A lunar base monitors its energy gain from solar panels and energy use

Question

A lunar base monitors its energy gain from solar panels and energy use across 8 equal 3-hour intervals over a 24-hour day (interval n = 1 to 8). The amount of energy gained (in kilowatt-hours) in interval n is given by:

Gained(n) = 400 – 100 * |n – 4|

The amount of energy used (in kilowatt-hours) in interval n is given by:

Used(n) = 120 + 30 * min(n, 9 – n)

If A is the total energy gained over the full day, and B is the maximum net surplus (Gained – Used) during any single interval, select one value for each from the list below.

mvgkannan · Answer

I think option for B should be 250 rather than 240.

poojaarora1818 · Answer

Solution:

Gained  
 Used 
 Net Surplus 
 
  1 
 400-100*3=100 
 120+30*min(1,8)=150 
 100-150=-50 
 
  2 
 400-100*2=200  
 120+30*min(2,7)=180  
 200-180=20 
 
  3 
 400-100*1=300  
 120+30*min(3,6)=210  
 300-210=90 
 
  4 
 400-100*0=400  
 120+30*min(4,5)=240 
 400-240=160 
 
  5 
 400-100*1=300  
 120+30*min(5,4)=240 
 300-240=60 
 
  6 
 400-100*2=200  
 120+30*min(6,3)=210 
 200-210=-10 
 
  7 
 400-100*3=100  
 120+30*min(7,2)=180  
 100-180=-80 
 
  8 
 400-100*4= 0  
 120+30*min(8,1)=150  
 0-150=-150

If we add the total of gained, then we will get 1600, and the maximum net surplus any single day, as per the calculation, is the 4th interval, which is 160, and anything above 160 is 240. I am not sure about the second choice.

SKaur3 · Answer

I got a similar solution to  poojaarora1818 , but I am also unsure about B.

Yosemite98 · Answer

https://gmatclub.com/forum/download/file.php?id=84331 Not sure about B, with a quick graph it is super easy to understand that the major difference will be at n=4 which indeed is 160 GMAT-Club-Forum-smat33zi.png

sarthak19203 · Answer

Can you please take a look at the OA for B? i'm not really sure how did we come up with 240, because minimum 400-120 =280 and then we would atleast have 30 subtracted from it (which is not consistent honestly, because to subtract 30 via used, we would also be getting a -300 factor on the gained. The maximum that i can see is at n=4, and that would require the factor to be 160.

Bismuth83 · Answer

1. Given two functions over n = 1 to 8, Gained(n) and Used(n), we need to  figure out the sum of all the Gained(n) and the maximum of Gained(n) - Used(n) .

2. Let's first write all the values of the functions in a chart:

n 
 Gained(n) 
 Used(n) 
 Gained(n) - Used(n) 
 
  1 
 100 
 150

2 
 200 
 180

3 
 300 
 210

4 
 400 
 240

5 
 300 
 240

6 
 200 
 210

7 
 100 
 180

8 
 0 
 150

3. The sum of all Gained(n) is equal to  100 + 200 + 300 + 400 + 300 + 200 + 100 + 0 = 2 * 600 + 400 = 1600 .

4. Now, let's take the difference of Gained(n) and Used(n):

n 
 Gained(n) 
 Used(n) 
 Gained(n) - Used(n) 
 
  1 
 100 
 150 
 -50 
 
  2 
 200 
 180 
 20 
 
  3 
 300 
 210 
 90 
 
  4 
 400 
 240 
 160 
 
  5 
 300 
 240 
 60 
 
  6 
 200 
 210 
 -10 
 
  7 
 100 
 180 
 -80 
 
  8 
 0 
 150 
 -150

The maximum out of all the values is 160.

5. Our answer will be:   D1 - 1600 and D2 - 160  .

Bismuth83 · Answer

Thank you for noticing this! The answer choices have been changed to include 160.

A lunar base monitors its energy gain from solar panels and energy use

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My Rewards

	Gained	Used	Net Surplus
1	400-100*3=100	120+30*min(1,8)=150	100-150=-50
2	400-100*2=200	120+30*min(2,7)=180	200-180=20
3	400-100*1=300	120+30*min(3,6)=210	300-210=90
4	400-100*0=400	120+30*min(4,5)=240	400-240=160
5	400-100*1=300	120+30*min(5,4)=240	300-240=60
6	400-100*2=200	120+30*min(6,3)=210	200-210=-10
7	400-100*3=100	120+30*min(7,2)=180	100-180=-80
8	400-100*4= 0	120+30*min(8,1)=150	0-150=-150

n	Gained(n)	Used(n)	Gained(n) - Used(n)
1	100	150
2	200	180
3	300	210
4	400	240
5	300	240
6	200	210
7	100	180
8	0	150