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# On a certain scale of intensity, each increment of 10 in mag

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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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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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08 Oct 2009, 03:00
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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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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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10 Mar 2011, 10:43
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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 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.

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10 Mar 2011, 11:08
Let intensity @ 125 be x
135-10x
145-100x
155-1000x
165-10000x

Ans: "E"
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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

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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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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On a certain scale of intensity, each increment of 10 in mag [#permalink]

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23 Aug 2017, 15:52
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. 1000
E. 10 000

I did the problem algebraically, but then checked by assigning an intensity value for magnitude 125. The latter made more sense to me.

Let magnitude of 125 have a corresponding intensity value of 2.

125 = 2

135 = 2 * 10 (tenfold increase in intensity) = 20

145 = 200
155 = 2000
165 = 20000

Intensity corresponding to a magnitude of 165 is 20,000/2 = 10,000 times the intensity corresponding to a magnitude of 125.

Whatever works.

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

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25 Aug 2017, 10:59
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. 1000
E. 10 000

Since 125 is 40 away from 165, the intensity of 165 compared to 125 is 10 x 10 x 10 x 10 = 10,000.

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Re: On a certain scale of intensity, each increment of 10 in mag   [#permalink] 25 Aug 2017, 10:59
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