5 to 8 questionsfrom this unit.The question types1. Arrangement of correct mathematical operations.2. Filling the correct mathematical operations.3. Mathematical-Logical problemsWhen you are solving these questions you should follow basic mathematical operations in acorrect sequence. The correct sequence of solving mathematical problems is known by“BODMAS Rule” or “VBODMAS Rule”.V – Vinculum B – Brackets O – Of D – Division M – Multiplication A – AdditionS – Subtraction.For Examples1. Which interchange of signs will make the following equation correct? 25 − 5 + 32 ÷ 4 × 6 = 13. (a) − and ÷ (b) × and ÷ (c) – and + (d) × and –Answer:- (a)2. Select the correct combination of mathematical signs to replace the * signs and to balance the given equation. 5 * 5 * 3 * 4 * 64 * 4. (a) ÷×−= + (b) ×−= +÷ (c) +×−=÷ (d) ×−=÷+Answer:- (c)
• 10. 3. Some equations are solved on the basis of certain system on the same basis find out the correct answer for the unsolved equation. 1 × 2 × 5 = 125, 2 × 1 × 6 = 216, 3×4×3=? (a) 125 (b) 216 (c) 512 (d) 343Answer:- (d)SYLLOGISMSyllogism: The literal meaning of syllogism is ‘Conclusion’ or ‘inference’. The questions in thesyllogism are in the form of statements (premises) followed by Conclusion (proposition) and thestudents are asked to find the correct conclusion on the basis of the statements.The general form of statements and conclusions in the syllogism is; All As are Bs. Some As are Bs. No A is a B.The graphical representation of the statements1. All As are Bs.2. Some As are Bs.3. No A is a B.
• 11. For ExampleDirection (Q. 1 – 2): In each questions below there are two/three statements followed by twoconclusions I and II. Assuming both the statements true, you have to decide which of the twoconclusions logically follows the statements and then give your answer(a) If only conclusion I follows.(b) If only conclusion II follows.(c) If either conclusion I or conclusion II follows.(d) If neither conclusion I nor conclusion II follows.(e) If both conclusion I and conclusion II follow.1. Statements: Some pens are pencils. Some pencils are books. Conclusions: I. Some pens are books. II. All books are pens.Answer:- (d)2. Statements: All bottles are bags. All bags are buses. Conclusions: I. Some bags are bottles. II. All bags are bottles.Answer:- (a)NON-VERBALThe questions from this unit can be of following types;1. Pictorial pattern based questions.2. Embedded figure questions.3. Mirror Image questions.4. Paper Cutting problems.5. Relationship based problems.
• 12. For Example1. Which answer figure will complete the pattern in the question figure? (SSC 2011) Question figure: Answer figures:Answer: (D) will complete the pattern2. Which of the answer figures is exactly the mirror image of the question figure, when the mirror is held on the line MN? (SSC 2002) Question figure: (a) (b) (c) (d)Answer: (d)3. Select the answer figure in which the question figure is hidden/embedded. Question figure: (a) (b) (c) (d)Answer: (d)
• 13. QUANTITATIVE APTITUDEHCF & LCMSome Important conceptsFactors and Multiples: If a number m divides the number n with remainder = 0, then m is calledthe factor of n and n is called multiple of m.Highest Common Factor (HCF): The HCF is the highest common factor for two or morenumbers.For Example, if we have two numbers 30 and 20, then the factors of 30 are 1, 2, 3, 5, 6, 10, 15and 30. And the factors of 20 are 1, 2, 4, 5, 10 and 20. Here 10 is the highest common factor.Least Common Multiple (LCM): The LCM is the lowest number which is exactly divisible by thegiven two or more than two numbers. For example, if we have two numbers 20 and 30, thenthe multiples of 20 are 20, 40, 60, 80………. And the multiples of 30 are 30, 60, 90, 120………..Here 60 is the lowest number which is exactly divisible by both 20 and 30. Therefore 60 is therequired LCM. Product of two numbers = Product of their LCM and HCF For example, the LCM for 20 and 30 is 60 and HCF for 20 and 30 is 10. Here 20 × 30 = 60 × 10 = 600.Co-primes: Two numbers are co-primes if their HCF = 1.HCF and LCM of Fraction: 𝐇𝐂𝐅 𝐨𝐟 𝐧𝐮𝐦𝐞𝐫𝐚𝐭𝐨𝐫𝐬 𝐋𝐂𝐌 𝐨𝐟 𝐧𝐮𝐦𝐞𝐫𝐚𝐭𝐨𝐫𝐬HCF = 𝐋𝐂𝐌 𝐨𝐟 𝐃𝐞𝐧𝐨𝐦𝐢𝐧𝐚𝐭𝐨𝐫𝐬 and LCM = 𝐇𝐂𝐅 𝐨𝐟 𝐃𝐞𝐧𝐨𝐦𝐢𝐧𝐚𝐭𝐨𝐫𝐬For ExampleThe L.C.M. of two numbers is 189 and H.C.F. of two numbers is 9.If one of the number is 63then other number is: (a) 27 (b) 28 (c) 29 (d) 30Answer:- (a)
• 14. RATIO & PROPORTIONRatio: A ratio is a comparison of two numbers (quantity in the same unit). It is written as, a: b =a = a ÷ b, where a and b are two number (quantity).bIn a ratio a : b, a and b are the terms of the ratio; ‘a’ is called the antecedent and ‘b’ is called theconsequent.The word ‘antecedent’ literally means ‘that which goes before’ and the word consequentliterally means ‘that which goes after’.Compound Ratio: Compounded ratio is a product of two or more ratios.Example: Find the ratio compounded of the three ratios:2 : 3, 3 : 4 and 7 : 11 2 × 3 ×7 7Solution: the required ratio is =3 ×4 ×11 = 22 . 1 1Inverse Ratio: If a : b is a given ratio, then a ∶ b or b : a is called its inverse ratio or reciprocalratio.1. If the antecedent (a) = the consequent (b), the ratio is called the ratio of equality, such as 2 : 2.2. If the antecedent (a) > the consequent (b), the ratio is called the ratio of greater inequality, such as 3 : 2.3. If the antecedent (a) < the consequent (b), the ratio is called the ratio of less inequality, such as 2 : 3. a cProportion: A proportion expresses the equality of two ratios. e.g. b = dOr a : b = c : d or a : b :: c : d.In a proportion in the form of a : b :: c : d the first and the last terms are called the extremesand the second and the third terms or the middle terms are called as the mean terms. Whenfour quantities are in proportion, we can write it in the mathematical form as a : b :: c : d a c = d  ad = bcb
• 15. For Example, 2 1If Rs. 510 be divided among A, B, C in such a way that A gets 3 of what B gets and B gets 4 ofwhat C gets, then their shares are respectively : (a) Rs. 120, Rs. 240, Rs. 150 (b) Rs. 60, Rs. 90, Rs. 360 (c) Rs. 150, Rs. 300, Rs. 60 (d) None of theseAnswer:- (b)PERCENTAGEPercentage: The term “per cent” means “for every hundred”. A fraction whose denominators is100 is called a percentage and the numerator of the fraction is called the rate per cent. It isdenoted by the symbol %. x 10 1Here x % =100 . For example 10% =100 = 10 .Similarly, fraction can be changed in the form of percentage when we multiply them by 100. 1 1For Example 4 = 4 × 100 = 25%.To decrease a number by a given %: 𝟏𝟎𝟎−𝐑𝐚𝐭𝐞Multiply the numbers by the factor 𝟏𝟎𝟎To find the % increase of a number: 𝐓𝐨𝐭𝐚𝐥 𝐢𝐧𝐜𝐫𝐞𝐚𝐬𝐞 𝐅𝐢𝐧𝐚𝐥 𝐯𝐚𝐥𝐮𝐞−𝐈𝐧𝐢𝐭𝐢𝐚𝐥 𝐯𝐚𝐥𝐮𝐞% increase = × 𝟏𝟎𝟎 = × 𝟏𝟎𝟎 𝐈𝐧𝐢𝐭𝐢𝐚𝐥 𝐯𝐚𝐥𝐮𝐞 𝐈𝐧𝐢𝐭𝐢𝐚𝐥 𝐯𝐚𝐥𝐮𝐞To find the % decrease of a number: 𝐓𝐨𝐭𝐚𝐥 𝐝𝐞𝐜𝐫𝐞𝐚𝐬𝐞 𝐅𝐢𝐧𝐚𝐥 𝐯𝐚𝐥𝐮𝐞−𝐅𝐢𝐧𝐚𝐥 𝐯𝐚𝐥𝐮𝐞% decrease = × 𝟏𝟎𝟎 = × 𝟏𝟎𝟎 𝐈𝐧𝐢𝐭𝐢𝐚𝐥 𝐯𝐚𝐥𝐮𝐞 𝐈𝐧𝐢𝐭𝐢𝐚𝐥 𝐯𝐚𝐥𝐮𝐞For Example1. In an election between two candidates, 75% of the voters casted their votes, out of which 2% of the votes were declared invalid. A candidate got 9261 votes which were 75% of the total valid votes. Find the total number of votes enrolled in that election. (SSC 2003) (a) 15000 (c) 16800 (b) 16000 (d) 17000
• 16. Answer:- (c) Let the total enrolled votes are x. Then, Number of votes cast = 75 % of x. Validvotes = 98 % of (75 % of x) ∴ 75 % 0f [98 % of (75 % of x)] = 92612. If 50% of (x – y) = 30% of (x + y), then what percent of x is y? (a) 25 % (b) 50 % (c) 75 % (d) 100 % 50 30Answer:- (a) here 50% of (x – y) = 30% of (x + y)  100 x − y = 100 (x + y) 1 10(x – y) = 6(x + y)  10x – 6x = 6y + 10y  4x = 16y  4 x = y  25 % of x = y. 1Because4 = 25%AVERAGEAverage is defined as the sum of n different numerical values divided by n. 𝐒𝐮𝐦 𝐨𝐟 𝐧 𝐝𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐭 𝐍𝐮𝐦𝐞𝐫𝐢𝐜𝐚𝐥 𝐕𝐚𝐥𝐮𝐞Average = 𝐧 𝐓𝐨𝐭𝐚𝐥 𝐃𝐢𝐬𝐭𝐚𝐧𝐜𝐞Average Speed = 𝐓𝐨𝐭𝐚𝐥 𝐒𝐩𝐞𝐞𝐝If a person covers half of his journey at a speed of x km/h and the next half at the speed of y 𝟐𝒙𝒚km/h, then the average speed during the whole journey is . 𝒙+ 𝒚 𝒘𝟏𝒙𝟏 + 𝒘𝟐𝒙𝟐Weighted Average of x1, x2 where weight is w1, w2 = . 𝒘𝟏 + 𝒘𝟐Average of different groupsIf the average of a group of n items is a and the average of another group of m items is m, then 𝐧𝐚 + 𝐦𝐛the combined average = 𝐧 + 𝐦 .For Example1. The average of a non-zero number and its square is 5 times the number. The number is: (SSC 2003) (a) 9 (b) 17 (c) 29 (d) 295Answer:- (a) Let the number be x. Then
• 17. x2+ x = 5x  x 2 + x = 10x  x 2 + x − 10x = 0  x x − 9 = 0 ∴ x = 9. 2SURDS AND INDICESLaws of Indices: (a) am × an = am + n am (b) a n = am – n (c) (am) n = a m n (d) (a b)m = a m b m a m am (e) b = bm 0 (f) a = 1Laws of Surds: 1 n n (a) n a = an (d) a =a n n n m n mn (b) ab = a × b (e) a= a n n m n (c) n a = n a (f) a = am b bFor Example, 11. Find the value of ( 81) 4 . (a) 3 (b) 3 (c) 9 3 (d) 3 1 1 1 1 1 1Answer: - (b) ( 81) 4 = (3 × 3 × 3 × 3 )2 × 4 = (34 )2 × 4 = (3)2 = 3. 1 −42. (− ) 3 =? 64 (a) 254 (b) -254 (c) 256 (d) -256 1 −4 1 (−4) 1Answer:- (c) (− ) 3 = (− 4)3 × 3 = (− 4)−4 = (−4)4 = 256. 64PROFIT & LOSS
• 18. Some important terms for Profit and Loss:Cost Price: It is the price at which an article is purchased. The cost price is generally abbreviatedas C.P.Selling Price: It is the price at which an article is sold. The selling price is generally abbreviatedas S.P.Profit or Loss: It is the difference of selling price (SP) and cost price (CP). If the difference ispositive, then it is called a profit and if the difference is negative, then it is called a loss.Important Formulae1. SP = CP + Profit = CP – Loss Gain × 1002. Gain % = C.P. Loss × 1003. Loss % = C.P. (100 + Gain %) (100 − Loss %)4. SP = × CP = × CP 100 100 100 1005. CP = (100 + Gain × SP = (100 −Loss × SP %) %)For Example1. Arjun buys an old car for Rs 1, 12, 000 and spends 8000 on its repairs. If he sells the Car for Rs 1,80,000, his gain percent is: (a) 50 (c) 52 (b) 51 (d) 53Answer:- (a) C.P. = Rs. (1,12,000 + 8,000) = Rs. 1,20,000, S.P = 1,80,000, 60,000Gain % = ( 1,20,000 × 100) =50%2. If loss is 1/4 of S.P., the loss percentage is: (a) 25% (b) 20% (c) 35% (d) 40% a 5a a 4Answer:- (b) S.P. = a, Loss = a/4, CP. =a + 4 =  Loss% =( × × 100) % = 20% 4 4 5aINTEREST CALCULATIONImportant Facts and Formulae related to Interest calculation
• 19. Principal: It is the money which is borrowed or lend out for certain period. Generally, Principalis denoted with P.Interest: It is the cost of holding others money for a certain period of time.Rate of Interest: It is the rate at which the interest is calculated. The rate of interest is theamount of interest which is calculated at every Rs. 100 for a fixed period. Generally for 1 year. P ×R ×T S.I.× 100 S.I.× 100 S.I.× 100Simple Interest (S. I.) = ,P= ,R= ,T= 100 R×T P×T P×R R nCompound Interest (C. I.) = P 1 + 100 −P (When interest is compounded annually) R 2n 2Compound Interest (C. I.) = P 1 + 100 −P (When interest is compounded Half yearly)When rates of interest are different for different years, for example R1%, R2% and R3% for 1st,2nd and 3rd year respectively. 1 R 2 R R3Then Compound Interest =P 1 + 100 1 + 100 1 + 100 − P.For Example1. Find the compound interest on Rs. 160, 000 in 2 years at 10 % per annum. The interest being compounded half yearly. (a) 32000 (b) 33000 (c) 34000 (d) 34481 10 2×2 2Answer:- (d) The required C.I. = 160, 000 1 + − 160, 000 100 21 21 21 21= 160, 000 − 160, 000 = 194, 481 − 160, 000 = 34, 481 20 20 20 202. The difference between the compound interest and simple interest on a certain sum at 10% per annum for 2 years is Rs. 200. Find the principal. (a) Rs. 10, 000 (b) Rs. 15, 000 (c) Rs. 20, 000 (d) Rs. 25, 000Answer:- (c) Let the required principal be P. then
• 20. 10 2 P × 10 × 2 121 1 1P 1 + 100 −P− = 200  P − 1 − 5 = 200  P = 200 100 100 100Therefore, P = 20, 000
• 21. TIME, SPEED & DISTANCEImportant facts and formulaeDistance = (Speed × Time) DistanceSpeed = Time DistanceTime = Speed 5x km/h = x × 18 m/s 18x m/s = x × km/h 5 Total DistanceAverage Speed = Total SpeedIf a person covers half of his journey at a speed of x km/h and the next half at the speed of y 𝟐𝒙𝒚km/h, then the average speed during the whole journey is . 𝒙+ 𝒚In water, the direction along the stream is called downstream. And the direction against thestream is called upstream.If the speed of a boat in still water is a km/h and the speed of the stream is b km/h, then:Speed downstream = (a + b) km/hSpeed upstream = (a + b) km/hIf the speed downstream is p km/h and the speed upstream is b km/h, then: 𝟏Speed in still water = 𝟐 (𝐚 + 𝐛) km/h 𝟏Rate of stream = 𝟐 (𝐚 − 𝐛) km/hFor Example1. A man travelled from the village to the post-office at the rate of 25km/h and walked back at the rate of 4 km/h. If the whole journey took 5 hours 48 minutes. Find the distance of the post-office from the village. (SSC, 2004) (a) 10 km (c) 20 km (b) 15 km (d) 25 km
• 22. 2 × 25 × 4 200Answer: - (c) The average speed for the whole journey = = km/h. the total time is 25 + 4 29 48 4 29 200 295 60 = 5 5 = h. ∴ the total distance = × = 40 km. Therefore the distance from village 5 29 5 40to the post office = = 20 km. 22. In what time will a train 100 m long, cross an electric pole, if its speed be 144 km/h? (a) 2.5 seconds (c) 3.5 seconds (b) 3 seconds (d) 4 seconds 5Answer: - (a) The speed in m/s = 144 × 18 = 40 m/s and the train has to cross its own length to 100cross a pole. The length of the train is 100 m. therefore the required time is = 2.5 s 40TIME & WORKSome Important Facts and Formulae1. If m persons can complete a work in p days, then the same work will be completed by n m persons in n × p days. 12. If P can complete a piece of work in A days, then P’s one day work = A of the complete work 1 and if P can do A of work in 1 day, then P can complete the work in A days.3. If A can complete a piece of work in m days and B can complete the same work in n days, 1 1 m+n then the 1 day work of both of them together = +n = of the work and both of m mn mn them together complete the work in m+n .4. If A is twice as good a workman as B, then ratio of work done by A and B = 2 : 1 and the ratio of times taken by both of them to finish the work = 1 : 2.For Example1. Aryan takes 5 hours to do a job, and Aryaman takes 6 hours to do the same job. How long should it take if both Aryan and Aryaman, working together but independently, to do the same job? 9 (a) 3 days (c) 211 days 8 (b) 211 days (d) 4 daysAnswer: - (b)
• 23. 12. A can finish a work in 12 days and B can do the same work in 3 the time taken by A. Then, working together, what part of same work they can finish in a day? 1 1 (a) (c) 5 3 1 1 (b) (d) 4 2Answer:- (c)ELEMENTARY ALGEBRABelow are some important formulae which are very useful to solve questions.1. (a + b)2 = a2 + b2 + 2ab2. (a – b)2 = a2 + b2 – 2ab3. a2 – b2 = (a + b)(a – b)4. (a + b)3 = a3 + 3a2b + 3ab2 + b35. (a − b)3 = a3 − 3a2b + 3ab2 − b36. a3 + b3 = (a + b)(a2 – ab + b2)7. a3 − b3 = (a − b)(a2 + ab + b2)8. a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) 1 = 2(a + b + c)[(a – b)2 + (b – c)2 + (c – a)2]9. (a + b + c)3 = a3 +b3 + c3 + 3(a + b)(b + c)(c + a)10. If (a + b + c) = 0 or a = b = c, then a3 + b3 + c3 = 3abc.11. am × an = am + n12. (am)n = am n13. (a b)m = a m b m 𝑎 𝑚14. = x m – n. 𝑎𝑛The questions in the SSC exam are generally based on the addition, subtraction, multiplicationand division of algebraic expressions. Some questions are also asked from quadratic equations.For Example,1. If for two real constants a and b, the expression ax3 + 3x2 – 8x + b is exactly divisible by (x + 2) and (x – 2). Find the value of a and b.Answer: - the expression is exactly divisible by (x + 2) and (x – 2). Thereforea(−2)3 + 3(−2)2 – 8(−2) + b = 0  −8a + 3×4 + 16 + b = 0  8a – b = 28 ………(1)Again, a(2)3 + 3(2)2 – 8(2) + b = 0  8a + 3×4 − 16 + b = 0  8a + b = 4 ………(2)
• 24. After adding (1) and (2)8a – b + 8a + b = 28 + 4  16a = 32 ∴ a = 2.After putting the value of a in (1)8 × 2 – b = 28  16 – 28 = b ∴ b = − 12.Hence, a = 2 and b = −12. 1 4a2. If a + a = 2, find the value of 3a 2 + 4a − 3. (a) 0 (b) 1 (c) 2 (d) −1 1 4aAnswer: - (b) Solve a + a = 2 and put the value in 3a 2 + 4a − 3.3. If a2 + b2 + 2b + 1 = 0, then find the value of a34 + b38. (a) 1 (b) 0 (c) −1 (d) 2Answer: - (a) Here a = 0 and b = −1, then a34 + b38 = 1
• 25. GEOMETRYConceptsTriangles: Triangle is a polygon which has three sides. The sum of all the angles of a triangle is180°.Types of Triangles:1. Acute angle triangle: It is a triangle in which all the angles are less than 90°.2. Obtuse angle triangle: It is a triangle in which one angle is greater than 90°.3. Right angle triangle: It is a triangle in which one angle is equal to 90°.4. Equilateral triangle: It is a triangle in which all the sides are equal and the all angles are equal to 60°.5. Isosceles triangle: It is a triangle in which two sides are equal and angle opposite the equal sides are also equal.6. Scalene triangle: It is a triangle in which all the sides are unequal.General Properties of Triangles:1. Sum of the length of any two sides of a triangle is always greater than the third side.2. Difference between the lengths of any two sides of a triangle is always less than the third side.3. The side opposite to the greatest angle will be the greatest similarly the side opposite to the smallest angle will be the smallest. a b c4. Sine rule in a triangle: sinA = sinB = sinC = 2R, where a, b and c are the side of a triangle and A, B and C are the corresponding angles of the side. R is the circum radius.5. Cosine rule in a triangle: a2 = b2 + c2 – 2bc cosA.6. The exterior angle is equal to the sum of two interior angles which are not adjacent to it.Important terms1. Median: A line joining the mid-point of a side of a triangle to the opposite vertex is called a median. The median divides a triangle into two parts of equal area and the point where all the medians meet is called the centroid of the triangle. The centroid divides each median in the ratio 2 : 1.2. Altitudes: Altitudes are the perpendiculars drawn from any vertex to the opposite sides. All the altitudes meet at the orthocentre of the triangle. The angle made by any side at the orthocentre and vertical angle is together equal to 180°. Here,  A +  BOC = 180°.
• 26. 3. Perpendicular Bisectors: It is a line which bisects a side and at the same time perpendicular to the side. The point at which the perpendicular bisectors of the sides meet is called the circumcentre of the triangle. The circumcentre of a triangle circumscribes the triangle and the circle is unique for the triangle. The angle made by a side of a triangle at the circumcentre is double to the opposite angle. Here, 2  A =  BOC.Circles: A circle is a set of points in a plane which are equidistant from a point O. The point O iscalled centre of the circle and the distance from the centre O is called radius which is generallyrepresented as r.Properties1. There can be only one circle passing through three or more non-collinear points in a plane.2. If two circles intersect in two points then the line through the centres is the perpendicular bisector of the common chord.For Example1. ABCD is a square. M is the midpoint of AB and N is the mid-point of BC. DM and AN are joined and they meet at O. Then which of the following is correct? (a) OA : OM = 1 : 2 (b) AN = MD (c)  ADM =  ANB (d)  AMD =  BANAnswer: - (b)2. AB = 8 cm and CD = 6 cm are two parallel chords on the same side of the centre of a circle. The distance between them is 1 cm. The radius of the circle (in cm) is (SSC) (a) 5 (b) 4 (c) 3 (d) 2Answer: - (a)
• 27. 3. The circumcentre of a triangle ABC is O. If  BAC = 85° and  BCA = 75°, then the value of  OAC is (a) 40° (b) 60° (c) 70° (d) 90°Answer: - (a)4. AB is a chord of a circle of radius 7 cm. The chord subtends a right angle at the centre of the 22 circle. Find the area of the minor segment in sq cm. Use 𝛑 = 7 . (a) 7 (b) 24.5 (c) 14 (d) 38.5Answer:- (c)MENSURATIONBelow are some important formulae which are very useful to solve questions.1. Area of a rectangle = Length × Breadth2. In a rectangle, (Diagonal)2 = (Length)2 + (Breadth)2.3. Perimeter of a rectangle = 2(Length + Breadth). 14. Area of square = (Side)2 = 2 (Diagonal)2.5. Perimeter of a square = 4 × Side.6. Area of 4 walls of a room = 2(Length + Breadth) × height7. Area of a parallelogram = Base × Height. 18. Area of a rhombus = 2 (Product of Diagonals). 39. Area of an equilateral triangle = 4 (Side)2.10. Perimeter of an equilateral triangle = 3(Side). a +b +c11. Area of a triangle = s s − a s − b s − c , where s = 2 .12. Area of a circle = 𝛑r2, where r is the radius of the circle.13. Circumference of a circle = 2𝛑r 2πrθ14. Arc AB on a circle = 360° , where AOB = 𝛉 and O is the centre. πr 2 θ15. Area of sector AOB = 360° .16. Volume of a cube = a3, where a is a side or edge.17. Whole suface area of a cube = 6a2.
• 28. 18. Diagonal of a cube = √3 a.19. Volume of a cuboid = lbh, where l = length, b = breadth and h = height.20. Whole suface area of a cuboid = 2(lb + bh + hl).21. Diagonal of a cuboid = l2 + b 2 + h2 .22. Volume of a cylinder = 𝛑r2h.23. Curved surface area of a cylinder = 2𝛑rh.24. Total surface area of a cylinder = (2𝛑rh + 𝛑r2). 425. Volume of a sphere = 3 πr 3 .26. Surface area of a sphere = 4𝛑r2. 227. Volume of a hemisphere = 3 πr 3 .28. Curved surface area of a hemisphere = 2𝛑r2.29. Whole surface area of a hemisphere = 3𝛑r2.30. Right circular Cone, I. Slant height l = r 2 + h2 , where r = radius of base and h = height. 1 II. Volume of a cone = 3 πr 2 h. III. Curved surface area of a cone = 𝛑rl = 𝛑r r 2 + h2 . IV. Total surface area of a cone = (𝛑rl + 𝛑r2). 131. Slant surface area of a pyramid = 2 (Perimeter of the base)×(Slant height). 132. Total surface are of a pyramid = 2 (Perimeter of the base)×(Slant height) + area of base. area of the base33. Volume of pyramid = ×Height. 3For Example1. There is a pyramid on a base which is a regular hexagon of side 2a cm. If every slant edge of 5a this pyramid is of length 2 cm, then the volume of this pyramid is (SSC) (a) 3a3 cm3 (b) 3 2a3 cm3 (c) 3 3a3 cm3 (d) 6a3 cm3 3 5a 2Answer:- (c) Here area of base = 6 × 4 (2a)2 = 6 3a2 and height = − 2a 2 = 2 25a 2 9a 2 3a 6 3a2 × 3a = − 4a2 = = , then the volume of a pyramid = 3 3a. 4 4 2 3 22. The area of the four walls of a room is 660 m2 and its length is twice its breadth. If the height of the room is 11 m, then the area of its floor (in m2) is (a) 120 (c) 200 (b) 150 (d) 330
• 29. Answer:- (c) Let b be the Breadth, l be the Length and h be the Height of the room. 2 × (l + b) ×h =660  2 × (2b + b) × 11 = 660  2 × 3b × 11 = 660  66b = 660 ∴ b = 10, then l = 200.Therefore, area of the floor = 20 × 10 = 200.3. A cylindrical rod of iron whose height is five times its radius is melted and cast into spherical balls each of half the radius of the cylinder. the number of such spherical balls is (a) 5 (b) 30 (c) 10 (d) 25Answer:- (b) Let the given radius of the cylinder be r and the height be 5r then the volume of r 4 r 3the cylinder =  r2(5r) = 5  r3. Again the volume of one sphere whose radius is 2 = 3 π 2 =4 r3 πr 3 5πr 3 π = . Therefore the number of spheres in the volume 5  r3 = πr 3 = 303 8 6 6TRIGONOMETRYPythagoras Theorem: In a right-angled triangle,h2 = p2 + b2 Here, h is the hypotenuse, p is the perpendicular and b is the base of the rightangledTrigonometric Ratios: the trigonometric ratios are calculated according to the angles. Thehypotenuse is the largest side of a right angled triangle, the perpendicular is the opposite sideof the angle and the base is the adjacent side.For example,Here, for angle  R, PR = hypotenuse, PQ = perpendicular and QR = base. P Q R AThe Trigonometric ratios are Sin Ѳ p Tan Ѳ p Sec Ѳ h h b b Cos Ѳ b Cot Ѳ b Cosec Ѳ h h p p
• 30. The important values of Trigonometric ratios Ѳ Sin Cos Tan Cot Sec Cosec 30® 1 3 1 √3 2 2 2 2 3 3 45® 1 1 1 1 √2 √2 2 2 60® 3 1 √3 1 2 2 2 2 3 3Some important formulae 1 1 1Sin Ѳ = cosec Ѳ, cos Ѳ = sec Ѳ, tan Ѳ = cot Ѳ, sin (90® - Ѳ) = cos Ѳ, tan (90®- Ѳ) = cot Ѳ,Sec (90® - Ѳ) = cosec Ѳ, cos(90®- Ѳ) = sin Ѳ, cot (90® - Ѳ) = tan Ѳ,Cosec (90® - Ѳ) = sec Ѳ, Sin2Ѳ + Cos2Ѳ = 1, Sec2Ѳ – Tan2Ѳ = 1, Cosec2Ѳ – Cot2Ѳ = 1.Angle of Elevation: If a person looking at an object whichis placed above the line of his eye. Then, the angle whichis the made by the object’s line with the eye and thehorizontal line is called the angle of elevation.Angle of Depression: If a person looking down at anobject which is placed below the level of his eye. Then,the angle which is made at his eye by the object’s linewith the horizontal line of his eye is called the angle ofdepression.For Example 3 5sin Ѳ+2cos Ѳ1. If tan Ѳ = 4, then the value of 4 cos 2 Ѳ−3 sin 2 Ѳ is (a) 1 (b) 0 115 (c) 37
• 31. 110 (d) 23 3 3 4Answer:- (c) Here tan 𝛉 = 4, then sin 𝛉 = 5 and cos 𝛉 = 5 from the Pythagoras Theorem.2. The value of tan 5°. tan 15°. tan 35°. tan 55°. tan 75°. tan 85° is (a) 1 (b) 0 (c) 0.5 (d) 0.75Answer:- (a) Here we can change tan 5° = tan (90° − 85°) = cot 85° and tan 𝛉 × cot 𝛉 = 1.3. A ladder leaning against a wall makes an angle of 60° with the ground. If the length of the ladder is 30m, find the height of the wall up to the ladder. (a) 15 (b) 15√3 (c) 30 (d) 30√3Answer:- (b) The diagram can be,DATA INTERPRETATIONThe questions in this unit can be asked from four different types of data.1. Histogram2. Frequency polygon3. Bar-diagram4. Pie-chartFor ExampleQuestion (1 to 5): The line diagram shows the cost of production and profit of five companiesfor the year 2011-12. (The figures are in 000).
• 32. 500 400 300 COST OF PRODUCTION 200 PROFIT 100 0 A B C D E1. The ratio of profits of company B to D is: (a) 2:3 (c) 4:3 (b) 3:4 (d) 3:22. The profit of company C is what percentage of the cost of production of company E? (a) 20% (c) 30% (b) 25% (d) 35%3. The cost of production of company B is how many times of company C profit? (a) 2.5 (c) 3.5 (b) 3 (d) 44. Which company has the maximum percentage of profit? (a) A (c) C (b) B (d) D5. What is the average profit of all the companies? (a) 50,000 (c) 70,000 (b) 60,000 (d) 80,000Answers:-1. The profit of B is 50,000 and D is 75,000 therefore the ratio is 2:3.2. The profit of C is 1,00,000 which is 25% of Es cost of production which is 4,00,000.3. The cost of B is 3,50,000 which 3.5 times the profit of C profit which is 50,000.4. The company C has a 40% profit.5. The total profit of all the companies are 75,000+50,000+1,00,000+75,000+1,00,000 = 4,00,000, therefore the average will be (4,00,000)/5= 80,000.
• 33. ENGLISH LANGUAGESYNONYMSSynonyms: The word comes from ancient Greek words ‘syn’ and ‘onoma’ where ‘syn’ means‘with’ and ‘onoma’ means ‘name’. Synonyms are the word which has similar meanings.Two words are said to be synonymous when they have similar meanings. The words caste andclass is synonymous because both the words have similar meaning.For Example,Q. Find the word which is synonymous to Deny. (SSC 2011) (a) Regain (b) Refuse (c) Repair (d) ReduceAnswer: The synonyms for Deny are Contradict, Refuse, Reject etc. here we have Refuse as anoption, which means the right answer is (b).Q. Find the word which is synonymous to Deposit. (a) Degrade (b) Dethrone (c) Place (d) RemovalAnswer:- (c)ANTONYMSAntonyms: The word “antonym “comes from ancient Greek words ‘anti’ and ‘onoma’ where‘anti’ means opposite and ‘onoma’ means name. The literal meaning of antonyms is theopposite name which means the word which has opposite meaning. For example; Day – Night,Long – Short, Up – Down, Small – Large, etc.For ExampleQ. Find the opposite word to Concur. (SSC 2011) (a) Disagree (c) Disarrange (b) Disappear (d) Discourage
• 34. Answer: The synonyms for Concur are agree, cooperate, combine etc. Therefore the antonymfor ‘agree’ is ‘disagree’. Hence right answer is (a).Q. Find the opposite word to Condensation. (a) Abridgment (c) Compression (b) Broadening (d) ConcentrationAnswer: - (b)IDIOMS AND PHRASESIdioms and phrases refer to commonly used groups of words in English. They are used inspecific situations and often used in an idiomatic, rather than a figurative sense. Idioms areoften full sentences. Phrases, however, are usually made up of a few words and are used as agrammatical unit in a sentence.For Example,1. Ram used very ugly words against his kind uncle; he threw down the gauntlet before him. (a) he abused and insulted him (b) he threw the challenge (c) he behaved as if he was a very great and important person (d) he put several conditions for negotiationAnswer: - (b)2. He always cuts both ends (a) Work for both sides (b) Inflicts injuries on others (c) Argues in support of both sides of the issue (d) Behaves dishonestlyAnswer: - (c)ONE WORD SUBSTITUTIONOne word substitution is the words that replace a group of words or sentences without creatingor changing the exact meaning of sentences. These words generally bring compression in anykind of writing.There are lots of words in English language that can be used effectively in place of complexsentences or words to make writing to the point without losing the meaning of the context.For Example,
• 35. 1. One who eats too much (a) Foodie (b) Glutton (c) Eater (d) Food lovingAnswer: - (b) glutton2. A book published after the death of its author (a) Posthumous (b) Anonymous (c) Synonymous (d) MysteriousAnswer: - (a) Posthumous
• 36. SPELLING TESTSpelling test is basically devised to test the vocabulary power and the candidate ability to writethe words with correct spellings.Question asked is such section generally required to choose the correct spelt word or the mis-spelt out of the alternatives given.For Example, find the mis – spelt word.1. (a) Comission (b) Commisson (c) Comession (d) CommissionAnswer: (d)2. (a) Liutenenat (b) Lieutanent (c) Lieutenant (d) LeiutanentAnswer: (c)INCORRECT SENTENCESDirections: In questions no. 1 to 2, a part of the sentence is underlined. Below are givenalternatives to the underlined part at (a), (b) and (c) which may improve the sentence. Choosethe correct alternative. In case no correction is needed, your answer is (d).1. All the allegations levelled against him were found to be baseless. (a) levelled for (b) level with (c) level against (d) no correctionAnswer:- (c)2. Last Sunday I went to the market and bought spectacles. (a) two spectacles (b) a pair of spectacles (c) a spectacle (d) no correctionAnswer:- (b)
• 37. SENTENCE RECONSTRUCTIONSentence reconstruction is the grammatical arrangement of words in sentences, phrasestructure, and syntax. The sentences are presented in a jumbled manner and the students aresupposed to arrange it chronologically.Here we are giving some examples. Students are advised to go through it carefully.1. There is a fashion now-a-days P. as an evil Q. who is born with a silver spoon R. to bewail poverty S. and to pity the youngman in his mouth (a) PSRQ (b) RPSQ (c) RSQP (d) SPRQAnswer:- (b)2. Though the government claims P. it has failed to arrest Q. the rate of inflation is down R. or the decrease S. the rise in prices 6. in the per capita income (a) PQRS (b) PQSR (c) PSQR (d) QPSRAnswer:- (d)SPOTTING THE ERRORDirections: Read each sentence to find out whether there is any grammatical mistake/error init. The error if any will be in one part of the sentence. Mark the number of the part with erroras your answer. If there is no error, mark the last option.1. (a) My brother lived at the top / (b) of an old house / (c) which attic had been / (d) converted into a flat. / (e) No error.Answer: - (c) Replace ‘which’ by ‘whose’.