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# Fundamentals of Quantitative Aptitude

Tips to prepare for Fundamentals of Quantitative Aptitude
Number Properties
LCM, HCF
Divisibility
Fractions & Decimals,
square
Square Roots
cyclicity
with shortcut tricks

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### Fundamentals of Quantitative Aptitude

1. 1. QUANTITATIVE APTITUDE Pawan Mishra
2. 2. FUNDAMENTAL S
3. 3. Important Concepts and Formulas - Numbers  Number Sets and Properties of Numbers   Counting Numbers (Natural numbers) : 1, 2, 3 ...   Whole Numbers : 0, 1, 2, 3 ...   Integers : -3, -2, -1, 0, 1, 2, 3 ...   Rational Numbers  Rational numbers can be expressed as ab where a and b are integers and b≠0 Examples: 112, 42, 0, −811 etc.  All integers, fractions and terminating or recurring decimals are rational numbers.
4. 4. Irrational Numbers Any number which is not a rational number is an irrational number. In other words, an irrational number is a number which cannot be expressed as ab where a and b are integers. For instance, numbers whose decimals do not terminate and do not repeat cannot be written as a fraction and hence they are irrational numbers. Example : π, 2√, (3+5√), 43√ (meaning 4×3√), 6√3 etc Please note that the value of π = 3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59230 78164 06286 20899 86280 34825 34211 70679... We cannot π as a simple fraction (The fraction 22/7 = 3.14.... is just an approximate value of π)
5. 5. Real Numbers Real numbers include counting numbers, whole numbers, integers, rational numbers and irrational numbers. Surds Let a be any rational number and n be any positive integer such that a√n is irrational. Then a√n is a surd. Example : 3√, 10−−√6, 43√ etc Please note that numbers like 9√, 27−−√3 etc are not surds because they are not irrational numbers Every surd is an irrational number. But every irrational number is not a surd. (eg : π , e etc are not surds though they are irrational numbers.)
6. 6. ADDITION, SUBTRACTIONAND MULTIPLICATION RULES FOR EVENAND ODD NUMBERS
7. 7. Addition Rules for Even and Odd Numbers The sum of any number of even numbers is always even The sum of even number of odd numbers is always even The sum of odd number of odd numbers is always odd
8. 8. Subtraction Rules for Even and Odd Numbers The difference of two even numbers is always even The difference of two odd numbers is always even
9. 9. Multiplication Rules for Even and Odd Numbers The product of even numbers is always even The product of odd numbers is always odd If there is at least one even number multiplied by any number of odd numbers, the product is always even
10. 10. 3STEPMULTIPLICATIONTRICK-A SHORTCUTMETHOD
11. 11. Step 1 First step is same as conventional method, here we multiply 2 with 2.
12. 12. Step 2 This is an interesting step. Now multiply last digit first value and first digit of second value and vice-versa. Then we add outcomes. But we need the last number that is 8 here.
13. 13. Step 3 This is the last step, in this step we do multiplication ten's digit of both value and add the remainder from previous calculation. That's it, we completed the calculation in 3 steps instead of six steps.
14. 14. We can use this method for multiplication of three or even four digit numbers but time management is really important in IBPS exam and other recruitment exams so for longer calculations, estimation is the best trick. I will post an article about how to do long calculations using estimation and result is 95% accurate which is enough to arrive at answer. As two of the readers namely Rahul and Ansh have requested me to use this technique in longer calculations multiplications. I am updating this article.
15. 15. Multiplication of 3 digit numbers In this example I will multiply 432 with 346. Now the 3 step multiplication method will become 5 step. This method can be used for 4 and even 5 digit numbers but as in bank exams there is lack of time available for calculations I recommend you to use approximation for long calculations.
16. 16. Step 1 Step 2 Step 3 Step 4 Step 5 In case you find any difficulty to understand the above multiplication method then ask your question in the comments. I will try to answer every query asap
17. 17. DIVISIBILITY
18. 18. Divisible By One whole number is divisible by another if the remainder we get after the division is zero. Examples 36 is divisible by 4 because 36 ÷ 4 = 9 with a remainder of 0. 36 is divisible by 6 because 36 ÷ 6 = 6 with a remainder of 0. 36 is not divisible by 5 because 36 ÷ 5 = 7 with a remainder of 1.
19. 19. Divisibility Rules By using divisibility rules we can easily find out whether a given number is divisible by another number without actually performing the division. This helps to save time especially when working with numbers.
20. 20. DivisibilityRule Description Examples Divisibility by 2 A number is divisible by 2 if the last digit is even. i.e., if the last digit is 0 or 2 or 4 or 6 or 8 Example1: Check if 64 is divisible by 2. The last digit of 64 is 4 (even). Hence 64 is divisible by 2 Example2: Check if 69 is divisible by 2. The last digit of 69 is 9 (not even). Hence 69 is not divisible by 2
21. 21. Divisibility by 3 A number is divisible by 3 if the sum of the digits is divisible by 3 Example1: Check if 387 is divisible by 3. 3 + 8 + 7 = 18. 18 is divisible by 3. Hence 387 is divisible by 3 Example2: Check if 421 is divisible by 3. 4 + 2 + 1 = 7. 7 is not divisible by 3. Hence 421 is not divisible by 3
22. 22. Divisibility by 4 A number is divisible by 4 if the number formed by the last two digits is divisible by 4. Example1: Check if 416 is divisible by 4. Number formed by the last two digits = 16. 16 is divisible by 4. Hence 416 is divisible by 4 Example2: Check if 481 is divisible by 4. Number formed by the last two digits = 81. 81 is not divisible by 4. Hence 481 is not divisible by 4
23. 23. Divisibility by 5 A number is divisible by 5 if the last digit is either 0 or 5. Example1: Check if 305 is divisible by 5. Last digit is 5. Hence 305 is divisible by 5. Example2: Check if 420 is divisible by 5. Last digit is 0. Hence 420 is divisible by 5. Example3: Check if 312 is divisible by 5. Last digit is 2. Hence 312 is not divisible by 5.
24. 24. Divisibility by 6 A number is divisible by 6 if it is divisible by both 2 and 3. Example1: Check if 546 is divisible by 6. 546 is divisible by 2. 546 is also divisible by 3. (Check the divisibility rule of 2 and 3 to find out this) Hence 546 is divisible by 6 Example2: Check if 633 is divisible by 6. 633 is not divisible by 2 though 633 is divisible by 3. (Check the divisibility rule of 2 and 3 to find out this) Hence 633 is not divisible by 6 Example3: Check if 635 is divisible by 6. 635 is not divisible by 2. 635 is also not divisible by 3. (Check the divisibility rule of 2 and 3 to find out this) Hence 635 is not divisible by 6 Example4: Check if 428 is divisible by 6. 428 is divisible by 2 but 428 is not divisible by 3.(Check the divisibility rule of 2 and 3 to find out this) Hence 428 is not divisible by 6
25. 25. Divisibility by 7 To find out if a number is divisible by 7, double the last digit and subtact it from the number formed by the remaining digits. Repeat this process until we get at a smaller number whose divisibility we know. If this smaller number is 0 or divisible by 7, the original number is also divisible by 7. Example1: Check if 349 is divisible by 7. Given number = 349 34 - (9 × 2) = 34 - 18 = 16 16 is not divisible by 7. Hence 349 is not divisible by 7 Example2: Check if 364 is divisible by 7. Given number = 364 36 - (4 × 2) = 36 - 8 = 28 28 is divisible by 7. Hence 364 is also divisible by 7 Example3: Check if 3374 is divisible by 7. Given number = 3374 337 - (4 × 2) = 337 - 8 = 329 32 - (9 × 2) = 32 - 18 = 14 14 is divisible by 7. Hence 329 is also divisible by 7. Hence 3374 is also divisible by 7.
26. 26. Divisibility by 8 A number is divisible by 8 if the number formed by the last three digits is divisible by 8. Example1: Check if 7624 is divisible by 8. The number formed by the last three digits of 7624 = 624. 624 is divisible by 8. Hence 7624 is also divisible by 8. Example2: Check if 129437464 is divisible by 8. The number formed by the last three digits of 129437464 = 464. 464 is divisible by 8. Hence 129437464 is also divisible by 8. Example3: Check if 737460 is divisible by 8. The number formed by the last three digits of 737460 = 460. 460 is not divisible by 8. Hence 737460 is also not divisible by 8.
27. 27. Divisibility by 9 A number is divisible by 9 if the sum of its digits is divisible by 9. (Please note that we can apply this rule to the answer again and again if we need) Example1: Check if 367821 is divisible by 9. 3 + 6 + 7 + 8 + 2 + 1 = 27 27 is divisible by 9. Hence 367821 is also divisible by 9. Example2: Check if 47128 is divisible by 9. 4 + 7 + 1 + 2 + 8 = 22 22 is not divisible by 9. Hence 47128 is not divisible by 9. Example3: Check if 4975291989 is divisible by 9. 4 + 9+ 7 + 5 + 2 + 9 + 1 + 9 + 8 + 9= 63 Since 63 is big, we can use the same method to see if it is divisible by 9. 6 + 3 = 9 9 is divisible by 9. Hence 63 is also divisible by 9. Hence 4975291989 is also divisible by 9.
28. 28. Divisibility by 10 A number is divisible by 10 if the last digit is 0. Example1: Check if 2570 is divisible by 10. Last digit is 0. Hence 2570 is divisible by 10. Example2: Check if 5462 is divisible by 10. Last digit is not 0. Hence 5462 is not divisible by 10
29. 29. Divisibility by 11 To find out if a number is divisible by 11, find the sum of the odd numbered digits and the sum of the even numbered digits. Now substract the lower number obtained from the bigger number obtained. If the number we get is 0 or divisible by 11, the original number is also divisible by 11. Example1: Check if 85136 is divisible by 11. 8 + 1 + 6 = 15 5 + 3 = 8 15 - 8 = 7 7 is not divisible by 11. Hence 85136 is not divisible by 11. Example2: Check if 2737152 is divisible by 11. 2 + 3 + 1 + 2 = 8 7 + 7 + 5 = 19 19 - 8 = 11 11 is divisible by 11. Hence 2737152 is also divisible by 11. Example3: Check if 957 is divisible by 11. 9 + 7 = 16 5 = 5 16 - 5 = 11 11 is divisible by 11. Hence 957 is also divisible by 11. Example4: Check if 9548 is divisible by 11. 9 + 4 = 13 5 + 8 = 13 13 - 13 = 0 We got the difference as 0. Hence 9548 is divisible by 11.
30. 30. Divisibility by 12 A number is divisible by 12 if the number is divisible by both 3 and 4 Example1: Check if 720 is divisible by 12. 720 is divisible by 3 and 720 is also divisible by 4. (Check the divisibility rules of 3 and 4 to find out this) Hence 720 is also divisible by 12 Example2: Check if 916 is divisible by 12. 916 is not divisible by 3 , though 916 is divisible by 4.(Check the divisibility rules of 3 and 4 to find out this) Hence 916 is not divisible by 12 Example3: Check if 921 is divisible by 12. 921 is divisible by 3. But 921 is not divisible by 4.(Check the divisibility rules of 3 and 4 to find out this) Hence 921 is not divisible by 12 Example4: Check if 827 is divisible by 12. 827 is not divisible by 3. 827 is also not divisible by 4.(Check the divisibility rules of 3 and 4 to find out this) Hence 827 is not divisible by 12
31. 31. Divisibility by 13 To find out if a number is divisible by 13, multiply the last digit by 4 and add it to the number formed by the remaining digits. Repeat this process until we get at a smaller number whose divisibility we know. If this smaller number is divisible by 13, the original number is also divisible by 13. Example1: Check if 349 is divisible by 13. Given number = 349 34 + (9 × 4) = 34 + 36 = 70 70 is not divisible by 13. Hence 349 is not divisible by 349 Example2: Check if 572 is divisible by 13. Given number = 572 57 + (2 × 4) = 57 + 8 = 65 65 is divisible by 13. Hence 572 is also divisible by 13 Example3: Check if 68172 is divisible by 13. Given number = 68172 6817 + (2 × 4) = 6817 + 8 = 6825 682 + (5 × 4) = 682 + 20 = 702 70 + (2 × 4) = 70 + 8 = 78 78 is divisible by 13. Hence 68172 is also divisible by 13. Example4: Check if 651 is divisible by 13. Given number = 651 65 + (1 × 4) = 65 + 4 = 69 69 is not divisible by 13. Hence 651 is not divisible by 13
32. 32. Divisibility by 14 A number is divisible by 14 if it is divisible by both 2 and 7. Example1: Check if 238 is divisible by 14 238 is divisible by 2 . 238 is also divisible by 7. (Please check the divisibility rule of 2 and 7 to find out this) Hence 238 is also divisible by 14 Example2: Check if 336 is divisible by 14 336 is divisible by 2 . 336 is also divisible by 7. (Please check the divisibility rule of 2 and 7 to find out this) Hence 336 is also divisible by 14 Example3: Check if 342 is divisible by 14. 342 is divisible by 2 , but 342 is not divisible by 7.(Please check the divisibility rule of 2 and 7 to find out this) Hence 342 is not divisible by 12 Example4: Check if 175 is divisible by 14. 175 is not divisible by 2 , though it is divisible by 7.(Please check the divisibility rule of 2 and 7 to find out this) Hence 175 is not divisible by 14 Example5: Check if 337 is divisible by 14. 337 is not divisible by 2 and also by 7 (Please check the divisibility rule of 2 and 7 to find out this) Hence 337 is not divisible by 14
33. 33. Divisibility by 15 A number is divisible by 15 If it is divisible by both 3 and 5. Example1: Check if 435 is divisible by 15 435 is divisible by 3 . 435 is also divisible by 5. (Please check the divisibility rule of 3 and 5 to find out this) Hence 435 is also divisible by 15 Example2: Check if 555 is divisible by 15 555 is divisible by 3 . 555 is also divisible by 5. (Please check the divisibility rule of 3 and 5 to find out this) Hence 555 is also divisible by 15 Example3: Check if 483 is divisible by 15. 483 is divisible by 3 , but 483 is not divisible by 5. (Please check the divisibility rule of 3 and 5 to find out this) Hence 483 is not divisible by 15 Example4: Check if 485 is divisible by 15. 485 is not divisible by 3 , though it is divisible by 5. (Please check the divisibility rule of 3 and 5 to find out this) Hence 485 is not divisible by 15 Example5: Check if 487 is divisible by 15. 487 is not divisible by 3 . It is also not divisible by 5 (Please check the divisibility rule of 3 and 5 to find out this) Hence 487 is not divisible by 15
34. 34. Divisibility by 16 A number is divisible by 16 if the number formed by the last four digits is divisible by 16. Example1: Check if 5696512 is divisible by 16. The number formed by the last four digits of 5696512 = 6512 6512 is divisible by 16. Hence 5696512 is also divisible by 16. Example2: Check if 3326976 is divisible by 16. The number formed by the last four digits of 3326976 = 6976 6976 is divisible by 16. Hence 3326976 is also divisible by 16. Example3: Check if 732374360 is divisible by 16. The number formed by the last three digits of 732374360 = 4360 4360 is not divisible by 16. Hence 732374360 is also not divisible by 16.
35. 35. Divisibility by 17 To find out if a number is divisible by 17, multiply the last digit by 5 and subtract it from the number formed by the remaining digits. Repeat this process until you arrive at a smaller number whose divisibility you know. If this smaller number is divisible by 17, the original number is also divisible by 17. Example1: Check if 500327 is divisible by 17. Given Number = 500327 50032 - (7 × 5 )= 50032 - 35 = 49997 4999 - (7 × 5 ) = 4999 - 35 = 4964 496 - (4 × 5 ) = 496 - 20 = 476 47 - (6 × 5 ) = 47 - 30 = 17 17 is divisible by 17. Hence 500327 is also divisible by 17 Example2: Check if 521461 is divisible by 17. Given Number = 521461 52146 - (1 × 5 )= 52146 -5 = 52141 5214 - (1 × 5 ) = 5214 - 5 = 5209 520 - (9 × 5 ) = 520 - 45 = 475 47 - (5 × 5 ) = 47 - 25 = 22 22 is not divisible by 17. Hence 521461 is not divisible by 17
36. 36. Divisibility by 18 A number is divisible by 18 if it is divisible by both 2 and 9. Example1: Check if 31104 is divisible by 18. 31104 is divisible by 2. 31104 is also divisible by 9. (Please check the divisibility rule of 2 and 9 to find out this) Hence 31104 is divisible by 18 Example2: Check if 1170 is divisible by 18. 1170 is divisible by 2. 1170 is also divisible by 9. (Please check the divisibility rule of 2 and 9 to find out this) Hence 1170 is divisible by 18 Example3: Check if 1182 is divisible by 18. 1182 is divisible by 2 , but 1182 is not divisible by 9. (Please check the divisibility rule of 2 and 9 to find out this) Hence 1182 is not divisible by 18 Example4: Check if 1287 is divisible by 18. 1287 is not divisible by 2 though it is divisible by 9. (Please check the divisibility rule of 2 and 9 to find out this) Hence 1287 is not divisible by 18
37. 37. Divisibility by 19 To find out if a number is divisible by 19, multiply the last digit by 2 and add it to the number formed by the remaining digits. Repeat this process until you arrive at a smaller number whose divisibility you know. If this smaller number is divisible by 19, the original number is also divisible by 19. Example1: Check if 74689 is divisible by 19. Given Number = 74689 7468 + (9 × 2 )= 7468 + 18 = 7486 748 + (6 × 2 ) = 748 + 12 = 760 76 + (0 × 2 ) = 76 + 0 = 76 76 is divisible by 19. Hence 74689 is also divisible by 19 Example2: Check if 71234 is divisible by 19. Given Number = 71234 7123 + (4 × 2 )= 7123 + 8 = 7131 713 + (1 × 2 )= 713 + 2 = 715 71 + (5 × 2 )= 71 + 10 = 81 81 is not divisible by 19. Hence 71234 is not divisible by 19
38. 38. Divisibility by 20 A number is divisible by 20 if it is divisible by 10 and the tens digit is even. (There is one more rule to see if a number is divisible by 20 which is given below. A number is divisible by 20 if the number is divisible by both 4 and 5) Example1: Check if 720 is divisible by 20 720 is divisible by 10. (Please check the divisibility rule of 10 to find out this). The tens digit = 2 = even digit. Hence 720 is also divisible by 20 Example2: Check if 1340 is divisible by 20 1340 is divisible by 10. (Please check the divisibility rule of 10 to find out this). The tens digit = 2 = even digit. Hence 1340 is divisible by 20 Example3: Check if 1350 is divisible by 20 1350 is divisible by 10. (Please check the divisibility rule of 10 to find out this). But the tens digit = 5 = not an even digit. Hence 1350 is not divisible by 20 Example4: Check if 1325 is divisible by 20 1325 is not divisible by 10 (Please check the divisibility rule of 10 to find out this) though the tens digit = 2 = even digit. Hence 1325 is not divisible by 20
39. 39. WHAT ARE FACTORS OF A NUMBER AND HOW TO FIND IT OUT?
40. 40. Factors of a number If one number is divisible by a second number, the second number is a factor of the first number The lowest factor of any positive number = 1 The highest factor of any positive number = the number itself Example The factors of 36 are 1, 2, 3, 4, 6, 9 12, 18, 36 because each of these numbers divides 36 with a remainder of 0
41. 41. How to find out factors of a number Write down 1 and the number itself (lowest and highest factors). Check if the given number is divisible by 2 (Reference: Divisibility by 2 rule) If the number is divisible by 2, write down 2 as the second lowest factor and divide the given number by 2 to get the second highest factor Check for divisibility by 3, 4,5, and so on. till the beginning of the list reaches the end
42. 42. Example1: Find out the factors of 72 Write down 1 and the number itself (72) as lowest and highest factors. 1 . . . 72 72 is divisible by 2 (Reference: Divisibility by 2 Rule). 72 ÷ 2 = 36. Hence 2nd lowest factor = 2 and 2nd highest factor = 36. So we can write as 1, 2 . . . 36, 72 72 is divisible by 3 (Reference: Divisibility by 3 Rule). 72 ÷ 3 = 24 . Hence 3rd lowest factor = 3 and 3rd highest factor = 24. So we can write as 1, 2, 3, . . . 24, 36, 72
43. 43. 72 is divisible by 4 (Reference: Divisibility by 4 Rule). 72 ÷ 4 = 18. Hence 4th lowest factor = 4 and 4th highest factor = 18. So we can write as 1, 2, 3, 4, . . . 18, 24, 36, 72 72 is not divisible by 5 (Reference: Divisibility by 5 Rule) 72 is divisible by 6 (Reference: Divisibility by 6 Rule). 72 ÷ 6 = 12. Hence 5th lowest factor = 6 and 5th highest factor = 12. So we can write as
44. 44. 72 is not divisible by 7 (Reference: Divisibility by 7 Rule) 72 is divisible by 8 (Reference: Divisibility by 8 Rule). 72 ÷ 8 = 9. Hence 6th lowest factor = 8 and 6th highest factor = 9. Now our list is complete and the factors of 72 are 1, 2, 3, 4, 6, 8, 9 12, 18, 24, 36, 72
45. 45. Example2: Find out the factors of 22 Write down 1 and the number itself (22) as lowest and highest factors 1 . . . 22 22 is divisible by 2 (Reference: Divisibility by 2 Rule). 22 ÷ 2 = 11. Hence 2nd lowest factor = 2 and 2nd highest factor = 11. So we can write as 1, 2 . . . 11, 22 22 is not divisible by 3 (Reference: Divisibility by 3 Rule). 22 is not divisible by 4 (Reference: Divisibility by 4 Rule). 22 is not divisible by 5 (Reference: Divisibility by 5 Rule). 22 is not divisible by 6 (Reference: Divisibility by 6 Rule). 22 is not divisible by 7 (Reference: Divisibility by 7 Rule). 22 is not divisible by 8 (Reference: Divisibility by 8 Rule). 22 is not divisible by 9 (Reference: Divisibility by 9 Rule). 22 is not divisible by 10 (Reference: Divisibility by 10 Rule). Now our list is complete and the factors of 22 are 1, 2, 11, 22
46. 46. Important Properties of Factors If a number is divisible by another number, then it is also divisible by all the factors of that number. Example : 108 is divisible by 36 because 106 ÷ 38 = 3 with remainder of 0. The factors of 36 are 1, 2, 3, 4, 6, 9 12, 18, 36 because each of these numbers divides 36 with a remainder of 0. Hence, 108 is also divisible by each of the numbers 1, 2, 3, 4, 6, 9, 12, 18, 36.
47. 47. WHAT ARE PRIME NUMBERS AND COMPOSITE NUMBERS?
48. 48. Prime Numbers A prime number is a positve integer that is divisible by itself and 1 only. Prime numbers will have exactly two integer factors. Examples: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, etc. Please note the following facts Zero is not a prime number because zero is divisible by more than two factors. Zero can be divided by 1, 2, 3 etc. (0 ÷ 1 = 0, 0÷ 2 = 0 ...) One is not a prime number because it does not have two factors. It is divisible by only 1
49. 49. Composite Numbers Composite numbers are numbers that have more than two factors. A composite number is divisible by at least one number other than 1 and itself. Examples: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, etc. Please note that zero and 1 are neither prime numbers nor composite numbers. Every whole number is either prime or composite, with two exceptions 0 and 1 which are neither prime nor composite
50. 50. WHAT ARE PRIME FACTORIZATION AND PRIME FACTORS ?
51. 51. Prime factor Prime factorization he factors which are prime numbers are called prime factors Prime factorization of a number is the expression of the number as the product of its prime factors Example 1: Prime factorization of 280 can be written as 280 = 2 × 2 × 2 × 5 × 7 = 23 × 5 × 7 and the prime factors of 280 are 2, 5 and 7 Example 2: Prime factorization of 72 can be written as 72 = 2 × 2 × 2 × 3 × 3 = 23 × 32 and the prime factors of 72 are 2 and 3
52. 52. How to find out prime factorization and prime factors of a number Repeated Division Method : In order to find out the prime factorization of a number, repeatedly divide the number by the smallest prime number possible(2,3,5,7,11, ...) until the quotient is 1.
53. 53. Example 1: Find out Prime factorization of 280 2*140=280 2*2*70=280 2*2*2*35=280 2*2*2*5*7=280 Hence, prime factorization of 280 can be written as 280 = 2 × 2 × 2 × 5 × 7 = 23 × 5 × 7 and the prime factors of 280 are 2, 5 and 7
54. 54. Example 2: Find out Prime factorization of 72 2*36=72 2*2*18=72 2*2*2*9=72 2*2*2*3*3=72 Hence, prime factorization of 72 can be written as 72 = 2 × 2 × 2 × 3 × 3 = 23 × 32 and the prime factors of 72 are 2 and 3
55. 55. Important Properties Every whole number greater than 1 can be uniquely expressed as the product of its prime factors. For example, 700 = 22 × 52 × 7
56. 56. Multiples Multiples of a whole number are the products of that number with 1, 2, 3, 4, and so on Example : Multiples of 3 are 3, 6, 9, 12, 15, ... If a number x divides another number y exactly with a remainder of 0, we can say that x is a factor of y and y is a multiple of x For instance, 4 divides 36 exactly with a remainder of 0. Hence 4 is a factor of 36 and 36 is a multiple of 4
57. 57. WHAT IS LEAST COMMON MULTIPLE (LCM) AND HOW TO FIND LCM
58. 58. Least Common Multiple (LCM) Least Common Multiple (LCM) of two or more numbers is the smallest number that is a multiple of all the numbers Example: LCM of 3 and 4 = 12 because 12 is the smallest multiple which is common to 3 and 4 (In other words, 12 is the smallest number which is divisible by both 3 and 4) We can find out LCM using prime factorization method or division method
59. 59. How to find out LCM using prime factorization method Step1 : Express each number as a product of prime factors. Step2 : LCM = The product of highest powers of all prime factors
60. 60. Example 1 : Find out LCM of 8 and 14 Step1 : Express each number as a product of prime factors. 8 = 23 14 = 2 × 7 Step2 : LCM = The product of highest powers of all prime factors Here the prime factors are 2 and 7 The highest power of 2 here = 23 The highest power of 7 here = 7 Hence LCM = 23 × 7 = 56
61. 61. Example 2 : Find out LCM of 18, 24, 9, 36 and 90 Step1 : Express each number as a product of prime factors 18 = 2 × 32 24 = 23 × 3 9 = 32 36 = 23 × 32 90 = 2 × 5 × 32 Step2 : LCM = The product of highest powers of all prime factors Here the prime factors are 2, 3 and 5 The highest power of 2 here = 23 The highest power of 3 here = 32 The highest power of 5 here = 5 Hence LCM = 23 × 32 × 5 = 360
62. 62. How to find out LCM using Division Method (shortcut) Step 1 : Write the given numbers in a horizontal line separated by commas. Step 2 : Divide the given numbers by the smallest prime number which can exactly divide at least two of the given numbers. Step 3 : Write the quotients and undivided numbers in a line below the first. Step 4 : Repeat the process until we reach a stage where no prime factor is common to any two numbers in the row. Step 5 : LCM = The product of all the divisors and the numbers in the last line.
63. 63. Example 1 : Find out LCM of 8 and 14 Hence Least common multiple (L.C.M) of 8 and 14 = 2 × 4 × 7 = 56
64. 64. Example 2 : Find out LCM of 18, 24, 9, 36 and 90  Hence Least common multiple (L.C.M) of 18, 24, 9, 36 and 90 = 2 × 2 × 3 × 3 × 2 × 5 = 360
65. 65. WHAT IS HIGHEST COMMON FACTOR (HCF) OR GREATEST COMMON MEASURE (GCM) OR GREATEST COMMON DIVISOR (GCD) AND HOW TO FIND IT OUT ?
66. 66. Highest Common Factor(H.C.F) or Greatest Common Measure(G.C.M) or Greatest Common Divisor (G.C.D) Highest Common Factor(H.C.F) or Greatest Common Measure(G.C.M) or Greatest Common Divisor (G.C.D) of two or more numbers is the greatest number which divides each of them exactly. Example : HCF or GCM or GCD of 60 and 75 = 15 because 15 is the highest number which divides both 60 and 75 exactly. We can find out HCF using prime factorization method or division method
67. 67. How to find out HCF using prime factorization method Step1 : Express each number as a product of prime factors. (Reference: Prime Factorization and how to find out Prime Factorization) Step2 : HCF is the product of all common prime factors using the least power of each common prime factor.
68. 68. Example 1 : Find out HCF of 60 and 75 (Reference: Prime Factorization and how to find out Prime Factorization) Step1 : Express each number as a product of prime factors. 60 = 22 × 3 × 5 75 = 3 × 52 Step2 : HCF is the product of all common prime factors using the least power of each common prime factor. Here, common prime factors are 3 and 5 The least power of 3 here = 3 The least power of 5 here = 5 Hence, HCF = 3 × 5 = 15
69. 69. Example 2 : Find out HCF of 36, 24 and 12 Step1 : Express each number as a product of prime factors. (Reference: Prime Factorization and how to find out Prime Factorization) 36 = 22 × 32 24 = 23 × 3 12 = 22 × 3 Step2 : HCF is the product of all common prime factors using the least power of each common prime factor. Here 2 and 3 are common prime factors. The least power of 2 here = 22 The least power of 3 here = 3 Hence, HCF = 22 × 3 = 12
70. 70. Example 3 : Find out HCF of 36, 27 and 80 Step1 : Express each number as a product of prime factors. (Reference: Prime Factorization and how to find out Prime Factorization) 36 = 22 × 32 27 = 33 80 = 24 × 5 Step2 : HCF = HCF is the product of all common prime factors using the least power of each common prime factor. Here you can see that there are no common prime factors. Hence, HCF = 1
71. 71. How to find out HCF using prime factorization method - By dividing the numbers (shortcut) Step 1 : Write the given numbers in a horizontal line separated by commas. Step 2 : Divide the given numbers by the smallest prime number which can exactly divide all of the given numbers. Step 3 : Write the quotients in a line below the first. Step 4 : Repeat the process until we reach a stage where no common prime factor exists for all of the numbers. Step 5 :We can see that the factors mentioned in the left side clearly divides all the numbers exactly and they are common prime factors. Their product is the HCF
72. 72. Example 1 : Find out HCF of 60 and 75 We can see that the prime factors mentioned in the left side clearly divides all the numbers exactly and they are common prime factors. no common prime factor is exists for the numbers came at the bottom. Hence HCF = 3 × 5 =15
73. 73. Example 2 : Find out HCF of 36, 24 and 12 We can see that the prime factors mentioned in the left side clearly divides all the numbers exactly and they are common prime factors. no common prime factor is exists for the numbers came at the bottom. Hence HCF = 2 × 2 × 3 = 12.
74. 74. Example 3 : Find out HCF of 36, 24 and 48 We can see that the prime factors mentioned in the left side clearly divides all the numbers exactly and they are common prime factors. no common prime factor is exists for the numbers came at the bottom. Hence HCF = 2 × 2 × 3 = 12.
75. 75. HOW TO FIND OUT HCF USING DIVISION METHOD (SHORTCUT)
76. 76. To find out HCF of two given numbers using division method Step 1: Divide the larger number by the smaller number Step 2: Divisor of step 1 is divided by its remainder Step 3: Divisor of step 2 is divided by its remainder. Continue this process till we get 0 as remainder. Step 4: Divisior of the last step is the HCF.
77. 77. To find out HCF of three given numbers using division method Step 1: Find out HCF of any two numbers. Step 2: Find out the HCF of the third number and the HCF obtained in step 1 Step 3: HCF obtained in step 2 will be the HCF of the three numbers
78. 78. To find out HCF of more than three numbers using division method In a similar way as explained for three numbers, we can find out HCF of more than three numbers also
79. 79. Example 1 : Find out HCF of 60 and 75
80. 80. Hence HCF of 3556 and 3224 = 4
81. 81. How to calculate LCM and HCF for fractions Least Common Multiple (L.C.M.) for fractions LCM for fractions = LCM of Numerators HCF of Denominators Example 1: Find out LCM of 1/2, 3/8, 3/4 LCM = LCM (1, 3, 3) / HCF (2, 8, 4)=32 Example 2: Find out LCM of 2/5, 3/10 LCM = LCM (2, 3) / HCF (5, 10)=65
82. 82. Highest Common Multiple (H.C.F) for fractions HCF for fractions = HCF of Numerators / LCM of Denominators Example 1: Find out HCF of 3/5, 6/11, 9/20 HCF = HCF (3, 6, 9) / LCM (5, 11, 20)=3220 Example 2: Find out HCF of 4/5, 2/3 HCF = HCF (4, 2) / LCM (5, 3)=215
83. 83. How to calculate LCM and HCF for Decimals Step 1 : Make the same number of decimal places in all the given numbers by suffixing zero(s) in required numbers as needed. Step 2 : Now find the LCM/HCF of these numbers without decimal. Step 3 : Put the decimal point in the result obtained in step 2 leaving as many digits on its right as there are in each of the numbers.
84. 84. Example1 : Find the LCM and HCF of .63, 1.05, 2.1 Step 1 : Make the same number of decimal places in all the given numbers by suffixing zero(s) in required numbers as needed. i.e., the numbers can be written as .63, 1.05, 2.10 Step 2 : Now find the LCM/HCF of these numbers without decimal. Without decimal, the numbers can be written as 63, 105 and 210 . LCM (63, 105 and 210) = 630 HCF (63, 105 and 210) = 21 Step 3 : Put the decimal point in the result obtained in step 2 leaving as many digits on its right as there are in each of the numbers. i.e., here, we need to put decimal point in the result obtained in step 2 leaving two digits on its right. i.e., the LCM (.63, 1.05, 2.1) = 6.30 HCF (.63, 1.05, 2.1) = .21
85. 85. HOW TO COMPARE FRACTIONS?
86. 86. Type 1 : Fractions with same denominators. Compare 3/5 and 1/5 These fractions have same denominator. So just compare the numerators. Bigger the numerator, bigger the number. 3 > 1. Hence 3/5>1/5 Example 2: Compare 2/7 and 3/7 and 8/7 These fractions have same denominator. So just compare the numerators. Bigger the numerator, bigger the number. 8 > 3 > 2. Hence 8/7>3/7>2/7
87. 87. Type 2 : Fractions with same numerators. Example 1: Compare 3/5 and 3/8 These fractions have same numerator. So just compare the denominators. Bigger the denominator, smaller the number. 8 > 5. Hence 3/8<3/5 Example 2: Compare 7/8 and 7/2 and 7/5 These fractions have same numerator. So just compare the denominators. Bigger the denominator, smaller the number. 8 > 5 > 2. Hence 7/8<7/5<7/2
88. 88. Type 3 : Fractions with different numerators and denominators. Example 1: Compare 3/5 and 4/7 To compare such fractions, find out LCM of the denominators. Here, LCM(5, 7) = 35 Now , convert each of the given fractions into an equivalent fraction with 35 (LCM) as the denominator. The denominator of 3/5 is 5. 5 needs to be multiplied with 7 to get 35. Hence, 3/5=3×7/5×7=2135 The denominator of 4/7 is 7. 7 needs to be multiplied with 5 to get 35. Hence, 4/7=4×5/7×5=2035 21/35>20/35 Hence, 3/5>4/7 Or Convert the fractions to decimals 3/5=.6 4/7=.5... (Need not find out the complete decimal value; just find out up to what is required for comparison. In this case the first digit itself is sufficient to do the comparison) .6 > .5... Hence, 3/5>4/7
89. 89. CO-PRIME NUMBERS OR RELATIVELY PRIME NUMBERS
90. 90. Two numbers are said to be co-prime (also spelled coprime) or relatively prime if they do not have a common factor other than 1. i.e., if their HCF is 1. Example1: 3, 5 are co-prime numbers (Because HCF of 3 and 5 = 1) Example2: 14, 15 are co-prime numbers (Because HCF of 14 and 15 = 1) A set of numbers is said to be pair wise co-prime (or pair wise relatively prime) if every two distinct numbers in the set are co-prime
91. 91. Example1 : The numbers 10, 7, 33, 13 are pair wise co-prime, because HCF of any pair of the numbers in this is 1. HCF (10, 7) = HCF (10, 33) = HCF (10, 13) = HCF (7, 33) = HCF (7, 13) = HCF (33, 13) = 1. Example2 : The numbers 10, 7, 33, 14 are not pair wise co-prime because HCF(10, 14) = 2 ≠ 1 and HCF(7, 14) = 7 ≠ 1. If a number is divisible by two co-prime numbers, then the number is divisible by their product also. Example 3, 5 are co-prime numbers (Because HCF of 3 and 5 = 1) 14325 is divisible by 3 and 5. 3 × 5 = 15 Hence 14325 is divisible by 15 also If a number is divisible by more than two pair wise co-prime numbers, then the number is divisible by their product also.
92. 92. Example1 : The numbers 3, 4, 5 are pair wise co-prime because HCF of any pair of numbers in this is 1 1440 is divisible by 3, 4 and 5. 3 × 4 × 5 = 60. Hence 1440 is also divisible by 60 Example2 The numbers 3, 4, 9 are not pair wise co-prime because HCF (3, 9 ) = 3 ≠ 1 1440 is divisible by 3, 4 and 9. 3 X 4 X 9 = 108. However 1440 is not divisible by 108 as 3, 4, 9 are not pair wise co-prime
93. 93. Important Points to Note on LCM and HCF Product of two numbers = Product of their HCF and LCM. Example LCM (8, 14) = 56 HCF (8, 14) = 2 LCM (8, 14) × HCF (8, 14) = 56 × 2 = 112 8 × 14 = 112 Hence LCM (8, 14) × HCF (8, 14) = 8 × 14
94. 94. SQUARING
95. 95. GENERAL SHORTCUT METHOD FOR FINDING SQUARES
96. 96. You can find square of any number in the world with this method. Let’s say the number is two digit number. i.e. AB. So B is units digit and A is tens digit. Step 1: Find Square of B Step 2: Find 2×A×B Step 3: Find Square of A Let’s take an example.
97. 97. We want to find square of 37. Step 1: Find square of 7. Square of 7 = 49. So write 9 in the answer and 4 as carry to the second step. Step 2: Find 2×(3×7) 2 × (3 × 7) = 42. 42 + 4(Carry) = 46. Write 6 in the answer and 4 as a carry to the third step. Step 3: Find square of 3 Square of 3 = 9 9 + 4(Carry) = 13. Write 13 in the answer. So the answer is 1369.
98. 98. Now, If the number is of three digit i.e. ABC Here C is unit’s digit, B is ten’s digit and A is hundredth digit. Step 1: Find Square of C Step 2: Find 2 × (B × C) Step 3: Find 2 × (A × C) + B2 (NOTE: You may observe that in odd number of digit case, we are multiplying end two digits with 2 (here: A and C) and squaring single digit (here B). Step 4: Find 2 × (A × B) (NOTE: You may observe that whenever there are double digits, we are multiplying it with 2. And whenever there is single digit, we are squaring it.) Step 5: Find square of A (NOTE: Here is single digit, so we are squaring it.) Let’s take an example.
99. 99. Find square of 456. Step 1: Find square of 6. Square of 6 = 36. So write 6 in the answer and 3 as a carry to the second step. Step 2: Find 2 × (5 × 6) 2 × (5 × 6) = 60 60 + 3(Carry) = 63 Write 3 in the answer and 6 as a carry to the third step. Step 3: Find 2 × (4 × 6) + 52 2 × (4 × 6) + 52 = 73 73 + 6(Carry) = 79 Write 9 in the answer and 7 as a carry to the fourth step. Step 4: Find 2 × (4 × 5) 2 × (4 × 5) = 40 40 + 7 = 47 Write 7 in the answer and 4 as a carry to the fifth step. Step 5: Find square of 4 Square of 4 = 16 16 + 4(Carry) = 20 Write 20 in the answer. So 4562 = 207936.
100. 100. Now, If the number is of four digit i.e. ABCD Here D is unit’s digit, C is ten’s digit, B is hundredth digit and A is thousands digit. Step 1: Find Square of D Step 2: Find 2 × (C × D) Step 3: Find 2 × (B × D) + C2 (NOTE: You may observe that in odd number of digit case, we are multiplying end two digits with 2 (here: B and D) and squaring single remaining digit (here C). Step 4: Find 2 × (A × D) + 2 × (B × C) (NOTE: You may observe that where ever there is even digits, we are multiplying end two digits with 2 + remaining two digits with 2.) Step 5: Find 2 × (A × C) + B2 (NOTE: You may observe that in odd number of digit case, we are multiplying end two digits with 2 (here: A and C) and squaring single remaining digit (here B). Step 6: Find 2 × (A × B) (NOTE: You may observe that here even digits so we are multiplying them with 2, and no remaining digits so we are not adding anything.) Step 7: Find square of A Let’s take an example.
101. 101. Square Of The Numbers With Unit Digit As 5 This method is to find square of the numbers which has unit’s digit as 5. i.e.: 25, 45, 65, etc. You can find square of these numbers by three easy steps. Step 1: Multiply ten’s digit with its next number. Step 2: Find square of unit’s digit. i.e.: Square of 5. Step 3: Write answers of step 1 and step 2 to together or side by side. Let’s take examples.
102. 102. Find square of 35 Step 1: Multiply ten’s digit with its next number. 3 × ( 3 + 1 ) = 3 × 4 = 12 Step 2: Find square of unit’s digit. i.e.: Square of 5. Square of 5 = 25 Step 3: Write answers of step 1 and step 2 to together. Answer = 1225
103. 103. Find square of 65 Step 1: Multiply ten’s digit with its next number. 6 × ( 6 + 1 ) = 6 × 7 = 42 Step 2: Find square of unit’s digit. i.e.: Square of 5. Square of 5 = 25 Step 3: Write answers of step 1 and step 2 to together. Answer = 4225
104. 104. Find square of 95 Step 1: Multiply ten’s digit with its next number. 9 × ( 9 + 1 ) = 9 × 10 = 90 Step 2: Find square of unit’s digit. i.e.: Square of 5. Square of 5 = 25 Step 3: Write answers of step 1 and step 2 to together. Answer = 9025
105. 105. Find square of 115 Step 1: Multiply ten’s digit with its next number. 11 × ( 11 + 1 ) = 11 × 12 = 132 (Note: We are taking whole 11 as a ten’s digit.) Step 2: Find square of unit’s digit. i.e.: Square of 5. Square of 5 = 25 Step 3: Write answers of step 1 and step 2 to together. Answer = 13225
106. 106. Find square of 215 Step 1: Multiply ten’s digit with its next number. 21 × ( 21 + 1 ) = 21 × 22 = 462 (Note: We are taking whole 21 as a ten’s digit.) Step 2: Find square of unit’s digit. i.e.: Square of 5. Square of 5 = 25 Step 3: Write answers of step 1 and step 2 to together. Answer = 46225
107. 107. Square Of The Numbers In 50s This method is used to find square of the numbers in 50s i.e. numbers from 51 to 59. You can find square of these numbers in three simple steps. Step 1: Add 25 to the unit’s digit. Step 2: Square the unit’s digit. Step 3: Write the answers of step 1 and step 2 together or side by side. Let’s take an example.
108. 108. Square of 56 Step 1: Add 25 to the unit’s digit 6 + 25 = 31 Step 2: Square the unit’s digit 62 = 36 Step 3: Write the answers of step 1 and step 2 together. Answer = 3136
109. 109. Square of 59 Step 1: Add 25 to the unit’s digit 9 + 25 = 34 Step 2: Square the unit’s digit 92 = 81 Step 3: Write the answers of step 1 and step 2 together. Answer = 3481
110. 110. Square of 53. Step 1: Add 25 to the unit’s digit. 3 + 25 = 28 Step 2: Square the unit’s digit. 32 = 9 Step 3: Write answers of step 1 and step 2 together. Answer: 2809 (NOTE: Whenever square of unit’s digit is on only single digit then we are adding 0 before it.)
111. 111. Square of 52 Step 1: Add 25 to the unit’s digit 2 + 25 = 27 Step 2: Square the unit’s digit 22 = 4 Step 3: Write the answers of step 1 and step 2 together. Answer = 2704 (NOTE: Whenever square of unit’s digit is on only single digit then we are adding 0 before it.)
112. 112. Square The Number If You Know Square Of Previous Number This method is to find square of the number if you know square of the previous number. You can find answers in three simple steps. Step 1: Find square of the previous number which is known. Step 2: Multiply the number being squared by 2 and subtract 1. Step 3: Add Step 1 and Step 2 Let’s take some examples.
113. 113. Find square of 31. Step 1: Find square of previous number (30) which is known. 302 = 900 Step 2: Multiply the number being squared (31) by 2 and subtract 1. (31 × 2) – 1 = 62 – 1 = 61 Step 3: Add Step 1 and Step 2 900 + 61 = 961
114. 114. Find square of 26. Step 1: Find square of previous number (25) which is known. 252 = 625 Step 2: Multiply the number being squared (31) by 2 and subtract 1. (26 × 2) – 1 = 52 – 1 = 51 Step 3: Add Step 1 and Step 2 625 + 51 = 676
115. 115. Find square of 81. Step 1: Find square of previous number (80) which is known. 802 = 6400 Step 2: Multiply the number being squared (31) by 2 and subtract 1. (81 × 2) – 1 =162 – 1 = 161 Step 3: Add Step 1 and Step 2 6400 + 161 = 6561
116. 116. SHORTCUT METHOD FOR FINDING CUBE
117. 117. If a2 = b, b√=a (i.e., square root of b is a.) · Square Root Examples 22=4and4√=2 52=25and25−−√=5 If a3 = b, b√3=a (i.e., cube root of b is a.) Cube Root Examples 23=8and8√3=2 53=125and125−−−√3=5 If a2 = b, b√=a (i.e., square root of b is a.) If a3 = b, b√3=a (i.e., cube root of b is a.)
118. 118. BASICS CONCEPTS AND FORMULAS - SIMPLIFICATION
119. 119. Order of Operations - BODMAS Rule What is BODMAS Rule? BODMAS rule defines the correct sequence in which operations are to be performed in a given mathematical expression to find its value. In BODMAS, B = Bracket, O = Order (Powers, Square Roots, etc.) DM = Division and Multiplication (left-to-right) AS = Addition and Subtraction (left-to-right)
120. 120. This means, to simplify an expression, the following order must be followed. Do operations in Brackets first, strictly in the order (), {} and [] Evaluate exponents (Powers, Roots, etc.) Perform division and multiplication, working from left to right. (division and multiplication rank equally and done left to right). Perform addition and subtraction, working from left to right. (addition and subtraction rank equally and done left to right).
121. 121. In India, the term BODMAS is used whereas in USA the acronym PEMDAS is used. The full form of PEMDAS is "Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction". Both are correct and only difference is that these are used in different parts of the world. Some other variations which are used across the world to represent the same concept are BIDMAS, ERDMAS, PERDMAS and BPODMAS.
122. 122. Examples 12 + 22 ÷ 11 × (18 ÷ 3)2 - 10 = 12 + 22 ÷ 11 × 62 - 10 (∵ Brackets first) = 12 + 22 ÷ 11 × 36 - 10 (∵ exponents) = 12 + 2 × 36 - 10 = 12 + 72 - 10 (∵ division and multiplication, left to right) = 84 - 10 = 74 (∵ Addition and Subtraction, left to right) 4 + 10 - 3 × 6 / 3 + 4 = 4 + 10 - 18/3 + 4 = 4 + 10 - 6 + 4 (∵ division and multiplication, left to right) = 14 - 6 + 4 = 8 + 4 = 12 (∵ Addition and Subtraction, left to right)
123. 123. Modulus of a Real Number For any real number x, the absolute value or modulus of x is denoted by |x| and is defined as |x|={x−xif x≥0if x<0 Hence, the Modulus(absolute value) of x is always either positive or zero, but never negative Example |8| = |-8| = 8
124. 124. FRACTIONS
125. 125. A fraction is an expression that indicates the quotient of two quantities. Examples of fractions: etc. A fraction has two parts, Numerator and Denominator. Numerator is the number at the top of the fraction and denominator is the number at the bottom of the fraction. For example, in the fraction ¼ , 1 is numerator and 4 is denominator The denominator of a fraction cannot be zero
126. 126. Type of Fractions i. Common Fraction A common fraction (also known as Vulgar fraction and simple fraction) is a fraction in which both numerator and denominator are integers (As with other fractions, the denominator cannot be zero) Examples: 2/3 , 7/12 , −6/5 i. Decimal Fraction A decimal fraction is a fraction in which denominator is an integer power of ten. (The term decimals are commonly used to refer decimal fractions). Generally, a decimal fraction is expressed using decimal notation and its denominator is not mentioned explicitly Examples: 1/10 = .1 1/100 = .01 17/100 = .17 4/10000 = .0004 121/100 = 1.21 Decimal fractions can also be expressed using scientific notation with negative exponents Example: 4.193 × 10−7 is a decimal fraction which represents .0000004193
127. 127.  Generally, a zero is added to the left of the decimal point if a whole number is absent to draw the attention to the fact that the number is a decimal fraction. For example, .1 and 0.1 represents the same  Annexing Zeros to the extreme right of a decimal fraction Annexing zeros to the extreme right of a decimal fraction does not change its value. Examples 0.4 = 0.40 = 0.400 = 0.4000, etc. 1.21 = 1.210 = 1.2100 = 1.21000, etc  If the numerator and denominator contain the same number of decimal places, we can remove the decimal sign Example 1.2/4.8=12/48=1/4 0.03/0.24=3/24=1/8  Conversion of a Decimal into Common Fraction Put 1 in the denominator under the decimal point and annex with it as many zeros as is the number of digits after the decimal point. Now, remove the decimal point and reduce the fraction to its lowest terms. Examples 0.5 = 5/10=1/2 .75 = 75/100=3/4 1.25 = 125/100=5/4
128. 128. IMPORTANT CONCEPTS AND FORMULAS - ALGEBRA (INCLUSIVE OF SURDS AND INDICES )
129. 129.  a(b+c)=ab+ac(Distributive Law)  (a+b)2=a2+2ab+b2  (a−b)2=a2−2ab+b2  (a+b)2+(a−b)2=2(a2+b2)  (a+b)3=a3+3a2b+3ab2+b3=a3+3ab(a+b)+b3  (a−b)3=a3−3a2b+3ab2−b3=a3−3ab(a−b)−b3  a2−b2=(a−b)(a+b)  a3+b3=(a+b)(a2−ab+b2)  a3−b3=(a−b)(a2+ab+b2)  an−bn=(a−b)(an−1+an−2b+an−3b2+ ... +bn−1)  (a+b+c)2=a2+b2+c2+2(ab+bc+ca)  (a+b)(a+c)=a2+(b+c)a+bc
130. 130.  an=a.a.a ... (n times)  am.an...ap=am+n+...+p  am.an=am+n  am/an=am−n  (am)n=amn=(an)m  (ab)n=anbn  (a/b)n=an / bn  a−n=1/an  an=1/a−n  a0=1 where a∈ R, a ≠0  ap/q=ap−−√q  if am=an where a≠0 and a≠±1, then m=n  if an=bn where n≠0, then a=±b
131. 131.  (x+a)2=x2+2ax+a2  (x−a)2=x2−2ax+a2  (x+a)(x+b)=x2+(a+b)x+ab  x2−a2=(x−a)(x+a)  (xn+1) is completely divisible by (x + 1) when n is odd  (xn+an) is completely divisible by (x + a) when n is odd  (xn−an) is completely divisible by (x−a) for every natural number n  (xn−an) is completely divisible by (x+a) when n is even
132. 132. SURDS AND IMPORTANT PROPERTIES
133. 133. Surds Let a be a rational number and n be a positive integer such that a√n is irrational. Then a√n is called a surd of order n. Examples : 3√ is a surd of order 2 7√3 is a surd of order 3 43√ is a surd of order 2 Please note that numbers like 9√, 27−−√3 etc are not surds because they are not irrational numbers Every surd is an irrational number. But every irrational number is not a surd. (eg : π , e etc are not surds though they are irrational numbers.)
134. 134. Quadratic Surds A surd of order 2 is called a quadratic surd Examples : 2√,3√,(3+5√), etc.
135. 135. Laws of Indices: i. am * an = am+n ii. am/an = am-n iii. (am)n =amn iv. (ab)n = anbn v. (a/b)n = an/bn vi. a0= 1
137. 137. A. Quadratic Equations A quadratic equation is a second degree univariate polynomial equation. A quadratic equation can be written as (general form or standard form of quadratic equation) ax2+bx+c=0 where x is a variable, a, b and c are constants and a≠0 Example : x2+5x+6=0 (Please note that if a=0, equation becomes a linear equation) The solutions of a quadratic equation are called its roots.
138. 138. B. How to Solve Quadratic Equations There are many methods to solve Quadratic equations. Quadratic equations can be solved by factoring, completing the square, graphing, Newton's method, and using the quadratic formula. We can go through some of the popular methods for solving quadratic equations here.
139. 139. How to Solve Quadratic Equations By Factoring Step 1: Write the quadratic equation in standard form. i.e. , bring all terms on the left side of the equal sign and 0 on the right side of the equal sign. Step 2: Factor the left side of the equation Step 3: Equate each factor to 0 and solve the equations Factoring is one of the fastest ways for solving quadratic equations. The concept will be clear from the following examples.
140. 140. Example1 : Solve the Quadratic Equation x2+4x−11=x−1 Step 1 : Write the quadratic equation in standard form. x2+4x−11=x−1 ⇒x2+4x−11−x+1=0 ⇒x2+3x−10=0 Step 2 : Factor the left side of the equation. We know that x2+(a+b)x+ab=(x+a)(x+b). We will use the same concept for factoring. Assume x2+3x−10=x2+(a+b)x+ab We need to find out a and b such that a + b = 3 and ab = -10 a = +5 and b = -2 satisfies the above condition. Hence x2+3x−10=(x+5)(x−2) x2+3x−10=0 ⇒(x+5)(x−2)=0 Step 3 :Equate each factor to 0 and solve the equations (x + 5)(x - 2) = 0 => (x + 5) = 0 or (x - 2) = 0 => x = -5 or 2 Hence, the solutions of the quadratic equation x2+4x−11=x−1 are x = -5 and x = 2. (In other words, x = -5 and x = 2 are the roots of the quadratic equation x2+4x−11=x−1
141. 141. Example2 : Solve the Quadratic Equation x2−7x+10=0 This equation is already in the standard form. Hence let's go to step 2 Step 2 : Factor the left side of the equation. Here we need to find out a and b such that a + b = -7 and ab = +10 a = -5 and b = -2 satisfies the above condition. Hence x2−7x+10=(x−5)(x−2) Hence, x2+3x−10=0 ⇒(x−5)(x−2)=0 Step 3 :Equate each factor to 0 and solve the equations (x - 5)(x - 2) = 0 => (x - 5) = 0 or (x - 2) = 0 => x = 5 or 2 Hence, the solutions of the quadratic equation x2−7x+10=0 are x = 5 and x = 2. (In other words, x = 5 and x = 2 are the roots of the quadratic equation x2−7x+10=0)
142. 142. Example3 : Solve the Quadratic Equation 3x2−14x+8=0 This equation is already in the standard form. Hence let's go to step 2 Step 2 : Factor the left side of the equation. Here we need to follow a slightly different approach for factoring because here the coefficient of x2 ≠ 1 (whereas In example 1 and example 2, the coefficient of x2 was 1) Hence, to factor 3x2−14x+8, we need to do the followings 1. Product of the second degree term and the constant. i.e., 3x2×8=24x2 2. We got product as 24x2 and have middle term as -14x. From this , it is clear that we can take -12x and -2x such that their sum is -14x and product is 24x2 Hence, 3x2−14x+8=0 = 3x2−12x−2x+8=3x(x−4)−2(x−4)=(x−4)(3x−2) Hence, 3x2−14x+8=0 ⇒(x−4)(3x−2)=0 Step 3 :Equate each factor to 0 and solve the equations (x - 5)(x - 2) = 0 => (x - 4) = 0 or (3x - 2) = 0 => x = 4 or 23. Hence, the solutions of the quadratic equation 3x2−14x+8=0 are x = 4 and x = 23 (In other words, x = 4 and x = 23 are the roots of the quadratic equation 3x2−14x+8=0
143. 143. Example4 : Solve the Quadratic Equation 6x2−17x+12=0 This equation is already in the standard form. Hence let's go to step 2 Step 2 : Factor the left side of the equation. coefficient of x2 ≠ 1 Hence, to factor 6x2−17x+12, we need to do the followings 1. Product of the second degree term and the constant. i.e., 6x2×12=72x2 2. We got product as 72x2 and have middle term as -17x. From this , it is clear that we can take -8x and -9x such that their sum is -17x and product is 72x2 Hence, 6x2−17x+12=6x2−8x−9x+12=2x(3x−4)−3(3x−4)=(3x−4)(2x−3) Hence, 6x2−17x+12=0 ⇒(3x−4)(2x−3)=0 Step 3 :Equate each factor to 0 and solve the equations (3x - 4)(2x - 3) = 0 => (3x - 4) = 0 or (2x - 3) = 0 => x = 43 or 32. Hence, the solutions of the quadratic equation 6x2−17x+12=0 are x = 43 and x = 32. (In other words, x = 43 and x = 32 are the roots of the quadratic equation 6x2−17x+12=0
144. 144. (SPECIAL FACTORIZATION EXAMPLES)
145. 145. Example5 : Solve the Quadratic Equation x2−9=0 This equation is already in the standard form. Hence let's go to step 2 Step 2 : Factor the left side of the equation. Here,x2−9 is in the form a2−b2 where a = x and b = 3 We know that a2−b2=(a−b)(a+b)We will use the same concept for factoring this. x2−9=x2−32=(x−3)(x+3) Hence, x2−9=0 ⇒(x−3)(x+3)=0 Step 3 :Equate each factor to 0 and solve the equations (x - 3)(x + 3) = 0 => x = 3 or -3 Hence, the solutions of the quadratic equation x2−9=0 are x = 3 and x = -3 (In other words, x = 3 and x = -3 are the roots of the quadratic equation x2−9=0
146. 146. Example6: Solve the Quadratic Equation x2+6x+9=0 This equation is already in the standard form. Hence let's go to step 2 Step 2 : Factor the left side of the equation. Here,x2+6x+9 is in the form a2+2ab+b2 where a = x and b = 3 We know that a2+2ab+b2=(a+b)2 We will use the same concept for factoring this. x2+6x+9=x2+2×x×3+32=(x+3)2 Hence, x2+6x+9=0 ⇒(x+3)2=0 Step 3 : (x+3)2=0 => x + 3 = 0 => x = -3 Hence, the solution of the quadratic equation x2+6x+9=0 is x = -3 (In other words, x = -3 is the root of the quadratic equation x2+6x+9=0
147. 147. Example7 : Solve the Quadratic Equation x2−6x+9=0 This equation is already in the standard form. Hence let's go to step 2 Step 2 : Factor the left side of the equation. Here,x2−6x+9 is in the form a2−2ab+b2 where a = x and b = 3 We know that a2−2ab+b2=(a−b)2 We will use the same concept for factoring this. x2−6x+9=x2−2×x×3+32=(x−3)2 Hence, x2+6x+9=0 ⇒(x−3)2=0 Step 3 : (x−3)2=0 => x - 3 = 0 => x = 3 Hence, the solution of the quadratic equation x2−6x+9=0 is x = 3 (In other words, x = 3 is the root of the quadratic equation x2−6x+9=0
148. 148. CYCLICITY OF NUMBERS
149. 149. Types of questions based on cyclicity of numbers There are mainly 3 categories of questions which fall under cyclicity of numbers, that include How to find units digit of ab How to find units digit of ab * cd * ef How to find units digit of abc Let's see how to solve these questions using the concept of cyclicity of numbers, with examples.
150. 150. Find units digit of ab Given ab, units place digit of the result depends on units place digit of a and the divisibility of power b. Consider powers of 2 As we know, 21 =2 22 = 4 23 = 8 24 = 16 25 = 32 26 = 64 27 = 128.. and so on What do you observe here? We can see that the units place digit for powers of 2 repeat in an order: 2, 4, 8, 6. So the "cyclicity" of number 2 is 4 (that means the pattern repeats after 4 occurrences) and the cycle pattern is 2, 4, 8, 6. From this you can see that to find the units place digit of powers of 2, you have to divide the exponent by 4. Let's check the validity of above formula with an example.
151. 151. Find the units place digit of 299? Using the above observation of cyclicity of powers of 2, divide the exponent by 4 99/4 gives reminder as 3. That means, units place digit of 299 is the 3rd item in the cycle which is 8.
152. 152. Shortcuts to solve problems related to units place digit of ab Case 1: If b is a multiple of 4 If a is an even number, ie: 2, 4, 6 or 8 then the units place digit is 6 If a is an odd number, ie: 1, 3, 7 or 9 then the units place digit is 1 Case 2: If b is not a multiple of 4 Let r be the reminder when b is divided by 4, then units place of ab will be equal to units place of ar Let's call it the general rule of cyclicity. Using these rules you can solve all the problems related to cyclicity of numbers.
153. 153. Number ^1 ^2 ^3 ^4 Cyclicity 2 2 4 8 6 4 3 3 9 7 1 4 4 4 6 4 6 2 5 5 5 5 5 1 6 6 6 6 6 1 7 7 9 3 1 4 8 8 4 2 6 4 9 9 1 9 1 2
154. 154. Find units digit for numbers of the form ab * cd Question of this type is similar to first category. First find the unit digit of ab and cd separately. Let the answers be x and y Then unit digit of ab * cd = units digit of x * y
155. 155. Find units digit of abc Questions of this type have to be approached on a case by case basis. Case 1: If cyclicity of units place digit of a is 4 then we have to divide the exponent of a by 4 and find out the remainder. Depending on the value of remainder we can apply the general rule of cyclicity given above and reach the solution. Case 2: If cyclicity of units place digit of a is 2, only extra information we need to find is if the exponent will be even or odd. Then we can apply the general rule of cyclicity given above and reach the solution. Let's see application of these rules with help of examples
156. 156. Find the units place digit of 24344 Here cyclicity of units place digit is 4 (Units place digit is 2, from the above table we can see the cyclicity of 2 is 4). Hence case 1 is applicable. Now we have to find the remainder when exponent of 2 is divided by 4, that is the remainder when 4344 is divided by 4. Remainder of 4344/4 = Remainder of (44 – 1)44/4 Using the binomial theorem, (as explained in number system tutorial) we can see that there is only one term in the expansion of (44 – 1)44 which is not divisible by 4. The term is 144/4 Remainder of 144/4 = 1 Now we can apply the general rules of cyclicity, (since reminder is 1, case 2 of general rule of cyclicity is applicable) which says, units place of 24344 = units place of 21 = 2.
157. 157. Find the units place digit of 293945 Here cyclicity of units place digit of a, that is cyclicity of 9 is 2. Hence case 2 is applicable Next step is we need to find if 3945 is even or odd 3945 will always result in an odd number, because we are multiplying an odd number 39, odd number of times (45). Now we can apply the general rule of cyclicity described in the first section. General rule says, first check if exponent is divisible by 4. Since the exponent here is odd, it's not divisible by 4. Again according to general rule, if exponent is not divisible by 4, find the reminder when exponent is divided by 4. Since the exponent is odd here, the possible reminders when it's divided by 4 are 1 and 3. This means, the units place digit of number will be 1st or 3rd element in the cyclicity of units place digit of a. In this case 1st and 3rd elements of 9 (units place digit of 29) is 9. Hence units place digit of 2293945 is 9.

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Tips to prepare for Fundamentals of Quantitative Aptitude Number Properties LCM, HCF Divisibility Fractions & Decimals, square Square Roots cyclicity with shortcut tricks

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