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18 Apr 2025, 09:54
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
74% (02:33) correct
26%
(02:25)
wrong
based on 126
sessions
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Not Attempted Yet
A private investigator charges a total of t dollars for the first 7 days of any investigation and 0.3t dollars in additional fees for each additional day of investigation. Did a specific investigation cost more than $3,000?
(1) The investigator charged a total of 4t dollars for the investigation.
(2) If the investigation had lasted for a total of 20 days, the investigator would have charged $900 more than what he did.
1000112321.jpg [ 1.74 MiB | Viewed 959 times ]
(1) The investigator charged a total of 4t dollars for the investigation.
(2) If the investigation had lasted for a total of 20 days, the investigator would have charged $900 more than what he did.
Attachment:
1000112321.jpg [ 1.74 MiB | Viewed 959 times ]
18 Apr 2025, 10:06
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It looks like you had a typo in the set up so I have modified and changed & to be t for the initial investigation cost
18 Apr 2025, 10:13
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From question, We have investigation cost as fixed charge + variable charge(if investigation days >7 )
Let investigation duration be d days
Total Cost = Fixed + Variable cost = t + 0.3t(d - 7)
Statement A:
Total Cost = 4t => t + 0.3t(d-7) = 4t => d = 17
total cost > 4000 -> We don't know ,because we still don't know t
Statement B:
Total Cost(20 Days) - Total Cost(actual d days) = 900
(t + 0.3t(20-7)) - (t+0.3t(d-7)) = 900 ->one equation, 2 variables (t,d)
Can't infer total cost > 4000
Both Statement A & B:
From Statement A, d = 7
We can put d in equation of statement 2 ,to get t =>From t ,we can calculate actual cost
So, Statement A& B together are sufficient
Let investigation duration be d days
Total Cost = Fixed + Variable cost = t + 0.3t(d - 7)
Statement A:
Total Cost = 4t => t + 0.3t(d-7) = 4t => d = 17
total cost > 4000 -> We don't know ,because we still don't know t
Statement B:
Total Cost(20 Days) - Total Cost(actual d days) = 900
(t + 0.3t(20-7)) - (t+0.3t(d-7)) = 900 ->one equation, 2 variables (t,d)
Can't infer total cost > 4000
Both Statement A & B:
From Statement A, d = 7
We can put d in equation of statement 2 ,to get t =>From t ,we can calculate actual cost
So, Statement A& B together are sufficient
