www.jagranjosh.comSSC Higher Secondary Exam

(Preparation Booklet)

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

Mensuration.......................................................................................................................... 28

Trigonometry ........................................................................................................................ 30

Data Interpretation............................................................................................................... 32

English 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

INTRODUCTION

SSC 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 Lower Division Clerk. SSC Combined Higher Secondary Level (10+2) Examination 2012 is going to be conducted on 21 October 2012 & 28 October 2012. To be eligible for SSC Combined Higher Secondary Level (10+2) Examination 2012, the candidate must have passed 12th Standard or equivalent or higher examination from a recognized Board or University. In addition, the candidate’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 is followed 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 be negative marking of 0.25 marks for each wrong answer.

Syllabus for Objective Paper

General Intelligence 

Semantic 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, Figural Series, Critical Thinking, Problem Solving, Emotional Intelligence, Word Building, Social Intelligence, Coding and de‐coding, Numerical operations

English Language

Spot 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 Sentence parts, Shuffling of Sentences in a passage, Cloze Passage, Comprehension Passage

Quantitative Aptitude

1. Arithmetic

Number Systems: Computation of Whole Number, Decimal and Fractions, Relationshipbetween numbers

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 work

2. Algebra

Basic 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 Equations

3. Geometry

Familiarity with elementary geometric figures and facts:

Triangle and its various kinds of centres viz. Centroid, In‐centre, Orthocentre, CircumcentreCongruence and similarity of triangles

Circle and its chords, tangents, angles subtended by chords of a circle, common tangents to twoor more circles

4. Mensuration

Triangle, Quadrilaterals, Regular Polygons (sum of the internal angles of a polygon), Circle, RightPrism, Right Circular Cone, Right Circular Cylinder, Sphere, Hemispheres, Rectangular Parallelepiped, Regular Right Pyramid with triangular or square Base

5. Trigonometry

Trigonometry (for acute angles 0 with 0≤0≤90): Trigonometric ratios, Degree and RadianMeasures, Standard Identities like sin20 + Cos20=1 etc

Complementary Angles, Heights and Distances (simple problems only)

6. Statistical Charts

Use of Tables and Graphs: Histogram, Frequency polygon, Bar‐diagram, and Pie‐chart

General Awareness

History, Culture Geography, Economic Scene, General policy and Scientific Research

GENERAL INTELLIGENCE

ANALOGY

Analogy: 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 form of 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) Rink 

Answer: ‐ (c) Court. Here Cricket is played at Pitch and Tennis is played at Court.

CLASSIFICATION

Classification: 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 Examples

1.   (a) 36(b) 125(c) 144(d) 121

Answers: ‐ (b) All except 125 are squares.

SERIES

Series: 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 wrong term in the series.

For Example

1. 6, 11, 18, 27, 38, ?.(a) 49(b) 50

(c) 51(d) 52

Answer: ‐ (c) The series is +5, +7, +9……

CODING­DECODING

Coding: 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 language how PEN be written?

Answer:‐ In the above example the letters of BOOK is coded on the basis of their position in the English alphabet. The letters B=2, O = 15, K = 11. In the same way we can encode PEN as 16514.

SEQUENCE

The 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. Triangle

Answer: ‐ The right order for these words will be Line – Angle – Triangle – Rectangle –Pentagon. Therefore the answer is 1, 2, 5, 4, and 3.

DIRECTION TEST

In this unit, the problems are generally formulated with successive follow‐up of the directions and the candidates are required to find the final direction, the final position with respect to the initial position or the distance between the final position and initial position.On paper, we can draw the directions as;

(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 position

We can draw his path as;His Home 

For Example, A starts walking towards East, turns right, again turn right, turns left, again turns left, turns left, again turns left. In which direction is A walking now?

(a) East(b) West(c) North(d) South

Answer:‐ (b)

PUZZLES

The 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 the father of R because S and R are sisters. Therefore Q is the uncle of R. because Q is the father’s brother 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 tallerthan Satish. Who is taller than Rajesh but shorter than Anil? (SSC 2011)

Answer: It is given that Kishore>Satish>Rajesh and Kishore>Anil>Satish. It means the sequence of height is Mohan>Kishore>Anil>Satish> Rajesh. Therefore, the person who is taller thanRajesh but shorter than Anil is Satish.

MATHEMATICAL PUZZLES

Introduction: 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 SSC exams. After analyzing previous papers, it is clear that in the exam you can get 5 to 8 questions from this unit.

The question types

1. Arrangement of correct mathematical operations.2. Filling the correct mathematical operations.3. Mathematical‐Logical problems

When you are solving these questions you should follow basic mathematical operations in a correct 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 – Addition

S – Subtraction.

For Examples

1. 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)

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) 343

Answer:‐ (d)

SYLLOGISM

Syllogism: The literal meaning of syllogism is ‘Conclusion’ or ‘inference’. The questions in the syllogism 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 statements

1. All As are Bs.

2. Some As are Bs.

3. No A is a B.

For Example

Direction (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)

NON­VERBAL

The 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.

For Example

1. Which answer figure will complete the pattern in the question figure? (SSC 2011)

Question figure:

Answer figures:

Answer: (D) will complete the pattern

2. 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)

QUANTITATIVE APTITUDE

HCF & LCM

Some Important concepts

Factors 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 more numbers. 

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 the given 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.

Co­primes: Two numbers are co‐primes if their HCF = 1.

HCF and LCM of Fraction: 

HCF = 𝐇𝐂𝐅 𝐨𝐟 𝐧𝐮𝐦𝐞𝐫𝐚𝐭𝐨𝐫𝐬

𝐋𝐂𝐌 𝐨𝐟 𝐃𝐞𝐧𝐨𝐦𝐢𝐧𝐚𝐭𝐨𝐫𝐬and LCM = 

𝐋𝐂𝐌 𝐨𝐟 𝐧𝐮𝐦𝐞𝐫𝐚𝐭𝐨𝐫𝐬

𝐇𝐂𝐅 𝐨𝐟 𝐃𝐞𝐧𝐨𝐦𝐢𝐧𝐚𝐭𝐨𝐫𝐬

For Example

The L.C.M. of two numbers is 189 and H.C.F. of two numbers is 9.If one of the number is 63 then other number is:

(a) 27(b) 28(c) 29(d) 30

Answer:‐ (a)

RATIO & PROPORTION

Ratio: A ratio is a comparison of two numbers (quantity in the same unit). It is written as, a: b =a

b= a ÷ b, where a and b are two number (quantity).

In 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

Solution: the required ratio is =2 × 3 ×7

3 ×4 ×11=

7

22.

Inverse Ratio: If a : b is a given ratio, then1

a∶1

bor b : a is called its inverse ratio or reciprocal 

ratio. 

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.

Proportion: A proportion expresses the equality of two ratios. e.g. a

b=

c

d

Or 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 extremes and the second and the third terms or the middle terms are called as the mean terms. When

four quantities are in proportion, we can write it in the mathematical form as a : b :: c : d �a

b=

c

d� ad = bc

For Example,

If Rs. 510 be divided among A, B, C in such a way that A gets 2

3of what B gets and B gets

1

4of

what 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 these

Answer:‐ (b)

PERCENTAGE

Percentage: The term “per cent” means “for every hundred”. A fraction whose denominators is 100 is called a percentage and the numerator of the fraction is called the rate per cent. It isdenoted by the symbol %.

Here x % =x

100. For example 10% =

10

100=

1

10. 

Similarly, fraction can be changed in the form of percentage when we multiply them by 100. 

For Example1

4=

1

4× 100 = 25%.

To decrease a number by a given %:

Multiply the numbers by the factor𝟏𝟎𝟎−𝐑𝐚𝐭𝐞

𝟏𝟎𝟎

To find the % increase of a number:

% increase =𝐓𝐨𝐭𝐚𝐥 𝐢𝐧𝐜𝐫𝐞𝐚𝐬𝐞

𝐈𝐧𝐢𝐭𝐢𝐚𝐥 𝐯𝐚𝐥𝐮𝐞× 𝟏𝟎𝟎 =

𝐅𝐢𝐧𝐚𝐥 𝐯𝐚𝐥𝐮𝐞−𝐈𝐧𝐢𝐭𝐢𝐚𝐥 𝐯𝐚𝐥𝐮𝐞

𝐈𝐧𝐢𝐭𝐢𝐚𝐥 𝐯𝐚𝐥𝐮𝐞× 𝟏𝟎𝟎

To find the % decrease of a number:

% decrease =𝐓𝐨𝐭𝐚𝐥 𝐝𝐞𝐜𝐫𝐞𝐚𝐬𝐞

𝐈𝐧𝐢𝐭𝐢𝐚𝐥 𝐯𝐚𝐥𝐮𝐞× 𝟏𝟎𝟎 =

𝐅𝐢𝐧𝐚𝐥 𝐯𝐚𝐥𝐮𝐞−𝐅𝐢𝐧𝐚𝐥 𝐯𝐚𝐥𝐮𝐞

𝐈𝐧𝐢𝐭𝐢𝐚𝐥 𝐯𝐚𝐥𝐮𝐞× 𝟏𝟎𝟎

For Example

1. In an election between two candidates, 75% of the voters casted their votes, out of which2% of the votes were declared invalid. A candidate got 9261 votes which were 75% of thetotal valid votes. Find the total number of votes enrolled in that election. (SSC 2003)

(a) 15000(b) 16000

(c) 16800(d) 17000

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)] = 9261

2. If 50% of (x – y) = 30% of (x + y), then what percent of x is y?(a) 25 %(b) 50 %(c) 75 %(d) 100 %

Answer:‐ (a) here 50% of (x – y) = 30% of (x + y)� 50

100(x− y) =

30

100(x + y)

�10(x – y) = 6(x + y) �10x – 6x = 6y + 10y�4x = 16y � 1

4x = y �25 % of x = y.

Because1

4= 25%

AVERAGE

Average is defined as the sum of n different numerical values divided by n.

Average = 𝐒𝐮𝐦 𝐨𝐟 𝐧 𝐝𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐭 𝐍𝐮𝐦𝐞𝐫𝐢𝐜𝐚𝐥 𝐕𝐚𝐥𝐮𝐞

𝐧

Average Speed = 𝐓𝐨𝐭𝐚𝐥 𝐃𝐢𝐬𝐭𝐚𝐧𝐜𝐞

𝐓𝐨𝐭𝐚𝐥 𝐒𝐩𝐞𝐞𝐝

If a person covers half of his journey at a speed of x km/h and the next half at the speed of y

km/h, then the average speed during the whole journey is �𝟐𝒙𝒚

𝒙 + 𝒚�.

Weighted Average of x1, x2 where weight isw1, w2 = 𝒘𝟏𝒙𝟏 +𝒘𝟐𝒙𝟐

𝒘𝟏 +𝒘𝟐.

Average of different groups

If the average of a group of n items is a and the average of another group of m items is m, then

the combined average =𝐧𝐚 +𝐦𝐛

𝐧 +𝐦.

For Example

1. 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) 295

Answer:‐ (a) Let the number be x. Then

x2+ x

2= 5x �x2 + x = 10x�x2 + x − 10x = 0�x(x− 9) = 0 ∴ x = 9.

SURDS AND INDICES

Laws of Indices:

(a) am × an = am + n

(b)am

an= am – n 

(c) (am) n = a m n

(d) (a b)m = a m b m

(e) �a

b�m

=am

bm

(f) a0 = 1

Laws of Surds:

(a) √an

= a1

n

(b) √abn

= √an

× √bn

(c) �a

b

n=

√an

√bn

(d) �√an

�n= a

(e) �√anm

= √amn

(f) �√an�m= √am

n

For Example,

1. Find the value of (�81)1

4.

(a) 3

(b) √3

(c) √9

(d) √33

Answer: ‐ (b) (�81)1

4 = (3 × 3 × 3 × 3 )1

2×1

4 = (34 )1

2×1

4 = (3)1

2 =√3.

2. (−1

64)−

4

3  = ?

(a) 254(b) ‐254(c) 256(d) ‐256

Answer:‐ (c) (−1

64)−

4

3  = (−1

4)3 ×

(−4)

3 = (−1

4)−4 = (−4)4 = 256.

PROFIT & LOSS

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 is positive, then it is called a profit and if the difference is negative, then it is called a loss.

Important Formulae

1. SP = CP + Profit = CP – Loss 

2. Gain % = �Gain × 100

C.P.�

3. Loss % = �Loss × 100

C.P.�

4. SP = (100 + Gain %)

100× CP = 

(100 − Loss %)

100× CP

5. CP = 100

(100 + Gain %)× SP = 

100

(100 −Loss %)× SP

For Example

1. Arjun buys an old car for Rs 1, 12, 000 and spends 8000 on its repairs. If he sells the Car forRs 1,80,000, his gain percent is:

(a) 50(b) 51

(c) 52(d) 53

Answer:‐ (a) C.P. = Rs. (1,12,000 + 8,000) = Rs. 1,20,000, S.P = 1,80,000, 

Gain % = (60,000

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%

Answer:‐ (b) S.P. = a, Loss = a/4, CP. =a + a

4= 5a

4�Loss% =( 

a

4×

4

5a× 100) % = 20%

INTEREST CALCULATION

Important Facts and Formulae related to Interest calculation

Principal: It is the money which is borrowed or lend out for certain period. Generally, Principal is 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 the amount of interest which is calculated at every Rs. 100 for a fixed period. Generally for 1 year.

Simple Interest (S. I.) =�P × R × T

100�, P =�

S.I.× 100

R × T�, R =�

S.I.× 100

P × T�, T = �

S.I.× 100

P × R�

Compound Interest (C. I.) = P �1 +R

100�n

− P (When interest is compounded annually)

Compound Interest (C. I.) = P�1 +�R

2�

100�

2n

− 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.

Then Compound Interest =P �1 +R1

100� �1 +

R2

100� �1 +

R3

100� − P.

For Example

1. Find the compound interest on Rs. 160, 000 in 2 years at 10 % per annum. The interestbeing compounded half yearly.

(a) 32000(b) 33000(c) 34000(d) 34481

Answer:‐ (d) The required C.I. = 160, 000 �1 +�10

2�

100�

2×2

− 160, 000

= 160, 000 �21

20� �

21

20� �

21

20� �

21

20� − 160, 000 = 194, 481 − 160, 000 = 34, 481

2. 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, 000

Answer:‐ (c) Let the required principal be P. then

P �1 +10

100�2

− P−P × 10 × 2

100= 200 � P �

121

100− 1−

1

5� = 200�P �

1

100� = 200

Therefore, P = 20, 000

TIME, SPEED & DISTANCE

Important facts and formulae

Distance = (Speed × Time)

Speed = Distance

Time

Time =Distance

Speed

x km/h = �x ×5

18�m/s

x m/s = �x ×18

5� km/h

Average Speed = Total Distance

Total Speed

If a person covers half of his journey at a speed of x km/h and the next half at the speed of y

km/h, then the average speed during the whole journey is �𝟐𝒙𝒚

𝒙 + 𝒚�.

In 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/h

Speed upstream = (a + b) km/h

If the speed downstream is p km/h and the speed upstream is b km/h, then:

Speed in still water = 𝟏

𝟐(𝐚+ 𝐛) km/h

Rate of stream = 𝟏

𝟐(𝐚 − 𝐛) km/h

For Example

1. A man travelled from the village to the post‐office at the rate of 25km/h and walked back atthe rate of 4 km/h. If the whole journey took 5 hours 48 minutes. Find the distance of thepost‐office from the village. (SSC, 2004)

(a) 10 km(b) 15 km

(c) 20 km(d) 25 km

Answer: ‐ (c) The average speed for the whole journey = 2 × 25 × 4

25 + 4=

200

29km/h. the total time is 

548

60= 5

4

5=

29

5h. ∴ the total distance =

200

29×

29

5= 40 km. Therefore the distance from village 

to the post office = 40

2= 20 km.

2. In what time will a train 100 m long, cross an electric pole, if its speed be 144 km/h?

(a) 2.5 seconds(b) 3 seconds

(c) 3.5 seconds(d) 4 seconds

Answer: ‐ (a) The speed in m/s = 144 × 5

18= 40 m/s and the train has to cross its own length to

cross a pole. The length of the train is 100 m. therefore the required time is 100

40= 2.5 s

TIME & WORK

Some Important Facts and Formulae

1. If m persons can complete a work in p days, then the same work will be completed by n

persons inm

n× p days.

2. If P can complete a piece of work in A days, then P’s one day work = 1

Aof the complete work

and if P can do1

Aof 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, 

then the 1 day work of both of them together = �1

m+

1

n� =

m+n

mnof the work and both of

them together complete the work inmn

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 ratioof times taken by both of them to finish the work = 1 : 2.

For Example

1. Aryan takes 5 hours to do a job, and Aryaman takes 6 hours to do the same job. How longshould it take if both Aryan and Aryaman, working together but independently, to do the same job?

(a) 3 days

(b) 28

11days

(c) 29

11days

(d) 4 days

Answer: ‐ (b)

2. A can finish a work in 12 days and B can do the same work in1

3the time taken by A. Then,

working together, what part of same work they can finish in a day?

(a)1

5

(b)1

4

(c)1

3

(d)1

2

Answer:‐ (c)

ELEMENTARY ALGEBRA

Below 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 + b3

5. (a − b)3 = a3 − 3a2b + 3ab2 − b3 

6. 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 + n

12. (am)n = am n

13. (a b)m = a m b m

14.𝑎𝑚

𝑎𝑛= x m – n.

The 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). Therefore 

a(−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)

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.

2. If a +1

a= 2, find the value of

4a

3a2 + 4a − 3. 

(a) 0(b) 1(c) 2(d) −1

Answer: ‐ (b) Solve a +1

a= 2 and put the value in

4a

3a2 + 4a − 3.

3. If a2 + b2 + 2b + 1 = 0, then find the value of a34 + b38.(a) 1(b) 0(c) −1(d) 2

Answer: ‐ (a) Here a = 0 and b = −1, then a34 + b38 = 1

GEOMETRY

Concepts

Triangles: 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.

4. Sine rule in a triangle: a

sinA=

b

sinB=

c

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 terms

1. 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 theorthocentre of the triangle. The angle made by any side atthe orthocentre and vertical angle is together equal to 180°.Here, � A + � BOC = 180°.

3. Perpendicular Bisectors: It is a line which bisects a side and at the same time perpendicular to theside. The point at which the perpendicular bisectors of the sides meet is called the circumcentre of the triangle. The circumcentreof a triangle circumscribes the triangle and the circle is unique for the triangle. The angle madeby a side of a triangle at the circumcentre isdouble 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 generally represented as r.

Properties

1. 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 Example

1. 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 = � BAN

Answer: ‐ (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) 2

Answer: ‐ (a)

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 

circle. Find the area of the minor segment in sq cm. Use 𝛑 =22

7.

(a) 7(b) 24.5(c) 14(d) 38.5

Answer:‐ (c)

MENSURATION

Below 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).

4. Area of square = (Side)2 = 1

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.

8. Area of a rhombus = 1

2(Product of Diagonals).

9. Area of an equilateral triangle =√3

4(Side)2.

10. Perimeter of an equilateral triangle = 3(Side).

11. Area of a triangle = �s(s− a)(s− b)(s− c), where s =a + b + c

2.

12. Area of a circle = 𝛑r2, where r is the radius of the circle.13. Circumference of a circle = 2𝛑r

14. Arc AB on a circle = 2πrθ

360°, where AOB = 𝛉 and O is the centre.

15. Area of sector AOB =πr2θ

360°.

16. Volume of a cube = a3, where a is a side or edge.17. Whole suface area of a cube = 6a2.

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 + b2 + h2.22. Volume of a cylinder = 𝛑r2h.23. Curved surface area of a cylinder = 2𝛑rh.24. Total surface area of a cylinder = (2𝛑rh + 𝛑r2).

25. Volume of a sphere =4

3πr3.

26. Surface area of a sphere = 4𝛑r2.

27. Volume of a hemisphere = 2

3πr3.

28. Curved surface area of a hemisphere = 2𝛑r2.29. Whole surface area of a hemisphere = 3𝛑r2.30. Right circular Cone,

I. Slant height l = √r2 + h2, where r = radius of base and h = height.

II. Volume of a cone = 1

3πr2h.

III. Curved surface area of a cone = 𝛑rl = 𝛑r√r2 + h2.IV. Total surface area of a cone = (𝛑rl + 𝛑r2).

31. Slant surface area of a pyramid =1

2(Perimeter of the base)×(Slant height).

32. Total surface are of a pyramid =1

2(Perimeter of the base)×(Slant height) + area of base.

33. Volume of pyramid = �area of the base

3� ×Height.

For Example

1. There is a pyramid on a base which is a regular hexagon of side 2a cm. If every slant edge of 

this pyramid is of length5a

2cm, then the volume of this pyramid is  (SSC)

(a) 3a3 cm3

(b) 3√2a3 cm3

(c) 3√3a3 cm3

(d) 6a3 cm3

Answer:‐ (c) Here area of base = 6 ×√3

4(2a)2 = 6√3a2 and height = ��

5a

2�2

− (2a)2 =

�25a2

4− 4a2 = �

9a2

4=

3a

2, then the volume of a pyramid =

6√3a2

3

× 3a

2

=  3√3a.

2. 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(b) 150

(c) 200(d) 330

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) 25

Answer:‐ (b) Let the given radius of the cylinder be r and the height be 5r then the volume of

the cylinder = � r2(5r) = 5� r3. Again the volume of one sphere whose radius is r

2= 4

3π �

r

2�3

= 

4

3πr3

8= πr3

6. Therefore the number of spheres in the volume 5� r3 = 

5πr3

πr3

6

= 30

TRIGONOMETRY

Pythagoras 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 rightangled

Trigonometric Ratios: the trigonometric ratios are calculated according to the angles. The hypotenuse is the largest side of a right angled triangle, the perpendicular is the opposite side of 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

The Trigonometric ratios are

Sin Ѳ p

hTan Ѳ p

bSec Ѳ h

b

Cos Ѳ b

h

Cot Ѳ b

p

Cosec Ѳ h

p

A

The important values of Trigonometric ratios

Ѳ Sin Cos Tan Cot Sec Cosec

30® 1

2√3

2

1

√3

√3 2

√3

2

45® 1

√2

1

√2

1 1 √2 √2

60® √3

2

1

2

√3 1

√3

2 2

√3

Some important formulae

Sin Ѳ =1

cosec Ѳ, cos Ѳ =

1

secѲ, tan Ѳ =

1

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

1. If tan Ѳ =3

4, then the value of

5sinѲ+2cosѲ

4 cos 2 Ѳ−3 sin 2 Ѳis

(a) 1 (b) 0

(c)115

37

(d)110

23

Answer:‐ (c) Here tan 𝛉 = 3

4, then sin 𝛉 = 

3

5and cos 𝛉 = 

4

5from 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) 0.75

Answer:‐ (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 theladder is 30m, find the height of the wall up to the ladder.(a) 15(b) 15√3(c) 30(d) 30√3

Answer:‐ (b) The diagram can be, 

DATA INTERPRETATION

The questions in this unit can be asked from four different types of data.

1. Histogram2. Frequency polygon3. Bar­diagram4. Pie­chart

For Example

Question (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').

1. The ratio of profits of company B to D is:

(a) 2:3(b) 3:4

(c) 4:3(d) 3:2

2. The profit of company C is what percentage of the cost of production of company E?

(a) 20%(b) 25%

(c) 30%(d) 35%

3. The cost of production of company B is how many times of company C profit?

(a) 2.5(b) 3

(c) 3.5(d) 4

4. Which company has the maximum percentage of profit?

(a) A(b) B

(c) C(d) D

5. What is the average profit of all the companies?

(a) 50,000(b) 60,000

(c) 70,000(d) 80,000

Answers:‐

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 E's 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.

0

100

200

300

400

500

A B C D E

COST OF PRODUCTION

PROFIT

ENGLISH LANGUAGE

SYNONYMS

Synonyms: 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) Reduce

Answer: 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) Removal

Answer:‐ (c) 

ANTONYMS

Antonyms: 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 the opposite name which means the word which has opposite meaning. For example; Day – Night, Long – Short, Up – Down, Small – Large, etc.

For Example

Q. Find the opposite word to Concur.   (SSC 2011) 

(a) Disagree

(b) Disappear 

(c) Disarrange

(d) Discourage

Answer: The synonyms for Concur are agree, cooperate, combine etc. Therefore the antonym for ‘agree’ is ‘disagree’. Hence right answer is (a).

Q. Find the opposite word to Condensation.

(a) Abridgment(b) Broadening

(c) Compression(d) Concentration

Answer: ‐ (b)

IDIOMS AND PHRASES

Idioms 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 negotiation

Answer: ‐ (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 dishonestly

Answer: ‐ (c)

ONE WORD SUBSTITUTION

One word substitution is the words that replace a group of words or sentences without creating or changing the exact meaning of sentences. These words generally bring compression in any kind of writing. 

There are lots of words in English language that can be used effectively in place of complex sentences or words to make writing to the point without losing the meaning of the context.

For Example,

1. One who eats too much(a) Foodie(b) Glutton(c) Eater(d) Food loving

Answer: ‐ (b) glutton

2. A book published after the death of its author(a) Posthumous(b) Anonymous(c) Synonymous(d) Mysterious

Answer: ‐ (a) Posthumous

SPELLING TEST

Spelling test is basically devised to test the vocabulary power and the candidate ability to write the 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) Commission

Answer: (d)

2.(a) Liutenenat(b) Lieutanent(c) Lieutenant(d) Leiutanent

Answer: (c)

INCORRECT SENTENCES

Directions: 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 correction

Answer:‐ (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 correction

Answer:‐ (b)

SENTENCE RECONSTRUCTION

Sentence 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 evilQ. who is born with a silver spoonR. to bewail povertyS. and to pity the youngman

in his mouth

(a) PSRQ(b) RPSQ(c) RSQP(d) SPRQ

Answer:‐ (b)

2. Though the government claimsP. it has failed to arrestQ. the rate of inflation is downR. or the decreaseS. the rise in prices6. in the per capita income(a) PQRS(b) PQSR(c) PSQR(d) QPSR

Answer:‐ (d)

SPOTTING THE ERROR

Directions: 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’.

2. (a) All companies must / (b) send its annual report to/ (c) its shareholders twenty one days /(d) before the annual general meeting./ (e) No error.

Answer: ‐ (a) Use ‘every’ in place of ‘all’.

3. (a) They agreed / (b) to repair the damage / (c)freely of charge / (d)No error

Answer: ‐ (c)

4. (a) Radha was trying for admission / (b) in the Science College / (c) even though her parentswanted/ (d) her to take up medicine. / (e) No error.

Answer: ‐ (c) Use ‘had wanted’ in place of ‘wanted’.

5. (a) My younger sister and / (b) I am interested / (c) in mathematics./ (d) No error

Answer: ‐ (b) Use ‘are’ in place of ‘am’.  

6. (a)  You should purchase/ (b) this  insurance  policy/  (c) as  the  company/  (d) offers  manybenefits./ (e) No error.

Answer: ‐ (e) No error.

7. (a) Having acquired some experience/(b) she is no longer/ (c) one of those who believes/ (d)every explanation she is given./(e) No error

Answer: ‐ (d) Replace ‘is’ with ‘has’

CLOZE TEST

Cloze test is the test of the ability to comprehend text in which the reader has to supply themissing words that have been removed from the text at regular intervals. 

It is basically a test for diagnosing reading ability; words are generally deleted from a prosepassage and the reader is required to fill in the blanks.

Directions (Q. 1‐1O): In the following passage there arc blanks, each of which has beennumbered. These numbers are printed below the passage and against each, five words aresuggested, one of which fits the blank appropriately. Find out the appropriate word in eachcase. 

Raju was orphaned at a very (1) age. He lost (2) of his parents because his family was (3) poor toafford any treatment. He was deeply affected by this. He decided to work hard and (4) on his own feet. Along with his studies he (5) part‐time in a restaurant and earned enough money inorder to (6) his studies. Being a brilliant student, he earned scholarships and later managed toget admission in a good medical college. He is now a (7) doctor and helps poor patient for (8).This shows that one can survive every (9) condition. Determine and hard work (10) to success inlife.

1. (a) youth (b) early (c) childhood (d) childish (e) recent2. (a) any (b) few (c) either  (d) couple (e) both3. (a) little (b) so  (c) too (d) some  (e) ample 4. (a) balance(b) establish (c) stand (d) erect (e) talk5. (a) worked (b) employed (c) busy (d) established(e) functioned6. (a) done (b) practice  (c) follow (d) pursue (e) proceed7. (a) recognise  (b) know (c) respected (d) worth (e) merit8. (a) favour  (b) less  (c) penny  (d) subsidy (e) free 9. (a) failed (b) adverse  (c) worse  (c) evil  (e) tragedy10. (a) result (b) follow (c) excel  (d) lead (e) urge

Answers:

1. (b)2. (e)

3. (c)4. (c)

5. (a)6. (d)

7. (c)8. (e)

9. (b)10. (d)