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Speed Mathematics

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Speed Mathematics  [#permalink]

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New post Updated on: 23 Nov 2011, 20:25
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This is for GMAT bees who have just started their preperation. This is taken from various books. Vedic Mathematics and Study materials from TIME, a premier institution that provide training for Indian Institute of Management CAT exam.
Shall be adding new techniques in speed mathematics as i go through various topics. So plz do keep up with this thread. If any of u has got more techniques plz do share it here.

To master the quants section, u need to have Knowledge in every area, speed in solving a question and approach in tackeling the test papers.
Speed is very important in GMAT because you are supposed to attend all the questions in GMAT and unattempted wuestions will lead to high penalty in the total score. I would like to demonstrate certain speed methodes of calculation which will be of great use to you in GMAT.
These techniques are drawn from different sources. i just thought of consolidating the useful techniques here.

Before getting into methodes of speed calculations, there are few aspects to be kept in mind. For getting the most out of speed calculation, u should be thorough in the following

Multiplication tables - (1*15) up to (20*15)
Squares up to 25 any higher square can be calculated easily.
Cubes upto 12
Powers of 2 - up to 12
Powers of 3 - upto 6
Reciprocals of numbers - upto 12
Compliments of 100 (i.e. the differance between 100 and the given two-digit number. eg- 25's compliment is 100-25 = 75)

Some ways of simplifying calculations

Multiplication by 5
For multiplication by 5, you should multiply the number to be multiplied by 10 and then divide it by 2.
eg: 6493 * 5 = 64930/2 = 32465

Multiplication by 25
Multiply the number to be multiplied by 100 and divide it by 4
eg: 6493 * 25 = 649300/4 = 162325

Multiplication by 125
Multiply the number by 1000 and divide by 8
eg: 6493 * 125 = 6493000/8 = 811625
( Alternatively, you can treat 125 as 100+25. So multiplication by 125 can be treated as multiplication by 100 and add to this number - 1/4th of itself because 25 is 1/4th of 100.)

Multiplication by 11
the rule is "for each digit add the right hand side and write the result as the corresponding figure in the product."
For the purpose of applying the rule, it will be easier if you assume that there is one "zero" on either side of the given number.
eg: 7469*11= 074690 (apply the above said rule) = 82159

Calculation of Squares
Getting the square of a number ending in 5 is simple. If the last digit of the number is 5, the last two digits of the square will be 25. Whatever is the earlier part of the number multiply it with one more than itself and that will be the first part of the answer. ( the second part of the answer will be 25 only)

35\(^2\)=1225.Here 3*4 = 12(first part) and the second part of the answer is always 25 so the answer is 1225

45\(^2\) = 2025 First part of th answer is 4 * 5=25 followed by 5*5=25

55\(^2\) = 3025

245\(^2\) = 24*25= 60025; First part - 24*25 = 600, last part - 5*5 = 25

Multiplying two numbers both of which are close to the same power of 10

Suppose we want to multiply 97 with 92. The power of 10 to which these two numbers are close is 100. Here 100 is called as the " base ". Write the two numbers with the differance from the base i.e., 100 (including the sign)
as shown below.

97 --> -3 ( 97 is 100-3 )
92 --> -8 ( 92 is 100-8 )

Then take the sum of the two numbers ( including their signs) along EITHER one of the two diagonals ( it will be same the same in both cases ).
Here diagonal sum is 97-8=92-3=89. This will form the first part of the answer.


The second part of the answer is the product ( along with the sign) of the difference from the power of 10 written for the two numbers, Here it is the profuct of -3 and -8 which is 24.

Now the last step is putting these two parts 89 and 24 together one next to the other.
Here the answer is 8924. That is 92*97=8924

Note:- The product of the two deviations should have asa many digits as the number of zeroes in the basae. In the above example the base is 100 having 2 zeroes so the product of -8 and -3 has 2 digits.
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Originally posted by cleetus on 29 Nov 2010, 04:03.
Last edited by cleetus on 23 Nov 2011, 20:25, edited 2 times in total.
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Re: Speed Mathematics  [#permalink]

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New post 29 Nov 2010, 04:23
Waiting for more........
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Re: Speed Mathematics  [#permalink]

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New post 29 Nov 2010, 15:27
Kool one............Thanks!!
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Re: Speed Mathematics  [#permalink]

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New post 02 Dec 2010, 11:07
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cleetus wrote:
Multiplication by 11
the rule is "for each digit add the right hand side and write the result as the corresponding figure in the product."
For the purpose of applying the rule, it will be easier if you assume that there is one "zero" on either side of the given number.
eg: 7469*11= 074690 (apply the above said rule) = 82159


Sorry but I find this explanation a bit ambiguous. I would've word it like that:

Step 1: write down the number -> 7469
Step 2: assume there are zeros on both sides of the number -> 074690
Step 3: starting from the right-most digit, add it to the one on the left and write the result as the corresponding figure in the product, starting with the units -> 0 + 9 = 9 => _ _ _ _ 9
Step 4: repeat Step 3 -> 9 + 6 = 15 => _ _ _ 5 9 (1 aside)
Step 5: repeat Step 3 -> 6 + 4 + 1 (recalling) = 11 => _ _ 1 5 9 (1 aside)
etc.

Nice post though. Good job!
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Re: Speed Mathematics  [#permalink]

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New post 02 Dec 2010, 11:13
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Thanks for ur comments. shall get back to this soon with more techniques. Plz make this thread alive by posting more easier techniques if anyone knows. Also do care to share this thread and site to your other GMAT friends.
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Re: Speed Mathematics  [#permalink]

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New post 02 Dec 2010, 11:19
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juukkk wrote:
cleetus wrote:
Multiplication by 11
the rule is "for each digit add the right hand side and write the result as the corresponding figure in the product."
For the purpose of applying the rule, it will be easier if you assume that there is one "zero" on either side of the given number.
eg: 7469*11= 074690 (apply the above said rule) = 82159


Sorry but I find this explanation a bit ambiguous. I would've word it like that:

Step 1: write down the number -> 7469
Step 2: assume there are zeros on both sides of the number -> 074690
Step 3: starting from the right-most digit, add it to the one on the left and write the result as the corresponding figure in the product, starting with the units -> 0 + 9 = 9 => _ _ _ _ 9
Step 4: repeat Step 3 -> 9 + 6 = 15 => _ _ _ 5 9 (1 aside)
Step 5: repeat Step 3 -> 6 + 4 + 1 (recalling) = 11 => _ _ 1 5 9 (1 aside)
etc.

Nice post though. Good job!


Thanks Juukkk. Explanation to Multiplication by 11 was ambiguous. I dint notice that in the beginning. Your explanation to the concept is better.
Thanks once again
+1 to you
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Re: Speed Mathematics  [#permalink]

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New post 02 Dec 2010, 17:40
Do you have any other speed calculations for the calculation of squares for any of the other integers?
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New post 04 Dec 2010, 20:40
i am travelling these days. shall get back to this thread soon with calculation of squares.
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New post 05 Dec 2010, 08:55
Very helpful! Thanks +1
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lol: waiting for you to get back home :P
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Re: Speed Mathematics  [#permalink]

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New post 29 Feb 2016, 11:44
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One nice and very general trick:

When you're multiplying two numbers together, be open to breaking down one or both of them into smaller values if that makes the multiplication quicker. For instance, if you're multiplying a large number by 9, multiply it by "10-1" instead:

894 * 9 = 894 * (10 - 1) = 8940 - 894 = 8046

Or if you're squaring a number that's very close to an easier value whose square you already know, try something like this:

301^2 = (300 + 1)^2 = 300^2 + 2*300*1 + 1^2 = 90000 + 600 + 1 = 90601

49^2 = (50 - 1)^2 = 2500 - 100 + 1 = 2401

Don't forget that you can change a value you're working with into a nicer form! The GMAT often gives you information that's actually pretty easy to work with, but that's intentionally presented in a harder or more complicated form than you technically need.
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Re: Speed Mathematics  [#permalink]

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New post 03 Mar 2016, 21:38
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Most of these "speed" math techniques are irrelevant for the GMAT. If you have to resort to these techniques then you are on the wrong track. For example, you will not be required to multiply by 125 on the GMAT. If there is a problem that ends up having a number being multiplied by 125, then somehow the terms will simplify by cancellation. The same is true for some of the other rules such as multiplication by 11, or squaring numbers such as 245.

Almost all of the GMAT problems are written in a way that grunt numerical work can be avoided. This is because there is no calculator on the exam, and the goal of the test writers is not to test one's numerical prowess. This is not to say that the problems aren't set in a way that one can get trapped in the numerical morass, you just have to spot the clean way and stay away from tedious calculations. This will typically involve factoring, cancelling, dividing, and in some cases approximating.

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Useful Formulae, Concepts & Tricks  [#permalink]

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New post 17 Jan 2018, 01:46
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Hello Fellow Aspirants,

This is my very first thread on GMAT Club, so if there are any mistakes or if any improvements can be made in the content then please comment below.

I am a maths enthusiast and I like to learn various methods to increase my calculation speed without depending on the calculator. I used to study Vedic mathematics in my high school and remember quiet a few techniques that I will be happy to share here.


Note: If such a thread is irrelevant/doesn't add anything useful to the community, then admin please let me know so that I can take this thread off ASAP.


My aim:
I wanted to make a thread where I could keep all the formulae, concepts and tricks at one place, so that I don't have to go through the various sources again and again. I want to share the same with the community.


1. How to find square root of a number


Example: Find the square root of 28376.

Step 1: Start by making pair of digits from the right until only 1 or no digit is left at the extreme left.
Our example would look like --> 2 83 76

Step 2: Find the number whose square is less than or equal to the extreme left one remaining digit or the pair of digit.
In our example since we have only 2 as the extreme left digit, we will search for a number whose square is less than or equal to 2. Clearly it's 1.
So, we have 1 (divisor) *1 (quotient)=1. Subtract 1 from 2. The remainder is 1. Now carry the next pair of digits, 83, besides 1 (just as you do in long division method).
We have quotient as 1 for now.

Step 3: Now add quotient to divisor. We get 1+1=2. Now think of a units digit D such that 2D*D<=183.
D=6 because 26(divisor)*6(quotient)=156<183 and 27*7=189>183.
Now subtract 156 from 183. We get remainder 27.
So now the quotient is 16.
Again carry the next pair of digits besides 27.

Step 4: Add 6(previous quotient) to 26(previous divisor). We get 32. The dividend this time is 2776.
Think of a digit D such that 32D*D<=dividend.
Clearly D=8; 328*8=2624<2776 and 329*9=2961>2776. Now subtract 2624 from 2776. We get 152.
Quotient = 168

Step 5: Now bring a pair of zeros besides the previous dividend such that a decimal comes into picture.
So dividend=15200
divisor=328(previous divisor)+8(previous quotient)=336
Again think of digit D such that 336D*D<=15200.
Clearly D=4; 3364*4=13456<15200 and 3365*5=16825>15200. Now subtract 13456 from 15200. We get 1744.
The quotient = 168.4

Thus continuing with this strategy you get quotient= 168.4517; The result can be verified via calculator!

The process might be looking tedious but it is simple long division which takes 10 -15 seconds max to find the desired value.

2. Find square of a number with units digit as 5


Example: \(75^2\)
Step 1: 5*5=25
Step 2: 7(tenth digit)*8(tenth digit+1)=56
Thus, \(75^2\)=5625; Result can be verified via calculator!

Hope the thread helps people is some way or the other.

*This is an ongoing thread in which I will add different variety of material including formulae/tricks/concepts*


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Re: Speed Mathematics  [#permalink]

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New post 18 Aug 2018, 03:10
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For a triangle inscribed inside a circle and with the base (or one side) of the triangle forming the diameter of the triangle and one of the vertices touching the circumference.

The angle formed at the vertex (touching the circumference ) is 90 degrees
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Re: Speed Mathematics &nbs [#permalink] 18 Aug 2018, 03:10
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