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shampoo
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(10^8 - 10^2) / (10^7 - 10^3)
= 10^2(10^6 - 1)/10^2(10^5-10)

10^6 - 1 --> insignificant, can be rounded off to 10^6

10^5 - 10 --> insignificant, can be rounded off to 10^5

Ans: 10
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10 it is

(10^8 - 10^ 2)/(10^7-10^3)
=> 10^2(10^6-1)/10^3(10^4-1)
=> (10^6-1)/(10^5-10)

Can be rounded to 10^6/10^5 = 10
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interesting approach...i didnt factor 10^2 or 10^3 out

i pretty much estimaed 10^8 is fairly large compared to 10^2 so i estimate this to 10^8

same thing with 10^7-10^3..is ~10^7

so then i get 10^8/10^7=~10
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I am getting B

Factor out 10^2[10^6-1]/[10^3[10^4-1]

which gives

10^-1[10^2]

which is approximately 10
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Doesnt' the rule of when bases are the same powers add applies here.

10 ^ 8 - 10 ^ 6 = 10 ^ 8-6 = 10 ^ 2

Or I got it all wrong...!!
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thanks everyone, OA is 10.

btw, what difficulty level this question is?
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What is the fastest way to solve this:
Attachments

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Given :

(10^8 - 10^2 ) / ( 10^7 - 10^3)

(10^8 - 10^2 ) approx 10^8
( 10^7 - 10^3) approx 10^7

therefore 10.
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The answer should be 10 only.

Lets take it this way:
(10^8-10^2)/(10^7-10^3) can be written as
10^2(10^6-1)/10^3(10^4-1) - taking 10^2 & 10^3 common from numerator and denominator respectively.
Now, 10^6>>1 so we can say that 10^6-1 is approximately equal to 10^6.
Similarly, 10^4>>1 so we can say that 10^4-1 is approximately equal to 10^4.

So, the expression now becomes : (10^2*10^6)/(10^3*10^4).

or, 10^8/10^7
or, 10.
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I have a slightly different approach which is easy and takes you 30 seconds. (However I also got it wrong on my GMATprep)

You can rewrite the term: 10^8 / 10^7 - 10^2 / 10^3

Now we know that two exponents with the same base (10) and which are divided by each other can be deducted from each other.

Therefore we get: 10^1 - 10^-1
This is the same as 10 - 1/10 = This is obviously closest to 10
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alimad
Doesnt' the rule of when bases are the same powers add applies here.

10 ^ 8 - 10 ^ 6 = 10 ^ 8-6 = 10 ^ 2

Or I got it all wrong...!!

You can only apply this rule if the two numbers are divided by each other.
I used this approach in my solution:

You can rewrite the term: 10^8 / 10^7 - 10^2 / 10^3

Now we know that two exponents with the same base (10) and which are divided by each other can be deducted from each other.

Therefore we get: 10^1 - 10^-1
This is the same as 10 - 1/10 = This is obviously closest to 10



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