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On a certain scale of intensity, each increment of 10 in mag [#permalink]
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07 Oct 2009, 20:46
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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? A. 40 B. 100 C. 400 D. 1000 E. 10 000
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Re: Scale of Intensity [#permalink]
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08 Oct 2009, 02:05
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. 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
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Re: Scale of Intensity [#permalink]
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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
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Re: Scale of Intensity [#permalink]
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08 Oct 2009, 03:13
bhushan252 wrote: can u explain in details 'each increment of 10 in magnitude represents a tenfold increase in intensity' this implies that whenever we add 10 to the magnitude the intensity increase by 10 times. Say we have a magnitude of x with its intensity being y. then x + 10 will have 10 times the intensity of x, which is y x 10 = 10y so basically we just multiply the intensity by 10 when 10 is added to the magnitude. If we add a 40 to the magnitude, which is 40/10 = 4 tens, we need to multiply the intensity by 10, 4times. so intensity = y x (10x10x10x10) = 10,000y
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On a certain scale of intensity, each increment of 10 in mag [#permalink]
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10 Mar 2011, 10:32
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?
A. 40 B. 100 C. 400 D. 1000 E. 10 000



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Re: magnitude [#permalink]
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10 Mar 2011, 10:43



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Re: magnitude [#permalink]
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10 Mar 2011, 11:08
Let intensity @ 125 be x 13510x 145100x 1551000x 16510000x Ans: "E"
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Re: Scale of Intensity [#permalink]
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18 Dec 2013, 10:02
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 Hence answer is E Hope it helps Cheers! J



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Re: On a certain scale of intensity, each increment of 10 in mag [#permalink]
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27 Feb 2014, 22:17
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!



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Re: On a certain scale of intensity, each increment of 10 in mag [#permalink]
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27 Feb 2014, 23:15
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! Yes, this is the logic.
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Re: On a certain scale of intensity, each increment of 10 in mag [#permalink]
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09 Mar 2015, 17:09
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Re: On a certain scale of intensity, each increment of 10 in mag [#permalink]
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10 Mar 2015, 02:50
so numerically speaking , initial +nX10=(10^n)xintial so 165=125+40=125+4x10=(10^4)125 in intensity hence 10000 (E)
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Re: On a certain scale of intensity, each increment of 10 in mag [#permalink]
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