GENERAL STUDIES PAPER

By X. KIANI

CHAPTER 1 NUMBERS CHAPTER 2 LCM AND HCF CHAPTER 3 PERCENTAGE CHAPTER 4 SIMPLE INTEREST CHAPTER 5 COMPOUND INTEREST CHAPTER 6 DECIMAL FRACTIONS CHAPTER 7 SQUARE ROOTS & CUBE ROOTS CHAPTER 8 PROFIT AND LOSS CHAPTER 9 MATHEMATICAL OPERATIONS: BASIC SIMPLIFICATION CHAPTER 10 AVERAGE CHAPTER 11 SURDS AND INDICES CHAPTER 12 PROBLEMS BASED ON AGE CHAPTER 13 RATIO AND PROPORTION CHAPTER 14 TIME, WORK AND WAGES CHAPTER 15 PIPE AND CISTERNS CHAPTER 16 TIME AND DISTANCE CONTENTS CHAPTER 17 BOATS AND STREAMS CHAPTER 18 AREA CHAPTER 19 VOLUME AND SURFACE AREA CHAPTER 20 LOGARITHMS CHAPTER 21 PROBLEM ON TRAINS CHAPTER 22 CALENDAR CHAPTER 23 CLOCKS CHAPTER 24 STOCKS AND SHARE CHAPTER 25 DISCOUNT CHAPTER 26 GAMES AND RACES CHAPTER 27 PROGRESSION CHAPTER 28 PROBABILITY CHAPTER 29 PERMUTATIONS AND COMBINATIONS CHAPTER 30 ALLIGATION OR MIXUTRE CHAPTER 31 PARTNERSHIP CHAPTER 32 ALGEBRAIC EXPRESSION AND OPERATIONS CHAPTER 33 GEOMETRY WE MAY ALSO WRITE THIS NUMBER AS 2465971385 = 2 × 109 + 4 × 108 + 6 × 107 + 5 × 106 + 9 × + 7 × 104 + 1 × 103 + 3 × 102 + 8 × 101 + 5 × HERE, THE PLACE VALUE OF VARIOUS DIGITS ARE DIGIT PLACE VALUE 5 × 100 = 8 × 101 = 3 × 102 = THUS, WE SEE THAT THE PLACE VALUE OF A DIGIT DEPENDS ON ITS LOCATION IN THE NUMBER FOR INSTANCE, TAKE THE DIGIT 5 IN THE ABOVE EXAMPLE ITS FACE VALUE REMAINS FIVE AT BOTH THE PLACES WHILE ITS PLACE VALUES AT TWO PLACES ARE 5 AND 5,000,000 RESPECTIVELY CLASSIFICATION OF NUMBERS NATURAL NUMBER : THESE ARE ALSO CALLED COUNTING NUMBERS AS THESE ARE THE NUMBERS WE USE FOR COUNTING PURPOSES IT IS REPRESENTED BY N : {1, 2, 3, 4 } WHOLE NUMBER : IT INCLUDES ALL NATURAL NUMBERS PLUS ZERO SO, WE CAN DENOTE IT BY W : {0, 1, 2, 3 } INTEGERS : IT INCLUDES ALL WHOLE NUMBERS ALONG WITH NEGA- TIVE NUMBERS IT IS REPRESENTED BY I : {–5, –4, –3, –2, –1, 0, 1, 2, 3,} THUS, WE SEE THAT WHOLE NUMBERS CONSIST OF POSITIVE INTE- GERS AND ZERO SIMILARLY, NATURAL NUMBERS CONSIST OF POSITIVE IN- TEGERS EVEN NUMBER : A NUMBER WHICH IS COMPLETELY DIVISIBLE BY IS CALLED AN EVEN NUMBER IN OTHER WORDS, SUCH NUMBERS HAVE AS A FACTOR WHEN THEY ARE WRITTEN AS PRODUCT OF DIFFERENT NUM- BERS FOR INSTANCE; 30 = 2 × 3 × 5, 42 = 2 × 3 × OTHER EXAMPLES OF EVEN NUMBERS ARE 2, 4, 6, 8, 10 (A NUM- BER IS SAID TO BE A FACTOR OR SUB-MULTIPLE OF ANOTHER WHEN IT DIVIDES THE OTHER EXACTLY FOR EXAMPLE, 5 AND 3 ARE FACTORS OF 15 ODD NUMBER : THESE NUMBERS ARE NOT COMPLETELY DIVISIBLE BY 2 IN OTHER WORDS, A NUMBER WHICH IS NOT EVEN IS AN ODD NUMBER FOR EXAMPLE, 1, 3, 5, 7, 9 IT MAY BE NOTED THAT ZERO IS AN EXCEPTION TO THIS EVEN-ODD CLASSIFICATION CYCLIC NUMBER : IT IS SUCH AN INTEGER OF N DIGITS WHICH ON BEING MULTIPLIED BY ANY NUMBER FROM 1 THROUGH N, GIVES A PROD- UCT CONTAINING THE SAME N DIGITS AS IN THE ORIGINAL NUMBER AND THESE DIGITS ARE IN THE SAME CYCLIC ORDER IN THE PRODUCT FOR EXAMPLE, 2 × 142857 = 285714; 3 × 142857 = 4 × 142857 = 571428; 5 × 142857 = FRACTIONS : A FRACTION IS A NUMBER WHICH REPRESENTS A RATIO OR DIVISION OF TWO INTEGERS IT IS EXPRESSED IN THE FORM P Q WHERE ‘P’ AND ‘Q’ ARE INTEGERS ‘P’ IS CALLED THE NUMERATOR AND ‘Q’ IS CALLED THE DENOMINATOR NUMBERS ARE A COLLECTION OF CERTAIN SYMBOLS OR FIGURES CALLED DIGITS THE MOST COMMON NUMBER SYSTEM IN USE IS DECIMAL SYSTEM WHICH HAS TEN SYMBOLS, EACH REPRESENTING A DIGIT THESE ARE 0, 1, 2, 3, 4, 5, 6, 7, 8 AND 9 A COMBINATION OF THESE FIGURES REPRESENTING A NUMBER IS CALLED A NUMERAL WE ALSO HAVE BINARY NUMBER SYSTEM IT USES ONLY 0 AND 1 THERE ARE OTHER NUMBER SYSTEMS TOO EVERY DIGIT IN A NUMBER HAS A FACE VALUE AND A PLACE VALUE THE FACE VALUE OF A DIGIT EQUALS THE VALUE OF THE DIGIT ITSELF, IRRESPECTIVE OF ITS PLACE IN THE NUMERAL FOR DETERMINING THE PLACE VALUE OF A DIGIT IN A GIVEN NUMBER WE BEGIN FROM THE EXTREME RIGHT AS UNIT’S PLACE, TEN’S PLACE, HUNDRED’S PLACE, THOUSAND’S PLACE AND SO ON IT HAS BEEN ILLUSTRATED WITH AN EXAMPLE GIVEN BELOW : THE NUMBER 2465971385 MAY BE REPRESENTED AS : BILLION TEN CRORE’S TEN LAC’S LAC’S TEN THOUSAND’S HUNDRED’S TEN’S UNIT’S CRORE’S (MILLION) THOUSAND’S NUMBERS CHAPTER A FRACTION CANNOT HAVE ZERO (Q ? 0) AS ITS DENOMINATOR SINCE DIVISION BY ZERO IS NOT DEFINED ZERO DIVIDED BY ANY INTEGER IS ALWAYS ZERO A FRACTION WITH 1 AS THE DENOMINATOR IS THE SAME AS THE WHOLE NUMBER WHICH IS ITS NUMERATOR FOR EXAMPLE, 1 IS 8; 5 IS 0 EQUIVALENT FRACTIONS : TWO FRACTIONS ARE SAID TO BE EQUIVA- LENT IF THEY REPRESENT THE SAME RATIO OR NUMBER SO, IF WE MULTI- PLY OR DIVIDE THE NUMERATOR AND DENOMINATOR OF A FRACTION BY THE SAME NON-ZERO INTEGER, THE RESULT OBTAINED WILL BE EQUIVA- LENT TO THE ORIGINAL FRACTION FOR EXAMPLE, ? ? IN A MULTIPLE-CHOICE TEST, THE ANSWER TO A PROBLEM THAT ONE OBTAINS NEED NOT NECESSARILY BE THE SAME AS GIVEN IN THE CHOICES, HOWEVER, ONE OF THE CHOICES MAY BE EQUIVALENT IN SUCH CASES ONE NEEDS TO EXPRESS THE ANSWER AS EQUIVALENT FRACTION TO FIND A FRACTION WITH A KNOWN DENOMINATOR EQUAL TO A GIVEN FRACTION : (I) WE DIVIDE THE DESIRED DENOMINATOR BY THE DENOMINATOR OF THE GIVEN FRACTION (II) THEN WE MULTIPLY THE RESULT OF (I) BY THE NUMERATOR OF THE GIVEN FRACTION WHICH GIVES THE NUMERATOR OF THE REQUIRED FRACTION EXAMPLE : FIND THE FRACTION WITH DENOMINATOR 20 WHICH IS EQUAL TO 3 SOL (I) 20 ÷ 5 = (II) 4 × 3 = ANSWER IS 20 TO REDUCE A FRACTION TO ITS LOWEST TERMS, WE CANCEL ALL THE FACTORS COMMON TO THE NUMERATOR AND DENOMINATOR REAL NUMBER : ANY MEASUREMENT CARRIED OUT IN THE PHYSI- CAL WORLD GIVES SOME MEANINGFUL FIGURE OR VALUE OR NUMBER THIS NUMBER IS CALLED REAL NUMBER IT CONSISTS OF TWO GROUPS : (I) RATIONAL NUMBER : A RATIONAL NUMBER CAN ALWAYS BE REP- RESENTED BY A FRACTION OF THE FORM P Q , WHERE P AND Q ARE INTE- GERS AND Q IS NOT EQUAL TO ZERO (Q ? 0) ALL INTEGERS AND FRAC- TIONS ARE RATIONAL NUMBERS FOR EXAMPLE, 5 13 ETC (II) IRRATIONAL NUMBER : AN IRRATIONAL NUMBER CAN’T BE EX- PRESSED IN THE FORM OF P Q , WHERE Q ? 0 FOR EXAMPLE, 3, 2 IT GIVES ONLY AN APPROXIMATE ANSWER IN THE FORM OF A FRACTION OR DECIMAL NUMBER THE DIGITS AFTER THE DECIMAL POINT ARE NON-ENDING THE SAME HOLDS TRUE FOR ? = 3 14 WHICH AGAIN IS IRRATIONAL ALTERNATIVELY, WE CAN SAY THAT AN INFINITE NON-RECURRING DECIMAL NUMBER IS AN IRRATIONAL NUM- BER PRIME NUMBER : A PRIME NUMBER IS A NUMBER WHICH HAS NO FACTOR BESIDES ITSELF AND UNITY, I E; IT IS DIVISIBLE ONLY BY ITSELF AND 1 BUT NOT BY ANY OTHER NUMBER FOR EXAMPLE : 2, 3, 5, 7, 11, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107 NOTE : (I) ‘2’ IS THE ONLY EVEN NUMBER WHICH IS PRIME (II) ALL PRIME NUMBERS OTHER THAN 2 ARE ODD NUMBERS BUT ALL ODD NUMBERS ARE NOT PRIME NUMBERS FOR EXAMPLE, 9 IS AN ODD NUMBER BUT IT IS NOT A PRIME NUMBER AS IT IS DIVISIBLE BY 3 COMPOSITE NUMBER : A COMPOSITE NUMBER IS ONE WHICH HAS OTHER FACTORS BESIDES ITSELF AND UNITY THUS, IT IS A NON-PRIME NUMBER FOR EXAMPLE, 4, 6, 9, 14, 15 ETC NOTE : (I) ‘1’ IS NEITHER PRIME NOR COMPOSITE (II) A COMPOSITE NUMBER MAY BE EVEN OR ODD THE NUMBER OF WAYS IN WHICH A NUMBER N CAN BE EXPRESSED AS PRODUCT OF TWO FACTORS WHICH ARE RELATIVELY PRIME TO EACH OTHER IS 2M–1, WHERE M IS THE NUMBER OF DIFFERENT PRIME FACTORS OF N FOR EXAMPLE, 540 = 22 × 33 × 5 ( = 4 × 27 × 5) ? M = 3(I E , 2, 3, 5) ? NO OF WAYS = 23–1 = I E , 20 × 27, 4 × 135, 108 × 5, 540 × 1 THE LARGEST PRIME NUMBER KNOWN SO FAR IS 22281–1 WHICH IS OF ABOUT 700 DIGITS CONSECUTIVE INTEGERS : THESE ARE SERIES OF NUMBERS DIFFER- ING BY 1 IN ASCENDING OR DESCENDING ORDER FOR EXAMPLE, 12, 13, 14, 15 SIMILARLY, EXAMPLES OF CONSECUTIVE EVEN NUMBERS ARE 4, 6, 8, 10 22, 24, 26, 28 AND SO ON EXAMPLES OF CONSECUTIVE PRIME NUMBERS ARE 7, 11, 13, 17 TEST FOR PRIME NUMBERS FOR NUMBERS LESS THAN 100, IT IS NOT VERY DIFFICULT TO DETER- MINE WHETHER IT IS A PRIME NUMBER OR NOT WE CAN ALSO CHECK IT UP FROM THE LIST GIVEN UNDER PRIME NUMBERS FOR TESTING ANY NUMBER GREATER THAN 100 WHETHER IT IS A PRIME NUMBER OR NOT : I WE TAKE THE NEAREST INTEGER LARGER THAN THE APPROXIMATE SQUARE ROOT OF THAT NUMBER SUPPOSE IT IS X II WE TEST THE DIVISIBILITY OF THE GIVEN NUMBER BY EVERY PRIME NUMBER LESS THAN X III IF THE NUMBER IS NOT DIVISIBLE BY ANY OF THEM, THEN IT IS A PRIME NUMBER, OTHERWISE IT IS A COMPOSITE NUMBER LET US TEST THE THREE NUMBERS (A) 331, (B) 481 AND (C) 881 (A) FOR 331, 18 × 18 = 324, 19 × 19 = 361, ? X = 19 THE PRIME NUMBERS LESS THAN 19 ARE 2, 3, 5, 7, 11, 13 AND 17 NOW TRY TO DIVIDE 331 BY THESE PRIME NUMBERS WE SEE THAT IS NOT COMPLETELY DIVISIBLE BY ANY OF THESE PRIME NUMBERS THEREFORE, 331 IS A PRIME NUMBER (B) FOR 481, X = 22 THE PRIME NUMBERS LESS THAN 22 ARE 2, 3, 5, 7, 11, 13, 17, 19 WE FIND THAT 481 IS DIVISIBLE BY 13 GENERAL STUDIES PAPERII THEREFORE, 481 IS NOT A PRIME NUMBER (C) FOR 881, X = THE PRIME NUMBERS LESS THAN 30 ARE 2, 3, 5, 7, 11, 13, 17, 19, 23 AND 29 WE SEE THAT 881 IS NOT DIVISIBLE BY ANY OF THE ABOVE PRIME NUMBERS THEREFORE, 881 IS A PRIME NUMBER ANY WHOLE NUMBER CAN BE WRITTEN AS A PRODUCT OF FACTORS WHICH ARE PRIME NUMBERS TO WRITE A NUMBER AS A PRODUCT OF PRIME FACTORS : I DIVIDE THE NUMBER BY 2 IF POSSIBLE; CONTINUE TO DIVIDE BY 2 UNTIL THE FACTOR YOU GET IS NOT DIVISIBLE BY 2 II DIVIDE THE RESULT FROM I BY 3 IF POSSIBLE; CONTINUE TO DI- VIDE BY 3 UNTIL THE FACTOR YOU GET IS NOT DIVISIBLE BY 3 III DIVIDE THE RESULT FROM II BY 5 IF POSSIBLE; CONTINUE TO DIVIDE BY 5 UNTIL THE FACTOR YOU GET IS NOT DIVISIBLE BY 5 IV CONTINUE THE PROCEDURE WITH 7, 11 AND SO ON, UNTIL ALL THE FACTORS ARE PRIMES FOR EXAMPLE, 504 = 2 × 2 × 2 × 3 × 3 × PERFECT NUMBER : IF THE SUM OF THE DIVISORS OF A NUMBER N EX- CLUDING N ITSELF IS EQUAL TO N, THEN N IS CALLED A PERFECT NUMBER FOR EXAMPLE, 6, 28, 496, FOR 6, DIVISORS ARE 1, 2, AND 6 : 1 + 2 + 3 = 1 + 2 + 4 + 7 + 14 = 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 = 8128 : 1 + 2 + 4 + 8 + 16 + 32 + 64 + 127 + 254 + 508 + 1016 + 2032 + 4064 = NOTE : THE SUM OF THE RECIPROCALS OF THE DIVISORS OF A PERFECT NUMBER INCLUDING THAT OF ITS OWN IS ALWAYS EQUALS 2 FOR EXAMPLE, FOR 28 THE FACTORS ARE 1, 2, 4, 7 AND 14 ? ? ? ? ? 28 ? ? ? ? A NUMBER CONTAINING A DECIMAL POINT IS CALLED A DECIMAL NUMBER 35 467 = (3 10 ) (5 10 ) (3 10) (5 1) ? ? ? ? =35 + MIXED NUMBER : A MIXED NUMBER CONSISTS OF A WHOLE NUM- BER AND A FRACTION FOR EXAMPLE, 3 5 IS A MIXED NUMBER THIS IS EQUIVALENT TO AND HENCE CAN BE WRITTEN AS ? ? HERE 3 IS THE WHOLE NUMBER AND 5 IS THE FRACTION CHANGING A MIXED NUMBER INTO AN EQUIVALENT FRACTION : A MIXED NUMBER CAN BE CHANGED INTO A FRACTION AS FOLLOWS : I MULTIPLY THE WHOLE NUMBER BY THE DENOMINATOR OF THE FRAC- TION II ADD THE NUMERATOR OF THE FRACTION TO THE RESULTANT PROD- UCT III THE EQUIVALENT FRACTION HAS NUMERATOR GIVEN BY II ABOVE AND DENOMINATOR SAME AS THAT OF THE FRACTIONAL PART OF THE GIVEN MIXED NUMBER IN THE ABOVE EXAMPLE : ? ? 3 ? ? ? ? WE CAN ALSO DO THE REVERSE OF IT, I E, WHEN THE NUMERATOR OF A GIVEN FRACTION IS GREATER THAN THE DENOMINATOR, WE CAN CHANGE IT INTO A MIXED NUMBER USING THE FOLLOWING PROCEDURE : I DIVIDE THE NUMERATOR BY THE DENOMINATOR II THE QUOTIENT, GIVES THE WHOLE NUMBER AND THE REMAINDER GIVES THE NUMERATOR OF THE FRACTIONAL PART THE DENOMINATOR OF THE FRACTIONAL PART WILL BE THE SAME AS THE DENOMINATOR OF THE GIVEN FRACTION FOR EXAMPLE, = SIGNED NUMBER : A NUMBER WHICH HAS EITHER A PLUS OR A MINUS SIGN ATTACHED AS PREFIX IS CALLED A SIGNED NUMBER SUCH NUMBERS ARE ALSO CALLED DIRECTED NUMBERS FOR EXAMPLE, + 3, – 6, –7, 5 , – 4 3 ETC IF NO SIGN IS GIVEN WITH A NUMBER, A PLUS SIGN IS ASSUMED AND VICE VERSA SINGED NUMBERS ARE OFTEN USED TO DISTINGUISH DIFFERENT CON- CEPTS FOR EXAMPLE, A PROFIT OF RS 25 MAY BE INDICATED BY + RS GENERAL STUDIES PAPERII WE NOTICE THAT THE OPERATION OF PLUS DOES NOT CHANGE THE SIGN OF THE NUMBER WHEREAS THE OPERATION OF MINUS CHANGES THE SIGN OF THE NUMBER IN MULTIPLICATION AND DIVISION, WHEN BOTH THE NUMBERS CARRY SIMILAR SIGN, WE GET POSITIVE SIGN IN THE RESULT, OTHERWISE, WE GET NEGATIVE SIGN IN THE RESULT 25 AND A LOSS OF RS 25 BY – RS 25 THUS, PLUS AND MINUS SIGNS HAVE OPPOSITE MEANINGS A HEIGHT OF 2500 METRES ABOVE THE SEA LEVEL MAY BE DENOTED BY +2500 METRES WHILE A DEPTH OF METRES BELOW THE SEA LEVEL IS DENOTED BY – 2500 METRES SIMI- LARLY, WE CAN WRITE FOR TEMPERATURE ABOVE AND BELOW 0°C SO, WE SEE THAT IN ALL CASES WE MUST HAVE A DATUM OR REFERENCE WITH RESPECT TO WHICH THE GIVEN QUANTITY IS MEASURED AND EXPRESSED IN THE ABOVE EXAMPLES, ZERO, COST PRICE, SEA LEVEL, 0°C ARE THE REFERENCE POINTS THE QUANTITY ON THE POSITIVE SIDE OF THE DA- TUM IS GIVEN PLUS SIGN AND ITS ABSOLUTE VALUE IS OBTAINED BY ADDING THE QUANTITY TO THE DATUM/REFERENCE VALUE ON THE OTHER HAND, THE OTHER (NEGATIVE) SIDE OF THE DATUM IS GIVEN MINUS SIGN AND ITS ABSOLUTE VALUE IS GIVEN BY SUBTRACTING THE GIVEN QUAN- TITY FROM THE DATUM WHICH, IN OTHER WORDS, IS EQUIVALENT TO ADD- ING THE QUANTITY WITH MINUS SIGN TO THE DATUM FOR EXAMPLE, 0 + 25 = + 25 = THUS, WE SEE THAT ADDING A NEGATIVE QUANTITY TO ANY NUM- BER IS EQUIVALENT TO SUBTRACTING ITS ABSOLUTE (POSITIVE) VALUE FROM THE GIVEN NUMBER YOU CAN IMAGINE NUMBERS TO BE ARRANGED ON A LINE CALLED A NUMBER LINE DRAW A LINE WHICH EXTENDS INDEFINITELY (I E , UPTO INFINITY) IN BOTH DIRECTIONS MARK A REFERENCE POINT ON THIS LINE AND CALL IT ZERO THE PORTION ON THE RIGHT OF ZERO IS CALLED POSITIVE SIDE AND THAT ON THE LEFT OF ZERO AS NEGATIVE SIDE (INSTEAD OF A HORI- ZONTAL LINE WE CAN ALTERNATIVELY DRAW A VERTICAL LINE IN WHICH CASE THE PORTION ABOVE ZERO IS POSITIVE AND THAT BELOW ZERO IS NEGA- TIVE) MARK ANOTHER POINT ON THE LINE TO THE RIGHT OF ZERO AND CALL IT ‘1’ THE POINT TO THE RIGHT OF 1 WHICH IS EXACTLY AS FAR AWAY FROM 1 AS 1 IS FROM 0 IS MARKED 2 THE POINT TO THE RIGHT OF 2 JUST AS FAR FROM 2 AS 1 IS FROM 0 (OR 2 IS FROM 1) IS CALLED 3 SIMILARLY, WE HAVE 4, 5, 6 AND SO ON THE POINT MID-WAY OR HALF-WAY BETWEEN 0 AND 1 IS CALLED 2 , THE POINT MIDWAY BETWEEN 1 AND 2 IS CALLED 12 OR 2 AND SO ON THE POINT HALFWAY BETWEEN 0 AND 2 IS CALLED 4 AND THE POINT MIDWAY BETWEEN 2 AND 1 IS CALLED 4 THUS ANY WHOLE NUMBER OR FRACTION CAN BE IDENTIFIED FOR EXAMPLE, 2 4 IT LIES BETWEEN 2 AND 3 DIVIDE THE DISTANCE BETWEEN 2 AND 3 INTO 4 EQUAL PARTS THE POINT AT THE THIRD MARK BETWEEN 2 AND 3 REPRESENTS 2 4 IF WE GO TO THE LEFT OF ZERO, THE SAME DISTANCES AS WE DID FROM 0 TO 1, THE POINT IS CALLED –1 IN A SIMILAR MANNER WE CAN REPRESENT – 2, –3, – 4, – 4 , – 2 , –1 , –2 4 AND SO ON THUS, ZERO IS NEITHER POSITIVE NOR NEGATIVE AND THEREFORE WE CAN WRITE, 0 = + 0 = – ANY NON-ZERO NUMBER IS EITHER POSITIVE OR NEGATIVE BUT CAN’T BE BOTH IT IS NOT ALWAYS NECESSARY TO TAKE ZERO AS DATUM/REFERENCE POINT DATUM OR REFERENCE POINT IS CHOSEN AS PER CONVENIENCE AND DEPENDS ON THE PROBLEM TO BE SOLVED THE SAME HOLDS TRUE FOR POSITIVE AND NEGATIVE TOO BUT ONCE CHOSEN, THEIR UNIFOR- MITY MUST BE MAINTAINED THROUGHOUT THE SOLUTION IT IS USUAL PRACTICE TO ASSIGN ZERO TO THE REFERENCE POINT/DATUM, WHATEVER ITS VALUE MIGHT BE FOR INSTANCE, WE TAKE GROUND LEVEL OF A PARTICULAR PLACE IN- STEAD OF SEA LEVEL AS DATUM FOR EXPRESSING THE HEIGHT OF A BUILD- ING SIMILARLY, THE REFERENCE FOR LOCAL TIME IS DIFFERENT IN DIFFERENT COUNTRIES MODULUS OR ABSOLUTE VALUE : THE MODULUS OR ABSOLUTE VALUE OF ANY SIGNED NUMBER (BE IT POSITIVE OR NEGATIVE) IS ALWAYS TAKEN AS POSITIVE AND IS EQUAL TO THE DISTANCE OF THE NUMBER FROM THE ZERO DATUM IT IS DENOTED BY WRITING THE NUMBER BETWEEN TWO VERTICAL LINES IN OTHER WORDS, IT DENOTES ONLY THE MAGNITUDE OF THE NUMBER IF A IS ANY SIGNED NUMBER, ITS MODULUS OR ABSOLUTE VALUE IS WRITTEN AS |A| IF A IS POSITIVE, |A| = A AND IF A IS NEGATIVE, |A| = –A SO, WHAT WE DO EFFECTIVELY IS SIMPLY DROP THE SIGN ATTACHED TO THE NUMBER, WHEN WE ARE TAKING ITS MODULUS OR ABSOLUTE VALUE FOR EXAMPLE, |+3| = + 3 = |–8| = – (–8) = + 8 = |+8| = + 8 = |–12| = –(–12) = + 12 = ABSOLUTE VALUE OF ZERO IS ALWAYS ZERO MATHEMATICAL OPERATIONS ON SIGNS : ( 20) (– 20) – – ( 20) – – (– 20) ?? ?? ? ? ? ? ?? ? ? ? ? ?? ? ( ) ( ) ( ) (–) – (–) ( ) – (–) (–) ( ) ( ) ( ) (–) – (–) ( ) – (–) (–) ?????? ? ?? ? ? ? ??? ? ? ???? ?????? ? ?? ? ? ??? ? ? ? ??? GENERAL STUDIES PAPERII SOLVED EXAMPLES EX 1 WHAT WILL BE THE DIGIT IN UNIT'S PLACE IN THE PRODUCT OF ? ? ? SOL 7 × 7 = 49, 7 × 7 × 7 = 343 AND 7 × 7 × 7 × 7 = ? DIGIT IN UNIT'S PLACE IN 7 IS 1 NOW, 735 = (74)8 × THEREFORE, DIGIT IN UNIT'S PLACE IN ? ? SIMILARLY ? ? ? ? AND 34 = 3 × 3 × 3 × 3 = ? ?? ? AND IN THE PRODUCT OF 11 × 11 × 11 55 TIMES, THE DIGIT IN UNIT'S PLACE IS 1 ? DIGIT IN UNIT'S PLACE IN THE PRODUCT OF ? ? I E 3 × 7 × 1 = 21 IS 1 EX 2 FIND THE SUM OF EVEN NUMBERS BETWEEN 1 AND 31 IS SOL REQUIRED NUMBER ARE 2, 4, 6 , 30 THIS IS AN A P CONTAINING 15 TERMS ? REQUIRED SUM = 30) ? ? EX 3 WHAT LEAST VALUE MUST BE ASSIGNED TO * SO THAT THE NUMBER 197*5462 IS DIVISIBLE BY 9 ? SOL LET THE MISSING DIGIT BE X ? SUM OF DIGITS = (1 + 9 + 7 + X + 5 + 4 + 6 + 2) = (34 + X) FOR (34 + X) TO BE DIVISIBLE BY 9, X MUST BE REPLACED BY HENCE THE DIGIT IN PLACE OF * MUST BE 2 EX 4 HOW MANY TERMS ARE THERE IN 2, 4, 8, 16 1024 ? SOL SINCE, 2, 4, 8, 16 1024 FORM A G P WITH A = 2 AND R = 2 = LET THE NUMBER OF TERMS BE N THEN N? ? ? ? N?? ? ? N – 1 = 9 OR N = EX 5 THE FIRST ODD NUMBER IS 1, THE SECOND ODD NUMBER IS 3, THE THIRD ODD NUMBER IS 5 AND SO ON THE 200TH ODD NUMBER IS SOL FIRST ODD NUMBER = SECOND ODD NUMBER = THIRD ODD NUMBER = ? N TH ODD NUMBER = 1 + (N – 1) 2 = 2N – ? 200TH ODD NUMBER = 2 × 200 – 1= 400 – 1 = EX 6 THE NUMBER OF DIGITS IN THE SQUARE ROOT OF 625686734489 IS THE NUMBER OF DIGITS IN 625686734489 IS 12 ? NUMBER OF DIGITS IN ITS SQUARE ROOT = PRACTICE EXERCISE 1 IN A THREE DIGIT NUMBER THE DIGIT IN THE UNIT’S PLACE IS TWICE THE DIGIT IN THE TEN’S PLACE AND 1 5 TIMES THE DIGIT IN THE HUNDRED’S PLACE IF THE SUM OF ALL THE THREE DIGITS OF THE NUMBER IS 13, WHAT IS THE NUMBER ? (A) (B) (C) (D) 2 TWO NUMBERS ARE LESS THAN THE THIRD NUMBER BY 50% AND 54% RESPECTIVELY BY HOW MUCH PER CENT IS THE SECOND NUMBER LESS THAN THE FIRST NUMBER ? (A) 3 IF (28)3 IS SUBTRACTED FROM THE SQUARE OF A NUMBER, THE ANSWER SO OBTAINED IS 1457 WHAT IS THE NUMBER ? (A) 4 THE DIFFERENCE BETWEEN A NUMBER AND 5 TH OF THE NUM- BER IS 30 THE NUMBER IS (A) (D) NONE OF THESE 5 WHAT IS THE GREATER OF TWO NUMBERS WHOSE PRODUCT IS 1092 AND THE SUM OF THE TWO NUMBERS EXCEEDS THEIR DIF- FERENCE BY 42? (A) 7 IF (12)3 IS SUBTRACTED FROM THE SQUARE OF A NUMBER, THE ANSWER SO OBTAINED IS 976 WHAT IS THE NUMBER ? (A) 8 IF THE SUM OF FOUR CONSECUTIVE EVEN NUMBERS IS 228, WHICH IS THE SMALLEST OF THE NUMBERS ? (A) (D) 4 9 THE PRODUCT OF TWO SUCCESSIVE POSITIVE INTEGERS IS 462 WHICH IS THE SMALLER INTEGER ? (A) 10 THE TOTAL NUMBER OF INTEGERS BETWEEN 200 AND 400, EACH OF WHICH EITHER BEGINS WITH 3 OR ENDS WITH 3 OR BOTH, IS (A) GENERAL STUDIES PAPERII 12 THE NUMBER OBTAINED BY INTERCHANGING THE DIGITS OF A TWO DIGIT NUMBER IS LESS THAN THE ORIGINAL NUMBER BY 18 IF SUM OF THE DIGITS IS 6, WHAT IS THE ORIGINAL TWO DIGIT NUM- BER? (A) 13 HOW MANY NUMBERS LESS THAN 1000 ARE MULTIPLES OF BOTH 10 AND 13 ? (A) 14 THE DIFFERENCE BETWEEN A TWO DIGIT NUMBER AND THE NUMBER OBTAINED BY INTERCHANGING THE POSITIONS OF ITS DIGITS IS 36 WHAT IS THE DIFFERENCE BETWEEN THE TWO DIGITS OF THAT NUMBER ? (A) (D) CANNOT BE DETERMINED 15 BY HOW MUCH IS TWO-FIFTH OF 200 GREATER THAN THREE-FIFTH OF 125? (A) 16 THE AVERAGE OF FOUR CONSECUTIVE EVEN NUMBERS IS ONE- FOURTH OF THE SUM OF THESE NUMBERS WHAT IS THE DIFFER- ENCE BETWEEN THE FIRST AND THE LAST NUMBER? (A) (D) CANNOT BE DETERMINED 17 IN A PARTY THERE ARE 120 MEMBERS TWO-THIRD OF THEM ARE MEN AND REST ARE WOMEN ALL THE MEMBERS ARE MARRIED EX- CEPT 12 WOMEN MEMBERS HOW MANY MARRIED WOMEN ARE THERE IN THE CLUB ? (A) 18 IF THE DIGITS OF A TWO DIGIT NUMBER ARE INTERCHANGED, THE NEWLY FORMED NUMBER IS MORE THAN THE ORIGINAL NUMBER BY 18, AND SUM OF THE DIGITS IS 8; THEN WHAT IS THE ORIGI- NAL NUMBER ? (A) (D) CANNOT BE DETERMINED 19 DIFFERENCE BETWEEN A TWO DIGIT NUMBER AND THE NUMBER OBTAINED BY INTERCHANGING THE DIGIT IS 27 IF THE SUM OF THE DIGIT IS 7, WHAT IS THE ORIGINAL NUMBER ? (A) (D) CANNOT BE DETERMINED 20 2 OF 4 OF A NUMBER IS 2 2 OF 10 WHAT IS THE NUMBER ? (A) 21 THE DIGIT IN THE UNIT’S PLACE OF A NUMBER IS EQUAL TO THE DIGIT IN THE TEN’S PLACE OF HALF THAT NUMBER AND THE DIGIT IN THE TEN’S PLACE OF THAT NUMBER IS LESS THAN THE DIGIT IN THE UNIT’S PLACE OF HALF OF THE NUMBER BY 1 IF THE SUM OF THE DIGITS OF THE NUMBER IS SEVEN, THEN WHAT IS THE NUM- BER ? (A) (D) DATA INADEQUATE 22 A CLASSROOM HAS EQUAL NUMBER OF BOYS AND GIRLS EIGHT GIRLS LEFT TO PLAY KHO-KHO, LEAVING TWICE AS MANY BOYS AS GIRLS IN THE CLASSROOM WHAT IS THE TOTAL NUMBER OF GIRLS AND BOYS PRESENT INITIALLY ? (A) CANNOT BE DETERMINED (B) 23 THE DIGITS OF A TWO-DIGIT NUMBER ARE IN THE RATIO 2 : 3 AND THE NUMBER OBTAINED BY INTERCHANGING THE DIGITS IS BIG- GER THAN THE ORIGINAL NUMBER BY 27 WHAT WAS THE ORIGI- NAL NUMBER ? (A) 24 THE NUMBER 0 121212 IN THE FORM IS EQUAL TO (A) (B) (C) (D) 25 TWO NATURAL NUMBERS ARE IN THE RATIO 3 : 5 AND THEIR PRODUCT IS 2160 THE SMALLER OF THE NUMBERS IS (A) ANSWER KEY 1 (B) 2 (D) 3 (D) 4 (A) 5 (C) 6 (B) 7 (D) 8 (B) 9 (C) 10 (C) 11 (C) 12 (C) 13 (D) 14 (A) 15 (C) 16 (B) 17 (A) 18 (C) 19 (A) 20 (C) 21 (A) 22 (D) 23 (D) 24 (C) 25 (A) EXPLANATORY NOTES 1 (B) LET THE TEN’S DIGIT BE X ? UNIT’S DIGIT = 2X AND HUNDRED’S DIGITS = X ACCORDING TO THE QUESTION, ? ? ? UNIT’S DIGIT = 6 AND HUNDRED’S DIGIT = 1 5 ? ? NUMBER = 2 (D) LET THE THIRD NUMBER BE 100 ? FIRST NUMBER = SECOND NUMBER = 100 – 54 = DECREASE = 50 – 46 = ? REQUIRED PERCENTAGE = 50 ? ? 3 (D) LET THE NUMBER BE X ACCORDING TO THE QUESTION, X2 – (28)3 = ? X2 – 21952 = ? X2 = 21952 + 1457 = ? X = ? 4 (A) LET THE NUMBER BE X ACCORDING TO THE QUESTION, X ? ???? 30 X ? ? ? 5 (C) LET THE NUMBERS BE X AND Y RESPECTIVELY ACCORDING TO THE QUESTION, X + Y – (X – Y) = ? 2Y = ? Y = 2 ? ? X = 21 ? 6 (B) LET THE NUMBER BE X ACCORDING TO THE QUESTION, (58 – 39)% OF X = ? X × 100 = ???X = ´ = ? 62% OF 1300 = 1300 × 7 (D) LET THE REQUIRED NUMBER BE X ACCORDING TO THE QUESTION, X2 – (12)3 = ? X2 – 1728 = ? X2 = 1728 + 976 = ? X = 2704 = 8 (B) ACCORDING TO THE QUESTION, X + X + 2 + X + 4 + X + 6 = ? 4X + 12 = ? 4X = 228 – 12 = ? X = 4 = ? THE LEAST EVEN NUMBER = 9 (C) SUPPOSE THE TWO CONSECUTIVE INTEGERS BE X AND X + 1 RESPECTIVELY ACCORDING TO QUESTION (X) × (X + 1) = ? X2 + X – 462 = 0 ? X2 + 22X – 21X – 462 = ? X (X + 22) – 21 (X + 22) = ? (X + 22) (X – 21) ? X = 10 (C) WHEN 3 LIES AT HUNDREDS PLACE ? TOTAL INTEGERS = 10 × 9 = WHEN 3 LIES AT UNIT’S PLACE TOTAL INTEGERS = WHEN 3 LIES AT UNIT’S AND HUNDRED’S PLACE TOTAL INTEGERS = ? TOTAL INTEGERS = 90 + 10 +10 = 11 (C) ? ? A B ? = A + B ? ? A B ? = A + B + AB HERE, A + B < A + B + 2 AB CLEARLY, A B ? < A B ? 12 (C) LET THE ORIGINAL NUMBER BE 10X + Y NUMBER OBTAINED BY INTERCHANGING THE DIGITS = 10Y + X ? 10X + Y – 10Y – X = ? 9X – 9Y = 18 ? X – Y = 2 (I) AGAIN, X + Y = (II) FROM EQUATIONS (I) AND (II), X = 4 AND Y = ? ORIGINAL NUMBER = 10 × 4 + 2 = 13 (D) THE NUMBER OF MULTIPLES OF 130 ARE OBTAINED BY DIVID- ING 1000 BY 130 THE QUOTIENT, I E 7 GIVES THE RESULT 14 (A) LET THE UNIT’S DIGIT BE X AND TEN’S DIGIT BE Y ALSO LET Y > X ? NUMBER = 10Y + X AND NUMBER OBTAINED BY INTERCHANGING THE DIGITS = 10X + Y ? 10Y + X – 10X – Y = ? 9Y – 9X = ? 9 (Y – X) = ? Y – X = 15 (C) REQUIRED DIFFERENCE = ? ? ? ? ? ? ? ? ? ? ? ? ? GENERAL STUDIES PAPERII ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = 80 – 75 = 16 (B) FOUR CONSECUTIVE EVEN NUMBERS ARE 2X, 2X + 2, 2X + AND 2X + 6 RESPECTIVELY ??REQUIRED DIFFERENCE = 2X + 6 – 2X = 17 (A) TOTAL MEMBERS IN THE PARTY = TWO-THIRDS OF THEM ARE MEN HENCE, ONE-THIRD OF THEM ARE WOMEN, I E , ? NUMBER OF MARRIED WOMEN = 40 – 12 = 18 (C) LET THE UNIT’S DIGIT BE Y AND TEN’S DIGIT BE X ? NUMBER = 10X + Y ? NEW NUMBER AFTER INTERCHANGE = 10Y + X ACCORDING TO THE QUESTION 10Y + X – 10X – Y = OR, 9 (Y – X) = OR, Y – X = (I) AGAIN, X + Y = (II) FROM EQUATIONS (I) AND (II) 2Y = OR, Y = [FROM EQUATION (I)] NUMBER = 10X + Y = 10 × 3 + 5 = 19 (A) LET THE NUMBER BE XY X + Y = 7 AND (10X + Y) – (10Y + X) = OR, 9X – 9Y = OR, X – Y = AND,X + Y = 2X = Y = ? THE REQUIRED NUMBER IS 52 20 (C) LET THE NUMBER IS X ACCORDING TO THE QUESTION OF OF X = 2 2 OF OR, X = 5 2 ? OR, X 5 10 ? 21 (A) 5 + 2 = 5 = 6 – THE NUMBER 52 SATISFIES ALL THE CONDITIONS GIVEN IN THE QUESTION 22 (D) LET THE NUMBER OF BOYS AND GIRLS INITIALLY EACH BE EQUAL TO X THEN 2 (X – 8) = X OR, 2X – 16 = X OR, X = ? TOTAL NUMBER OF BOYS AND GIRLS INITIALLY PRESENT = 2X = 2 × 16 = ALSO, (10Y + X) – (10X + Y) = OR, 9Y – 9X = OR, Y – 3 Y = 3 ? ? ? THE ORIGINAL NUMBER IS 69 24 (C) 0 121212 = 0 12 = ? 25 (A) LET THE NUMBERS BE 3X AND 5X RESPECTIVELY ? 3X × 5X = ? X2 = 3 5 ? ? = 12 × 12 ? X = ? SMALLER NUMBER = 3X = 3 × 12 = LCM AND HCF A NUMBER IS SAID TO BE MULTIPLE OF ANOTHER WHEN IT IS EX- ACTLY DIVISIBLE BY THE OTHER SIMILARLY, WE CAN DEFINE COMMON MULTIPLE AS THE NUMBER WHICH IS EXACTLY DIVISIBLE BY NUMBERS FOR WHOM IT IS COMMON FOR EXAMPLE, 10 IS A COMMON MULTIPLE FOR 2 AS WELL AS 5, 20, 30, 40 ETC ARE OTHER COMMON MULTIPLES OF 2 AND 5 FOR 2 AND 3, COMMON MULTIPLES ARE 6, 12, 18 AND SO ON FOR 4 AND 6, COMMON MULTIPLES ARE 12, 24, 36 AND SO ON LEAST COMMON MULTIPLE (LCM) LCM OF TWO OR MORE NUMBERS IS THE SMALLEST NUMBER WHICH IS COMMON MULTIPLE OF THE GIVEN NUMBERS IN OTHER WORDS, LCM OF GIVEN NUMBERS IS THE SMALLEST NUMBER WHICH IS EXACTLY DIVIS- IBLE BY EACH OF THEM IN THE ABOVE EXAMPLES, THE LCM FOR 2 AND 5 IS 2 AND 3 IS 4 AND 6 IS WE CAN FIND LCM BY TWO METHODS METHOD A : I WRITE EACH OF THE GIVEN NUMBERS AS PRODUCT OF PRIME FAC- TORS II FIND THE PRODUCT OF THE HIGHEST POWERS OF THE PRIME FAC- TORS WHICH WILL BE THE LCM NOTE : DON’T REPEAT ANY FACTOR WHILE WRITING THE PRODUCT IN STEP II EX 1 FIND THE LCM OF 27 AND 63 SOL 27 = 3 × 3 × 3 = 63 = 3 × 3 × 7 = 32 × (33 INCLUDES 32 AND THEREFORE, WE WILL NOT WRITE 32 WHILE FIND- ING THE PRODUCT FOR LCM) LCM = 33 × 7 = EX 2 FIND THE LCM OF 54 AND 21 SOL 54 = 2 × 21 = 3 × LCM = 2 × 33 × 7 = EX 3 FIND THE LCM OF 36, 56, 105 AND 108 SOL 36 = 22 × 56 = 23 × 105 = 3 × 5 × 108 = 22 × THE LCM MUST CONTAIN EVERY PRIME FACTORS OF EACH OF THE NUMBERS ALSO, IT MUST INCLUDE THE HIGHEST POWER OF EACH PRIME FACTOR WHICH APPEARS IN ANY OF THEM SO, IT MUST CONTAIN 23 OR IT WOULD NOT BE A MULTIPLE OF 56, IT MUST CONTAIN 33 OR IT WOULD NOT BE A MULTIPLE OF 108, IT MUST CONTAIN 5 OR IT WOULD NOT BE A MULTIPLE OF 105, AND IT MUST CON- TAIN 7 OR IT WOULD NOT BE A MULTIPLE OF 56 OR OF 105 THEREFORE, THE LCM = 23 × 33 × 5 × 7 = NOTE : 36 IS ITSELF A FACTOR OF 108 AND HENCE ANY MULTIPLE OF 108 MUST BE A MULTIPLE OF 36 TOO THEREFORE, WE COULD HAVE STRUCK OUT 36 METHOD B : THIS IS A QUICKER METHOD TO FIND THE PRIME FACTORS AND HENCE LCM IN THIS METHOD THERE CAN BE MORE THAN ONE ARRANGEMENTS FOR THE SAME NUMBERS I WRITE THE GIVEN NUMBERS IN A ROW AND STRIKE OUT THOSE NUMBERS WHICH ARE FACTORS OF ANY OTHER NUMBER IN THE SET II WRITE THE FACTOR ON THE LEFT HAND SIDE WHICH CAN DIVIDE MAXIMUM OF THE NUMBERS III WRITE IN THE NEXT ROW THE QUOTIENTS OBTAINED AND ALSO THOSE NUMBERS (AS THEY ARE) WHICH ARE NOT DIVISIBLE BY THAT FACTOR YOU CAN STRIKE OUT FROM ANY ROW 1, IF IT APPEARS IV REPEAT STEPS II AND III UNTIL WE GET A ROW WHERE NO TWO NUMBERS HAVE A COMMON FACTOR OR DIVISOR, I E , ALL THE NUMBERS IN THE ROW ARE PRIME TO EACH OTHER, THOUGH INDIVIDUALLY THEY MAY NOT BE PRIME NUMBERS V MULTIPLY ALL THE FACTORS/DIVISORS AND THE NUMBERS LEFT IN THE LAST ROW THE PRODUCT GIVES THE LCM OF THE GIVEN NUMBERS LET US NOW SEE HOW IT WORKS AND HOW SIMPLE IT IS EX 4 FIND THE LCM OF 36, 56, 105 AND 108 SOL 56, 105, 28, 105, 14, 105, 14, 35, 2, 5, 36 IS A FACTOR/SUBMULTIPLE OF 108 (36 × 3 = 108) AND HENCE WE STRIKE IT OFF 56 AND 108 ARE DIVISIBLE BY 2 SO WE WRITE 2 ON THE LEFT SIDE AND PERFORM STEP III NEXT FACTORS ARE 2, 3, AND 7 THEREAFTER, WE FIND THAT 2, 5 AND 9 LEFT IN THE LAST ROW HAVE NO COMMON DIVISOR I E , 2, 5, 9 ARE PRIME TO EACH OTHER, THOUGH ITSELF IS NOT A PRIME NUMBER SO, WE FIND THE PRODUCT OF 2, 2, 3, 7, 2, 5 AND 9 TO GET THE REQUIRED LCM LCM = 2 × 2 × 3 × 7 × 2 × 5 × 9 = WE MAY HAVE ANOTHER ARRANGEMENT AS GIVEN BELOW WHICH WILL SAVE MORE TIME ALSO, IN ORDER TO SAVE TIME WE CAN FIND THE PRODUCT FROM THE TABLE ITSELF INSTEAD OF WRITING MULTIPLICATION SEPARATELY AS ALL THE NUMBERS TO BE MULTIPLIED LIE ON THE PERIPH- ERY OF THE TABLE AS INDICATED BY ARROW BELOW THE BASIC LOGIC REMAINS THE SAME AS EXPLAINED EARLIER CHAPTER , , , , , , , , , LCM = 4 × 3 × 7 × 2 × 5 × 9 = 7560 THE BASIC PRINCIPLE FOLLOWED IN BOTH THE METHODS OF FINDING LCM IS TO ENSURE THAT THE FACTORS WHICH ARE COMMON TO TWO OR MORE NUMBERS DO NOT GET REPEATED IN THE FINAL PRODUCT NOTE : THE LCM OF A SET OF NUMBERS WHICH ARE PRIME RELATIVE TO EACH OTHER IS EQUAL TO THEIR PRODUCT THIS IS SO BECAUSE SUCH A SET OF NUMBERS DO NOT HAVE ANY COMMON FACTOR THE NUMBERS INDIVIDUALLY MAY NOT BE PRIME NUMBERS EX 5 FIND THE LCM OF 3, 4, AND 7 SOL HERE, 4 ITSELF IS NOT A PRIME NUMBER BUT THE THREE GIVEN NUMBERS 3, 4 AND 7 ARE PRIME TO EACH OTHER AS NONE OF THEM HAVE ANY FACTOR COMMON TO THE OTHER SO, LCM = 3 × 4 × 7 = EX 6 FIND THE LCM OF 4, 6 AND 7 SOL HERE, 4 AND 6 ARE NOT PRIME TO EACH OTHER AS THEY HAVE 2 AS COMMON FACTOR AND THIS SHOULD NOT GET REPEATED IN THE PRODUCT FOR FINDING LCM SO, WE CAN USE EITHER OF THE TWO METH- ODS TO FIND LCM WHICH ENSURES NON-REPETITION OF SUCH COM- MON FACTORS IN THE PRODUCT METHOD A : 4 = 6 = 2 × 7 = ? LCM = 22 × 3 × 7 = METHOD B , , , , ? LCM = 2 × 2 × 3 × 7 = LCM OF MORE THAN TWO NUMBERS LCM OF THREE NUMBERS IS EQUAL TO THE LCM OF (LCM OF ANY TWO NUMBERS THE REMAINING THIRD NUMBER) FOR FOUR NUMBERS THIS CAN BE EXTENDED TO LCM OF (LCM OF ANY THREE NUMBERS AND FOURTH NUMBER) AND SO ON WE CAN ALSO BREAK FOUR NUMBERS IN TWO SUBGROUPS OF TWO NUMBERS EACH AND TAKE THEIR LCMS SEPARATELY AND THEREAFTER, WE TAKE THE LCM OF THESE TWO LCMS EX 7 FIND THE LC