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Learning Material. For Examination For Recruitment of Clerical Cadre Personnel INSTITUTE OF BANKING PERSONNEL SELECTION, MUMBAI Post Box No. 8587, IBPS House, Near Thakur Polytechnic, 90´ D. P. Road, Off. Western Express Highway, Kandivali (East), MUMBAI 400 101.

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  1. Learning Material For Examination For Recruitment of Clerical Cadre Personnel INSTITUTE OF BANKINGPERSONNEL SELECTION, MUMBAI Post Box No. 8587, IBPS House, Near Thakur Polytechnic, 90´ D. P. Road, Off. Western Express Highway, Kandivali (East), MUMBAI 400 101

  2. NATURE OF THE WRITTEN EXAMINATION The Selection process for Clerical Cadre consists of the following : 1. Written Examination 2. Personal Interview This presentation provides information only about the first part i.e., the Written Examination : Written Examination : The written examination will comprise of objective type of tests as follow : The details of the tests are given below : Sr. No. Test Questions Max. marks (i) General Awareness 40 40 (ii)General English 40 40 (iii) Quantitative Aptitude 40 40 (iv) Reasoning Ability 40 40 (v) Marketing / Computer 40 40 Except the Test of General English all the questions in the tests will be printed in Hindi and English. For these objective type tests, you will be given a composite time of 2 Hours 15 Minutes. Candidate has to pass in each test separately.

  3. Passing Marks : The candidate MUST pass in each of the objective type tests. The passing marks in each of the five tests will be decided by the Bank on the basis of the performance of all the competing candidates taken together in each test subject to a minimum required level. Candidates are also required to score a minimum of 40% ( 35% for SC/ST/PWD/XS Category) on aggregate. Candidates who pass in each test and on aggregate and rank sufficiently high on the basis of the aggregate marks in the objective tests as stated above will be called for Personal Interview. Bank reserves the right to change this cut-off. The number of such candidates would depend upon the number of vacancies to be filled in. Final selection will be on the basis of candidate’s performance in the written examination, and Personal Interview. Sample Questions : Below are given some sample questions for each of these tests. The type of questions are only illustrative and not exhaustive. In actual test you may find questions on some or all these types and also questions on the type not mentioned here. HOW TO INDICATE YOUR ANSWER : How to solve an objective type question is explained in detail in the “Information Handout” which is usually sent to you along with the call letter. See the illustration of an Objective Type question given below : Q.11. 86 + 57 + 278 = ?   (1)  431 (2)  421 (3)  411 (4)  321 (5)  None of these Here, you have to find out the sum of the three numbers 86, 57 and 278. When these three numbers are added you get 421. Among the alternatives 421 is given as the second one. Hence, your answer to this question is alternative number (2). In an objective type question, only one of these given alternatives is correct and you are supposed to give the number of that alternative as your answer. If you give more than one alternative number as your answer, you will not get any credit.

  4. You have to indicate your answer on a separate answersheet. Unlike the answer sheets you have handled for the school or college examinations, this is a special kind of answersheet, designed for an Optical Mark Reader (OMR). OMR helps to meet the need for rapid and accurate test scoring of your answer sheets. OMR reads data in the form of codes indicated by HB pencil only. OMR reads HB pencil codes by measuring the reflectivity of the areas marked by you. Therefore, accuracy of scoring depends on the use of HB pencil and the method of darkening codes. It is very important to remember that OMR is very sensitive to improperly coded/handled/torn answer sheets. Therefore, to facilitate scoring by OMR you have to be careful enough to keep your answer sheet clean, unwrinkled, unfolded, and intact. The method of indicating the answer is explained in the “Information Handout” which will be sent to you along with the call letter. Please study the handout carefully. While marking your answers on the answer sheet you have to darken the appropriate ovals. You should note the following points in this context : (a) You have to use HB pencil to indicate your answers. HB pencils are ordinary writing pencils easily available in the market. Do not use too soft pencils like 2B/BB, 3B or ink pens/ball point pens to indicate your answers on the answer sheet. Also do not use very hard pencils like H, 2H, 3H etc., as use of these pencils may damage the answer sheet. Further, OMR may not be able to sense the very light codes. (b) If, for example, the answer to Q. 11 is (2), then it has to be shown as follows : Q.11 . 1 2 3 4 5 Note that the oval number 2 against question number 11 is darkened with pencil. (c) If you have skipped a question you should mark the answer against the correct question number while answering the next question. (d) Darkening the ovals need not be artistic but you should darken the ovals completely. Also see that you are not darkening any portion outside the oval. (e) If you need to change an answer you should completely erase the previous answer with a good quality eraser and darken the correct oval. While erasing you should not leave any smudge. Keep your eraser clean in order to avoid smudging the answer sheet.

  5. NOTE FOR TEST OF REASONING ABILITY This is a test to see how well you can think and analyse. These questions will be printed in both Hindi and English versions. (A) VERBAL QUESTIONS Reasoning in a verbal medium depends upon the ability to understand words/directions given in the questions. You should also be able to analyse relationships among the various things expressed through words. Further, if you are able to discriminate the various aspects expressed through different words, you will be able to identify the correct answer easily. Ability to read quickly without missing important details will help. The examples of verbal questions given in this section are not exhaustive. However, the examples given here will help you to familiarize with the nature of verbal questions. Although for each example a method to solve is provided, it may be possible to solve the problem in some other way. The verbal questions may be of verbal-analogies, classification, coding, letter-series and following directions etc. Some of these types are discussed here. (1) VERBAL-ANALOGIES Each question of this type has one pair of related words, and one word from a second pair. Five answer-words are given for completing the second pair. You have to select a word completing the second pair, so that it expresses a relationship similar to that expressed in the first pair.

  6. Here is an example. Example: Q. ‘Table’ is related to ‘Wood’ in the same way as ‘Shoe’ is related to ____ ?(1) Cobbler (2) Chappal (3) Hide (4) Leather (5) Shop Explanation : Notice the first pair, “Table is related to Wood”. In this pair, the second word is a “raw material” of which the first word (Table) is made. Now look at the first word of the second pair, i.e., Shoe. You have to complete, “Shoe is related to _?_”. To complete this pair you must express the same relationship as identified in the first pair, so that the second word in the second pair becomes “raw material” of which the first word (Shoe) is made. In order to select the correct word, we will have to consider each answer-word and check if it expresses best the required relationship. With the first answer-word the second pair will be “Shoe : Cobbler”. This pair does not express the relationship indicated in the first pair. In this case Cobbler is the one who cobbles or mends shoes. With the second answer-word the second pair will be “Shoe: Chappal”. The word Chappal is a synonym of Shoe, hence, does not express the relationship we are looking for. With the third answer-word the second pair will be “Shoe : Hide”. The word Hide means skin of an animal. Since, the shoe is made by using animal skin, this could be the choice. In order to confirm correctness of this answer-word we must check if the other answer-words express the required relationship in a better way or not. With the fourth answer-word the second pair becomes “Shoe : Leather”. The word Leather means a tanned or dressed skin of an animal. The shoe is usually made of an animal skin after it is tanned. Hence, ‘Leather’’ is a better answer-word than ‘Hide’, expressing the relationship reflected in the first pair. After, checking with the fifth answer-word, however, we will be able to finalise this as a correct answer. If firth answer word used, we will have the second pair “Shoe : Shop”. Shop is the place (and not the raw material), hence, is not at all the correct word. From the above discussion it follows that “Shoe : Leather” is the correct pair, expressing “Object : raw material” relationship. The word leather is the fourth answer-word. Hence, for the given question, 4 is the correct answer.

  7. Here is another example : Q. ‘Mountain’ is related to ‘Hillock’ in the same way as ‘Tree’ is related to –––––– ? (1) Jungle (2) Bush (3) Plant (4) Forest (5) Trunk Explanation : Notice the relationship between Mountain and Hillock. It is LARGE to SMALL. Remember that in such a case when you select the second word for the second pair the order expressed must be the same in both the pairs, i.e., the second word should stand for the smaller version of the thing given in the first word, and not otherwise. In the first pair ‘Mountain’ is a high hill, whereas, ‘Hillock’ is a small portion slightly raised above the ground. In the second pair the first work is ‘Tree’. If this is taken as LARGE (TALL) version of plant we should have a second word representing a SMALL (SHORT) version of plant. If we try to pair the word ‘Tree’ with each of the given answer-words we find that only the word ‘Bush’ (i.e. a plant which grows just above the ground level) fits well. The pair “Tree : Bush” expresses the same kind of relationship represented by the pair “Mountain : Hillock”. Since, the correct word Bush is numbered (2), 2 is the correct answer. From the two examples given above you might have learnt that the Analogy Questions test the ability to recognize relationships among words, or the concept they represent and to recognize when the relationships become parallel. Relationships could be of kind, size, degree, purpose, make etc. Therefore recognise the correct kind of the relationship for the complete pair given, first. Then, on the basis of the recognised relationship try out the possibility of each of the five possible pairs for parallelness. Select the pair which has a relationship closest to that you have recognised. (2) LETTER‑SERIES In this type of questions, a series made up of letters in Roman alphabet is given. You have to find out the rule by which is series is formed. On the basis of the rule, you have to decide as to what would be the letter/s in the place of question mark.

  8. An example of this type of question is given below : Example : Q. Which of the following will come in the place of the question mark (?) in the following letter series ?QP SM UJ ? YD (1) WG (2) WH (3) VG (4) VH (5) None of these Explanation : To find out the rule underlying the given letter series it is useful to keep in front of you all the Roman alphabet arranged, serially and numbered 1 to 26 as given below : A B C D E F G H I J K L M N O P Q R S T U V W X Y Z 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 Keeping this arrangement in front of you, you can complete the following table : Pair No. First Number in Difference Second Number in Difference Letter Alphabet Letter Alphabet First Q 17 P 16 +2 -3 Second S 19 M 13 +2 -3 Third U 21 J 10 +2 -3 Fourth ? W 23 ? G 7 +2 -3 Fifth Y 25 D 4

  9. Look at the numbers of every first letter of each of the pairs. These numbers are in increasing order. Further, the difference in the two consecutive numbers is two. From this we can easily find out the number of the alphabet that should appear first in the fourth i.e. the missing pair. The number is 23 and the corresponding alphabet is W . Now look at the numbers of every second letter of the pairs. They are in decreasing order and the difference in the two consecutive numbers is three. From this we can easily work out the second alphabet in the fourth missing pair. The number we get is 7 and the corresponding alphabet is G . From the discussion above the pair of letters that should appear at the question mark is WG . In the five answers provided WG appears as the first answer. Hence , 1 is the correct answer. To save your time you may memorize the number/ rank of each alphabet and use it whenever demanded by the question. An alternate way to solve the problem is presented below. STEP-I First of all consider only the first letters of the given pairs. When these letter are placed in sequence, they will appear as Q S U ? Y STEP-II See the direction in which the letters are picked up from Roman alphabet-i.e. whether from left to right of from right to left. You will observe that the letters are picked up from left to right, as indicated below. A ‑‑‑‑‑ Q R S T U V W X Y ‑‑‑‑‑‑ Z Left ––––––––––––––––––––––––––––––––––––––– Right STEP ‑ III Now find out the number of letter skipped while picking up the letters, to from the given series. After Q the letter S is picked up by skipping letter R. After S the letter T is skipped and the next letter U is picked up.

  10. Obviously the letter immediately following U is V and that should be skipped. After skipping V we will pick up letter W. Let us confirm whether exactly one letter gets skipped between the identified letter W and the next letter picked up i.e. Letter Y. It can be seen that only letter X has been skipped; hence, W is the correct choice as a first letter of the missing pair. STEP ‑ IV Follow the same steps with the second letters of the given pairs. P M J ? D is the sequence obtained. The way these letters are picked up, is indicated below : A ‑‑‑‑‑ D E F G H I J K L M N O P ‑‑‑‑‑ Z Left –––––––––––––––––––––––––––––––––––––––––––––––– Right You would have observed that letters are picked up from right to left of the Roman alphabet. Further, after a particular letter two letters from the alphabet are skipped and the next letter is picked up. Therefore, we have to skip two letters in the alphabet in the right to left direction and pick up the next letter. By doing so we get G as a second letter of the missing pair. We can confirm correctness of the identified letter by checking if two letters get skipped between G and D. Indeed, F and E are the only two letters skipped between G and D. Hence, G is the correct choice as the second letter of the missing pair. Therefore, the missing pair is WG only. The correct answer therefore is ‘1’ representing WG. (3) THE ODD ONE OUT (CLASSIFICATION) In this type of questions five words representing objects, professions, concepts etc. are given. Four of these are alike in some way or hold some common property, thus forming a group. The remaining one is different and can not go with the four. You have to find out the different word/thing and number representing that would be the correct answer. Here is an example. Example. Q. Four of the following five alike in a certain way and so form a group. Which is the one that does not belong to the group ? (1) Sun (2) Earth (3) Uranus (4) Saturn (5) Jupiter

  11. Explanation : Study the five words given above. All these (objects) are found in the sky. However, Earth, Uranus, Saturn and Jupiter are Planets. They revolve about the sun and reflect the Sun’s Light. These are not the characteristics of ‘Sun’. Hence, Sun does not belong to the group of the other four planets and is odd one out. Since, Sun is given as 1, that is the correct answer. (4) LETTER OR DIGIT CODING In this type of questions, a set of letters (words) or digits is presented in its original and also coded (disguised) form. You have to mentally analyse the coding system or rule. Applying this rule you have to find out the coded form of the other set of letters (word) or digits. Here is an example : Q. In a certain code ‘POWDER’ is written as ONVCDQ. How would ‘BELONG’ be written in that code ? (1) AFKPNH (2) CFMPOH (3) ADKNMF (4) CDMNOF (5) None of these Explanation : In order to identify the correct coded form of the word BELONG; we have to study the method used in deriving the coded form of POWDER. Again, we can use the number of each of the letters used in the NORMAL and CODED form of the word and study the pattern in the numbers. The numbers could be noted down by referring to the previous slide and entered in the table given below : Letters in the first word in the first pair P O W D E R Number in the alphabet 16 15 23 4 5 18 –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– Letter in the coded form O N V C D Q Number in the alphabet 15 14 22 3 4 17 Observe the number of the letter in the original word in the first place and the number of the corresponding coded form. The number of the coded letter is one less than that of the original letter. Therefore the coded form of any letter

  12. in the original word becomes just the preceeding letter in the alphabet. While the coded form of the word BELONG, this rule we will have to follow We will follow the observed rule in the following steps. Step 1 Writing letters in the given word B ELO N G Step 2 Writing the number of the letter in the alphabet 2 5 12 15 14 7 Step 3 Writing the new number by subtracting one from the earlier number 1 4 11 14 13 6 Step 4 Replacing the number with the letters A D K N M F Thus, by following the identified rule we obtained, ‘ADKNMF’ as the coded form of word ‘BELONG’. Matching the obtained coded form with those given as answers we find that answer 3 is ‘ADKNMF’. Therefore, 3 is the correct answer of the question under discussion. Similarly, when digits are given you can first find out the rule used for coding and then by applying the rule to the digits given as normal you can derive its coded form. (5) FOLLOWING DIRECTIONS In this type of questions you are provided serially arranged letters in the Roman alphabet. With the help of this and the directions given, you have to identify the letter asked for. Here is an example. Q. Which of the following letters will be fifth to the right of the twelfth letter from the left end of the following alphabet ? A B C D E F G H I J K L M N O P Q R S T U V W X Y Z (1) O (2) R (3) I (4) J (5) None of these To solve this type of question is important to know the convention followed with respect to the LEFT and RIGHT. Observing the alphabets kept in front of you, as on this page, A is to your left-hand i.e. the left and Z is to your right hand. i.e. the right direction, whenever it appears in these type of questions. Now before attempting the solution read the question completely. This will help you to decide the sequence of the steps required for the solution. In the present case we have to take the following steps :

  13. Step 1 Go to left end of the alphabet, (i.e. A) Step 2 From A count the 12th letter (i.e. L) Step 3 From L in the direction of Z, count 5th letter skipping L, (i.e. ‘Q’) Thus, Q is the required answer. Since this is not provided in the first four answer choices, we have to select the answer ‘None of these’. This being the fifth answer our correct answer to the given question is ‘5’. Let us now see the nature of the pictorial questions or the non-verbal questions. (B) PICTORIAL QUESTIONS (NON–VERBAL) Rest of the questions will be of pictorial form. These questions are given in the form of diagrams/figures/symbols. Generally a set of five diagrams is given as ‘Problem Figures’. You have to find out the relationship between these figures and apply it to a new situation. The types of questions asked in this test can be classified into ‑‑‑‑ (1) Series, (2) Classification and (3) Analogies. (1) SERIES Each question of this series type has two sets of figures. The set labeled ‘Problem Figures’ is on the left hand side consisting of five or more figures. The set labeled as ‘Answer Figures’ is on the right hand side and consists of five figures numbered 1, 2, 3, 4 and 5. In ‘Series’ type questions, there is some systematic change occurring from left to right in the ‘Problem Figures’. You have to find out the nature of this change and decide as to what would be the next figure, if the same change continues. Then you have to pick out the one from the ‘Answer Figures’ which can be sixth or next in the row of ‘Problem Figures.’ here is an example of a ‘Series’ question : Q. Problem Figures Answer Figures OBSERVATIONS ANSWER In the given problem figures the only element is an In the position next to that shown in the last problemarrow. In each successive figure, the arrow-head is figure the arrow would be lying towards its anticlockwise tilting more towards the left or in the anticlockwise direction and its head will be pointing towards the right. direction, at an angle of 450. This description perfectly matches will the answer figure 4, which is the correct answer.

  14. Reasons for rejecting the other answer figures : Answer figure No.Reason for Rejection 1, 2 & 5 Arrow is tilted much more than the sequence indicates. 3 Arrow head is titled in clockwise direction. The changes which occur in figures in different questions could be of various types. Some of these are : a. Addition – Some element is added in the successive figures. b. Subtraction – An element is deleted from the successive figures. c. Substitution – A new element replaces an element already there in every successive figure. d. Rotation – The elements are rotated clockwise or anticlockwise across the figures. e. Rearrangement – The elements are rearranged in each successive figure. f. Change in element – The elements in each successive figure are modified or changed. g. Combination – Change involving two or more of the above Note : An element refers to a line, a dot, a circle, or any symbol which appears in the Problem Figures or Answer Figures. These types are discussed below with the help of an example. (a) Addition : Q. Problem Figures Answer Figures OBSERVATION ANSWER In each successive problem figure a single line is See the last problem figure. The figure next in getting attached to the earlier line at a right angle, in the the series will have a line added parallel to the clockwise direction, making a rectangle. After completing horizontal line on the top. The figure resultingthe rectangle a second round of addition begins outside from this addition will perfectly match with the earlier rectangle, repeating the earlier sequence of answer figure 1, which is the correct answer.addition of a line.

  15. Answer figure No. Reason for Rejection 2 Addition of a line is in the anticlockwise direction. 3 Three lines are added instead of one. 4 Without adding a line the initial line is rotated. 5. The addition of lines is not as per the requirement of the problem figures. (b) Subtraction Q. Problem Figures Answer Figures OBSERVATIONS ANSWER In the first problem figure eight dots are placed at Last problem figure is the result of removal of a equal distance from each other, outside the circle. In dot from the upper left hand corner of the referenceeach of the successive figures one dot gets removed square. The next alternate position for removal ofRemoval of the dot starts from the one at the lower left the dot is the lower left hand corner, where there ishand corner of the reference square. In the successive no dot. The dot closer to this position in the anti- figures an alternate dot is removed in the anticlockwise clockwise direction is the one closer to the lowerdirection . line of the reference square. This dot will get removed in the figure next in the series. After removing this dot the resulting figure will match with answer figure 1, which is the correct answer. .

  16. Reasons for rejecting the other answer figures : Answer figure No. Reason for Rejection 2 The dot in the clockwise direction is removed. 3 & 4 Placement of the dots is not in conformity with the trend indicated in the problem figures. 5 Instead of removal there has been addition of a dot. (c) Substitution : Q. Problem Figures Answer Figures OBSERVATIONS ANSWER Each problem figure has a pair of geometrical shapes. See the last problem figure. The smaller shape is The first one is larger, the next to it is smaller. The a rectangle. Next figure in the series will have an larger shape is an enlargement of the smaller shape in enlarged rectangle, taking the position of larger the preceding problem figure. The smaller shape when shape. The smaller shape introduced will be a introduced is new in the set. shape which has not appeared in the earlier problem figures. This description makes only answer figure 1 as the correct answer. Reasons for rejecting the other answer figures : Answer figure No. Reason for Rejection 2 & 5 Larger shape is not a rectangle. 3 Shape in the place of smaller shape is not small enough. 4 Smaller shape in this figure has already appeared in the problem figures.

  17. (d) Rotation : Q. Problem Figures Answer Figures OBSERVATIONS ANSWER Each problem figure has an upright pentagon with a ‘X’ In the last problem figure the ‘X’ mark is closer mark outside it, closer to one of its five sides. The size to the baseline of the upright pentagon. In the and the position of the pentagon remains unchanged in figure next in the series the ‘X’ mark will move the series. However the ‘X’ mark moves to the adjacent to the right side of the upright pentagon side in the anticlockwise direction, in the successive Hence, answer figure 4 is the correct answer. problem figures. Reasons for rejecting the other answer figures : Answer figure No.Reason for Rejection 1 The ‘X’ mark has skipped one side. 2 The pentagon is inverted. 3 The pentagon is replaced by a square. 5 Size of the pentagon is changed.

  18. (e) Rearrangement : Q. Problem Figures Answer Figures OBSERVATIONS ANSWER Each problem figure has four elements viz. a dot, Study the position of the four elements in the a square, a circles, and a triangle. The position of each last problem figure. Next positions of these of these elements undergo following changes in the elements in the series will be as described successive problem figures : below : – Dot moves from one corner of the reference – Dot will be at the lower left hand corner of thesquare to the other, in the anticlockwise direction reference square. – Square moves from one corner of the reference – Square will be at the upper left hand corner ofsquare to the other in the clockwise direction. of the reference square. – Circle oscillates diagonally from upper left hand – Circle will be at the lower right hand corner ofcorner to the lower right hand corner of the the reference square. reference square – Triangle is initially at the centre of the reference – Triangle will be near the centre of the right square. From the centre it moves to the left side hand side of the reference square.by half the width of the reference square, then tothe right side by full width of the reference square These positions are matching with those inand returns to the initial position. The movement answer figure 2, which is the correct answer is repeated in the given sequence.

  19. Reasons for rejecting the other answer figures. Answer figure No.Reason for Rejection 1 Dot has skipped one corner. 3 Dot has rotated clockwise instead of anticlockwise. 4 Square has rotated anticlockwise instead of clockwise. 5 Triangle is inverted. (f) Change in element : Q. Problem Figures Answer Figures OBSERVATIONS ANSWER The first problem figure is made up of eight straight The last problem figure has two alternate lines lines. In each successive problem figure a single curved outward and two alternate lines incurved. straight line gets curved. For the first line the curve The next figure in the series will have the fifth in outward. The next adjacent line gets incurved. straight line in the anticlock-wise direction curved This process proceeds in the anticlockwise direction. outward and two lines incurved. This description matches perfectly with the answer figure 3, which is the correct answer. Reasons for rejecting the other answer figures. Answer figure No. Reason for Rejection 1, 2, 4 & 5 Three lines are incurved instead of two.

  20. (g) Combination : In this type of questions the changes taking place in the problem figures could be of more than one of the above types. An example of such a type is discussed below : Q. Problem Figures Answer Figures OBSERVATIONS ANSWER First problem figure has a hexagon at the centre of the See the last problem figure. Next in the series, reference square. There are eight dots placed at equal the dot near the centre of the lower side of the distance from each other, outside the hexagon andreference square will be removed. closer to the boundary of the reference square. In each Simultaneously, a line from the centre joining successive problem figure following two changes take the vertex at the 5 O’clock position will get place. added. The resulting figure will perfectly match with the answer figure 2, which is the correct answer. – The Dot removed first is the one at the twelve O’clock position. The next removal is in the clockwise direction for the adjacent dot. – A single line joining one vertex and the centre is added. The vertex at 9 O’clock position, then 11 O’clock position. 7 O’clock position, 1 O’clock position..... is selected, in the given order for adding the line.

  21. Reasons for rejecting the other answer figures. Answer figure No. Reason for Rejection 1 The position of the newly added line is wrong. 3 & 5 Two dots are removed instead of one. 4 Two lines are added instead of one. (2) CLASSIFICATION In this type of questions five figures, numbered 1, 2, 3, 4 and 5 are given as Problem Figures as well as Answer Figures. Four of these figures have something in common to form a class. One of the five figures does not share the common feature/s observed in the four figures. Thus, one of the five figures, is the ‘odd-one-out’. You are supposed to identify the figure which does not belong to the group of four figures, forming a class. Example of this type is given below : Q. PROBLEM‑CUM‑ANSWER FIGURES OBSERVATIONS ANSWER Each figure given above has four segments. Further Four segments of the fifth figure are of unequal except the figure numbered as ‘5’, all the figures have area unlike the other four figures in the group. segments of equal area. Hence, figure 5, is the correct answer. (3) ANALOGY In these questions there are two sets of figures. The problem Figures and the Answer Figures. The Problem Figures are presented in two units. The first unit contains two figures and the second unit contains one figure and a question mark. You are supposed to find out which one of the Answer Figures should be in place of the question mark.

  22. Q. Problem Figures Answer Figures OBSERVATIONS ANSWER The first unit has two figures. One figure is a dark circle The logic is simple. One dark shape is paired and the other has three rings. The three rings form a with three similar hollow shapes in an inverted triangular pattern of an inverted triangle. Therefore, three hollow squares forming an inverted triangular pattern given as answer figure 2, is the correct answer. Reasons for rejecting the other answer figures. Answer figure No. Reason for Rejection 1 One of the hollow shapes is dissimilar. 3 & 5 Pattern is of circles instead of squares. 4 Two lines are added instead of one. Precautions to be taken : In the test of Reasoning Ability, while attempting the pictorial/non‑verbal questions, the candidate has to grasp the logic in the problem figures correctly. In order to solve these questions you should note the following points : 1. Make sure that you study the problem figures carefully to arrive at the above logic. On the basis of this logic select one answer figure after studying the given answer figures. 2. Sometimes, you may need to take help of the answer figures also (see’(c) Substitution’ explained earlier) to decide the correct answer.

  23. 3. In the questions of the ‘Series’ type, some candidates make the mistake of deciding the answer figure only on the basis of the observations made on the fourth and the fifth problem figures. 4. Also remember that it being an important test with respect to the marks allotted to it you should try to solve correctly as many number of questions (both verbal and non‑verbal) as possible. 5. Out of the total time of 135 minutes given to you, you may allot roughly 30 minutes to attempt this test. ‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑

  24. NOTE FOR TEST OF GENERAL ENGLISH What does this test measure ? This section deals with the test measuring your knowledge of English language. English language ability will be tested through questions/items on functional grammar, reading comprehension, context based vocabulary etc. There will be no questions on English literature or on poetic expressions. The different types of questions which are used in this test can be classified as (1) Spotting the Error (2) Sentence Completion (3) Reading Comprehension (4) Sentence Construction (5) Rearrangement of Sentences each of which measures one or more areas of language abilities. (1) SPOTTING THE ERROR Candidate’s familiarity with the conventions and grammatical rules of standard written English is tested in this type of questions. Emphasis here will be on assessing knowledge of correct expression. A correct sentence should be grammatically and structurally correct. Each question in this section, is divided into four parts and each is numbered. You have to decide whether there is any error in the sentence and find out in which of the parts the error exists, if there is any. If the sentence is correct, your answer is “5” i.e., “No error”. Study the example given below : Q.1. Last week / Ramesh and Haresh / do the work / which was pending. No error (1) (2) (3) (4) (5) Note that the sentence has been divided into four parts and each part is numbered. The sentence is about the completion of pending work two persons did the previous week. The error is in the verb used. The verb should denote an act that is already completed. The verb should read as “did” and the correct sentence should be “Last week / Ramesh and Haresh / did the work / which was pending”. Since the error is in the third part, the correct answer is “3”. Now try the next example.

  25. Q.2. Before they left / the office / they switched off / all the lights. No error (1) (2) (3) (4)    (5) In the above sentence you will note that there is no grammatical error. It is also meaningful. Therefore this question has no error and your answer should be “5” which is “No error”. (2) SENTENCE COMPLETION This type of questions measure your ability to recognise words and phrases that both logically and grammatically complete the sentence. A sentence is given with a word missing in it. The missing word will be indicated by a blank. You have to find the most appropriate word from the given alternatives. In deciding which of the five words best fills the blank space, you must consider the context provided by the sentence. Given below are two items for practice : Q.3. We were so late, we ----------- had time to catch the train. (1) nearly (2) almost (3) simply (4) not (5) hardly “So late” sets the tone of the sentence. Because the people were late they just managed to get into the train and the only word that conveys this meaning is “hardly”, which is denoted by (5). “Nearly” and “simply” could have been the right choices if the sentence was to read as “missed the train” after the blank. “Almost” and “not” are not correct grammatically. Thus, correct answer is “5”. Q.4. Raju’s shirt has a pattern of boats all ---------- it. (1) on (2) over (3) down (4) round (5) with In the above example, the only word which grammatically fits is “over”. So “2” is the answer. You may also be given a passage like the one given below : Instructions :  In the following passage there are blanks, each of which has been numbered. These numbers are printed below the passage and against each, five words are suggested, one of which fills the blank appropriately in the context of the whole passage. Find out the appropriate word. My father waved me goodbye and the bus (5). The person sitting (6) to me was a Government Engineer (7) to Hyderabad, (8) inspect the roads. Q.5. (1) going (2) started (3) arrived (4) stopped (5) travelling

  26. Q.6. (1) next (2) besides (3) near (4) side (5) neighbor Q.7. (1) coming (2) arriving (3) going (4) visiting (5) flying Q.8. (1) to (2) for (3) was (4) so (5) then In the above passage you have to fill up the blanks in such a way that the whole passage becomes meaningful in the context of the entire passage. It is not sufficient that the sentence alone is meaningful and grammatically correct. In the first blank the word which would fit in the blank (5) is “started”. Even though “arrived” or “stopped” might be grammatically correct it is not meaningful since his father was waving goodbye. The other words viz., “going” and “travelling” are not grammatically correct and also not meaningful. Thus, “2” is the correct answer. In the next blank (6) only “next” fits properly. This is decided by the preposition “to” which comes after the blank. It is unidiomatic to use “near” or “besides” with the preposition “to”. To use the words “side” and “neighbour” one has to change the sentence and therefore they are not correct answers. Hence, correct answer is “1”. The blank (7) has to be filled by keeping the entire passage in view. Since the passage is about a bus journey “flying” is a wrong answer. The description of the bus journey is of someone travelling to Hyderabad and therefore “coming” is a wrong alternative. “Arriving” and “visiting” are grammatically incorrect. Therefore the correct answer is “going”, i.e. alternative “3”. For the blank (8) “to” is the correct answer. To use “was” and “then” is ungrammatical. The other words “so” and “for” are meaningless. Thus, the correct answer is “1”. (3) SENTENCE CONSTRUCTION The questions of this type assess the ability to comprehend, organise and structure sentences. The question consists of a set of five words or groups of words. These five words have to be arranged to make a meaningful sentence. Note that all the five words should be used and each word should be used only once. The alphabet in the bracket preceding the word is the code given to the word. Each choice given represents a certain way in which the words can be arranged. You have to find which of the arrangements forms a sentence which is meaningful and grammatically correct. See the following illustration and try to solve it : Q.9. (A) NOW (B) REPLACE (C) THE (D) PLEASE (E) BOOK (1) DBAEC (2) BDCEA (3) DBCEA (4) DEACB (5) DACEB

  27. You should mentally rearrange the words so that a meaningful sentence is formed. You may take one word as a starting point and rearrange various other words to see whether they will form a meaningful sentence e.g., you may try starting with the word “book” as the first word and rearrange the other words. By referring to the alternatives given you may also be able to decide which of the words to use first as a reference point to arrive at the correct answer. Needless to say, you should be able to do this rearrangement in your mind itself. In the above illustration, the correct answer will be “Please replace the book now”. The correct order of letters therefore would be “DBCEA”. The correct answer therefore will be “3”. (4) READING COMPREHENSION Questions on reading comprehension measure the ability to understand, analyse and apply information and concepts presented in the written form. All questions are to be answered on the basis of what is stated or implied in the given passage. Reading comprehension, therefore evaluates your ability to : • understand words and statements in the given passage • understand the logical relationships between points and concepts in the given passage • draw inference from facts and statements in the given passage. Guidelines for answering the Reading Comprehension Test Given below are some guidelines which would be of use to you in answering satisfactorily the questions set on the passage. (1) Answer all questions on the basis of what is stated or implied in the passage itself. Even when you do not agree with what the author of the passage is saying, do not let your opinions or knowledge and information influence your judgment of what the author is saying. (2) Read the questions carefully, making sure that you understand what is being asked. If need be refer back to the passage for finding the answer. (3) Read all the alternatives carefully. Never assume that you have selected the best answer without first reading all the alternatives. (4) Remember that understanding is the critical factor in reading comprehension. (5) Remember that certain words in the passage are printed in bold or underlined to help you locate them while answering some of the questions.

  28. An example of this type of questions is given below : Q.10-14. Read the following passage carefully and answer the questions given below it. Certain words in the passage are printed in bold to help you locate them while answering some of the questions : Malaria is always associated with damp and marshy land. This is not because the land is damp but because still water is the breeding place of the mosquito, which begins its life as a larva living in water. Malaria does not frequently occur in dry desert countries. We should destroy mosquitoes to prevent their breeding in still water. This can be done by draining all ponds and pools, and by keeping them covered in the breeding season with a film of kerosene oil, which kills the larvae. Q.10. Where is malaria not very common ? (1) In cold countries (2) In dry countries     (3) In countries having a lot of rain (4) In hot countries (5) Not mentioned in the passage In the above illustration alternative “2” is correct. It is stated in the passage that mosquitoes carry malaria and mosquitoes breed in damp conditions. Dry countries are the places where damp and marshy conditions are absent. A country is called a “dry country” because the country is not only hot but also without rains. An alternative that comes closest to this is “in hot countries”. However, this cannot be selected because it does not mean that it does not rain in hot countries and consequently there is no dampness. The chances of damp conditions prevailing in countries having lot of rains is very high, so alternative “3” also cannot be a right choice. Same thing is true about alternative “cold countries”. The fifth alternative cannot be the right choice because in the passage the conditions under which mosquitoes breed and do not breed is mentioned. So the right answer is “2”. Q.11. What is the breeding place of the mosquito ? (1) Flowing water (2) Shallow water (3) Dirty water (4) Deep water (5) Still water For the above question you should refer back to the passage. The second sentence in the passage says “....... because still water is the breeding place of the mosquito”. Therefore the correct answer to the above question is “still water” i.e., “5” is the answer. Q.12. What is the use of kerosene oil in preventing Malaria ? (1) It kills the fully grown mosquitoes (2) It cleans the pools and ponds (3) It kills the developing mosquitoes (4) It helps in burning the things around the ponds (5) It purifies the air

  29. Referring to the passage you will find that the correct answer is given in the last sentence of the passage. It says “....... by keeping them covered in breeding season with a film of kerosene oil which kills the larvae”. A larva as you know is a developing mosquito and therefore among the given alternatives “3” is correct. You may have questions like this also. Q.13. Which of the following words is most SIMILAR in meaning of the word DRAINING as used in the passage ? (1) Depleting (2) Discharging (3) Emptying (4) Straining (5) Cleaning “Draining” means letting water off. So “Emptying” is the right choice, as the one with similar meaning of “draining”. Hence, “3” is the answer. Q.14. Which of the following words is most OPPOSITE in meaning of the word STILL as used in the passage ? (1) Noisy (2) Flowing (3) Living (4) Yet (5) Steady “Still” here means without movement or motion. So the choice “Flowing” is appropriate, as the one opposite in meaning of “still”. Hence, the answer is “2”. (5) REARRANGEMENT OF SENTENCES Another set of questions, in which the ability to understand what is read and to extract information is assessed, is discussed below. These questions also test your ability to organise your throughts and ideas in a suitable sequence. Questions of this type require the candidates to rearrange the given sentences in the proper sequence so as to form a meaningful paragraph. Q.15-19. Arrange the following five sentences A, B, C, D & E in the proper sequence so as to form a meaningful paragraph; then answer the questions given below them : A. Do not use the lift B. Assemble in the garden and wait for further instructions C. Fire ! There is a fire in the office D. Instead proceed down by the main staircase, in a calm and orderly fashion. E. Will all staff please vacate the building immediately ? Q.15. Which of the following should be the first sentence ? (1) A (2) B (3) C (4) D (5) E

  30. Q.16. Which of the following should be the second sentence ? (1) A (2) B (3) C (4) D (5) E Q.17. Which of the following should be the third sentence ? (1) A (2) B (3) C (4) D (5) E Q.18. Which of the following should be the fourth sentence ? (1) A (2) B (3) C (4) D (5) E Q.19. Which of the following should be the fifth sentence ? (1) A (2) B (3) C (4) D (5) E While trying the above questions you should first read all the five sentences and see what is the context in which the whole sentence has been given. In the above it is a fire which has broken out in an office. Obviously announcement or alarm regarding the fire should appear first, i.e., ‘C’, should be the first sentence in the paragraph. It is followed by a request to all the staff members to vacate the building immediately. Therefore the sentence “E” should follow sentence “C”. Had sentence “C” been only “there is a fire in the office” the rearrangement would have been with “C” following “E”. However, since “C” appears as a cry this comes first. The sentence “do not use the lift “follows “C” & “E”. This is followed by the sentence “D” i.e., “Instead proceed down by the main staircase, in a calm and orderly fashion”. To put this sentence earlier would not be correct because of the word “instead”. The word “instead” indicates an alternative way of doing and therefore has to have some action which is prohibited as has been given in sentence “A”. Sentence “B” “Assemble in the garden and wait for further instructions” is the last sentence because it tells people what they should do once they have vacated the office. In the proper sequence the paragraph would read as follows : C. Fire ! There is a fire in the office. E. Will all staff please vacate the building immediately ? A. Do not use the lift. D. Instead proceed down by the main staircase, in a calm and orderly fashion. B. Assemble in the garden and wait for further instructions. The correct sequence of the sentence therefore would be C, E, A, D, & B. Once you have got the sequence the questions can be answered easily; e.g., the answer to question number (15) would be “3” and so on. Do not answer the questions till you have sorted out the sequence of the complete paragraph.

  31. NOTE FOR TEST OF QUANTITATIVE APTITUDE This test measures candidates’ ability to solve simple arithmetical computation problems alongwith problems related to arithmetical reasoning (verbal problems) with speed and accuracy. All the problems in this test involve basic operations like addition, subtraction, multiplication, division, percentage, squares and square roots with whole numbers as well as vulgar fractions and decimals. All these arithmetical operations are included in the middle school syllabus and at best in the high school syllabus. You may refer to these books for revising your knowledge. Reviewing essential concepts Some of the basic ideas have been discussed in detail here. They should help you solve the simple arithmetical problems. Some more ideas which need only a brief discussion have been presented in the explanation of the problems. (1) COMPOUND NUMBERS A number, which has other divisors than 1 and the number itself, is called a compound number. Example All the divisors of 12 are, 12 : 1, 2, 3, 4, 6, 12. There are many divisors of 12 other than 1 and 12. Thus, 12 is a compound number. 4, 9, 15, 28, 64, 1000 are a few other compound numbers. (2) PRIME NUMBERS (1) All the divisors of 11 are, 11 : 1, 11. There are no divisors of 11 other than 1 and 11. (2) All the divisors of 19 are, 19 : 1, 19. There are no divisors of 19 other than 1 and 19. A number of which the only divisors are 1 and the number itself is called a prime number. Thus, 11 and 19 are prime numbers. 2, 3, 29, 37, 97 are a few other prime numbers. The number 1 is neither a compound number nor a prime number.

  32. (3) FACTORS OF THE NUMBER We can express each compound number as a product of two or more of its divisors. If any number is expressed as a product of the divisors then the divisors are called factors of the number. Thus, 12 = 2 × 6 or 12 = 2 × 2 × 3. It means 2 and 6 are factors of 12. That factor of a number which is also a prime number is called a prime factor. In the expression 12 = 2 × 6, 6 is not a prime factor, but 2 is a prime factor. In the expression 12 = 2 × 2 × 3, each factor is a prime factor. Each compound number can be expressed as a product of prime factors. (4) FACTORISATION The phrase ‘factorising a number’ means expressing it as a product of its prime factors. We will study how to factorise the given compound number. Example : 1    Factorise 12 Explanation Method 2 is a factor of 12 as 12 is exactly divisible by 2, and 12 = 2 x 6 12 ÷ 2 = 6. = 2 x 2 x 3 12 = 2 × 6. Now, 2 is a factor of 6 and 6 ÷ 2 = 3. 6 = 2 × 3 12 = 2 × 2 × 3. In this arrangement, each factor of 12 is a prime factor. Hence, the process of factorization of 12 is complete. Example : 2 Factorise 30. We can factorise 30 by using vertical arrangement, which is another way of factorising compound number :- Explanation Method Using the test of 2, 2 is a factor of 30 and 30 ÷ 2 = 15. 2 30 Using the test of 3, 3 is a factor of 15 and 15 ÷ 3 = 5. 3 15 5 is a factor of 5 and 5 ÷ 5 = 1. 5 5 1

  33. In this arrangement the operation of factorisation has to be carried out till we obtain 1 as a factor. Here, the product of all the numbers in the left hand column is 30. 30 = 2 × 3 × 5. Further, each factor here is a prime factor. Remember the following facts whenever you wish to factorise a number : (1) It is convenient to find the smallest factor first. (2) Check whether the same factor occurs again. (3) Next, find factors greater than the preceding ones. (4) Lastly, make sure that all the factors found are prime factors. The ideas explained so far are useful while solving the problems on fractions, square roots etc. (5) CROSS-MULTIPLICATION Since = (their value is ½ or 0.5), we say that and represent the same fraction. The product of the numerator 5 of the first fraction and the denominator 6 of the second fraction is 30. Similarly the product of the numerator 3 of the second fraction and the denominator 10 of the first fraction is also 30. We observe that these two products are the same. The numbers are taken as indicated by the cross arrow and then they are multiplied. Therefore, we call this operation as cross-multiplication. This operation can be utilised to find out missing value of either the numerator or the denominator of one of the two equivalent fractions given.

  34. ARITHMETICAL COMPUTATIONS Some questions in this test will be on arithmetical computations. Let us now discuss the various types of computation problems you may come across in this test. (A) ADDITION AND SUBTRACTION OF WHOLE AND DECIMAL NUMBERS : Q.1. 209 + 541 = ?(1) 740 (2) 749 (3) 750 (4) 850 (5) None of these This is a simple addition of whole numbers. The sum of the two numbers given is 750. Hence, alternative number (3) is the correct answer.Q.2. 7350 – 5941 = ? (1) 1311 (2) 1409 (3) 2409 (4) 2419 (5) None of these This is a simple subtraction of whole numbers. If we subtract 5941 from 7350, the number in place of the question mark (?) will be 1409 which is alternative number (2). Hence, alternative number (2) is the correct answer. In questions on addition and subtraction of numbers with decimal fractions, even if the decimal fractions in the given sums be arranged horizontally, it is more convenient to arrange them vertically before solving the sum. If the number of digits on the right hand side of the decimal point is the same in two or more decimal fractions, they are said to be equiplace decimal fractions. If the number of places on the right hand side of the decimal point is not the same in the given fractions, they have to be converted into equiplace decimal fractions before carrying out their addition or subtraction. Observe the following examples : Example : Q.3. 10.09 + 0.013 (1) 10.22 (2) 10.103 (3) 10.113 (4) 10.13 (5) None of these Explanation Method These decimal fractions are not equiplace. In the first 10.090 number, there are 2 digits after the decimal point and in + 0.013 the second number there are 3 digits. let us write number 10.09 as 10.090 to make 3 digits after the 10.103 decimal point in the first number. Thus, the correct answer is 10.103, which is alternative number (2). (Note : 10.09 = 10.090. The value of the original number which has a decimal point in it does not change by placing one zero at the extreme right.)

  35. Let us take one example of subtraction. Example : Q.4. 4.35 – 2.425 = ? (1) 2.935 (2) 1.935 (3) 1.925 (4) 1.825 (5) None of these Explanation Method In this example, we can write the number 4.350 4.35 as 4.350 to get equiplace decimal fractions. + 2.425 1.925 Thus, the correct answer is 1.925, which is alternative number (3). (B) MULTIPLICATION OF WHOLE AND DECIMAL NUMBERS Example Q.5. 432 × 25 = ? (1) 2160 (2) 8020 (3) 10800 (4) 1007550 (5) None of these Explanation Method This is an example of multiplication of whole numbers. 432 × 25 10800 Thus, the correct answer is 10800, which is alternative number (3). To find the product of two decimal fractions. (1) First, the multiplication should be carried out without considering the decimal point in the decimal fractions. (2) Then the decimal places in the two given numbers should be added together. The decimal point in the answer should be so placed that the number of digits to the right of the decimal point is equal to the total of the decimal places in the two given numbers. If the number of digits in the product is less than the number required, write as many zeros as may be needed to make up the total, in the proper places in the product and then place the decimal point.

  36. Let us understand the actual procedure with the help of following example. Example Q.6. 3.6 × 0.06 = ? (1) 18.6 (2) 2.16 (3) 21.6 (4) 1.86 (5) None of these Explanation Method The number of digits to the right of the decimal point in the 36 3.6 multiplicand is one. The number of digits to the right of the × 6 × 0.06 decimal point in the multiplier is two. If we consider both the 216 0.216 multiplier as well as the multiplicand the total number of digits The obtained answer 0.216 is not in the to the right of the decimal point become one plus two, i.e. three. four alternatives given above. Hence, In the product 216, which is the answer, the decimal point is alternative number (5) is the correct placed in such a way that the number of digits to the right of answer. the decimal point is equal to this total, that is, three. (C) DIVISION OF WHOLE AND DECIMAL NUMBERS : Example : Q.7. 504 ÷ 36 = ? (1) 1 (2) 8 (3) 12 (4) 14 (5) None of these Explanation Method Here, the division is a two-digit number. So let us 14 first consider a number formed by taking the first two digits 36 ) 504 of the dividend and divide it by 36. -36 Thus divide 50 by 36. 144 Now, let us consider the first digit of the dividend and -144 that of the divisor, starting from the extreme left and 000 determine what the quotient could be. If 5 is divided by 3 Quotient = 14. Remainder = 0. Since thethe quotient is 1. 36 x 1 = 36 . 36 is less than 50 or 36 < 50. remainder is zero the process of division is Hence, the quotient here is 1. complete. Hence, the correct answer is 14,The example may be completely solved by using this method. which is alternative number (4).

  37. In division of numbers with decimal fraction there may be three possibilities. 1. There is no decimal point in divisor and there is decimal point in dividend. 2. There is decimal point both in divisor and in dividend. 3. There is no decimal point in dividend and there is decimal point in divisor. Let us take an example of first type i.e., when there is no decimal point in divisor and there is decimal point in dividend. Example Q.8. 4.8 ÷ 16 = ? (1) 3 (2) 30 (3) 300 (4) 0.3 (5) None of these Explanation Method Here, although the divisor is a two-digit number, the 0.3 integer part of the divisor is one-digit number. So we 16) 4.8 have to first consider the number 4 and divide it by 16. - 0 48 Thus, divide 4 by 16. We get zero quotient. Now, let us - 48 consider the digit to the right of the decimal point in the 00 dividend. That digit is 8. When we will set down this 8 to get number 48, we will place the decimal point after zero, in the quotient. Quotient = 0.3 Remainder = 0 Thus, divide 48 by 16. We get 3 as the quotient, Which we write after the decimal point in the quotient. Since, we, Hence the correct answer is 0.3 get the remainder 0, the process of division is complete. which is alternative number (4).

  38. Remember to place a decimal point in the partial quotient obtained, as soon as the number to the immediate right of the decimal point in the dividend is set down for division. Now, let us consider an example of second type i.e., having decimal point both in dividend as well as in divisor. Example : Q.9. 0.48 ÷ 1.6 = ? (1) 3 (2) 30 (3) 300 (4) 0.3 (5) None of these First the divisor should be converted into an integer. Explanation Method In the divisor 1.6, there is only one digit to the right of the 0.48 0.48 decimal point. So, if we move the decimal point one place 1.6 1.6 x 10 to the right we will obtain the integer 16. So, we multiply 1.6 by 10 orally, and obtain the integer 16. Now, if we 0.48 x 10 = 4.8 multiply the denominator only by 10, the value of the 1.6 x 10 16 original fraction 0.48 will change. In order to retain the 1.6 Now this problem is reduced to original value of the fraction or for obtaining equivalent the problem where there is no fraction, we have also to multiply the dividend by 10. decimal point in the divisor. We do so orally and move the decimal point in the Hence, one can solve this fraction 0.48 one place to the right. problem as done in the Q.8. The correct answer will be 0.3 as in the earlier example, which is alternative number (4). Now, let us consider an example of third type i.e., when there is no decimal point in dividend and there is decimal point in divisor.

  39. Example : Q.10. 48 ÷ 1.6 = ? (1) 3 (2) 30 (3) 300 (4) 0.3 (5) None of these Explanation Method First, the divisor should be converted into an integer. 48 48 As explained in the earlier example we will multiply 1.6 1.6 1.6 x 10 by 10 to obtain the integer 16. To follow the rule for obtaining equivalent fractions, we have also to multiply 48 x 10 480 the dividend by 10. We do so by placing one zero to the 1.6 x 10 = 16 extreme right of 48. Now this problem is reduced to the problem Where there is no decimal point in the divisor. Hence, one can solve this problem using the method explained for solving the problem in Q. 7. The quotient will be 30. Hence, the correct answer will be 30, which is Alternative number (2). (Precautions with decimal point : While adding, subtracting, multiplying or dividing any number with decimal point, one should be very careful about the digits and placement of decimal point at appropriate place in the answer.)

  40. (D) INVERSE OPERATIONS : ADDITION, SUBTRACTION, MULTIPLICATION AND DIVISION OF NUMBERS. For solving the question of this type we will follow the properties of the equation mentioned earlier under the title “Reviewing essential concepts”. Example : Q.11. 435 + ? = 650 (1) 1085 (2) 1015 (3) 285 (4) 215 (5) None of these Method 435 + ? = 650 435 + ? – 435 = 650 – 435 (Subtracting 435 from both sides) Therefore ? = 215 Hence, the correct answer is 215, which is alternative number (4). Example : Q.12. 460 – ? = 284 (1) 276 (2) 186 (3) 286 (4) 176 (5) None of these Method 460 – ? = 284 460 – ? + ? = 284 + ? (Adding ? to both sides) 460 = 284 + ? 460 – 284 = 284 + ? – 284 (Subtracting 284 from both sides) 176 = ? or ? = 176 Hence, the correct answer is 176, which is alternative number (4). Similarly, questions of multiplications and divisions can be solved. Q.13. 216 × ? = 2592 (1) 559872 (2) 11 (3) 12 (4) 10 (5) None of these

  41. Method 216 × ? = 2592 216 x ? = 2592 (Dividing both sides by 216) 216 216 ? = 2592 216 Therefore ? = 12 Hence, the correct answer is 12, which is alternative number (3). Example : Q.14. 14 = 2 ? 5 (1)   (2)   (3)    (4) 35 (5) None of these Method 2 × ? = 14 × 5 (After cross-multiplication, see the explanation given under the title ‘Reviewing essential concepts’) (Dividing both sides by 2) ? = 35 Hence, the correct answer is 35, which is alternative number (4).

  42. (E) MIXED OPERATIONS An expression may contain many terms and several operations of different kinds such as addition, subtraction, multiplication, and division. In such cases, there are certain conventions to be followed as to the order in which these operations are performed. (1) The operations are carried out proceeding from left to right. (2) When there are no brackets in the given expression, out of the four kinds of operations (viz., addition, subtraction, multiplication, and division), multiplication or division is to be carried out first, proceeding from the left to the right. In 8 + 4 ÷ 2, the operation 4 ÷ 2 is carried out first, we get the answer 10. In 20 – 8 × 2, the operation 8 × 2 is carried out first, we get the answer 4. (3) When there are no brackets in the given expression, after carrying out multiplication or division; addition or subtraction are to be carried out, proceeding from the left to the right. In 6 + 7 – 4, the operation 6 + 7 is carried out first, we get the answer 13. Next the operation 13-4 carried out, we get the answer 9. In 22 – 15 + 3 – 4, the operation 22 – 15 is carried out first, we get 7. Next, the operation 7 + 3 is carried out, we get 10. In the end, the operation 10 – 4 is carried out, we get the answer 6. In 72 ÷ 12 – 3 × 2, the operation 72 ÷ 12 is carried out first, we get 6. Next the operation 3 × 2 is carried out, we get 6. In the end, the operation 6 – 6 is carried out, we get the answer 0. (4) When brackets are used, the operations inside the brackets are to be carried out first, then the remaining operations are to be carried out in the order mentioned above. In 32 ÷ (6 + 2) × 2 – 1, the operation 6 + 2 (inside the brackets) is carried out first, we get 8. Next, the operation 32 ÷ 8 is carried out, we get 4. After this, the operation 4 × 2 is carried out, we get 8. The last operation carried out is 8 – 1, we get the answer 7. In short, the operations should be carried out in the following order : 1. Brackets 2. Multiplication or division 3. Addition or subtraction

  43. Let us consider examples of mixed operations you may come across in the test. Example : Q.15. 84 + 48 ÷ 12 – 6 × 3 = ? (1) 276 (2) 15 (3) 70 (4) 92 (5) None of these Method Using the conventions mentioned under “mixed operations” the question will become : 84 + \ 48 ÷ 12 / – \ 6 × 3 / = ? 1st operation 2nd operation 48 ÷ 12 = 46 × 3 = 18 Replacing the obtained values we get the expression 84 + 4 – 18 = ? ln 84 + 4 – 18, the operation 84 + 4 is carried out first, we get 88. The operation 88 – 18 is carried out in the last, we get the answer 70. The correct answer is 70, which is alternative number (3). Now, let us solve another question given below : Example : Q. 16. (1) 5 (2) (3) 21 (4) 20 (5) None of these Method To solve , it should be taken into two parts i.e., (1) Numerator and (2) Denominator. Solve the two parts separately by following the conventions mentioned above. Numerator is \ 12 × 7 / – 4   12 × 7 = 84 84 – 4 = 80 Denominator is 8 × 2 = 16

  44. After putting the numerator and the denominator together, the expression becomes ? = 5 Thus, the correct answer is 5, which is alternative number (1). (F) ADDITION AND SUBTRACTION OF FRACTIONS : When the denominator of the fractions is the same, the simple rule for addition and subtraction of fractions is to add or subtract (as the case may be) the numerators of the fractions to get the numerator of the answer. The denominator of the answer will be common factor. See the following example : Q.17. (1)   (2)   (3)   (4)    (5)  None of these Method = (Common factor 8 will be the denominator) = (Numerators of both the fractions are added) Thus, the correct answer is which is alternative number (4). To add or subtract fractions with different denominators, it is necessary to convert the given fractions into fractions with a common denominator. Let us see the example given below : Q.18. (1)   (2)   (3)   (4)   (5)  None of these Explanation Method To convert and into fractions with a common denominator, we will find out a smallest number which is multiple of 5 as well as 6. We can do this by finding out factors of 5 and 6. Factors of 5 = 1 × 5, i.e., 1 and 5 (Denominators are made common) Factors of 6 = 2 × 3, i.e., 2 and 3

  45. Explanation Method Therefore, denominator common to both and =will be 1 × 5 × 2 × 3, i.e., 30. The denominator of first fraction is 5. If we multiply this by 6 we will (Common factor 30, will be the obtain 30 as the denominator. However, we will also denominator of the answer)have to multiply the numerator 3 of the first fraction by 6 to keep its original value unchanged. In effect we (Numerators of both the fractions are added)are multiplying by i.e. 1. Similarly, the denominator of the second fraction is 6. Thus, the correct answer is , which is In order to make it 30, we will have to multiply it by 5. alternative number (1)To keep the original value of the fraction unchanged we have to multiply the numerator also by 5. Example : Q.19. (1)   (2)   (3)   (4)   (5) None of these Explanation Method To convert and into fractions with a common denominator, we will find out a smallest number which is multiple of 5 as well as 3. We can do this by (Denominators are made common)finding out factors of 5 and 3. Factors of 5 = 1 × 5, i.e., 1 and 5 Factors of 3 = 1 × 3, i.e., 1 and 3 (Common factor 15 will be the denominator Therefore, denominator common to both and of the answer) will be 1 × 5 × 3, i.e. 15. (Numerators of both the fractions are added) We then convert and into fractions with Since, the numerator in the answer is greater than thea common denominator 15 and solve the denominator we can express it as problem as indicated in the earlier example.

  46. Because, Thus, Is the correct answer, which is alternative number (1). In the same way questions based on subtraction of fractions can also be solved. Let us take an example of subtraction. Example : Q.20. = ? (1)   (2)   (3)   (4)   (5) None of these Explanation Method To convert and into fractions with a common = denominator, we will find out a smallest number = (Denominators are made common) multiple of 6 as well as 8. We can do this by finding out factors of 6 and 8. Factors of = 6 = 2 × 3, i.e. 2 and 3. Factors of 8 = 2 × 2 × 2 i.e. 2, 2, and 2. Therefore, = (Common factor 24 will be the denominator common to both will be denominator of the answer) 2 × 3 ×2 × 2 i.e. 24. We then convert and into = ( Subtraction is carried out for the fractions with a common denominator 24 and solve numerators of the fractions) the problem. Thus, is the correct answer, which is alternative number (2).

  47. (G) MULTIPLICATION AND DIVISION OF FRACTIONS : In multiplication of fractions, write the product of their numerators as the numerator and the product of their denominators as the denominator of the answer. Here, cancel the common factor/s, if any, before writing the product Example : Q.21. (1) (2) (3) (4) (5) None of these Explanation Method We can do it in two ways (1) Both the numerator and the denominator, have a common factor 5, we can cancel it andwrite the fraction in the reduced form. We can cancel the common factor even (2) before finding the product. The answerobtained is the same. To divide a fraction by another fraction, we must multiply the The correct answer is , which is alternative former by the reciprocal of the later number (1)

  48. Let us take the following example. Example : Q.22. = ? (1) 49 (2)   (3)   (4)  4 (5) None of these Explanation Method To find out the reciprocal of a given fraction the number in its numerator and denominator should be made to interchange their places. Therefore, the reciprocal of will be . and are reciprocals of each other, because their product ( ) is 1. The correct answer is , which is alternative number (3) (H) PERCENTAGE : A percent is a fraction with a denominator of 100. Instead of writing the denominator, we write the symbol %. Thus, to write a percent as a fraction, drop the % sign and write a fraction with the original number as the numerator and 100 as denominator. If a decimal is needed, this fraction can be very easily converted to a decimal. Example : One percent = 1% = = 0.01.

  49. Let us solve some examples on percentage. Q.23. 15% of 840 = ? (1) 126 (2) 12600 (3)  1200 (4) 12000 (5) None of these To find given percent value of given number as in the above question we can make use of decimals. 15% of 840 = ? = 15% can be written as or 0.15 or = 0.15 × 840 = 126 Thus, 126 is the correct answer, which is alternative number (1). Now let us take another example of inverse nature. Example : Q.24. 15% of ? = 60 (1) 4 (2) 900 (3) 40 (4) 90 (5) None of these To solve this, as done in the earlier case, 15% can be written as 0.15. Thus, the question will be 0.15 of ? = 60 0.15 × ? = 60 ? = (Dividing both sides by 0.15) = (Following the rules of division for decimals explained in Q.9.) = 400 In the alternatives 400 is not given, therefore ‘None of these’ i.e., alternative number (5) is the correct answer.

  50. (I) SQUARES AND SQUARE ROOTS : When any number is multiplied to itself, the product obtained is called as square of that number. e.g. Square of 7 is 7 × 7 = 49. Here 49 is the square of 7. So 7 is the square root of 49. The sign is used to denote the square root of a number. means the square root of 49. In other words, is a number whose square is 49. Take another example –– = 15, Square root of 225 is 15. The number of prime factors of a perfect square is even. These factors can be so grouped into pairs that each pair contains the same factor twice. In order to find the square root of a number, we just have to take one factor from each pair. The product of these factors is the required square root. Let us solve the following example : Example : Q.25. (1) 24 (2) 26 (3) 34 (4) 36 (5) None of these As 576 is divisible by 2 and 3, we can find out if 576 is also divisible by squares of 2 and 3 i.e. 4 and 9. 9 576 As explained in the earlier pages under the title ‘Reviewing essential concepts’ 4 66 we have found out the factors of 576. 4 16 576 = 9 × 4 × 4 × 4 4 4 We can now write these factors in terms of prime factors and pair them. 1 576 = 3 × 3 × 2 × 2 × 2 × 2 × 2 × 2 We can now have two groups with same factors. 576 = (3 × 2 × 2 × 2) (3 × 2 × 2 × 2) = (24) (24) So, = 24 Thus, the correct answer is 24, which is alternative number (1).

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