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22 Sep 2025, 04:33
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R: 21
S: 10
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
19% (02:21) correct
81%
(03:31)
wrong
based on 95
sessions
History
Date
Time
Result
Not Attempted Yet
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.
| R | S | |
| 10 | ||
| 12 | ||
| 14 | ||
| 15 | ||
| 19 | ||
| 21 |
ShowHide Answer
Official Answer
R: 21
S: 10
30 Sep 2025, 01:30
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Bunuel
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, 12, or 14.
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
General Discussion
23 Sep 2025, 07:23
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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
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
Bunuel
