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Bunuel
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meanup
Bunuel
If 80 lamps can be lighted, 5 hours per day for 10 days for $21.25, then the number of lamps, which can be lighted 4 hours daily for 30 days, for $6.50, is

(A) 100
(B) 120
(C) 150
(D) 160
(E) 180

Dear Bunuel, Kindly review this question. If 120 is the answer then the second value should be $76.50 (Reverse approach)
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Edited. Thank you!
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­Ans (B)

\(\frac{80*5*10}{21.25}\) = \(\frac{n*4*30}{76.5}\)

\(n\) = \(\frac{80*5*10*76.5}{21.25*4*30}\) = \(120\)­
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meanup
\(\frac{5*10*21.25}{80}=\frac{4*30*76.50}{x}\)

\(x= 120\)
­meanup the equation you have created does not solve for the number of lamps recquired and will not result in \(x=120\). 

By multiplying the hours by the days by the price, and then dividing it by the number of lamps you are working out what the price is for a single lamp which stays on for that specific time frame. To make this method work you need to set them to both work for the same amount of time: 

80 lamps, lit 5 hours per day for 10 days for $21.25: Lamps = 80, hours worked = 50, cost = 21.25 (1)
x number of lamps, lit for 4 hours per day for 30 days, for $76.50: Lamps = x, hours worked = 120, cost 76.5 (2)

The LCM time would be 600: 
(1) Lamps = 80, hours worked = 50*12, cost= 21.25*12
(2)
 Lamps = x, hours worked = 120*5, cost= 76.5*12

Now one can use the same equation as you did above [ie. (time*price)/lamps]: \(\frac{(50*12*21.25*12)}{80}=\frac{(120*5*76.5*5)}{x}\)
Solving for x will result in 120, but it muddies things. 

The easiest way would be to solve for the number of lamps required is to set up the equation to be for price per hour worked. As an equation that would be: (cost)/(number of lamps*hours)

\(\frac{21.25}{80*50} = \frac{76.5}{120*x}\)

\(21.25x = \frac{76.5*50*80}{120}\) Multiply through by 4

\(85x = \frac{76.5*50*80}{30}\)

\(x = \frac{(76.5*80)}{51}\)

\(x = 1.5*80\)

\(x = 120\)


ANSWER B ­
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Nidzo

meanup
\(\frac{5*10*21.25}{80}=\frac{4*30*76.50}{x}\)

\(x= 120\)
­meanup the equation you have created does not solve for the number of lamps recquired and will not result in \(x=120\). 

By multiplying the hours by the days by the price, and then dividing it by the number of lamps you are working out what the price is for a single lamp which stays on for that specific time frame. To make this method work you need to set them to both work for the same amount of time: 

80 lamps, lit 5 hours per day for 10 days for $21.25: Lamps = 80, hours worked = 50, cost = 21.25 (1)
x number of lamps, lit for 4 hours per day for 30 days, for $76.50: Lamps = x, hours worked = 120, cost 76.5 (2)

The LCM time would be 600: 
(1) Lamps = 80, hours worked = 50*12, cost= 21.25*12
(2)
 Lamps = x, hours worked = 120*5, cost= 76.5*12

Now one can use the same equation as you did above [ie. (time*price)/lamps]: \(\frac{(50*12*21.25*12)}{80}=\frac{(120*5*76.5*5)}{x}\)
Solving for x will result in 120, but it muddies things. 

The easiest way would be to solve for the number of lamps required is to set up the equation to be for price per hour worked. As an equation that would be: (cost)/(number of lamps*hours)

\(\frac{21.25}{80*50} = \frac{76.5}{120*x}\)

\(21.25x = \frac{76.5*50*80}{120}\) Multiply through by 4

\(85x = \frac{76.5*50*80}{30}\)

\(x = \frac{(76.5*80)}{51}\)

\(x = 1.5*80\)

\(x = 120\)


ANSWER B ­
­Thanks, I recall now that the question was incorrectly posted and the solution was based on that assumption where I backtrack to find the missing information. 

Also, it was 3 years ago I don't remember much, what I did and how I did. So, your answer is correct.
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Can anyone help me tackle this question?

Here's how far I got >

80 lamps * 50 hours = 4000 "work"

4000 "work" = $21.25

Next, x lamps * 120 hours = $76.50

At this point, I wanted to find the value of $1 or 1 unit of work but I'm thoroughly confused and might be wrong
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saynchalk
If 80 lamps can be lighted, 5 hours per day for 10 days for $21.25, then the number of lamps, which can be lighted 4 hours daily for 30 days, for $76.50, is

(A) 100
(B) 120
(C) 150
(D) 160
(E) 180

Can anyone help me tackle this question?

Here's how far I got >

80 lamps * 50 hours = 4000 "work"

4000 "work" = $21.25

Next, x lamps * 120 hours = $76.50

At this point, I wanted to find the value of $1 or 1 unit of work but I'm thoroughly confused and might be wrong
4,000 hours cost $21.25. Hence, 1 hour costs $21.25/4,000.
120x hours cost $76.50. Hence, 1 hour costs $76.50/(120x).

Equating the above gives 21.25/4,000 = 76.50/(120x), which yields x = 120.

Answer: B.
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Bunuel

saynchalk
If 80 lamps can be lighted, 5 hours per day for 10 days for $21.25, then the number of lamps, which can be lighted 4 hours daily for 30 days, for $76.50, is

(A) 100
(B) 120
(C) 150
(D) 160
(E) 180

Can anyone help me tackle this question?

Here's how far I got >

80 lamps * 50 hours = 4000 "work"

4000 "work" = $21.25

Next, x lamps * 120 hours = $76.50

At this point, I wanted to find the value of $1 or 1 unit of work but I'm thoroughly confused and might be wrong
4,000 hours cost $21.25. Hence, 1 hour costs $21.25/4,000.
120x hours cost $76.50. Hence, 1 hour costs $76.50/(120x).

Equating the above gives 21.25/4,000 = 76.50/(120x), which yields x = 120.

Answer: B.
­Thanks Bunuel!!! :D :D
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