Tiêu đề MATH basic tricks.pdf ✅

NUMBER SYSTEM 1. Method to multiply 2-digit number. (i) AB × CD = AC / AD + BC / BD 35 × 47 = 12 / 21 + 20 / 35 = 12 / 41 / 35 = 1645 (ii) AB × AC = A2 / A (B + C) / BC 74 × 76 = 72 / 7(4 + 6) / 4 × 6 = 49 / 70 / 24 = 49 / 70 / 24 = 5624 (iii) AB × CC = AC / (A + B)C / BC = 35 × 44 = 3 × 4 / (3 + 5) × 4 / 5 × 4 = 12 / 32 / 20 = 12 / 32 / 20 = 1540 2. Method to multiply 3-digit no. ABC × DEF = AD / AE + BD / AF + BE + CD / BF + CE / CF 456 × 234 = 4 × 2 / 4 × 3 + 5 × 2 / 4 × 4 + 5 × 3 + 6 × 2 / 5 × 4 + 6 × 3 / 6 × 4 = 8 / 12 + 10 / 16 + 15 + 12 / 20 + 18 / 24 = 8 / 22 /43 / 38 / 24 = 106704 3. If in a series all number contains repeating 7. To find their sum, we start from the left multiply 7 by 1, 2, 3, 4, 5 & 6. Look at the example below. 777777 + 77777 + 7777 + 777 + 77 + 7 = ? = 7 × 1 / 7 × 2 / 7 × 3 / 7 × 4 / 7 × 5 / 7 × 6 = 7 / 14 / 21 / 28 / 35 / 42 = 864192 4. 0.5555 + 0.555 + 0.55 + 0.5 = ? To find the sum of those number in which one number is repeated after decimal, then first write the number in either increasing or decreasing order. Then -find the sum by using the below method. 0.5555 + 0.555 + 0.55 + 0.5 = 5 × 4 / 5 × 3 / 5 × 2 / 5 × 1 = 20 / 15 / 10 / 5 = 2.1605 5 Those numbers whose all digits are 3. (33)2 = 1089 Those number. in which all digits are number is 3 two or more than 2 times repeated, to find the square of these number, we repeat 1 and 8 by (n – 1) time. Where n ® Number of times 3 repeated. (333)2 = 110889 (3333)2 = 11108889 6. Those number whose all digits are 9. (99)2 = 9801 (999)2 = 998001 (9999)2 = 99980001 (99999)2 = 9999800001 7. Those number whose all digits are 1. A number whose one’s, ten’s, hundred’s digit is 1 i.e., 11, 111, 1111, .... In this we count number of digits. We write 1, 2, 3, ..... in their square the digit in the number, then write in decreasing order up to 1. 112 = 121 1112 = 12321 11112 = 1234321


S-4 101 Shortcuts in Quantitative Aptitude 22. Divisible by 7 : We use osculator (– 2) for divisibility test. 99995 : 9999 – 2 × 5 = 9989 9989 : 998 – 2 × 9 = 980 980 : 98 – 2 × 0 = 98 Now 98 is divisible by 7, so 99995 is also divisible by 7. 23. Divisible by 11 : In a number, if difference of sum of digit at even places and sum of digit at odd places is either 0 or multiple of 11, then no. is divisible by 11. For example, 12342 ̧ 11 Sum of even place digit = 2 + 4 = 6 Sum of odd place digit = 1 + 3 + 2 = 6 Difference = 6 – 6 = 0 \ 12342 is divisible by 11. 24. Divisible by 13 : We use (+ 4) as osculator. e.g., 876538 ̧ 13 876538: 8 × 4 + 3 = 35 5 × 4 + 3 + 5 = 28 8 × 4 + 2 + 6 = 40 0 × 4 + 4 + 7 = 11 1 × 4 + 1 + 8 = 13 13 is divisible by 13. \ 876538 is also divisible by 13. 25. Divisible by 17 : We use (– 5) as osculator. e.g., 294678: 29467 – 5 × 8 = 29427 27427: 2942 – 5 × 7 = 2907 2907: 290 – 5 × 7 = 255 255: 25 – 5 × 5 = 0 \ 294678 is completely divisible by 17. 26. Divisible by 19 : We use (+ 2) as osculator. e.g: 149264: 4 × 2 + 6 = 14 4 × 2 + 1 + 2 = 11 1 × 2 + 1 + 9 = 12 2 × 2 + 1 + 4 = 9 9 × 2 + 1 = 19 19 is divisible by 19 \ 149264 is divisible by 19. 27. HCF (Highest Common factor) There are two methods to find the HCF– (a) Factor method (b) Division method (i) For two no. a and b if a < b, then HCF of a and b is always less than or equal to a . (ii) The greatest number by which x, y and z completely divisible is the HCF of x, y and z. (iii) The greatest number by which x, y, z divisible and gives the remainder a, b and c is the HCF of (x –a), (y–b) and (z–c). (iv) The greatest number by which x, y and z divisible and gives same remainder in each case, that number is HCF of (x–y), (y–z) and (z–x). (v) H.C.F. of , a c b d and H.C.M. of (a, c, e) L.C.M. of (b, d, f) e f = 28. LCM (Least Common Multiple) There are two methods to find the LCM– (a) Factor method (b) Division method (i) For two numbers a and b if a < b, then L.C.M. of a and b is more than or equal to b. (ii) If ratio between two numbers is a : b and their H.C.F. is x, then their L.C.M. = abx.