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The "luminous flux," or perceived brightness, of a light source is
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03 Jul 2015, 02:29
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Re: The "luminous flux," or perceived brightness, of a light source is
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03 Jul 2015, 03:02
Bunuel wrote: The "luminous flux," or perceived brightness, of a light source is measured in lumens and is inversely proportional to the square of the distance from the light. If a light source produces 200 lumens at a distance of 3 meters, at what distance will the light source produce a luminous flux of 25 lumens?
A. 6 meters B. \(\sqrt{72}\) meters C. 9 meters D. 24 meters E. 72 meters Flux is Inversely Proportional to \((Distance)^2\) i.e. \(F_1 / F_2 = (D_2 / D_1)^2\) \(F_1 = 200\) \(D_1 = 3\) \(F_2 = 25\) so, \(200 / 25 = (D_2 / 3)^2\) i.e. \(8 = (D_2 / 3)^2\) i.e. \(2\sqrt{2} = (D_2 / 3)\) i.e. \(D_2 = 6\sqrt{2} = \sqrt{72}\) Answer: Option B
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Re: The "luminous flux," or perceived brightness, of a light source is
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03 Jul 2015, 03:47
If one thing x is directly proportional to another thing y, then x = ky, for some constant k. If one thing x is inversely proportional to another thing y, then x = k/y for some constant k. So here if brightness b is inversely proportional the the square of distance d, then b = k/d^2, for some constant k. We know that when d=3, we have b=200, so 200 = k/3^2, and k = 1800. So the equation relating brightness and distance is b = 1800/d^2. Thus when b = 25, we have: 25 = 1800/d^2 d^2 = 1800/25 d = √(1800/25) d = √2 √900 / 5 = (√2)*30/5 = 6√2 I really dislike that they've written the correct answer as √72; that is not simplified, and a root in an answer choice should not be written that way. A test taker with good habits, who simplifies roots immediately, will get to the answer 6√2, and won't see their answer among the choices. That's really not fair  a test taker shouldn't be punished for doing math the right way.
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Re: The "luminous flux," or perceived brightness, of a light source is
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Re: The "luminous flux," or perceived brightness, of a light source is
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03 Jul 2015, 04:42
Bunuel wrote: The "luminous flux," or perceived brightness, of a light source is measured in lumens and is inversely proportional to the square of the distance from the light. If a light source produces 200 lumens at a distance of 3 meters, at what distance will the light source produce a luminous flux of 25 lumens?
A. 6 meters B. \(6\sqrt{2}\) meters C. 9 meters D. 24 meters E. 72 meters Solution  Given that, Flux ∝ 1/distance^2 F1/F2 = d2/d1 > 200/25 = d^2/9 > d^2=72 > d=\(\sqrt{36*2}\)>d=\(6\sqrt{2}\) meters ANS B
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Re: The "luminous flux," or perceived brightness, of a light source is
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03 Jul 2015, 04:50
Bunuel wrote: The "luminous flux," or perceived brightness, of a light source is measured in lumens and is inversely proportional to the square of the distance from the light. If a light source produces 200 lumens at a distance of 3 meters, at what distance will the light source produce a luminous flux of 25 lumens?
A. 6 meters B. \(6\sqrt{2}\) meters C. 9 meters D. 24 meters E. 72 meters \(l=\frac{constant*1}{d^2}\) \(200=\frac{constant*1}{3^2}\) \(constant = 200*9=1800\) for 25 lumens.. \(25=\frac{1800*1}{d^2}\).. \(d=6\sqrt{2}\) ans B
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Re: The "luminous flux," or perceived brightness, of a light source is
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03 Jul 2015, 05:30
Bunuel wrote: The "luminous flux," or perceived brightness, of a light source is measured in lumens and is inversely proportional to the square of the distance from the light. If a light source produces 200 lumens at a distance of 3 meters, at what distance will the light source produce a luminous flux of 25 lumens?
A. 6 meters B. \(6\sqrt{2}\) meters C. 9 meters D. 24 meters E. 72 meters 2 ways to solve this question: Given that the luminous flux varies inversely with square of the distance. So we can write the equation as: \(L=k/d^2\), had it be varying directly instead of inversely, we would have written \(L=kd^2\) Coming back to the question, We are given that for initial condition, L=200, d=3. Here, we can either substitute these values to get the value of k in the equation above or keep it as it is. If we do calculate the value of k, we get k=1800 lumens*\(m^2\) Then, we will use this value of k and L=25 to find the value of d as : \(d^2 = k/L\) Thus, d= 72^0.5=6*2^0.5 If you don want to calculate the value of k, \(L1=k/d1^2\) and \(L2=k/d2^2\) Dividing the 2, \(\frac{L1}{L2}=\frac{d2^2}{d1^2}\) Substitute, L1=200, L2=25, d1=3, We get d2=72^0.5=6*2^0.5



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Re: The "luminous flux," or perceived brightness, of a light source is
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03 Jul 2015, 10:33
F= k/d^2 200=K/9 K = 1800 So f = 1800/25 = 72 Option E
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Re: The "luminous flux," or perceived brightness, of a light source is
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03 Jul 2015, 10:44
adityadon wrote: F= k/d^2
200=K/9 K = 1800
So f = 1800/25 = 72
Option E You did not take the square root as f is actually f^2 in the equation you have written above.



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Re: The "luminous flux," or perceived brightness, of a light source is
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04 Jul 2015, 10:35
OPtion B is correct l=constant/d2
200=constant∗132 constant=200∗9=1800
for 25 lumens.. 25=1800∗1d2.. d=6√2



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Re: The "luminous flux," or perceived brightness, of a light source is
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06 Jul 2015, 05:48
Bunuel wrote: The "luminous flux," or perceived brightness, of a light source is measured in lumens and is inversely proportional to the square of the distance from the light. If a light source produces 200 lumens at a distance of 3 meters, at what distance will the light source produce a luminous flux of 25 lumens?
A. 6 meters B. \(6\sqrt{2}\) meters C. 9 meters D. 24 meters E. 72 meters MANHATTAN GMAT OFFICIAL SOLUTION:Because the intensity of the light source and the square of the distance are inversely proportional, you can write the product of the “before” intensity and distance squared and the product of the “after” intensity and distance squared. Then set these two products equal to each other: \(I_1*(d_1)^2 = I_2*(d_2)^2\) \((200 \ lumens)*(3 \ meters)^2 =(25 \ lumens)*(d_2)^2\) \((d_2)^2 = \frac{(200 \ lumens)*(3 \ meters)^2}{25}\) \(d_2 = 6\sqrt{2}\) Answer: B.
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Re: The "luminous flux," or perceived brightness, of a light source is
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23 Apr 2017, 22:38
Bunuel wrote: The "luminous flux," or perceived brightness, of a light source is measured in lumens and is inversely proportional to the square of the distance from the light. If a light source produces 200 lumens at a distance of 3 meters, at what distance will the light source produce a luminous flux of 25 lumens?
A. 6 meters B. \(6\sqrt{2}\) meters C. 9 meters D. 24 meters E. 72 meters Wouldn't the easiest way to do this problem be to set up two equations 200(x) = 9 25(x) = D 200 is 8 times larger than 25 so you just multiply 9 by 8 and take the square root I initially got 72 but even still i think that this method is easier as long as you don't forget the square root.



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Re: The "luminous flux," or perceived brightness, of a light source is
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28 Jan 2019, 03:53
Bunuel wrote: The "luminous flux," or perceived brightness, of a light source is measured in lumens and is inversely proportional to the square of the distance from the light. If a light source produces 200 lumens at a distance of 3 meters, at what distance will the light source produce a luminous flux of 25 lumens?
A. 6 meters B. \(6\sqrt{2}\) meters C. 9 meters D. 24 meters E. 72 meters 200 is inversely proportional to 1/(root 3)^2 => 200 (3)^(1/2)*2 = constant 25 is inversely proportional to x^2 => 25 (x)^2 = constant 25 x^2 = 200 *3 x^2 = 24 x = \(6\sqrt{2}\) meters B
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Re: The "luminous flux," or perceived brightness, of a light source is
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30 Jan 2019, 18:52
Please suggest if my method is correct?
Lumens = LU Luminious Flux = LF Distance = d Multiplier = x
LU=1/3^2 * X 200 = 1/9 * X X=1800
To find for 25 25=1/d^2 *1800 25d^2=1800 D=√72 D=6√2




Re: The "luminous flux," or perceived brightness, of a light source is
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