Some More Useful Tricks for SSC

#dailyquizadda #SquareTricksandmore #ssc


1. Squaring a number with lightning fast speed

       Well you might have come across various tricks to find the square of a number, but this trick is the simplest and the fastest.
Firstly memorize squares of numbers upto 25. I think many of you might be knowing it already, but for those who don't, please mug it. Now this trick will help you in squaring numbers upto 125 and trust me, for SSC you need this much only.

(a) For numbers close to 50

      (i) Numbers less than 50






(ii) Numbers greater than 50

              The process is almost same -  Compute (N - 50)
              Add this difference to 25
              E.g. 56*56=
               1. 56 - 50 = 6
               2. Last two digits = 6 * 6 = 36
               3. First two digits = 25 + 6 = 31
                    Therefore, 56 * 56 = 3136

(b) Numbers close to 100

     (i) Numbers less than 100

      (ii) Numbers greater than 100
              The process is almost same -  Compute (N - 100)
              Add this difference to N
              E.g. 107*107=
              1. 107 - 100 = 7
              2. Last two digits = 7 * 7 = 49
              3. First three digits = 107 + 7 = 114
          Therefore, 107 * 107 = 11449

2. Squaring a number containing only '1'. E.g. 111, 1111, 1111111, etc.        Just count the number of ones. Like in 1111, there are 4 ones. So the square of 1111 is 1234321 (go from 1 to 4 and then back). Similarly, the square of 111111 = 12345654321

3. Finding product of two numbers of the form ab and ac, such that b+c=10. E.g. product of 43 and 47Last two digits = 3*7 = 21
First two digits = 4*(4+1) = 20
Hence 43*47 = 2021

4. Finding product of two numbers close to 100

(a) 103*109

Last two digits = 3*9 = 27

First three digits = (103 + 9) or (109 + 3) = 112

So 103*109 = 11227

(b) 92*97

Last two digits = 8*3 = 24  [100 - 92 = 8 and 100 - 97 = 3]

First two digits = (92 - 3) or (97 - 8) = 89

Hence 92*97 = 8924

Bonus Trick :

In SSC CGL, there is a very common question of nested square roots and there are total 3 cases for it (addition, subtraction and multiplication)

Case 1

In such questions, you should avoid forming quadratic equations because that will be time consuming. Here is the trick : Break the number written inside the square root into n*(n + 1) form. So 12 = 3*4

Answer is (n+1), i.e., 4

Similarly, for 72 (written as 8*9), the the answer will be 9; for 20(written as 4*5) the answer will be 5, and so on. In CGL questions, the number will always break into n*(n + 1) form.

Case 2

If instead of '+' you have '-' (above image), then

72 = 8*9

The answer is n, instead of n+1. So the above expression evaluates to 8.

If instead of + you have * (multiplication), then the answer is the number itself. E.g. - 

Case 3

Simple, isn't it? :)

Direct formulae:




 Special Right-angled Triangle:



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