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# solving fraction's power

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Senior Manager
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12 Nov 2010, 16:33
How do we solve questions like ( 0.4)^5 or (.25)^4.
Any quick way or approach, please share.
Thanks
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12 Nov 2010, 18:11
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Its always better to be thorough with the following. This will help u in doing quick calculations. If u are proficient in this, then u can do speed calculations.
Multiplication tables - (1*15) up to (20*15)
Squares up to 25
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)

.4$$^5$$ = ?
2$$^10$$ = 4$$^5$$ = 1024.
.4 has one decimal place. so when its multiplied 5 times, u need to put 5 decimal places to the left of 1024.

same way do .25$$^4$$
25*25 = 625
625*2 = 25$$^4$$ = 390625
now count the decimal points. Ans=0.390625

U are supposed to know the techniques for speed mathematics.
Once u master those techniques, then u can do a lot of lengthy calculations like multiplication of a 3 digit number with another 3 digit number, squaring a three digit number and the like in few seconds.
Shall start a new post on speed mathematics soon. if there is already a thread on that, then somebody plz attach the link here

Cleetus
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12 Nov 2010, 18:23
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shrive555 wrote:
How do we solve questions like ( 0.4)^5 or (.25)^4.
Any quick way or approach, please share.
Thanks

I do agree with cleetus... It is good to be comfortable with things he mentioned...
I would like to add here that try and be comfortable with using fractions rather than decimals. I know it looks easier with the decimals but the calculation involved are terrible.
e.g. $$0.4 = \frac{2}{5}, so (\frac{2}{5})^5 = \frac{32}{3125}$$

or $$0.25 = \frac{1}{4}, so (1/4)^4 = \frac{1}{(2^8)} = \frac{1}{256}$$

Very very very slim chance that you will need to convert these fractions to decimals to arrive at the correct answer in GMAT...
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12 Nov 2010, 18:44
That sounds good !! Thanks

calculation with speed !!!
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28 Feb 2016, 15:04
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Re: solving fraction's power   [#permalink] 28 Feb 2016, 15:04
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