# Problem Solving (Fated to earn money) Question that has no explanation yet

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Hi, I came up with a word problem that I know the answer to but I have no idea why the solution is what it is or how to reason through it. I posted this on another forum before, but I did not get the answer (explanation) I was looking for.

Joe has investments in Company A, Company B, and Company C.

Joe is fated to earn $25.00 from Company A within 2 days from now. Joe is fated to earn$45.00 from Company B within 3 days from now.

Joe is fated to earn $100.00 from Company C within 5 days from now. Joe is fated to earn no more than$26.00 from Company C and Company B on day 1 (1 day from now).

Joe is fated to earn at least $14.00 from Company A and Company C on day 2 (2 days from now). Joe has to earn twice the amount of money on the first day than the second day from Companies A, B, and C and twice the amount of money on the second day than the third day from Companies A, B, and C. This can be expressed algebraically as Joe earning x money on day 3 (3 days from now), 2x money on day 2 (2 days from now), and 4x money on day 1 (1 day from now). Joe can earn whatever amount of money (that satisfies the other conditions) from Companies A, B, and C on day 4 and day 5 (4 and 5 days from now). What is the lowest amount of money Joe can earn on day 1 (1 day from now) from Companies A, B, and C? Explain your reasoning. P.S. How come there doesn't seem to be good formulas to use for this question? #### Share this post ##### Link to post ##### Share on other sites I think you need to give us what you've got so far, because this sounds like homework. #### Share this post ##### Link to post ##### Share on other sites 1 hour ago, Yukang said: P.S. How come there doesn't seem to be good formulas to use for this question? Some of the relationships do not lend themselves to simple algebra, since they are conditional statements, and deal with inequalities. But some do. You can assign a variable to the amount of money each company earns on which day and set that equal to given amounts (e.g. A1 is the amount earned by company A on day 1, B2 is the amount of money B earns on day 2, etc.) There are relationships between some of these variables. Whether there is a single solution at the end of that, I'm not yet sure. In some problems you can make a graph and exclude certain answers, as they violate inequalities. #### Share this post ##### Link to post ##### Share on other sites Hi, I came up with the answer$44 through guess and check. Because the earnings are $26 for C & B on day 1, that leaves$18 for A on day 1 and 25-18=7 $7 dollars on day 2 for A. You want the contributions from Company C to be as low as possible so,$14 from C & A 2 days from now means the contributions from Company C is at least 14-7=7 $7. When setting the contributions for Company C in every other day to 0 until day 4 or 5, you get a total of$44 on day 1.

However, I have no idea how to solve the problem using step by step reasoning or even graphical analysis.

The problem is actually mine, and I tried to make it as hard as possible while not involving big numbers or difficult formulas.

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A1 + A2 = 25

B1 + B2 + B3 = 45

C1 + C2 + C3 +C4 +C5 = 100

(A1 + B1 +C1) = 2(A2 + B2 + C2) = 4(A3 + B3 + C3)

B1 + C1 <= 26

A2 + C2 >= 14

You could rewrite the third equation as C1 + C2 + C3 <= 100, in case that helps, since you don't need the C4 and C5 variables

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1 hour ago, swansont said:

A1 + A2 = 25

B1 + B2 + B3 = 45

C1 + C2 + C3 +C4 +C5 = 100

(A1 + B1 +C1) = 2(A2 + B2 + C2) = 4(A3 + B3 + C3)

B1 + C1 <= 26

A2 + C2 >= 14

You could rewrite the third equation as C1 + C2 + C3 <= 100, in case that helps, since you don't need the C4 and C5 variables

Thanks, that looks about right. Looks like 8 equations with 9 unknowns after you throw away the C4 & C5 variables? Is 8 equations with 9 unknowns solvable? I thought you need 9 equations? This is assuming the 3 way equal equation can be arranged into 3 equations:

I think the 3 way equal equations may be reversed in their coefficient though

4(A1 + B1 +C1) = 2(A2 + B2 + C2)

2(A2 + B2 + C2) = (A3 + B3 + C3)

4(A1 + B1 +C1) = (A3 + B3 + C3)

How can you solve 8 equations with 9 unknowns? (A1, A2, A3, B1, B2, B3, C1, C2, C3)

I know from guess and check A1 = 7, A2= 18, A3 = 0, B1 = 26, etc. (and the rest can be filled out), but isn't it strange there are only 8 equations?

Edited by Yukang

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49 minutes ago, Yukang said:

Thanks, that looks about right. Looks like 8 equations with 9 unknowns after you throw away the C4 & C5 variables? Is 8 equations with 9 unknowns solvable? I thought you need 9 equations? This is assuming the 3 way equal equation can be arranged into 3 equations:

For an exact solution, yes. But you are asking for a maximum value. There are an infinite number of values that satisfy the conditions, and you're picking the biggest one.

49 minutes ago, Yukang said:

I think the 3 way equal equations may be reversed in their coefficient though

4(A1 + B1 +C1) = 2(A2 + B2 + C2)

2(A2 + B2 + C2) = (A3 + B3 + C3)

4(A1 + B1 +C1) = (A3 + B3 + C3)

You said the day 1 income was twice as big as day 2, so (A1 + B1 +C1) = 2(A2 + B2 + C2)

49 minutes ago, Yukang said:

How can you solve 8 equations with 9 unknowns? (A1, A2, A3, B1, B2, B3, C1, C2, C3)

I know from guess and check A1 = 7, A2= 18, A3 = 0, B1 = 26, etc. (and the rest can be filled out), but isn't it strange there are only 8 equations?

You will get a range of values for solutions, rather than a single value.

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54 minutes ago, swansont said:

For an exact solution, yes. But you are asking for a maximum value. There are an infinite number of values that satisfy the conditions, and you're picking the biggest one.

You said the day 1 income was twice as big as day 2, so (A1 + B1 +C1) = 2(A2 + B2 + C2)

You will get a range of values for solutions, rather than a single value.

Ok, my bad about day 1 income being twice as big as day 2, you're right. The question asks for the minimum on day 1, so I guess I look for the minimum of the range of values? I never done this type of math before. Are you saying all 9 variables will have a range of values for solutions and the answer would be the minimum of A1 + B1 + C1?

You helped a lot, thanks, but I still don't know what to do to reason through it 100%. There is only one answer right, because it asks for the minimum on day 1?

I also have no idea how to find a range of solutions for 8 equations with 9 unknowns, although I know how to find solutions for 9 equations with 9 unknowns.

Edited by Yukang

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