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Quant#2

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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by beat_gmat_09 » Thu Nov 25, 2010 9:16 am
10^2 (10^6 - 1)
_____________

10^3 (10^4 - 1)

10^6 - 1 approx = 10^6 same for 10^4

10^6
= _____

10^5

= 10
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by anirudhbhalotia » Thu Nov 25, 2010 10:38 pm
beat_gmat_09 wrote:10^2 (10^6 - 1)
_____________

10^3 (10^4 - 1)

10^6 - 1 approx = 10^6 same for 10^4

10^6
= _____

10^5

= 10

Didn't get it.

How do you expand/solve 10^8 - 10^2? Isnt this = 10^6?

Similarily10^7 - 10^3 = 10^4?
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by beat_gmat_09 » Fri Nov 26, 2010 12:00 am
anirudhbhalotia wrote:
beat_gmat_09 wrote:10^2 (10^6 - 1)
_____________

10^3 (10^4 - 1)

10^6 - 1 approx = 10^6 same for 10^4

10^6
= _____

10^5

= 10

Didn't get it.

How do you expand/solve 10^8 - 10^2? Isnt this = 10^6?

Similarily10^7 - 10^3 = 10^4?
No. 10^8 - 10^2 !=10^6 ; Is 10^2 - 10 = 10 ? No. 10^2 = 100 so 100-10 = 90
Check out indices rules, you'll need to use these rules.

If m and n are positive integers, and then
(i) a^m a^n = a^(m + n)
(ii) [Quotient Law]
(iii) (a^m)^n = a^(mn)
(iv) (ab)^m = a^m . b^m
(v) a^0= 1
(vi) a^(m/n) = (a^m)^1/n


10^8 is quite a huge figure and subtracting 100 is approximately the same value - 10^8
moreover the question asks you the closest value not the exact/accurate value, so you can use approximation.
Hope this helps.
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by purnimaksingh » Thu Dec 16, 2010 9:23 am
To make it simple
10^2(10^6-1)
---------------
10^3(10^4-1)


(10^3*2-1^2)
------------------
10(10^2*2 -1^2)


use the formula (a^2-b^2)= (a-b)(a+b)
therefore numrator will be (10^3-1)(10^3+1)
and dr will be (10^2-1)(10^2+1)


so to simplify

round off 10^3-1 to 10^3 and so on

therefore we will get

(10^3-1)(10^3+1)
-------------------------
10(10^2-1)(10^2+1)

to make it approx round off +1 and -1

so we get 10^3*10^3
-------------------- = 10
10*10^2*10^2


hope this make sense
_____________

10^3 (10^4 - 1)

10^6 - 1 approx = 10^6 same for 10^4

10^6
= _____

10^5

= 10[/quote]


Didn't get it.

How do you expand/solve 10^8 - 10^2? Isnt this = 10^6?

Similarily10^7 - 10^3 = 10^4?[/quote]
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by purnimaksingh » Thu Dec 16, 2010 9:23 am
To make it simple
10^2(10^6-1)
---------------
10^3(10^4-1)


(10^3*2-1^2)
------------------
10(10^2*2 -1^2)


use the formula (a^2-b^2)= (a-b)(a+b)
therefore numrator will be (10^3-1)(10^3+1)
and dr will be (10^2-1)(10^2+1)


so to simplify

round off 10^3-1 to 10^3 and so on

therefore we will get

(10^3-1)(10^3+1)
-------------------------
10(10^2-1)(10^2+1)

to make it approx round off +1 and -1

so we get 10^3*10^3
-------------------- = 10
10*10^2*10^2


hope this make sense
_____________

10^3 (10^4 - 1)

10^6 - 1 approx = 10^6 same for 10^4

10^6
= _____

10^5

= 10[/quote]


Didn't get it.

How do you expand/solve 10^8 - 10^2? Isnt this = 10^6?

Similarily10^7 - 10^3 = 10^4?[/quote]
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by goyalsau » Sun Dec 19, 2010 2:29 am
anirudhbhalotia wrote:Image


This one as per my understanding and calculation answer was 10 factor 2.

But actual answer is 10.

How?


To make it simple
10^2(10^6-1)
---------------
10^3(10^4-1)


10^6 = 100,00,00 ,,,,,,,,, 10^4 = 100,00

10^6 - 1 = 999,999 ,,,,,, 10^4 - 1 = 999,9

999,999
------------------ { Cancel by 9 }
10(999,9)

111,111
-------------
10 ( 111,1)

1111 * 100 = 111,100 ....... Numerator can be written as 1111 * 100 { i know 11 is left that is very small and we can neglect that as problem asks for the approximation }

1111 * 100
-------------
10 ( 111,1)

100
-----
10

Answer is 10........... ( :wink: )
Saurabh Goyal
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by turbohaq » Tue Jan 18, 2011 4:11 pm
10^2(10^6-1)
___________ =
10^3(10^4-1)



(10^6-1)
_________ =
10(10^4-1)


99999.9
________ ~
9999


100000
________ =
10000


10 :)
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by gmat1011 » Fri Jan 21, 2011 3:17 am
A million has 6 zeros

For numerator 100 mill minus 100
For denominator 10 mill minus 1000

The 100 and 1000 will hardly make any significant dents in the large numbers

So Q is really 10^8/10^7

=> 10
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