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06 Dec 2023, 18:59
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C
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Difficulty:
75%
(hard)
Question Stats:
63% (02:59) correct
37%
(03:04)
wrong
based on 428
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At a certain farm , all of the 240 animals are either cows or pigs, and 180 of the animals have been fed today. If less than 1/3 of the animals are cows, and less than 1/3 of the pigs have not been fed today, what is the greatest number of cows that could have been fed today?
A) 53
B) 61
C) 72
D) 79
E) 108
A) 53
B) 61
C) 72
D) 79
E) 108
06 Dec 2023, 20:16
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Total number of Animals = 240
Total Number of Cow < 80
Therefore, total number of Pigs > (240-80) = 160 (Fed Pigs + Unfed Pigs)
Since, 1/3 of the pigs have not been fed today:
Number of unfed Pigs < 160/3 = 53.33
Since number of unfed pigs have to be a whole number,
Unfed Pigs <=53
Since, FedPigs + Unfed Pigs >160
Fed Pigs > (160-53) = 107
Since it is given that total number of fed animals = 180 =FedCows + FedPigs
180 - FedCows > 107
Therefore, FedCows < 73
Therefore Max possible number of FedCows = 72.
Total Number of Cow < 80
Therefore, total number of Pigs > (240-80) = 160 (Fed Pigs + Unfed Pigs)
Since, 1/3 of the pigs have not been fed today:
Number of unfed Pigs < 160/3 = 53.33
Since number of unfed pigs have to be a whole number,
Unfed Pigs <=53
Since, FedPigs + Unfed Pigs >160
Fed Pigs > (160-53) = 107
Since it is given that total number of fed animals = 180 =FedCows + FedPigs
180 - FedCows > 107
Therefore, FedCows < 73
Therefore Max possible number of FedCows = 72.
General Discussion
07 Dec 2023, 18:16
Kudos
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Draw the double set matrix to help you visualize the following:
The keys to maximizing the number of fed cows are to:
1) maximize the total number of cows and fed animals, and
2) minimize the other components in those rows and columns; in other words, minimize the number of unfed cows and fed pigs.
There are 240 animals on the farm, and less than a third are cows. So the number of cows is below 80. But we want to maximize the total number of cows, because we want to eventually max the number of fed cows. So set the number of total cows to 79, the maximum allowed by the problem.
That leaves 161 pigs on the farm.
If less than 1/3 of those pigs are unfed, that means more than 2/3 were fed. But we want to minimize the number of pigs that were fed, because that will allow us to max out the number of cows that were fed. Remember, the problem tells us that 180 of the animals on the farm were fed, and we want as many of them to be cows in this problem.
2/3 of 161 is just over 107, so the least possible value for unfed pigs is 108. If there are 108 pigs that were fed, that leaves room for 72 cows to have been fed.
Answer is C
The keys to maximizing the number of fed cows are to:
1) maximize the total number of cows and fed animals, and
2) minimize the other components in those rows and columns; in other words, minimize the number of unfed cows and fed pigs.
There are 240 animals on the farm, and less than a third are cows. So the number of cows is below 80. But we want to maximize the total number of cows, because we want to eventually max the number of fed cows. So set the number of total cows to 79, the maximum allowed by the problem.
That leaves 161 pigs on the farm.
If less than 1/3 of those pigs are unfed, that means more than 2/3 were fed. But we want to minimize the number of pigs that were fed, because that will allow us to max out the number of cows that were fed. Remember, the problem tells us that 180 of the animals on the farm were fed, and we want as many of them to be cows in this problem.
2/3 of 161 is just over 107, so the least possible value for unfed pigs is 108. If there are 108 pigs that were fed, that leaves room for 72 cows to have been fed.
Answer is C
