Whenever we do a calculation, there is always a trade-off between speed and accuracy. Here are some tricks derived from Vedic Maths that will help you do faster calculations without compromising accuracy in your competitive exams like JEE Main and JEE Advanced.

## Tips for Faster Calculations

**1. Squaring a number ending with 5**

Multiply the rest of the number leaving the 5 in the unit digit with its successive number and write the result with 25 in the end.

For e.g. 45^{2} can be written as = (4 * (4+1)) 25 = (4*5) 25 = 2025

**2. Difference between two consecutive natural numbers’ square is the sum of the two numbers.** (n+1)^{2 }– n^{2 }= n + (n+1).

Use this fact to find the square of a number if square of its previous natural number is known.

For e.g. 46^{2} = 45^{2} + 45 + 46. We know 45^{2} from the technique (1).

**3. Tip for finding any square**:

- Choose a base closer to the number whose square is to be found.
- Find the deficiency if the number is smacker to its base.
- Add the deficiency to the number.
- Multiply the result with the base.
- Add the product of the square of the deficiency with the result of of the above point.

For e.g. 98^{2}.

- Choose 100 as base.
- Deficiency = 98-100 = -2
- Number + Deficiency = 98 + (-1) = 96
- Multiplying result with base = 96*100 = 9600
- Adding above result with square of deficiency = 9600 + (-2)
^{2}=**9604**which is the result 98^{2}

**4. To multiply by 5, it is easier to add a zero at the end (*10) and divide the result by 2.**

**5. To find the complement of a number (difference from the next highest power of 10)**

Subtract all but the unit digits of the number from 9 and subtract the unit place from 10.

For e.g. to know complement of 32,056 (its difference from 100,000), we do:

9-3 = 6; 9-2 = 7; 9-0 = 9; 9-5 = 4; 10-6 = 4;

The result is **67,944**. Easier than borrowing them 1’s?

**6. To Multiply a 2 digit number by 11, for e.g. 35, write it as 3( )5. Inside the ( ), the sum of the two digits goes in. In this case, 3+5 = 8, hence 8 goes inside ( ). So, 35*11 = 382.**

In case the sum of the two digits is a two digit number, keep the unit digit and add one to the digit preceding the ( ). For e.g. if we are to multiply 48 by 11, 4( )8. Inside ( ), we have 4+8 which is 12. So we place 2 and add 1 to 4 which precedes the ( ). Hence the result is **528**.

The above were just some simple tricks you can apply to for faster calculations. With more practice, one can be sure of the answer and need not verify it by conventional methods.

This is just the tip of the iceberg in the world of Vedic Maths.