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On a certain scale of intensity, each increment of 10 in mag [#permalink]
07 Oct 2009, 19:46

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On a certain scale of intensity, each increment of 10 in magnitude represents a tenfold increase in intensity. On this scale, an intensity corresponding to a magnitude of 165 is how many times an intensity corresponding to a magnitude of 125?

Re: Scale of Intensity [#permalink]
08 Oct 2009, 01:05

amitgovin wrote:

On a certain scale of intensity, each increment of 10 in magnitude represents a tenfold increase in intensity. On this scale, an intensity corresponding to a magnitude of 165 is how many times an intensity corresponding to a magnitude of 125?

A. 40 b. 100 c. 400 d. 1,000 e. 10,000

this seems like it should be straight forward but I think that I'm missing something. please explain. thanks.

Re: Scale of Intensity [#permalink]
08 Oct 2009, 02:00

jax91 wrote:

amitgovin wrote:

On a certain scale of intensity, each increment of 10 in magnitude represents a tenfold increase in intensity. On this scale, an intensity corresponding to a magnitude of 165 is how many times an intensity corresponding to a magnitude of 125?

A. 40 b. 100 c. 400 d. 1,000 e. 10,000

this seems like it should be straight forward but I think that I'm missing something. please explain. thanks.

let 125 be of intensity x

so 125 + 10 =135 = 10x

135 + 10 = 145 = 10 (10x) = 100x

145 + 10 = 155 = 10 (100x) = 1000x

155 + 10 = 165 = 10 (1000x) = 10,000x

can u explain in details _________________

Bhushan S. If you like my post....Consider it for Kudos

On a certain scale of intensity, each increment of 10 in mag [#permalink]
10 Mar 2011, 09:32

On a certain scale of intensity, each increment of 10 in magnitude represents a tenfold increase in intensity. On this scale, an intensity corresponding to a amgnitude of 165 is how many times an intensity corresponding to 125?

On a certain scale of intensity, each increment of 10 in magnitude represents a tenfold increase in intensity. On this scale, an intensity corresponding to a amgnitude of 165 is how many times an intensity corresponding to 125? A/ 40 B/ 100 C/ 400 D/ 1000 E/ 10 000

Increase of 40 in magnitude corresponds to 10^4 increase in intensity:

If intensity for 125 is x then for 135 it'll be 10*x, for 145 it'll be 10*10*x=10^2*x, for 155 it'll be 10*10*10*x=10^3*x and for 165 it'll be 10*10*10*10*x=10^4*x.

Re: Scale of Intensity [#permalink]
18 Dec 2013, 09:02

amitgovin wrote:

On a certain scale of intensity, each increment of 10 in magnitude represents a tenfold increase in intensity. On this scale, an intensity corresponding to a magnitude of 165 is how many times an intensity corresponding to a magnitude of 125?

A. 40 b. 100 c. 400 d. 1,000 e. 10,000

this seems like it should be straight forward but I think that I'm missing something. please explain. thanks.

Increment of 10 between 165 and 125 is 4

So then the increase in intensity will be 10^4 = 10,000

Re: On a certain scale of intensity, each increment of 10 in mag [#permalink]
27 Feb 2014, 21:17

Is this logic ok? When I looked at this I subtracted 125 from 165 to get 40. Then thinking that this is the difference in the magnitudes and that magnitude gives us how many tenfold to calculate (so we have 4 increments of 10 in the magnitude so that would tell us 10^4) to give us 4 0s in the answer choice.

I think this may be similiar logic to jldgr but I just wanted to confirm. Thanks for the help!

Re: On a certain scale of intensity, each increment of 10 in mag [#permalink]
27 Feb 2014, 22:15

Expert's post

amjet12 wrote:

Is this logic ok? When I looked at this I subtracted 125 from 165 to get 40. Then thinking that this is the difference in the magnitudes and that magnitude gives us how many tenfold to calculate (so we have 4 increments of 10 in the magnitude so that would tell us 10^4) to give us 4 0s in the answer choice.

I think this may be similiar logic to jldgr but I just wanted to confirm. Thanks for the help!

Re: On a certain scale of intensity, each increment of 10 in mag [#permalink]
09 Mar 2015, 16:09

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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