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For a light that has an intensity of 60 candles at its source, the intensity in candles, S, of the light at a point d feet from the source is given by the formula \(S=\frac{60k}{d^2}\), where k is a constant. If the intensity of the light is 30 candles at a distance of 2 feet from the source, what is the intensity of the light at a distance of 20 feet from the source?
(A) 3/10 candles
(B) 1/2 candles
(C) 1 1/3 candles
(D) 2 candles
(E) 3 candles
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(A) 3/10 candles
(B) 1/2 candles
(C) 1 1/3 candles
(D) 2 candles
(E) 3 candles
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SOLUTION
For a light that has an intensity of 60 candles at its source, the intensity in candles, S, of the light at a point d feet from the source is given by the formula S=60k/d^2, where k is a constant. If the intensity of the light is 30 candles at a distance of 2 feet from the source, what is the intensity of the light at a distance of 20 feet from the source?
(A) 3/10 candles
(B) 1/2 candles
(C) 1 1/3 candles
(D) 2 candles
(E) 3 candles
The intensity of the light is 30 candles (S) at a distance of 2 feet (d) from the source: \(30=\frac{60k}{2^2}\) --> \(k=2\).
Therefore the intensity of the light at a distance of 20 feet from the source is \(S=\frac{60*2}{20^2}=\frac{3}{10}\).
Answer: A.
_________________
For a light that has an intensity of 60 candles at its source, the intensity in candles, S, of the light at a point d feet from the source is given by the formula S=60k/d^2, where k is a constant. If the intensity of the light is 30 candles at a distance of 2 feet from the source, what is the intensity of the light at a distance of 20 feet from the source?
(A) 3/10 candles
(B) 1/2 candles
(C) 1 1/3 candles
(D) 2 candles
(E) 3 candles
The intensity of the light is 30 candles (S) at a distance of 2 feet (d) from the source: \(30=\frac{60k}{2^2}\) --> \(k=2\).
Therefore the intensity of the light at a distance of 20 feet from the source is \(S=\frac{60*2}{20^2}=\frac{3}{10}\).
Answer: A.
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General Discussion
WoundedTiger
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For a light that has an intensity of 60 candles at its source, the intensity in candles, S, of the light at a point d feet from the source is given by the formula S=60k/d^2, where k is a constant. If the intensity of the light is 30 candles at a distance of 2 feet from the source, what is the intensity of the light at a distance of 20 feet from the source?
Sol: For the given intensity S=30 candles at a distance of 2 feet we can find value of k as
30*d^2/60= k
30*4/60 so k=2
Now substituting the value of k and d=20 ft we get
S= 60*2/(20^2)
Or
120/400 = 3/10 candles
Ans A
Posted from my mobile device
Sol: For the given intensity S=30 candles at a distance of 2 feet we can find value of k as
30*d^2/60= k
30*4/60 so k=2
Now substituting the value of k and d=20 ft we get
S= 60*2/(20^2)
Or
120/400 = 3/10 candles
Ans A
Posted from my mobile device
arunspanda
Joined: 04 Oct 2013
Last visit: 31 Oct 2021
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Location: India
GMAT Date: 05-23-2015
GPA: 3.45
Products:
Posts: 127
For a light that has an intensity of 60 candles at its source, the intensity in candles, S, of the light at a point d feet from the source is given by the formula \(S=\frac{60k}{d^2}\), where k is a constant. If the intensity of the light is 30 candles at a distance of 2 feet from the source, what is the intensity of the light at a distance of 20 feet from the source?
(A) 3/10 candles
(B) 1/2 candles
(C) 1 1/3 candles
(D) 2 candles
(E) 3 candles
Using the formula \(S=\frac{60k}{d^2}\), the ratio between light intensity at 2 feet and 20 feet \(=\frac{Light Intensity At 20 Feet}{Light Intensity At 2 Feet}=\frac{60k}{20^2}/\frac{60k}{2^2}= \frac{1}{100}\)
Given that light intensity at 2 feet equals 30 candles, the intensity of light at 20 feet is equal to \(30(\frac{1}{100}) = \frac{3}{10} candles\)
Answer: (A)
(A) 3/10 candles
(B) 1/2 candles
(C) 1 1/3 candles
(D) 2 candles
(E) 3 candles
Using the formula \(S=\frac{60k}{d^2}\), the ratio between light intensity at 2 feet and 20 feet \(=\frac{Light Intensity At 20 Feet}{Light Intensity At 2 Feet}=\frac{60k}{20^2}/\frac{60k}{2^2}= \frac{1}{100}\)
Given that light intensity at 2 feet equals 30 candles, the intensity of light at 20 feet is equal to \(30(\frac{1}{100}) = \frac{3}{10} candles\)
Answer: (A)
Bunuel
30 = 60* k/4 => k =2
60*2/20*20 = 3/10 - Option A)
option A.
First we can calculate the value of k by putting S=30 and d=2 in
S=60k/d^2
From this k=2
Now we can calculate S=60*2/(20)^2
S=3/10
First we can calculate the value of k by putting S=30 and d=2 in
S=60k/d^2
From this k=2
Now we can calculate S=60*2/(20)^2
S=3/10
Bunuel
Hi, Bunuel. I chose C because I think candle can not be apart, that means the number of candle is integer. Why the number of candle can be fraction? Thanks in advance:)
It changed in the OG Quantitive 2017 where the choices are
A) 3/10 candle
B) 1/2 candle
C) 1 candle
D) 2 candles
E) 3 candles.
And the question remains the same.
Jez0612
Intensity is kind of a unit of a brightness, so it's not actual candles, so we can have an intensity in fractions.
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60k/2^2 = 30 => k=2
So if d=20 then 60*2/20*20 == 3/10
So if d=20 then 60*2/20*20 == 3/10
JeffTargetTestPrep
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Last visit: 05 Jan 2024
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Expert reply
Posts: 2,973
rlevochkin
We are given that for a light that has an intensity of 60 candles at its source, the intensity in candles, S, of the light at a point d feet from the source is given by the formula S=60k/d^2, where k is a constant.
For this direct variation problem, we first will use the given formula to solve for k, the constant of variation, and then we can answer the question.
30 = 60k/(2^2)
30 = 60k/4
120 = 60k
k = 2
We know that k is 2, and we use that fact to determine the intensity of the light at a distance of 20 feet from the source.
S = 60(2)/20^2
S = 120/400 = 12/40 = 3/10 candles
Answer: A
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WoundedTiger
HI !
Can someone please explain why Ks coeffficient ( 60 ) remains the same . Thats what threw me !
Hanzel101
Joined: 15 Jul 2016
Last visit: 23 Feb 2020
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Location: India
Schools: Oxford "21 (A) Cambridge "21 (A$)
GMAT 1: 690 Q48 V36
Argp
Hi
The light is the same in all 3 calculations and has an intensity of 60 candles at the source. It decreases as we move away from light, so d, the distance, in the formula changes.
Rest everything remains the same.
Maybe if we were dealing with some other light that has an intensity of 30 candles AT THE SOURCE, the 60 would have changed but then too, it would be mention that 60 is the luminosity related to candles at the source.
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