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 Post subject: Tricks for quick numeric computations
PostPosted: Sun Dec 27, 2009 9:45 pm 

Joined: Tue Nov 17, 2009 12:50 pm
Posts: 11
Numeric computation is an important part of GRE quant.
Each of has a few tricks, often based on basic maths to make computations quicker.
Lets share what we know individually.
I am sure we will have a nice list that applies to questions that appear in GRE like tests.


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 Post subject: Re: Tricks for quick numeric computations
PostPosted: Sun Dec 27, 2009 9:51 pm 

Joined: Tue Nov 17, 2009 12:50 pm
Posts: 11
Here is one.

9P4 = 9 x 8 x 7 x 6 x 5

Look for numbers like 5 (or 15 or 25).
These when multiplied with an even number result into a numbers whose multiplication tables you know.
In this case 6x5 = 30 .... 3 is easy to handle.

9x8x7x30

Now we have a choice of
a. (9x8)x(7x30) = 72 x 210 or
b. (9x8)x(7)x(30) = 72 x 7 x 30


Would prefer (b). It is unlikely that one knows the multiplication tables of 21.
7 is relatively easy.

(b) becomes : 504 x 3 x 10 = 15120


Last edited by GREedge Math Expert on Sun Dec 27, 2009 10:00 pm, edited 1 time in total.

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 Post subject: Re: Tricks for quick numeric computations
PostPosted: Sun Dec 27, 2009 9:59 pm 

Joined: Tue Nov 17, 2009 12:50 pm
Posts: 11
Here is one more.

If you come across (N x 25 ) or (N x 125) where N is a big number, say 45679.

So, 45679 x 25

Instead of multiplying by 25, one could do the following:

45679 x 25 = 45679 x 25 x 4 /4 = (45679/4) x (25x4)
= (45679/4) x 100

This method is good if you are comfortable with dividing a number by 4 rather that multiplying with 25.

= 11419.75 x 100 = 1141975

Similarly 45679 x 125 = 45679 x 125 x 8/8 = (45679/8) x (125 x8)
= (45679/8) x 1000
= 5709.875 x 1000 = 5709875
This method is good if you are comfortable with dividing a number by 8 rather that multiplying with 125.

Applies as well to N x 50 = (N/2) x 100.
45678/2 is less prone to error than computing 45678 x 5


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 Post subject: Re: Tricks for quick numeric computations
PostPosted: Tue Dec 29, 2009 4:17 pm 

Joined: Fri Nov 27, 2009 5:11 pm
Posts: 55
If you are asked to compute 499*501

Think ? :geek:
Can you solve this without direct multiplying?
:?:

Solution:
this can be written as (500-1)(500+1) ;)
So this is nothing but 500^2 - 1 = 249999


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 Post subject: Re: Tricks for quick numeric computations
PostPosted: Tue Jan 12, 2010 11:45 am 

Joined: Fri Nov 27, 2009 5:11 pm
Posts: 55
If you are dealing with a large calculation like 13^6. This can come while solving a problem of permutations.
Now you find that the answers are in the simplified form.

The examiner does want to test your calculation ability.

If you break 13^6 = 13^2 x 13^2 x 13^2 = 169 x 169 x 169.

Surely this will end in a 9 so the choice will be the answer with 9 in its units place.

Mostly there will be only one choice which will end in 9 and so you can directly make your choice.


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 Post subject: requests pls......
PostPosted: Fri Feb 05, 2010 1:42 am 

Joined: Thu Feb 04, 2010 10:18 pm
Posts: 4
sir, could u give the detailed concept about percentiles............it is rather new...and not clearly explained any where


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 Post subject: Re: Tricks for quick numeric computations
PostPosted: Sat Mar 06, 2010 1:37 pm 

Joined: Fri Nov 27, 2009 5:11 pm
Posts: 55
Square of numbers ending with 5

25 :[2x3] [25] = [6][25] = 625

45 : [4x5] [25] = [20][25] = 2025

a5 =[a x a+1 ] [25]


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 Post subject: Re: Tricks for quick numeric computations
PostPosted: Tue Mar 16, 2010 12:23 pm 

Joined: Fri Nov 27, 2009 5:11 pm
Posts: 55
sqrt (a^2 + h ) where h<<a^2

=a + h/(2a)

example sqrt(1.7) =sqrt(1.69 + 0.01)
we know 1.3^2 = 1.69

so, sqrt(1.7) =1.3 + (0.01)/2.6 approx = 1.3+ 0.01/2.5 = 1.3 +1/250= 1.3 + 4/1000 =  1.3 +  0.004 = 1.304

sqrt (a^ - h ) where h <<a^2
= a - h/(2a)


sqrt (630) =sqrt (625 + 5 ) = 25 +5/(50) = 25.01


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