## 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)

**LINK II :**

**Click Here**for Direct Download G-Drive (One Click Direct Download – wait a bit after clicking)

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

**Sharing is Caring, please share this page.**

Source: Qmaths

Not aggreed

This does interest me