QUANTITATIVE APTITUDE

NUMBER SYSTEM

PROF: RAVINDRA P. MANGRULKARAPTITUDE TRAINERPIBM PUNEQUANTITATIVE APTITUDE

REAL NUMBERS DECIMALS (FRACTIONS) + INTEGERS (NO DECIMALS) DECIMALS 1) FINITE OR TERMINATING DECIMALS (e.g. 4.2, 5.36, 6.274 SUCH NUMBERS CAN BE REPRESENED IN THE FORM OF P/Q, HENCE ARE RATIONAL NUMBERS).

2) a) IN FINITE OR NON-TERMINATING DECIMALS (e.g. 3.666, 4.2626, 5.427427427, 6.2718424242 SUCH NUMBERS CAN ALSO BE REPRESENED IN THE FORM OF P/Q, HENCE ARE RATIONAL NUMBERS). b) IRRATIONAL NUMBERS (e.g. 3,,e,5 CANT BE WRITTEN IN THE FORM OF P/Q)

REAL NUMBER

DECIMALS

2

INTEGERS NEGATIVE INTEGERS (-3,-2,-1) + ZERO (0) + POSITIVE INTEGERS (1,2,3)NATURAL NUMBERS = POSITIVE INTEGERS[UNITY(1) + PRIME NUMBERS(2,3,5,7,11,13,17) + COMPOSITE NUMBERS(4,6,8,9,10,12,14,15)]

WHOLE NUMBERS = NON-NEGATIVE INTEGERS (0,1,2,3,4,5,6)

NON-NEGATIVE NUMBERS = 0,1,2,,3,4,5.NON-POSITIVE NUMBERS = 0,-1,-2,-3,-4,-5.

NEITHER PRIME NOR COMPOSITE NUMBER = 1

NON-NEGATIVE NUMBERS = 3

SHORTCUT PROCESS

TO CHECK WHETHER A NUMBER N IS PRIME OR NOT TAKE THE SQUARE ROOT OF THE NUMBER.ROUND OF THE SQUARE ROOT TO THE IMMEDIATELY LOWER INTEGER. CALL THIS NUMBER Z. e.g. IF YOU HAVE TO CHECK FOR 181, ITS SQUARE ROOTWILL BE 13.__. HENCE, THE VALUE OF Z, IN THIS CASE WILL BE 13.CHECK FOR DIVISIBILITY OF THE NUMBER N BY ALL PRIME NUMBERS BELOW Z. IF THERE IS NO PRIME NUMBER BELOW VALUE OF Z WHICH DIVIDES N THEN THE NUMBER N WILL BE PRIME..

THE VALUE OF 4

CHECK FOR 239

THE VALUE OF 239 LIES BETWEEN 15 TO 16, HENCE THE VALUE OF Z IS 15.

PRIME NUMBERS LESS THAN 15 ARE 2,3,5,7,11 AND 13.

239 IS NOT DIVISIBLE BY ANY OF THESE. HENCE 239 IS A PRIME NUMBER.

THE CONCEPT OF GCD / HCF (GREATEST COMMON DIVISOR / HIGHEST COMMON FACTOR)EXAMPLE : FIND THE GCD OF 150,210,375

STEP 1: WRITING DOWN THE STANDARD FORM OF NUMBERS 150 = 5 * 5 * 3 * 2210 = 5 * 2 * 3 * 7375 = 5 * 5 * 3 * 5

STEP 2 : WRITING PRIME FACTORS COMMON TO ALL THE THREE NUMBERS IS 5 * 3

STEP 3 : HENCE THE HCF WILL BE 5 * 3 = 15 .

THE CONCEPT6

SHORTCUT METHOD

STEP 1 : FIRST FIND THE SMALLEST DIFFERENCE BETWEEN ANY OF THESE 2 NUMBERS ( 210 150 = 60).

STEP 2 : THUS, THE REQUIRED HCF WOULD BE A FACTOR OF THE NUMBER 60. THE FACTORS OF 60 ARE 1 * 60, 2 * 30, 3 * 20, 4 * 15, 5 * 12, 6 * 10.

STEP 3 : ANY 1 OF THESE 8 NUMBERS IS HCF. AS 375 IS ONE OF THE NUMBERS SO NUMBER ENDING WITH 5 HAS TO BE HCF. HENCE 15 IS THE ANSWER.

EXAMPLE 1 :

THE SIDES OF A HEXAGONAL FIELD ARE 216, 423, 1215, 1422, 2169 AND 2223 METERS. FIND THE GREATEST LENGTH OF TAPE THAT WOULD BE ABLE TO EXACTLY MEASURE EACH OF THESE SIDES WITHOUT HAVING TO USE FRACTIONS/PARTS OF THE TAPE?

SOLUTION : WE ARE REQUIRED TO FIND OUT THE HCF OF THE NUMBERS 216, 423, 1215, 1422, 2169 AND 2223.

EX.8

SOLUTION : WE ARE REQUIRED TO FIND OUT THE HCF OF THE NUMBERS 216, 423, 1215, 1422, 2169 AND 2223.

STEP 1: IN ORDER TO DO THAT WE FIRST FIND THE SMALLEST DIFFERENCE BETWEEN ANY TWO OF THESE NUMBERS. [ 2223 2169 = 54 ]

STEP 2 : FACTORS OF 54 ARE 1*54, 2*27, 3*18, 6*9.

STEP 3 : CHECK FOR HCF. 54 AND 18 CANT BE HCF BEING EVEN NUMBERS AND SO CANT DIVIDE ODD NUMBERS LIKE 423 AND 2223. 27 DOESNT DIVIDE 423. 9 DIVIDES 216,423,1215,1422 AND 2169. (NEED NOT TO CHEK FOR 2223 AS IT DIVIDES 2169).

HENCE 9 IS THE REQUIRED ANSWER.

EXAMPLE 2 :

A NURSERY HAS 363,429,693 PLANTS RESPECTIVELY OF 3 DISTINCT VARIETIES. IT IS DESIRED TO PLACE THESE PLANTS IN STRAIGHT ROWS OF PLANTS OF 1 VARIETY ONLY SO THAT THE NUMBER OF ROWS REQUIRED IS MINIMUM. WHAT IS THE SIZE OF EACH ROW AND HOW MANY ROWS WOULD BE REQUIRED?

SOLUTION : WE ARE REQUIRED TO FIND OUT THE HCF OF THE NUMBERS 363,429,693.

SOLUTION : WE ARE REQUIRED TO FIND OUT THE HCF OF THE NUMBERS 363,429,693.

STEP 1: IN ORDER TO THAT WE FIRST FIND THE SMALLEST DIFFERENCE BETWEEN ANY TWO OF THESE NUMBERS. [ 429 363 = 66]

STEP 2 : FACTORS OF 66 ARE 1*66, 2*33, 3*22, 6*11.

STEP 3 : CHECK FOR HCF. 66 CANT BE HCF BEING EVEN NUMBERS AND SO CANT DIVIDE ODD NUMBERS LIKE 363. 33 DIVIDES 363 HENCE AUTOMATICALLY DIVIDES 429 AND ALSO DIVIDES 693. HENCE 33 IS THE REQUIRED ANSWER FOR SIZE OF EACH ROW.

STEP4 : MINIMUM NUMBER OF ROWS REQUIRED = 363/33 +429/33 + 693/33 = 45ROWS

THE CONCEPT OF LCM ( LEAST COMMON MULTIPLE ) FIND THE LCM OF 150,210,375 STEP 1: WRITING DOWN THE STANDARD FORMOF NUMBERS 150 = 5 * 5 * 3 * 2210 = 5 * 2 * 3 * 7375 = 5 * 5 * 3 * 5 STEP 2 : WRITING PRIME FACTORS THAT APPEARS AT LEAST ONCE IN ALL THE THREE NUMBERS IS 5, 3, 2, 7. STEP 3 : RAISE EACH OF THE PRIME FACTOR TO THEIR HIGHEST AVAILABLE POWER 53*31*71*21 = 5250.

12

SHORTCUT METHOD

FIND THE LCM OF 9,10,12,15

STEP 1: WRITE DOWN THE CO-PRIME NUMBERS 9*10

STEP 2 : 12 = 2*2*3. SINCE 9*10 ALREADY HAS 2*3. SO MULTIPLY 9*10 BY REMAINING 2. HENCE WE NOW HAVE 9*10*2.

STEP 3 : NOW 15 = 3*5. SINCE 9*10*2 ALREADY HAS 5*3. HENCE FINAL ANSWER WILL BE 9*10*2

HCF AND LCM OF FRACTIONS

HCF OF 2 OR MORE FRACTIONS = HCF OF NUMERATORS LCM OF DENOMINATORS

(B) LCM OF 2 OR MORE FRACTIONS = LCM OF NUMERATORS HCF OF DENOMINATORS

IMPORTANT RULE :

HCF (N1, N2) * LCM (N1, N2) = (N1 * N2)

IMPORTANT RULE14

NUMBER AND SUM OF FACTORS

FIND THE SUM OF FACTORS AND NUMBER OF FACTORS OF 240

240 = 24*31*51= (20+21+22+23+24)(30+31)(50+51) = (1+2+4+8+16)(1+3)(1+5)= 31*4*6= 744

NUMBER OF FACTORS = 5*2*2= 20.

(20+21+22+23+24)(30) = 1+2+4+8+16 (20+21+22+23+24)(31) = 3+6+12+24+48(20+21+22+23+24)(51) = 5+10+20+40+80(20+21+22+23+24)(31 )(51) = 15+30+60+120+240

DIVISIBILITY15

DIVISIBILTY RULE

2 OR 5 : A NUMBER IS DIVISIBLE BY 2 OR 5 IF THE LAST DIGIT IS DIVISIBLE BY 2 OR 5. 12, 14, 64, 92, 132 88, 36, 70. 123486755448 ARE DIVISIBLE Y 2. 15, 20, 125, 140, 185, 225, 320 ARE DIVISIBLE BY 5.

2) 3 OR 9 : THE SUM OF DIGITS DIVISIBLE BY 3 OR 9 ARE DIVISIBLE BY 3 OR 9. 15, 12, 18, 27, 162 ARE DIVISIBLE BY 3. 18, 27, 162, 729, 1881 ARE DIVISIBLE BY 3 AND 9.

3) 4 : IF THE LAST 2 DIGITS OF A NUMBER ARE 0 OR ARE DIVISIBLE BY 4 THEN THE NUMBER IS DIVISIBLE BY 4. 20, 148, 1500, 136, 125478980 ARE DIVISIBLE BY 4. 4) 6 : IF THE NUMBER IS SIMULTANEOUSLY DIVISIBLE BY 2 AND 3 THEN THE NUMBER IS DIVISIBLE BY 6. 6, 12, 18, 24, 96, 576 ARE DIVIDED BY 2 AND 3 BOTH HENCE ARE DIVIDED BY 6.

DIVISIBILTY RULE16

5) 8 : IF THE LAST 3 DIGITS OF A NUMBER ARE 0 OR ARE DIVISIBLE BY 8 THEN THE NUMBER IS DIVISIBLE BY 8 . 824, 848, 15000, 1360, 1254789880 ARE DIVISIBLE BY 4. 6) 11 : A NUMBER IS DIVISIBLE BY 11 IF THE DIFFERENCE OF THE SUM OF DIGITS IN THE ODD PLACES AND IN THE EVEN PLACES IS 0 OR DIVISIBLE BY 11. 121, 1331, 423467, 56789392 ARE DIVISIBLE BY 11.

7) 12 : ALL THE NUMBERS DIVISIBLE BY 3 AND 4 SIMULTANEOUSLY ARE DIVISIBLE BY 12. 12, 24, 36, 1440, 8574849300 ARE DIVISIBLE BY 12.

8) 7 OR 13 : IF THE DIFFERENCE OF THE NUMBER OF ITS THOUSANDS AND THE REMAINDER IS DIVISIBLE BY 7 OR 13. e.g.- 473312 IS DIVISLBLE BY 7 SINCE 473 312 =161 IS DIVISIBLE BY 7.

UNIT DIGIT OF A NUMBEREXAMPLE : FIND THE UNIT DIGIT OF THE EXPRESSION: 785562*56256*971250 CYCLE OF 4 : 781 8 782 4 783 2 784 68N+1 8 8N+2 4785562 = 78 (4*1390 + 2) = (784)1390*782 WILL YIELD 4 AS A UNIT DIGIT.

561 6 562 6 563 6 564 656256 = 56 (4*64 ) = (564) 64 WILL YIELD 6 AS A UNIT DIGIT.

971 7 972 9 973 3 974 1971250 = 97(4*312+2 ) = (974) 312 *972 WILL YIELD 9 AS A UNIT DIGIT.

NOW MULTIPLY ALL LAST DIGITS 4*6*9 216HENCE THE ANSWER WILL BE 6.

UNIT DIGIT18

SUM OF SERIES

SUM OF FIRST N NATURAL NUMBERS LET THE SUM BE S = 1+ 2 +3 +4 +..+N IS GIVEN BY: S = [N(N+1)] / 2 .

2. SUM OF THE SQUARES OF FIRST N NATURAL NUMBERS LET THE SUM BE S = 12+22+32+42+..+N2 IS GIVEN BY: S = [N(N+1)(2N+1)] / 6 .

3. SUM OF THE CUBES FIRST N NATURAL NUMBERS LET THE SUM BE S = 13+23+33+43+..+N3 IS GIVEN BY: S = [N(N+1) / 2] 2 .

SUM OF SERIES19

SUM OF SERIES

SUM OF FIRST N ODD NATURAL NUMBERS LET THE SUM BE S = 1+3+5+..+(2N-1) IS GIVEN BY: S = N2. a) IF N IS ODD [(N+1)/2] 2 b) IF N IS EVEN [N/2] 2

2. SUM OF FIRST N EVEN NATURAL NUMBERS LET THE SUM BE S = 2+4+6+..+2N IS GIVEN BY: S = N(N+1). a) IF N IS ODD [(N/2){(N/2 )+1)/2}] b) IF N IS EVEN [(N-1)/2] [(N+1)/2]

THANK YOU

THANK YOU21