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Vedic Maths Tricks

Posted On 2015-06-28

Vedic Maths Tricks

Since ancient times, Maths has not been a favorite subject of many people. Even the simplest of the calculation involving large numbers take a lot of time, even when it is solved by an elder. Children of 4th or 5th grade are not able to tackle their Maths problems very well. For all such people there are Vedic Maths tricks available. Vedic Maths is a solution to all the calculation problems. Vedic Maths was founded by an Indian saint named tirthaji in the earlier 20th century. He coined a total of twelve formulas for basic calculations that reduce the time of calculating simple multiplication or division problems involving large digits. This article has some Vedic Maths tricks that will definitely help you deal with your Maths problems in a better and easier way.


1. To find the square root of a perfect square This method is very handy when it comes to finding square root of large perfect squares. For this, one has to memorize the square of first ten natural numbers. Make a table for it. In the table of squares of the numbers, you will notice that if the unit's digit in the number is 1 then its square root will end with 1 or 9. If the number ends in 4 then its square root will end with 2 or 8. Similarly, if the number ends with 9, its square root will end with either 3 or 7 and if the number ends with 6 its square root will end with 4 or. 6. In case the number has 5 as its unit's digit, the square root will also have 5 at the unit's place and same is the case with 0. If any number ends with a 0, then its square root will also end with a 0.


2. To find the square of a number ending with 5 This is a real simple technique to find the square of any number, big or small that ends with 5. It will take few seconds rather than a long time that is required when you do it with the usual multiplication method. All you need to do is write 25 as the last two digits of the square because we know that 5*5 equals 25. The other number left has to be multiplied by a number greater than it and the answer has to be written before the 25. For e.g. For finding the square of 35, firstly write 25 as the last two digits and then simply multiply 3 by one digit greater i.e. 4, which equals 12 and write 12 before 25 which is the required number. Isn't that easy?


3. Maths subtraction technique This Vedic technique can be applied to subtract any number from a given number of seconds. It’s quite fast and easy as there is no borrow taking as we were taught in our schools. Two basic methods used in this are By a one more than that one before and last from 10 and all from 9. These two sutras will be used here to solve problems. For subtracting two digits, first begin with the right side. If the digit on the top is smaller than the digit which is below, then we have to take the complement of that digit. The complement will be added in the top digit which will be the answer at unit's place. Next we have to increase the digit next to last by one simply by putting a dot on it. Follow this and complete the whole question. This trick does not require any borrowing as in the case with normal methods. With more practice, one can easily perform subtraction using this trick.


4. Finding the reciprocal To find a the value of x in an equation like x+1/x = 10/3, first we cross multiply to form a 2nd degree equation in x and then solve it to obtain the value of x but by applying the Vedic Maths trick, solving such equations will become much more easier. After some practice, you will be able to do these questions without even using pen and paper. You just have to write the fraction part on the right hand side just like the question on the left hand side. For e.g., consider the above given equation x+1/x = 10/3. In this case, instead of cross multiplying and complication the equation, you can simply write 10/3 as 3+1/3 and on comparing the equation on both sides, you will see that the value of x comes out to be 3. As simple as that. Just break the number and get the answer. Let’s take another example for better understanding. Let the question be (1/x+1) +(x+1/x) = 26/5. Just split 26/5 as 5+ (1/5) and the problem is solved. When you will compare the left and right hand side of the equation, you will see that x+1=5 which implies that x=4, the required answer.

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