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

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Math Expert
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The "luminous flux," or perceived brightness, of a light source is  [#permalink]

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03 Jul 2015, 03:29
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55% (hard)

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63% (01:54) correct 37% (01:50) wrong based on 246 sessions

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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

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Re: The "luminous flux," or perceived brightness, of a light source is  [#permalink]

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03 Jul 2015, 04:02
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. $$D_2 = 6\sqrt{2} = \sqrt{72}$$

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Re: The "luminous flux," or perceived brightness, of a light source is  [#permalink]

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03 Jul 2015, 04:47
1
1
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  [#permalink]

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03 Jul 2015, 04:50
IanStewart wrote:
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.

Edited. Thank you.
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Re: The "luminous flux," or perceived brightness, of a light source is  [#permalink]

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03 Jul 2015, 05:42
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  [#permalink]

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03 Jul 2015, 05:50
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  [#permalink]

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03 Jul 2015, 06:30
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  [#permalink]

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03 Jul 2015, 11:33
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  [#permalink]

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03 Jul 2015, 11:44
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  [#permalink]

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04 Jul 2015, 11:35
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  [#permalink]

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06 Jul 2015, 06:48
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:

$$I_1*(d_1)^2 = I_2*(d_2)^2$$

$$(200 \ lumens)*(3 \ meters)^2 =(25 \ lumens)*(d_2)^2$$

$$(d_2)^2 = \frac{(200 \ lumens)*(3 \ meters)^2}{25}$$

$$d_2 = 6\sqrt{2}$$

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Re: The "luminous flux," or perceived brightness, of a light source is  [#permalink]

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23 Apr 2017, 23:38
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  [#permalink]

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28 Jan 2019, 04:53
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  [#permalink]

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30 Jan 2019, 19:52
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
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The "luminous flux," or perceived brightness, of a light source is  [#permalink]

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06 Mar 2019, 18:20
Hello experts! chetan2u

Hello experts!

Could someone please clarify to me why is it wrong to solve it like the following?

$$\frac{200}{8}=\frac{k}{(3^2)(8)}$$

$$25=\frac{k}{72}$$

Then I didn't know what to do and picked E.

Thank you so much in advance!
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Re: The "luminous flux," or perceived brightness, of a light source is  [#permalink]

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06 Mar 2019, 19:55
jfranciscocuencag wrote:
Hello experts! chetan2u

Hello experts!

Could someone please clarify to me why is it wrong to solve it like the following?

$$\frac{200}{8}=\frac{k}{(3^2)(8)}$$

$$25=\frac{k}{72}$$

Then I didn't know what to do and picked E.

Thank you so much in advance!

Hi
You have to tell how you got to the equation.
$$\frac{200}{8}=\frac{k}{(3^2)(8)}$$
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Re: The "luminous flux," or perceived brightness, of a light source is  [#permalink]

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06 Mar 2019, 20:32
chetan2u wrote:
jfranciscocuencag wrote:
Hello experts! chetan2u

Hello experts!

Could someone please clarify to me why is it wrong to solve it like the following?

$$\frac{200}{8}=\frac{k}{(3^2)(8)}$$

$$25=\frac{k}{72}$$

Then I didn't know what to do and picked E.

Thank you so much in advance!

Hi
You have to tell how you got to the equation.
$$\frac{200}{8}=\frac{k}{(3^2)(8)}$$

chetan2u

Well, I just know that when variables are inversely proportional means that if one is divided then the other one must be multiplied, hence I also use the following expression:

$$x=\frac{1}{y}$$

But I don't understand why do we have to solve for k.
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Re: The "luminous flux," or perceived brightness, of a light source is  [#permalink]

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06 Mar 2019, 20:45
1
jfranciscocuencag wrote:
chetan2u wrote:
jfranciscocuencag wrote:
Hello experts! chetan2u

Hello experts!

Could someone please clarify to me why is it wrong to solve it like the following?

$$\frac{200}{8}=\frac{k}{(3^2)(8)}$$

$$25=\frac{k}{72}$$

Then I didn't know what to do and picked E.

Thank you so much in advance!

Hi
You have to tell how you got to the equation.
$$\frac{200}{8}=\frac{k}{(3^2)(8)}$$

chetan2u

Well, I just know that when variables are inversely proportional means that if one is divided then the other one must be multiplied, hence I also use the following expression:

$$x=\frac{1}{y}$$

But I don't understand why do we have to solve for k.

There are two statements..
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. So, l=k/m^2, There are three variables...Two values changing l and m, while k is a constant. When you are talking of proportionality, we have to take a constant k
If a light source produces 200 lumens at a distance of 3 meters, This gives two variables and so you can find the constant k
at what distance will the light source produce a luminous flux of 25 lumens...Here you have one variable l and value of constant k is known from top, so you can find distance or m
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Re: The "luminous flux," or perceived brightness, of a light source is  [#permalink]

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07 Mar 2019, 02:44
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

luminous = constant * 1/distance^2
so given
200 = k * 1/3^2
k = 1800
so
25 = 1800*1/d^2
d^2 = 1800/25 = 72 = 6√2
IMO B
Re: The "luminous flux," or perceived brightness, of a light source is   [#permalink] 07 Mar 2019, 02:44
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