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A farmer wants to allocate her expenditures to maximize her profits du
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28 Feb 2024, 22:26
28 Feb 2024, 22:26
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Select for Greatest returns the resource for which, according to the desired calculation from the model, each euro of spending beyond 100 euros is predicted to make the greatest contribution to profits (or least contribution to losses). And select for Least returns the resource for which, according to the desired calculation, each euro of spending beyond the minimum necessary amount is predicted to make the least contribution to profits (or greatest contribution to losses). Make only two selections, one in each column.
All Data Insight question: TPA [ Official Guide DI Review 2023-24]
A farmer wants to allocate her expenditures to maximize her profits during the coming year. Let C be her spending on cropland, F her spending on fertilizer, L her spending on labor, M her spending on machinery, and S her spending on seed. She has estimated that, if she allocates at least 100 euros to each of these categories, then her profit during the coming year will be (C – 97)(L – 92)(S – 95)(M – 87)(F – 90) euros. She wishes to calculate the benefits of expenditure greater than 100 euros in one these categories, if she were to spend exactly 100 euros in each of the others.Select for Greatest returns the resource for which, according to the desired calculation from the model, each euro of spending beyond 100 euros is predicted to make the greatest contribution to profits (or least contribution to losses). And select for Least returns the resource for which, according to the desired calculation, each euro of spending beyond the minimum necessary amount is predicted to make the least contribution to profits (or greatest contribution to losses). Make only two selections, one in each column.
| Greatest returns | Least returns | |
| Cropland | ||
| Fertilizer | ||
| Labor | ||
| Machinery | ||
| Seed |
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Official Answer
Greatest returns: Cropland
Least returns: Machinery
A farmer wants to allocate her expenditures to maximize her profits du
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08 Mar 2024, 01:06
08 Mar 2024, 01:06
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Since you are reading this post, its fair to assume that you have solved this question. Before I give you the video solution, I want you to tell me your experience with this question:
Now watch the solution and observe the following:
Would love to hear your thoughts on it.
- Did you read this question and got scared of the profit expression?
- Did you brave through the profit expression but reached the question stem and then gave up on this question?
- Or did you brave through the complex language and processed the question but ended up selecting the wrong answer?
Now watch the solution and observe the following:
- This question is the testament of the translate process skill – do not give up when you see dense language. Instead read that with the mindset of ‘owning the dataset’.
- This question exemplifies of drawing proper inferences and arriving at proper approach so that you can arrive at the correct answer.
Would love to hear your thoughts on it.
A farmer wants to allocate her expenditures to maximize her profits du
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06 Mar 2024, 22:06
06 Mar 2024, 22:06
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She has estimated that, if she allocates at least 100 euros to each of these categories, then her profit during the coming year will be (C – 97)(L – 92)(S – 95)(M – 87)(F – 90) euros.
Further all except one are 100, and this one is >100.
Let all be 100 => (C – 97)(L – 92)(S – 95)(M – 87)(F – 90)
(100 – 97)(100 – 92)(100– 95)(100 – 87)(100 – 90)
3 * 8 * 5 * 13 * 10
We have to make the above highest by increasing one of them, the one that is >100.
Let us check with extremes
3*(8*5*13*10) = 3*5200….if increased by 1, then 4*5200 = 20800
(3*8*5*10)*13= 1200*13….if increased by 1, then 14*1200 = 16800
So increasing C makes it the largest and increasing M makes it smallest.
Let us check logically too
Let us see what happens when we add some positive number x to each one by one.
A. 3 * 8 * 5 * 13 * 10 becomes (3+a) * 8 * 5 * 13 * 10 = 3 * 8 * 5 * 13 * 10 + a * 8 * 5 * 13 * 10
B. 3 * 8 * 5 * 13 * 10 becomes 3 * (a+8) * 5 * 13 * 10 = 3 * 8 * 5 * 13 * 10 + a * 3 * 5 * 13 * 10
C. 3 * 8 * 5 * 13 * 10 becomes 3 * 8 * (a+5) * 13 * 10 = 3 * 8 * 5 * 13 * 10 + a * 3 * 8 * 13 * 10
D. 3 * 8 * 5 * 13 * 10 becomes 3 * 8 * 5 * (a+13) * 10 = 3 * 8 * 5 * 13 * 10 + a * 3 * 8 * 5 * 10
E. 3 * 8 * 5 * 13 * 10 becomes 3 * 8 * 5 * 13 * (a+10) = 3 * 8 * 5 * 13 * 10 + a * 3 * 8 * 5 * 13
The colored portion is same for all, so concentrate on only the second part. In the second part 'a' replaces the number to which it is added. So, when a replaces smallest number 3, the product will be the maximum, and when a replaces the largest number 13, the product will be the least
Further all except one are 100, and this one is >100.
Let all be 100 => (C – 97)(L – 92)(S – 95)(M – 87)(F – 90)
(100 – 97)(100 – 92)(100– 95)(100 – 87)(100 – 90)
3 * 8 * 5 * 13 * 10
We have to make the above highest by increasing one of them, the one that is >100.
Let us check with extremes
3*(8*5*13*10) = 3*5200….if increased by 1, then 4*5200 = 20800
(3*8*5*10)*13= 1200*13….if increased by 1, then 14*1200 = 16800
So increasing C makes it the largest and increasing M makes it smallest.
Let us check logically too
Let us see what happens when we add some positive number x to each one by one.
A. 3 * 8 * 5 * 13 * 10 becomes (3+a) * 8 * 5 * 13 * 10 = 3 * 8 * 5 * 13 * 10 + a * 8 * 5 * 13 * 10
B. 3 * 8 * 5 * 13 * 10 becomes 3 * (a+8) * 5 * 13 * 10 = 3 * 8 * 5 * 13 * 10 + a * 3 * 5 * 13 * 10
C. 3 * 8 * 5 * 13 * 10 becomes 3 * 8 * (a+5) * 13 * 10 = 3 * 8 * 5 * 13 * 10 + a * 3 * 8 * 13 * 10
D. 3 * 8 * 5 * 13 * 10 becomes 3 * 8 * 5 * (a+13) * 10 = 3 * 8 * 5 * 13 * 10 + a * 3 * 8 * 5 * 10
E. 3 * 8 * 5 * 13 * 10 becomes 3 * 8 * 5 * 13 * (a+10) = 3 * 8 * 5 * 13 * 10 + a * 3 * 8 * 5 * 13
The colored portion is same for all, so concentrate on only the second part. In the second part 'a' replaces the number to which it is added. So, when a replaces smallest number 3, the product will be the maximum, and when a replaces the largest number 13, the product will be the least
