SERIES COMPLETION. Number Series

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3 SERIES COMPLETION Number series problems deal with numbers. While attempting to solve the question, you have to check the pattern of the series. Series moves with certain mathematical operations. You have to check the pattern. Type of questions asked in the examination : (i) Find the missing term(s). (ii) Find the wrong term(s). Number Series In this type of series, the set of given numbers in a series are related to one another in a particular pattern or manner. The relationship between the numbers may be Consecutive odd/even numbers, Consecutive prime / composite numbers, Squares/cubes of some numbers with/without variation of addition or substraction of some number, Sum/product/difference of preceding number(s), Addition/subtraction/multiplication/division by some number, and Many more combinations of the relationship given above. Directions : (1 to 13) Find the missing numbers : Ex 1. 3, 5, 7, 9, 11, 13, 15, 17,? (A) 14 (B) 19 (C) 15 (D) 21 (B) Each term has a common difference = + 2. Hence, next term = = 19. Ex 2. 2, 3, 5, 7, 11,?, 17 (A) 14 (B) 13 (C) 10 (D) 12 (B) The series is made up of consecutive prime numbers. Therefore, the missing term is 13. Ex 3. 1, 4, 9, 16, 25,? (A) 35 (B) 36 (C) 37 (D) 49 (B) Each term is a square of 1, 2, 3, 4 and so on 1 2 = 1, 2 2 = 4, 3 2 = 9, 4 2 = 16, 5 2 = 25. Hence, next term = 6 2 = 36. Ex 4. 2, 5, 10, 17,? (A) 24 (B) 25 (C) 26 (D) 27 (C) Each term is a square of 1, 2, 3, 4 and so on and 1 is added to it, i.e , (2) 2 +1, (3) 2 +1,...= 2, 5, 10, Hence, next term = (5) = 26. Ex 5. 2, 3, 10, 15, 26, 35,? (A) 48 (B) 51 (C) 49 (D) 50 (D) The series exhibits the pattern of n 2 + 1, n 2 1, alternately, n taking values 1, 2,... Ex 6. 1, 8, 9, 64, 25, 216,?,? (A) 49, 64 (B) 343, 64 (C) 49, 512 (D) 343, 512 (C) Odd positioned digits are squares of 1, 3, 5 and so on, i.e. 1 2 = 1, 3 2 = 9, 5 2 = 25 and so on. Similarly, even positioned digits are cubes of 2, 4, 6, etc., i.e. 2 3 = 8, 4 3 = 64, 6 3 = 216. Therefore, the next term would be 7 2 i.e. 49 and 8 3 = 512 respectively. Ex 7. 0, 7, 26,?, 124, 215 (A) 51 (B) 37 (C) 63 (D) 16 (C) Each term is a cube of 1, 2, 3, 4 and so on and 1 subtracted from it, i.e , 2 3 1, 3 3 1, 4 3 1, 5 3 1, Therefore, the term replacing the question mark would be = 64 1 = 63. Ex 8. 3, 4, 10, 33, 136,? (A) 240 (B) 430 (C) 685 (D) 820 (C) The terms of the series are, previous term 1 + 1, previous term 2 + 2, previous term and so on. Hence, the next term will be = = 685. Ex 9. 11, 15, 21, 29,? (A) 40 (B) 41 (C) 37 (D) 39 (D) This series consists of increasing numbers. The pattern is +4, +6, +8,... Ex 10. 3, 6, 18, 72, 360,? (A) 720 (B) 1080 (C) 1600 (D) 2160 (D) The sequence in the given series is 2, 3, 4, 5, 6. Hence, the missing number is =2160. Ex 11. 6, 12, 7, 11, 8, 10, 9,? (A) 8 (B) 9 (C) 11 (D)10 (B) Alternate series [Difference series] (i) 6, 7, 8, 9 (ii) 12, 11, 10,? Ex 12. 0, 5, 22, 57, 116,? (A) 205 (B) 216 (C) 192 (D) (A) Difference Difference Hence, the next term = 205 PAGE # 1

4 Ex , 158, 172, 182,? (A) 210 (B) 193 (C) 197 (D) 203 (B) = 7, The difference between 151 & 158 is seven (7) = 14, The difference between 158 & 172 is (14) = and so on, Missing term = = 193. Directions : (14 to 15) In each of the following questions, a number series is given. After the series, below it in the next line, a number is given followed by (P), (Q), (R), (S) and (T). You have to complete the series starting with the number given following the sequence of the given series. Then answer the question given below it. Ex (P) (Q) (R) (S) (T) Which number will come in place of (T)? (A) 1412 (B) 164 (C) 696 (D) 78 (A) Similarly (P) (Q) (R) (S) (T) Therefore, the number 1412 will come in place of (T). Ex (P) (Q) (R) (S) (T) Which number will come in place of (Q)? (A) 113 (B) 17 (C) 3912 (D) 8065 (A) Similarly, (P) (Q) (R) (S) Therefore, the number 113 will come in place of (Q). Direction : (16 to18) Find the wrong term : Ex. 16 2, 5, 9, 11, 14 (A) 2 (B) 5 (C) 9 (D) 11 (C) Series : + 3, + 3, + 3,... The next term is got by adding 3 in preceeding term = 5, = 8 9 is wrong term. Ex , 100, 1100, 11000, , (A) (B) (C) 100 (D) (D) Given series is : is wrong. The correct term is Ex. 18 2, 6, 11, 17, 23, 32, 41 (A) 6 (B) 17 (C) 23 (D) 32 (C) Given series is : 24 2, 6, 11, 17, 23, 32, Alphabet Series problems deals with alphabets and Alpha-Numeric. While attempting to solve the question, you have to check the pattern of the series. Type of questions asked in the examination : (i) Find the missing term(s). (ii) Find the wrong term(s). Alphabet Series In these types of questions, a series of single or pairs of groups of letters is given. The terms of the series form a certain pattern as regards the position of the letters in the English alphabet. Position of Alphabet : (i) Alphabet in order : (ii) Alphabet in reverse order : Z Y X W V U T S R Q P O N M L K J I H G F E D C B A PAGE # 2

5 Directions : (1 to 7) Find the missing term : Ex 1. A, C,?, G, I (A) E (B) D (C) F (D) H (A) Series consists of alternate letter in order. So, the missing term would be E. Ex 2. V, T, R,?, N,? (A) O,M (B) P,M (C) L,P (D) P,L (D) Given series consists of alternate letters in reverse order. So, the missing terms would be P and L. Ex 3. Ex 4. A, C, F,?, O (A) G (C) H (B) Hint (B) J (D) K A C F J O DC, DE, FE,?, HG, HI (A) FE (C) GF +2 (B) Hint D C D E 0 +2 Ex 5. CIR, GMV, KQZ, OUD,? (A) RYH (C) SZI 0 (B) FG (D) GH F E FG H G H I (B) SYH (D) SYI (B) There is a continuous difference of 4 letters between the first letter of each group, second letter of each group and third letter of each group. So the missing term would be SYH. Ex 6. ZSD, YTC, XUB, WVA,? (A) VWZ (B) UVW (C) VXY (D) UWZ (A) The first letter of each group is in continuation in backward direction. The second letter of each group is in continuation in forward direction. The third letter of each group is in continuation in backward direction. Therefore, the missing term would be VWZ. Ex 7. KTE, SBM, AJU, IRC,? (A) OZL (B) QYZ (C) QZL (D) QZK (D) First letter of each group differ by 8 letters. Second letter of each group differ by 8 letters. Third letter of each group differ by 8 letters. Therefore, the missing term would be QZK. Directions : (8 to 9) Find the wrong term (s) : Ex 8. Ex 9. DOU, EPV, FQW, GRX, HTY, ITZ (A) EPV (B) FQW (C) GRX (D) HTY (D) In every term first, second and third letter is in alphabetical order to its next term respectively. Fourth term is not following the same rule. Hence, HTY is the wrong term and should be replaced by HSY. ABC, DGJ, HMR, NTA, SBK, ZKV (A) DGJ (B) HMR (C) NTA (D) SBK (C) First letter of first, second, third,...terms is moved three, four, five,...steps forward respectively. Similarly, second letter is moved five, six, seven,...steps forward respectively and third letter is moved seven, eight, nine,...steps forward respectively. Hence, NTA is the wrong term and should be replaced by MTA. Alpha-Numeric Series A series in which both alphabets and number are used. Direction : (10) Find the missing term : Ex 10. F3X, H7U, J15R, L31O,? (A) M46L (B) N44L (C) N63L (D) N44M (C) The first letter of each term is moved two steps forward and the last letter is moved three steps backward to obtain the corresponding letters of the next term. The numbers form the sequence = 7, = 15, = 31, = 63. So, the missing term would be N63L. Letter Repeating Series These type of questions usually consists of a series of small letters which follow a certain pattern. However, some letters are missing from the series. These missing letters are then given in a proper sequence as one of the alternatives. Pattern of such questions is that some letters in sequence are missing. (i) The letters may be in cyclic order (clockwise or anti-clockwise). (ii) To solve a problem, we have to select one of the alternative from the given alternatives. The alternative which gives a sequence form of letters is the choice. Directions : (1 to 6) Which sequence of letters when placed at the blanks one after the other will complete the given letter series? Ex 1. a _ a b _ b a _ a _ a b (A) babb (B) abba (C) baba (D) aabb (A) we proceed step by step to solve the above series : Steps : 1. We have two letters a and b making the series. 2. The first blank space should be filled in by 'b' so that we have one a followed by one b. 3. Second blank space should be filled in by 'a' so that the same pattern followed till end. PAGE # 3

6 Ex 2. a _ cab _ a _ c _ b c (A) bbac (C) abba (B) abab (D) bcba (A) The series is x y z u v / y z u v x/ z u v x y/u v x y z Thus the letters are written in a cyclic order. Ex 3. Sol- Ex 4. Ex 5. Sol- (D) Series is abc/ abc/ abc/ abc. So, pattern abc is repeated. _ abb _ a _ baa _ b (A) baba (B) abba (C) aabb (D) aaab (C) Series is aabb/ aabb/ aabb. So, pattern aabb is repeated. ba _ cb _ b _ bab _ (A) acbb (B) bcaa (C) cabb (D) bacc (D) The series is b a b c/b a b c/b a b c So, pattern babc is repeated ab _ aa _ caab _ c _ abb _ c (A) bbcaa (B) bcbca (C) cabac (D) cbbac (D) Series is abc / aabc / aabbc / aabbcc Ex 6. bc _ b _ c _ b _ ccb (A) cbcb (B) bbcb (C) cbbc (D) bcbc Sol- (A) Series is bccb / bccb / bccb. So, pattern bccb is repeated Direction : (7 to 9) The series given below is based on the letter series, In the series, some letters are missing. Select the correct alternative. If more than five letters are missing, select the last five letters of the series. Ex 7. Ex 8. x _ xxy _ x _ xy _ yxx yy _ y (A) xyyyy (B) xxyyx (C) yxxyx (D) xyxyx (A) The pattern of series is xy/xxyy/xxxyyy... xyzu _ yz _ v uv _ (A) uvxyz (B) vuzyx (C) uvzyx (D) vuxyz Ex 9. abcd _ bc _ e de _ (A) deabc (B) edcba (C) decba (D) edabc (A) The series is a b c d e / b c d e a / c d e a b / b e a b c Thus the letters are written in a cyclic order. Directions : (10 to 11) There is a letter series in the first row and a number series in the second row. Each number in the number series stands for a letter in the letter series. Since in each of that series some term are missing you have to find out as to what those terms are, and answer the questions based on these as given below in the series. Ex 10. Ex 11. _ m i a x _ i r x a m a 4 _ 5 _ 7 3 _ 6 The last five term of the letter series are (A) r m x i a (B) x m r a i (C) x r m a i (D) r m i x a (D) a = 6, i = 5, m = 3, r = 4 and x = 7 the letter series runs as rmiax mirxa irmax rmixa. By taking the letter in the groups of five, we find that first letter of the first group (i.e. r) is the third letter of the second group and the last two letters have interchanged their positions. The same rule applies in others groups also. a _ h c _ n e _ h _ e a c _ 2 1 _ 4 3 _ The last five terms in the series are (A) (B) (C) (D) (B) By taking a = 2, c = 1, n = 4, h = 5 and e = 3, the numbers series runs as If first digit of a group of five digits is placed as the last digit, we obtain the second group of five digits and so on. Directions : (1 to 10) Find the missing term/number(s) : Missing Term in Figure Ex 1. In such type of questions, a figure, a set of figures, an arrangement or a matrix is given each of which bears certain characters, be it numbers, letters or a group or combination of letters or numbers, following a certain pattern. (A) 125 (B) 25 (C) 625 (D) 156 (C) Clearly (1 + 3) 2 = 16 (15 + 6) 2 = (21) 2 = 441 (10 + 5) 2 = (15) 2 = 225 missing number in figure] ( ) 2 = (25) 2 = 625. PAGE # 4

7 Ex 2. In the second diagram, (1 2 3) = 6 and the number on the sides twice as the number on the opposite vertex. 3 2 = 6, 2 2 = 4, 1 2 = 2 In the Third diagram, (7 4 5) = 140 and the number on the sides twice as the number on the opposite vertex. 7 2 = 14, 5 2 = 10, 4 2 = 8. Ex 3. (A) 64 (B) 36 (C) 34 (D) 60 (A) Moving clockwise, in every quarter region, value of numbers gets doubled. 2 2 = 4, 8 2 = 16, 16 2 = 32, 32 2 = 64, = 256. Ex ? 10 (A) 0 (B) 5 (C) 10 (D) 15 (B) In first figure, (14 + 6) 2 2 = 16 In second figure, (12 + 5) 3 1 = 14 In third figure, (11 + 4) 10 1 = 5 4 (A) 120 (B) 100 (C) 125 (D) 64 (C) In the first column, = 53 In the second column, = 90 So, missing number, = 125. Ex 4. B C O Q M N (A) A (C) G? S R (B) D (D) P (A) In each column, the sum of top & bottom letter is equal to the order of the middle letter in that column ? 8 Ex (A) 16 (B) 14 (C) 20 (D) 22 (B) In the first diagram, (3 4 6) = 72 and the number on the sides twice as the number on the opposite vertex. 3 2 = 6, 6 2 = 12, 4 2 = Ex (A) 262 (B) 622 (C) 631 (D) 631 (B) In first figure, ( ) = 551. In second figure, ( ) = 246. In third figure, missing number = ( ) = Ex ? (A) 5 (B) 19 (C) 27 (D) 89 (D) In first figure, (6 3) + (5 15) = = 93. In second figure, (4 8) + (18 + 1) = = 50. In third figure, missing number = (9 6) + (7 5) = = 89. Ex 10. Which one number can be placed at the sign of interrogation? (A) 5 (B) 6 (C) 8 (D) 9 (D) In first figure, = 93 In second figure, = 79 So, In third figure, x = 67, x = 9 42? 6? PAGE # 5

8 CLOCK TEST Important Facts Minute hand and hour hand coincides once in every hour. They coincide 11 times in 12 hours and 22 times in 24 hours. They coincide only one time between 11 to 1 O clock. at 12 O clock. Minute hand and hour hand are opposite once in every hour. They do it 11 times in 12 hours and 22 times in 24 hours. They opposite only one time between 5 to 7 O clock. at 6 O clock. Both hands (minute and hour) are perpendicular twice in every hour. 22 times in 12 hours and 44 times in 24 hours. In one minute, hour hand moves 1/2º and minute hand moves 6º. In one hour, hour hand moves 30º and minute hand moves 360º. In an hour, minute hand moves 55 minutes ahead of hour hand. Ex.1 Ex.2 Ex.3 At what time between 3 O Clock and 4 O Clock will the two hands coincide? At 3 O clock the distance between the two hands is 15 minutes when they coincide with each other the distance between the two hands will be 0 min. So, the time taken ( ) = 15 minutes. Minute hand is 55 min. ahead of hour hand in 60 min. Minute hand is 1 min. ahead of hour hand in 60 min. 55 Minute hand is 15 min. ahead of hour hand in = = 16 min Hence the right time is 4 16 minute past At what time between 2 O clock and 3 O clock will the two hands be opposite? At 2 O clock the distance between the two hands is 10 minutes. When they are at 30 minutes distance, they are opposite to each other. The time taken ( ) = 40 min. Minute hand is 55 min. ahead of hour hand in 60 min. Minute hand is 1 min. ahead of hour hand in 60 min. 55 Minute hand is 40 minutes ahead of hour hand in = = 43 min Hence, the right time is 43 min. past At what time between 4 O clock and 5 O clock will the hands are perpendicular? At 4 O clock the distance between the two hands is 20 min. When they are at 15 minutes distance, they are perpendicular to each other. Case-I When the time taken (20 15) = 5 min. Minute hand is 55 min. ahead of hour hand in 60 min. Minute hand is 5 min. ahead of hour hand in = = 5 min Hence, the right time is 5 min. past Case-II When the time taken ( ) = 35 min. Minute hand is 55 min. ahead of hour hand in 60 min. Minute hand is 35 min. ahead of hour hand in = = 38 min Hence, the right time is 38 min. past Mirror Image of Clock If the time is between 1 O clock to 11 O clock, then to find the mirror image, time is subtracted from 11 : 60. If the time is between 11 O clock to 1 O clock, then to find the mirror image, time is subtracted from 23 : 60. Ex.4 The time in the clock is 4 : 46, what is the mirror image? (11 : 60) (4 : 46) = 7 : 14. Ex.5 The time in the clock is 12 : 35, then find its mirror image. (23 : 60) (12 : 35) = 11 : 25. Angle between Two Hands Angle are of two types : Positive angle : It is obtained by moving from hour hand to minute hand moving in clockwise direction. Negative angle : It is obtained by moving from minute hand to hour hand. Ex.6 Ex.7 Both types of angles are 360º in total. If one angle is known, other can be obtained by subtracting from 360º. At 4 : 30, what is the angle formed between hour hand and minute hand? At 4 O clock angle between hour and min. hand is of 120º. In 30 min. minute hand make an angle of 180º. So, the resultant angle is 180º 120º = 60º. But in 30 min. hour hand will also cover an angle of 15º. Hence, the final angle between both hands is 60º 15º = 45º. A bus for Delhi leaves every thirty minutes from a bus stand. An enquiry clerk told a passenger that the bus had already left ten minutes ago and the next bus will leave at 9.35 A.M. At what time did the enquiry clerk give this information to the passenger? Bus leaves after every 30 minutes. The next bus will leave at 9 : 35 A.M. The last bus left at 9 : 35 0 : 30 = 9 : 05 A.M. but clerk said that bus had left 10 minutes earlier. 9 : : 10 = 9 : 15 A.M. PAGE # 6

9 CALENDAR TEST To find the day of the week on a mentioned date. Certain concepts are defined as under. An ordinary year has 365 days. In an ordinary year, first and last day of the year are same. A leap year has 366 days. Every year which is divisible by 4 is called a leap year. For example 1200, 1600, 1992, 2004, etc. are all leap years. For a leap year, if first day is Monday than last day will be Tuesday for the same year. In a leap year, February is of 29 days but in an ordinary year, it has only 28 days. Year ending in 00's but not divisiable by 400 is not considered a leap year. e.g., 900, 1000, 1100, 1300, 1400, 1500, 1700, 1800, 1900, 2100 are not leap years. Days Odd Days Similarly, 200 years = 10 odd days = 03 odd days years = = 1 odd day years = = 0 odd day (1 is added as is a leap year) Similarly, 800, 1200, 1600, 2000, 2400 years contain 0 odd days. After counting the odd days, we find the day according to the number of odd days. Sunday for 0 odd day, Monday for 1 odd day and so on as shown in the following table. Table : 1 (Odd days for week days) Sunday Monday Tuesday Wednesday Thursday Friday Saturday The day on which calendar started (or the very first day ) i.e., 1 Jan, 0001 was Monday. Calendar year is from 1 Jan to 31 Dec. Financial year is from 1 April to 31 March. Odd Days The number of days exceeding the complete number of weeks in a duration is the number of odd days during that duration. Every ordinary year has 365 days = 52 weeks +1 day. Ordinary year has 1 odd day. Every leap year 366 days = 52 weeks + 2 days. Leap year has 2 odd days. Odd days of 100 years = 5, Odd days of 200 years = 3, Odd days of 300 years = 1, Odd days of 400 years = 0. Explanation : 100 years = 76 ordinary years + 24 leap years ( The year 100 is not a leap year) = 76 odd days odd days = 124 odd days. Ordinary Year Table : 2 (Odd days for months in a year) Days Odd Days Leap year Days Odd Days January 31 3 January 31 3 February 28 0 February 29 1 March 31 3 March 31 3 April 30 2 April 30 2 May 31 3 May 31 3 June 30 2 June 30 2 Total 181 days 6 Total 182 days 0 July 31 3 July 31 3 August 31 3 August 31 3 September 30 2 September 30 2 October 31 3 October 31 3 November 30 2 November 30 2 December 31 3 December 31 3 Total 184 days 1 Total 184 days 2 M onths of years Total days Odd days Table : 3 (Odd days for every quarter) I st three months 1 Jan to 31 March 90 / 91 Ord. / Leap 6 / 0 Ord. / Leap II nd three m onths 1 Apr to 30 June III rd three months 1 July to 30 Sep. Iv th three months 1 Oct. to 31 Dec Odd day 1 Odd day 1 Odd day Total year 1 Jan to 31 Dec. 365 / 366 Ord. / Leap 1 / 2 Ord. / Leap Odd days = 124 = 5 odd days. 7 PAGE # 7

10 Ex 1. If it was Saturday on 17th December 1982 what will be the day on 22nd December 1984? Total number of odd days between 17 Dec.1982 to 17 Dec.1984 the number of odd days = 1+2 = 3. From 17 to 22 Dec. number of odd days = = 8 odd days = 1 odd day. Saturday + 1 odd day = Sunday. Ex 2. Find the day of the week on 18 July, Here 1600 years have 0 odd day...(a) Ex years have 5 odd days...(b) 75 years = (18 leap years + 57 ordinary years) = ( ) = 93 odd days = ( ) = 2 odd days....(c) Now, the number of days from 1 st January to 18 th July, 1776 = = ( ) days = 4 odd days...(d) Adding, A, B, C & D = = 4 odd days. Hence, the required day is Thursday. On what dates of October, 1975 did Tuesday fall? For determining the dates, we find the day on 1 st Oct, years have 0 odd days...(a) 300 years have 1 odd days...(b) 74 years have (18 leap years + 56 ordinary years) = 92 odd days = 1 odd days...(c) Days from 1 st January to 1 st Oct. = 1 st Jan to 30 June + 1 st July to 1 st Oct. = = 274 days. Odd days 274/7= 1 odd days....(d) Adding A, B, C & D = = 3 odd days. So, Wednesday falls on 1 st Oct. Hence, 7, 14, 21, 28 October will Tuesday fall. Ex 4. Which year will have the same calendar next to The calendar for 1995 and the required year will be the same if day on 1 st January of both the years is same. This is possible only if the total odd days from 1 st January 1995 to 31 st December of the previous year of required year is 0. Let the required year is 2006 then, we have 3 leap years (1996, 2000, 2004) and 8 ordinary years (1995, 1997, 1998, 1999, 2001, 2002, 2003, 2005) Total odd days = ( ) = 14 = 0 odd days. Hence, the required year is Ex 5. Prove that last day of a century cannot be Tuesday, Thursday or Saturday. 100 years have = 5 odd days. Last day of 1 st century is Friday. 200 years have = 10 odd days = 3 odd days. Last day of 2 nd century is Wednesday. 300 years have = 15 odd days = 1 odd day Last day of 3 rd century is Monday. 400 years have = = 21 = 0 odd days. So, the last day of 4 th century is Sunday. Since the order keeps on cycling, we see that the last day of the century cannot be Tuesday, Thursday or Saturday. PAGE # 8

11 VENN DIAGRAM 1. An object is called a subset of another object, if former is a part of latter and such relation is shown by two concentric circles. (i) Pencil, Stationary (ii) Brinjal, Vegetable (iii) Chair, Furniture It is very clear from the above relationship that one object is a part of other, and hence all such relationships can be represented by figure below Directions : (1 to 6) Each of these questions given below contains three group of things. You are to choose from the following four numbered diagrams, a diagram that depicts the correct relationship among the three groups of thing in each question. Ex 1. Moon, Earth, Universe (A) (B) Pencil Brinjal Chair Stationary Vegetable Furniture 2. An object is said to have an intersection with another object, when two objects share some thing in common. (i) Surgeon, Males (ii) Politicians, Indian (iii) Educated, Unemployed Surgeon Politician Educated Males Indian Unemployed All the three relationships given above have something in common as some surgeons can be male and some female, some politicians may be Indian and some may belong to other countries, educated may be employed and unemployed as well. And all the three relationships can be represented by figure above. 3. Two objects are said to be disjoint when neither one is subset of another nor they share anything in common. In other words, totally unrelated object fall under this type of relationship (i) Furniture, Car (ii) Copy, Cloth (iii) Tool, Shirt Ex 2. Ex 3. Ex 4. Ex 5. (C) (D) (A) Moon and Earth are the parts of universe and therefore are subsets of universe. India, Pakistan, Asia (A) (C) (B) (D) (A) India and Pakistan are the subsets of Asia. Batsman, Cricket, Stick (A) (C) (B) (D) (D) Batsman is a subset of Cricket and Stick is something unrelated to Cricket. Book, Pen, Pencil (A) (C) (B) (D) (C) Book, Pen, Pencil are neither subset of one another nor have anything in common. W hich of the following diagrams correctly represents the relationship among Tennis fans, Cricket players and Students. It is clear from the above relationship that both the objects are unrelated to each other, and hence can be represented diagrammatically as shown in figure above. From the above discussion we observe that representation of relationship of two objects is not typical if students follow the above points. But representation of three objects diagrammatically pose slight problem before the students. A variety of such relationship is being discussed in the following examples. (A) (C) (B) (D) (A) From the relationship given in the question, we observe that each of the objects carries something in common to one another. A Tennis fan can be a cricket player as well as student. Hence Diagram (A) represents this relationship. PAGE # 9

12 Ex.6 W hich of the following diagrams correctly represents the relationship among smokers, bidi smokers, cancer patients. Ex 11. School teachers who are neither married nor do live in joint families are represented by (A) F (B) C (C) B (D) A (A) (C) (B) (D) (B) Bidi smokers is a subset of smokers and cancer patient may be a smoker, bidi smoker and non-smoker. Hence third object shares a common relationship with first and second object as well. Directions : (7 to 11) In the following diagram, three classes of population are represented by three figures. The triangle represents the school teachers, the square represents the married persons and the circle represents the persons living in joint families. Ex 7. Ex 8. Ex 9. Ex 10. F E B C D A Married persons living in joint families but not working as school teachers are represented by (A) C (B) F (C) D (D) A Persons who live in joint families, are unmarried and who do not work as school teachers are represented by (A) C (B) B (C) E (D) D Married teachers living in joint families are represented by (A) C (B) B (C) D (D) A School teachers who are married but do not live in joint families are represented by (A) C (B) F (C) A (D) D : (Ex. 7 to EX. 11) EX 7. EX 8. EX 9. EX 10. EX 11. (C) Married persons living in joint families are presented by the region common to the square and the circle i.e., D and B. But, according to the given conditions, the persons should not be school teachers. So, B is to be excluded. Hence, the required condition is denoted by region D. (C) Persons living in joint families are represented by the circle. According to the given conditions, the persons should be unmarried and not working as school teachers. So, the region should not be a part of either the square of the triangle. Thus, the given conditions are satisfied by the region E. (B) Married teachers are represented by the region common to the square and the triangle i.e., B and C. But, according to the given conditions, the persons should be living in joint families. So, the required region should be a part of the circle. Since B lies inside the circle, so the given conditions are satisfied by the persons denoted by the region B. (A) As in the above question, married teachers are represented by B and C. But, here, the given conditions lay down that the persons should not be living in joint families. So, the required region should lie outside the circle. Since C lies outside the circle, so the given conditions are satisfied by the persons denoted by the region C. (A) School teachers are represented by the triangle. But according to the given conditions, persons are neither married nor do they live in joint families. So, the region should not be a part of either the square or the circle. Such a region is F. PAGE # 10

13 CODING-DECODING Coding-Decoding A code is a system of signals. Coding is, therefore, a method of transmitting a message between sender and receiver which cannot be understood or comprehended by a third person. The coding decoding test is set up to judge the candidate s ability to decipher a particular word/ message and break the code to decipher the message. In coding, actual alphabets/words/ terms/numbers are replaced by certain other alphabets/words/terms/symbols etc. according to a specific rule. To solve these type of questions we have to detect the rule and then answer the questions. Decoding : It is a method to find the meaning of something that has written in code. Letter-Letter Coding Ex 3. C A S T L E D C V X Q K If PAINTER is written in a code language as NCGPRGP, then REASON would be written as : (A) PCYQMN (B) PGYQMN (C) PGYUMP (D) PGYUPM (C) P A I N T E R N C G P R G P Similarly, Ex 1. In these type of questions, the letters in a word are replaced by certain other letters according to a specific rule to form its code. The candidate is required to detect the coding pattern / rule and answer the questions accordingly. If in any code language, KUMAR is coded as LVNBS, How is EMOTIONAL coded in that language. (A) FNQUJQBM (B) FNPUJPOBM (C) GNPUJPOBM (D) GNQUJQOBM Ex 4. R E A S O N P G Y U M P If in any code language NATIONAL is written as MZGRLMZO than how is JAIPUR written in that language. (A) QZRKFI (B) PZRKFI (C) QZRIFK (D) QARKFI (B) Similarly, N A T I O N A L M Z G R L M Z O 14+13=27 (A) 1+26 = = = = = = = 27 Similarly, E M O T I O N A L F N P U J P O B M J A I P U R Q Z R K F I Letter-Number Coding = = = = = = 27 Ex 2. If JAPAN is coded as KCSES, then the code for CASTLE will be - (A) DCIJOB (B) DCJKRD (C) DCKMSG (D) DCVXQK (D) J A P A N Similarly, K C S E S In these types of questions, either numerical code values are assigned to a word or alphabetical code letters are assigned to the numbers. Ex 5. In a certain code, if TREE is coded as 7100, FROG as 2159, how is FREE coded in that code? (A) 2100 (B) 3100 (C) 1003 (D) 1002 (A) T R E E F R O G Hence, F R E E PAGE # 11

14 Ex 6. In a certain code, C is coded as 0, E as 7, T as 4, I as 9, P as 1, R as 3, and U as 5. How is coded in that code? (A) PICTRUE (B) PICTURE (C) RICTPUE (D) PCTUREI (B) P I C T U R E Ex 7. Ex 8. If OX is coded as 39, what will be the code number for LION? (A) 20 (B) 25 (C) 38 (D) 50 (D) By their natural position in alphabet, O 15, X 24 So, OX = ( ) = 39 Hence, L I O N = ( ) = 50 If AJAY is written as 1117, then in same code NAMA would be written as: - (A) 5114 (B) 5411 (C) 5141 (D) 4511 (C) Hence A J A Y Substitution Coding Ex 9. Ex N A M A In these types of questions, some particular objects are assigned code names. Then a question is asked that is to be answered in the code language. If paper is called eraser, eraser is called bag, bag is called scale, scale is called pencil and pencil is called paper, what will a person write with? (A) Pencil (B) Paper (C) Eraser (D) Bag (B) A person will write with a pencil and a pencil is called paper. If water is called food, food is called tree, tree is called sky, sky is called wall, on which of the following does a fruit grow? (A) Water (B) Food (C) Tree (D) Sky (D) Clearly, a fruit grows on a tree. As given that tree is called sky, a fruit grows on sky. Puzzle Based Coding Ex 11. Ex 12. In this type of questions, some messages are given in the coded language and the code for a particular word or message is asked. To analyses such codes, any two messages bearing a common word are picked up. The common code word will thus represent that word. Proceeding similarly by picking up all possible combinations of two, the entire message can be decoded and the order for individual words found. In a certain code language 389 means run very fast. 964 means come back fast and 487 means run and come. Which digit in the language means come? (A) 7 (B) 9 (C) 4 (D) 8 (C) In the second and third sentences, common number is 4 and common code is come. Hence, number 4 stands for come. In a certain code language, ken poti means good morning, hu shang means come on, and hu ken sue means come for good. Which word in that language does mean for? (A) shang (B) ken (C) sue (D) hu (C) In the second and third statements, the common code word is come and the common word is hu. So, come means hu. In the first and third statements, the common code word is good and the common word is ken. So, good means ken. Thus, in third and above statements for means sue. Column Coding Ex 13. Decode the underlined letters in column I from the same row of choices provided under column II. Each small letter in column II stands for some capital letter in column I. However, the small letters in column II are not arranged in the same order as their corresponding letters in column I. The code is the same for all the terms in column I. Column Column 1. H N T B Z v b h n t 2. C T N Z B t h n w v 3. D N B Z C x h v t w 4. O H N T Z t b h i n 5. T Z O B K n i v e t From terms 1 & 2, in column (I), NTBZ is common. From col. (II) we have vhnt common. Hence H=b & C = w. From term-3, NBZC have already occurred and the code for D must be a letter which did not appear in the earlier codes. Thus, the codes for D is x. From term-4 HNTZ have occured earlier. The letter which has not occured earlier is the code for O, that is, i. From term-5 TZOB have occured earlier. The letter which has not occured earlier is the code for K, that is e. The code of underlined letters are given in the following table Letter H C D O K Code b w x i e PAGE # 12

15 Directions : (14 to 16) In column I below, some words are given. In column II, their codes are given but they are not arranged in the same order. Study both the columns and find out the code for the letter given in each of the following questions, from among the given alternatives. The code for a letter will be same throughout. Column I Column II (i) DRGEX (a) (ii) AXPRD (b) (iii) SDRKG (c) (iv) KLPSX (d) (v) SGPAK (e) (vi) PXDAG (f) (vii) GKSAE (g) Ex 14. What is the code used for the letter D? (A) 2 (B) 3 (C) 5 (D) 1 Ex 15. What is the code used for the letter P? (A) 8 (B) 5 (C) 3 (D) 2 Ex 16. What is the code used for the letter E? (A) 4 (B) 7 (C) 8 (D) 6 (14 to16) 14. (A) In statement (iii) and (vi), common letters are D and G and common code digit are 2 and 7. Hence, it is clear that D and G stand for 2 and 7 but not respectively. From statement (v), it is clear that the word has letter G and code 7 in its coding. Hence, code for G is 7 and D is (C) In statement (ii) and (iv), common letters are P and X and common code digit are 3 and 5. Hence, it is clear that P and X stand for 3 and 5 but not respectively. From statement (v), it is clear that the word has letter P and code 3 in its coding. Hence, code for P is (D) In statement (i) and (vii), common letters are G and E and common code digit are 6 and 7. Hence, it is clear that G and E stand for 6 and 7 but not respectively. From statement (vi), it is clear that the word has letter G and code 7 in its coding. Hence, code for E is 6. Directions : (17 to 19) In each questions there is a word written in capital letters with one letter underlined. For each letter in that word there is a code written in small letters. That code is denoted by either (A), (B), (C), (D) or (E) not in the same order. You have to find out the exact code for the underlined letter in the word. The number of that code is the answer. Please note that the same letter appearing in other word (s) may be coded differently. Ex 17. MAGIC (A) km (B) eg (C) ik (D) ce (E) oq (C) M(+2) o (+2)q oq, A(+2) c(+ 2) e Ex 18. ce, G (+2)i(+2)k ik, I (+2)k (+2)m km and C (+2) e (+2)g eg QUITE (A) hj (B) su (C) tv (D) pr (E) df (D) Q( 1) p(+2) r pr, U ( 1) t(+2) v tv Ex 19., I ( 1) h (+2) j hj, T ( 1) s (+2) u su and E ( 1) d(+2) f df. BLAST (A) i (B) e (C) w (D) p (E) d (D) B(+3) is e, L ( 3) is i,a(+3) is d S( 3) is p and T(+3) is w PAGE # 13

16 BLOOD-RELATIONS Blood Relations Problems on Blood Relations involve analysis of information showing blood relationship among members of a family. In the questions, a chain of relationship is given in the form of information and on the basis of these information relation between any two members of the chain is asked. Students are supposed to be familiar with the knowledge of different relationship in a family. Grandfather's son Grandmother's son Grandfather's only son Grandmother's only son Mother's or Father's mother Mother's or Father's father Grandfather's only daughter-in-law Grandmother's only daughter-in-law Mother's or Father's son Mother's or Father's daughter Mother's or Father's brother Mother's or Father's sister Husband's or wife's sister Husband's or wife's brother Son's wife Daughter's husband Brother's son or Sister's son Brother's daughter or Sister's daughter Uncle or Aunt's son or daughter Sister's husband Brother's wife Father or uncle Father or uncle Father Father Grandmother Grandfather Mother Mother Brother Sister Uncle Aunt Sister-in-law Brother-in-law Daughter-in-law Son-in-law Nephew Niece Cousin Brother-in-law Sister-in-law Remark : A relation on the mother s side is called maternal while that on the father s side is called paternal. Thus, mother s brother is maternal uncle while father s brother is paternal uncle. Note : To build a family tree, certain standard notations are used to indicate a relationship between the members of the family. + stand for male person. stand for female person. +/ male or female person. stand for married couple. +/ B +/ A +/ A +/ B Tree Diagram of Blood Relation Asuming Your Self as Male + A is Son or Daughter of B. B is uncle or aunt of A. PAGE # 14

17 Tree Diagram of Blood Relation Asuming Your Self as Female Direct-Relationship Relation Puzzle Ex 1. Ex 2. In these type of questions, around about description is given in the form of certain small relationship and you required to analyses the whole chain of relations and decipher the direct relationship between the persons concerned. Pointing towards a man in the photograph, Archana said, He is the son of the only son of my grandmother. How is that man related to Archana? (A) Cousin (B) Nephew (C) Brother (D) Son (C) Only son of Archana s grandfather means Archana s father & his son is Archana s brother. Pointing to a photograph, a lady tells Amit, "I am the only daughter of this lady and her son is your material uncle." How is the speaker related to Amit's father? (A) Sister-in-law (B) Wife (C) Either (A) or (B) (D) Neither (A) nor (B) (B) The lady who is talking to Amit is the daughter of the lady in the photograph. The son of that lady who is the brother of the lady who is talking to Amit. The brother of this lady is the maternal uncle of Amit. The lady is the mother of Amit & wife of Amit s Father. Ex 3. In these type of questions, mutual blood relations of more than two persons are mentioned. The candidate is required to analysis the given information, work out a family chart and then answer the given questions. Rohit and Rohan are brothers. Soniya and Sunita are sisters. Rohit s son is Sunita s brother. How is Rohan related to Soniya. (A) Father (B) Brother (C) Grand Father (D) Uncle (D) Rohit s son is Sunita s brother means Rohit is Sunita s father. Rohit and Rohan are brothers. Sunita and Soniya are sisters. So, Rohan is the uncle of Soniya. Directions : (4 to 6) P,Q,R,S,T,U,V & W are the family members. Q is the sister of V and V is the brother of R. P whose s father is W, is husband of T. S is the husband of Q and U is the son of V. P is the father of Q. Ex 4. How U is related with T? (A) Son (C) Grandson (B) Mother (D) Nephew Ex 5. How S is related with R? (A) Son (B) uncle (C) Brother-in-law (D) Brother Ex 6. How W is related with R? (A) Grand father (B) uncle (C) Son (D) Brother PAGE # 15

18 : (4 to 6) + W Father Wife P + Father Husband S + Sister Q Son + + T V U Brother + R Ex 7. A + B means A is the son of B, A B means A is the wife of B. A B means A is the brother of B, A B means A is the mother of B, A = B means A is the sister of B. Which of the following represents P is the maternal-uncle of Q? (A) R P Q (B) P R Q (C) P + R Q (D) P + R Q (B) P R P is brother of R. R Q R is mother of Q. P is maternal uncle of Q. 4. (C) U is son of V and V is son of T. U is grandson of T. 5. (C) S is husband of Q and Q is sister of R. S is brother -in- law of R. 6. (A) R is son of P and P is son of W. W is grandfather of R. Coded Relation In such questions, the relationship are represented by certain specific symbols such as +,,,,, etc. The candidate is then required to analyse some given codes to determine then relationship between a set of persons, or to express a given relationship in the coded form. Ex 8. A B means A is the sister of B, A B means A is the daughter of B, A B means A is the son of B. On the basis of this information you have to tell, how is P related to S in the relationship P Q R S (A) Brother (C) Grandson (D) According to the directions Q Son P + Daughter +/ S Sisters R Daughter (B) Son P is the son of the daughter of S. (D) Daughter s son PAGE # 16

19 SYLLOGISM Syllogism literally means to stay together, giving us the sense of putting two propositions or premises together in order to draw a logical conclusion from them. Venn-Diagram Representation of Four Propositions 1. Universal Affirmative (A) : All S are P. There are two possibilities to represent the relation between S and P given by universal affirmative proposition "All S are P". This can also be understood with the help of set theory. Case I. S = {a, b, c} P = {a, b, c, d} Case II. S = {a, b, c} P = {a, b, c} Case I is represented by Fig. (i) and Case II is represented by Fig. (ii). In both these cases, we see that every element of set S is also the element of set P. Hence, we can definitely say that the above two figures show "All S are P". 2. Universal Negative (E) : No S are P There are three possible representations given by Fig. (iv), Fig. (v) and Fig. (vi) depicting particular affirmative proposition "Some S are P". This can be supported with the help of following sets: Case I. S = {a, b, c, d} P = {c, d, e, f} Case II. S = {a, b} P = {a, b, c, d} Case III. S = {a, b, c, d} P = {a, b} In all the three cases we find that some of the elements of S are also the elements of set P. 4. Particular Negative (O) : Some S are not P There is only one possibility of drawing the relationship between S and P. S = {a, b, c} P = {d, e, f} From the above two sets, it is clear that none of the elements of S is the element of set P. 3. Particular Affirmative (I) : Some S are P The particular negative proposition "Some S are not P" can be represented with the help of three possible figures given in (vii), (viii) and (ix). Venn-diagram representations of the above propositions can be supported by way of following sets: Case I. S = {a, b, c} P = {c, d, e} Case II. S = {a, b, c, d} P = {c, d} Case Ill. S = {a, b, c} P = {d, e, f} In all the three cases, we find that there are some elements in set S which are not elements of set P. Hence all the cases along with the respective figures support the proposition "Some S are not P". PAGE # 17

20 Venn-Diagram Representation Immediate inferences drawn from each type of propositions (A, E, I, O). One of the important points to be noted while drawing inference from Venn-diagrams is that all possibilities of Venndiagrams should be taken in account. Let us now discuss each type of proposition in relation to the pictorial representation. 1. A-Type-All S are P It is clear from the A type of proposition that all S are contained in P. Therefore, circle representing S will be either inside or equal to circle representing P. However, in both the cases, conclusions (Some P are S) and (Some S are P) are true This case can be understood clearly by taking two sets in all possible ways. S P 5 S, P (i) No P is S. (ii) Some S are not P. (iii) Some P are not S. Any other immediate inference drawn from E- type proposition is not valid. 3. -Type -Some S are P This Proposition gives rise to many possible representations of Venn-diagrams and hence most of the inferences drawn therefrom are invalid and doubtful This relationship can be shown by following sets and respective Venndiagrams. (i) S = { 1, 2, 3, 4 } P = { 3, 4, 5, 6 } (i) S = {1,2,3} P = {1,2,3,4,5} (ii) S = {1,2,3} P = {1,2,3} The above cases show the all the possibilities of two sets S and P showing the relationship between each other as represented by the proposition. All S are P. Now in both the cases we see that set {2,3} is the part of set S and also of set P. Hence, it is clear that inference (Some S are P) is true from this relationship. Likewise set { 2,3} is the part of set P and also of set S. Therefore, it is also clear that inference (Some P are S) is true. Inference (Some P are not S) is not valid because it is true not from case (i) but false from case (ii). Inference (All P are S) is not valid because it is true from case (ii) and false from case (i). 2. E- Type-No S is P There is only one possibility of Venn-diagram representation of E-type proposition. The relationship can also be shown by two sets S = {1,2,3} and P = { 4, 5, 6 }. From these two sets, we see that set { 2, 3} is the part of set S but not of set P. It implies that inference (Some S are not P ) is true. Similarly, set {5,6} is the part of set P but not of set S. This means that inference (Some P are not S) is true. Therefore, on the basis of E-type proposition, we can draw following immediate inferences. Set {3,4} is the part of set S as well as set P, hence some S are P (ii) S = { 1, 2, 3, 4} P = { 1, 2} Set {1,2 } is the part of set S as well as set P, hence some S are P. (iii) S = { 1,2} P = { 1,2,3,4} PAGE # 18

21 Set { 1,2} is the part of set S as well as set P, hence some S are P. The above three combinations of sets and respective diagrams show the relationship between S and P as represented by I-type proposition. From all the possible combinations, it is clear that inference (Some P are S ) is true. Inference (Some S are not P} is true from combinations (i) and (ii). But it is not true from combinations (iii). Therefore, inference (Some S are not P) is not a valid inference drawn from the above proposition. 4. O-Type-Some S are not P From this proposition no immediate inference can be drawn. Let us discuss this proposition in the light of Venn-diagram representation. (i) S = {1,2,3,4} P = {3,4,5,6} Set {1,2} is the part of set S but not of set P, hence this shows the relationship represented by the proposition 'Some S are not P'. (ii) S = {1,2,3} P = {4,5,6} Set {1,2,3} is the part of set S but not of set P, hence denotes proposition 'Some S are not P'. On the basis of all possible combinations showing relationship between S and P, no valid inference can be drawn. Inference - Some S are P is true from case (i) and (iii) but not true from case (ii) and hence it is an invalid inference. Inference - Some P are not S is true from case (i) and (ii) but not true from case (iii) and hence it is an invalid inference. Students should note that if an inference is true, it has to comply with or follow all the possible pictorial representation of Venn-diagrams. Directions : (1 to 5) In these types of questions, two statements followed by two conclusions, I and II, are given. You have to take the given two statements to be true even if they seem to be at variance from commonly known facts. Read the conclusions and then decide which of the given conclusions logically follows from the two given statements, disregarding commonly known facts. Mark the answer (A) if only I follows (B) if only II follows (C) if both I and II follow (D) if neither I nor II follows Ex. 1 Statements : Some players are dancers All dancers are tall. Conclusions : I. Some players are tall. II. All players are tall. Sol: (A) P I P II Set {2, 3} is the part of set S but not of set P, hence this shows the relation represented by the proposition 'Some S are not P'. (iii) S = {1,2,3,4,5} P = {4,5} So some players (those who are dancers) are tall. Ex. 2 Statements : Anil is a good sportsman. Sportsmen are healthy. Conclusions: I. All healthy persons are sportsmen. II. Anil is healthy. (B) P I P II So Anil has to be healthy. PAGE # 19

22 Ex.3 Statements: Businessmen married only fair girls. (D) Conclusions: Anisha is fair. I. Anisha was married to a businessman. II. Anisha was not married to a businessman. (D) So grapes may or may not be apples. Ex. 5 Statements : All men are bachelors. Some men are educated. Conclusions: I. Some bachelors are educated. II. Some educated are bachelors. Sol: (C) So Anisha may or may not be married to a businessman. Ex. 4 Statements : All apples are oranges. Some oranges are grapes. Conclusions: I. Some apples are grapes. II. Some grapes are apples. So Here some bachelors are definitely educated. both conclusions follow. PAGE # 20

23 MATHEMATICAL OPERATIONS Mathematical Operations This section deals with questions on simple mathematical operation. There are four fundamental operations, namely : Addition i.e. + ; Subtraction i.e. ; Multiplication i.e., x; and Division i.e.,. There are also statements such as Less than i.e, < ;greater than i.e, > ; equal to i.e, = ; and not equal to i.e, etc. Such operations are represented by symbols different from the usual ones. The candidate has to make a substitution of real signs and solve the equation accordingly. While attempting to solve a mathematical expression, proceed according to the rule BODMAS that is, Brackets, Of, Division, Multiplication, Addition & Subtraction.We can perform addition or subtraction in any order. Ex 1. What is the value of (48 12) =? (A) 10 (B) 0.6 (C) 2 (D) 18 (D) given expression = (48 12) = (Solving Bracket) = (Solving Division) = ( S o l v i n g Multiplication) = 18 (Solving Addition) Ex 2. What is the value of =? (A) 12 (B) 58 (C) 122 (D) (A) Given expression = 7 = 84 7 = Problems - Solving by Substitution Ex 3. In these type of questions, you are provided with substitutes for various mathematical symbols or numerals followed by a questions involving calculations of an expression or choosing the correct / incorrect equations. The candidate is required to put in the real signs or numerals in the given equation and then solve the questions as required. If + means, means, means + and means, then what will be the value of =? (A) 20 (B) 52 (C) 12 (D) None of these (A) Using the correct symbols, = = 20 Ex 4. If L denotes x, M denotes, P denotes + and Q denotes, then 8 P 36 M 6 Q 6 M 2 L 3 =? 13 (A) 6 (C) (B) 6 1 (D) 5 (D) Using the correct symbols, Given expression = = = 5 Ex 5. It being given that : > denotes +, < denotes, + denotes, denotes =, = denotes less than and denotes greater than, find which of the following is a correct statement. (A) > 4 = < 1 (B) 3 > 2 > 4 = < 2 (C) 3 > 2 < < 2 (D) < < 3 (C) Using proper notations, we have : (A) given statement is < or 11 < 2, which is not true. 2 (B) given statement is < or 9 < 4, which is not true. (C) given statement is > or 1 > 0, which is true. (D) given statement is > or 2 5 > 0, which is not true. Interchange of signs and numbers Ex 6. Find out to sign to be interchanged for making the given equation correct = 10 (A) + and (B) + and (C) and (D) + and (B) By making the interchanges given in (A), the equation as = 10 or 109 = 10 which is false By making the interchanges given in (B), the equation as =10 or 10=10 which is true By making the interchanges given in (C), the equation as = 10 or 109 = 10 which is false By making the interchanges given in (D), the equation as = 10 or 89 = 10 which is false PAGE # 21