At a mountain observatory, astronomers set exposure times for two type

Question

At a mountain observatory, astronomers set exposure times for two types of targets. Each faint target requires R seconds per image, and each bright target requires a shorter exposure of S seconds per image. On Night 1, the team captured 35 images using a total of 625 seconds. On Night 2, they captured 140 images using a total of 1730 seconds. On both nights, both faint and bright targets were captured.

Select for R and S values for R and S that are jointly consistent with the information provided. Make only two selections, one in each column.

Select for  R  and  S  values for R and S that are jointly consistent with the information provided. Make only two selections, one in each column.

Bunuel · Accepted Answer

Night 1 average = 625/35 ≈ 17.9

Night 2 average = 1730/140 ≈ 12.4

So S must be below both averages and R must be above both. Since R > S, that narrows to R = 19 or 21, and S = 10 or 12.

Night 1 equation is xR + (35 - x)S = 625, where x is the number of faint images on Night 1.

First test R = 19. Substituting gives 19x + S(35 - x) = 625. For S = 10, 12, or 14, this does not give an integer solution for x, so R cannot be 19.

Now test R = 21. Substituting gives 21x + S(35 - x) = 625.

For S = 10, this gives x = 25, which is an integer solution.

For S = 12 or 14, no integer solution exists.

Therefore the only consistent pair is  R = 21  and  S = 10 .

I29-172

AbhishekP220108 · Answer

Let faint target be "a" and bright target be "b" and we know R>S

a1+b1=35
a1=35-b1.......i
Ra1+Sb1=625.....ii
R(35-b1)+Sb1=625

Simplifying this, we get

b1=(35R-625)/(R-S)

similarly
a2+b2=140
a2=140-b2.......iii
Ra2+Sb2=1730.......iv
R(140-b2)+Sb2=1730

Simplifying this, we get

b2=(140R-1730)/(R-S)

By looking at the information one thing we know that both b1 and b2 are integral values. So lets try by inserting the values
since R>S, lets try with larger R which is 21

inserting R1 in b1 we get

b1=(35*21-625)/21-S
b1=110/21-S

b2=(140*21-1730)/21-S
b2=1210/21-S

for both b1 and b2 to be integer if we insert S as 10, we got the value of b1 as 10 and b2 as 110. 
accordingly a1=25 and a2=30

Inserting these value of a1, b1, a2, and b2 in equation ii & iv our equation is justified.

Hope it helps

Vaishbab · Answer

I ended up having to resort to plugging in the values to find a possible answer to this.

But, initially, it was important to figure out the average exposure per image on each night. This helped eliminate a lot of combinations and made my work easier.

On night 1 :
x=number of faint targets & y=number of bright targets
which means,  xR + yS = 625  ----- (1)
and with a total of 35 images,  x + y = 35  ------ (2)

On night 2 :
a=number of faint targets & b=number of bright targets
therefore,  aR + bS = 1730  ----- (3)
&  a + b = 140  ----- (4)

On night 1, with a total of 625 seconds for 35 images, the average exposure per image was 17.86 seconds. For night 2, with 1730 seconds for 140 images, it was 12.35 seconds per image.

With this information, we can identify the R & S pairs. Both averages  have  to be between the R & S values. With this condition, we can eliminate most possibilities, barring: (R,S) = (19,10), (21,10), (19,12) & (21,12)

Plugging R=19 & S=19 and using equations (1) & (2) in night 1, I couldn't get integer values for x & y. So that one's eliminated.

Then I tried R=21 & S=10 using (1) & (2), I got x=25 & y=10. 
Tried the same R & S values for (3) & (4) and I got a=30 & b=110.

So, my pick is R=21 and S=10.

I would have never figured this out within 3 mins though.

Parag22jain · Answer

MY strategy is to skip this question as this will take 10 mins to solve by any method.

GMATCoachBen · Answer

Parag22jain Indeed, it is a good strategy for almost everyone to skip some questions on DI if they look particularly long and hard, and remember that we can always come back to them if we have extra time. It's essential to make sure we finish and get all of the lower hanging fruit, since we are penalized much more for missing easier questions. Students often don't realize that the difficulty fluctuates a lot throughout the test, even when you're scoring very high, so you can often get some questions at the end that aren't that hard. As an example, my data from a DI 90 practice test is attached; I got 2 easy, 9 medium, and 9 hard questions. Long and hard two-part analysis questions are often good candidates to skip. Statistically, this question is harder than any official focus edition practice test question, with only 21% accuracy. I was curious, so I looked at the 29 805+ TPA practice test questions; only one was below 30% accuracy, at 28%: https://gmatclub.com/forum/a-city-counc ... 26193.html The others are all in the 30's, 40's, or 50's for % accuracy. The advanced search here is a very useful tool: https://gmatclub.com/forum/advanced-search/ For practice questions, I recommend the "Official Guide" and "GMAT Prep (Classic)" tags for the source.

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