101 maths tricks and shortcuts pdf download

101 maths tricks and shortcuts pdf download

Today we are sharing a small maths tricks and shortcuts but very useful pdf (e-book) “101 shortcuts in quantitative Aptitude” (Maths). This book is very very helpful for all aspirants whether they want to learn short cut tricks or want to increase their speed in maths. This e-book is very useful for upcoming exams like ssc cgl, SBI clerk 2016, etc.
Name of Book :101 shortcuts in quantitative Aptitude
Format: PDF
No. Of Pages: 20
 
Size : less than 1 MB
Quality: High
   
LINK I : Click Here To Download 101 shortcuts in quantitative Aptitude in pdf for free (Google-Drive)
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LINK III : Click Here for Alternate/Similar File (One Click Direct Download) By Disha Publications
 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
 
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