Problem Solving

(10^8)– (10^2)/(10^-7) – (10^-3)

What is the approximate value of the equation

A. 1
B. 10
C. 10^2
D. 10^3
E. 10^4

The answer was B. I am wondering if any one can help me follow the logic because when I work through I do not get 10. I get something closer to 1
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Hi !
OA is indeed 10 :)


(10^8)– (10^2)/(10^-7) – (10^-3)

10^2(10^6-1)/10^3(10^4-1)

10^6-1/10(10^4-1)

assuming -1 to be close to the same

10^6/10^5

10 is the answer

Vishu

about (10^-7) – (10^-3)

i solved it,, i got ,,,

1/10^7 - 1/10^3

10^3 - 10^7/ 10^10

i dont know how you got,,,

(10^-7) – (10^-3) =

10^3(10^4-1).

????

(10^-7) – (10^-3)

here i am takign common factor has 10 ^3 (10^4-1)

so 10^4-1 is approx equal to 10^4 ....

hence this is base line logic for the post

Please let me know if still doubt

Vishu

Vishu,
If the denominator is

10 ^ 7 - 10^3 the what u did is right

but may be there is a typo in the problem since the problem reads

10 ^ -7 - 10 ^ -3 like Raun pointed out.


Let me know your thoughts

Yes.. I agree with cramya..

denominators have negative powers..

oh ya its a typoo i guess coz .....i remember to have solved similar kinda problem
sorry being ignorant ramya

vishu

[email protected] fyi please confirm... if its not a typo . i am on the same boat as othrs

Yes you are right the problem should read
10^8-10^2/10^7-10^3
sorry for for confusion

Vishu,
Thank you fo rthe reponse, but I do not see how 10^8 - 10^3 become 10^2(10^6-1). 10^6-1 is same as 10^6/10^1. Could you please explain the logic behind you factorization . I know that you substract exponents when you divide, but not when you add or substract. Could any one else provide another explanation . Thank you