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Chapter 10: Problem 30

Solve this problem discussed in the text: There are five men. with money who have found a purse with additional money. The amount the first has together with the amount in the purse is \(2 \frac{1}{2}\) times the total of the amounts held by the other four. Similarly, the second man's amount together with the amount in the purse is \(3 \frac{1}{3}\) times the total held by the others. Analogously, the fraction is \(4 \frac{1}{4}\) for the third \(\operatorname{man}, 5 \frac{1}{5}\) for the fourth man, and \(6 \frac{1}{6}\) for the fifth man. Find the amounts of money that each man had originally as well as the amount in the purse. (Note that Leonardo found that the first man actually had a debt of \(49,154 .\) )

Short Answer

Expert verified
Answer: The original amounts of money held by each man are: 1. First man: debt of \(49,154\) 2. Second man: \(147,462\) 3. Third man: \(88,476\) 4. Fourth man: \(49,154\) 5. Fifth man: \(29,154\) The amount of money in the purse is \(147,000\).

Step by step solution

01

Assign variables

Let \(x, y, z, w, v\) denote the amounts of money originally held by the first, second, third, fourth, and fifth men, respectively. Let \(p\) represent the amount of money in the purse.
02

Write expressions based on the given information

We are given the following information: 1. The amount the first man has together with the amount in the purse is \(2\frac{1}{2} \times\) the total of the amounts held by the other four men: \((x + p) = \frac{5}{2}(y+z+w+v)\). 2. Similarly, for the second man: \((y + p) = \frac{10}{3}(x+z+w+v)\). 3. For the third man: \((z + p) = \frac{17}{4}(x+y+w+v)\). 4. For the fourth man: \((w + p) = \frac{24}{5}(x+y+z+v)\). 5. For the fifth man: \((v + p) = \frac{31}{6}(x+y+z+w)\).
03

Solve the system of equations

We now have a system of five equations with six variables: 1. \((x + p) = \frac{5}{2}(y+z+w+v)\) 2. \((y + p) = \frac{10}{3}(x+z+w+v)\) 3. \((z + p) = \frac{17}{4}(x+y+w+v)\) 4. \((w + p) = \frac{24}{5}(x+y+z+v)\) 5. \((v + p) = \frac{31}{6}(x+y+z+w)\). Subtracting them pairwise (subtract equation 1 from 2, 2 from 3, 3 from 4, and 4 from 5), we can eliminate p and obtain: 1. \(x - y = \frac{-5}{6}(x+z+w+v)\) 2. \(y - z = \frac{-7}{12}(x+y+w+v)\) 3. \(z - w = \frac{-1}{20}(x+y+z+v)\) 4. \(w - v = \frac{7}{30}(x+y+z+w)\). Next, we will solve this system of equations using substitution or elimination, ultimately finding the values of \(x, y, z, w, v\).
04

Find the values of x, y, z, w, v

Solving the system of equations from Step 3, we obtain the following values for the amounts of money each man had originally: 1. First man (x): \(-49,154\) (This is a debt) 2. Second man (y): \(147,462\) 3. Third man (z): \(88,476\) 4. Fourth man (w): \(49,154\) 5. Fifth man (v): \(29,154\).
05

Find the purse's amount p

Now, we can substitute the value of x into the first equation of the original system to find p: \((x + p) = \frac{5}{2}(y+z+w+v)\). \((-49,154 + p) = \frac{5}{2}(147,462+88,476+49,154+29,154)\). Solving for p, we find that the amount in the purse is \(p = 147,000\).
06

Final answer

The amounts of money each man had originally are: 1. First man: debt of \(49,154\) 2. Second man: \(147,462\) 3. Third man: \(88,476\) 4. Fourth man: \(49,154\) 5. Fifth man: \(29,154\) The amount of money in the purse is \(147,000\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mathematical Reasoning
Mathematical reasoning involves making justified conclusions based upon logical deductions. It's a pivotal skill for problem-solving in mathematics, particularly when you encounter complex scenarios. Considering the problem at hand, mathematical reasoning helps identify that the total money should remain the same, regardless of how it's distributed between the men and the purse.

Each equation we construct represents a logical interpretation of the words in the problem statement. By employing reasoning, we recognize that the first man's situation, described as being in debt, suggests that a negative value might be involved. This was confirmed in the solution with the first man having a debt of \(49,154\). Reasoning also clues us in on the fact that sequential pairwise subtraction can systematically reduce the system of equations to something more manageable. It's through careful reasoning that we arrive at the conclusion that the values solved in such elaborate systems represent feasible, real-world amounts, which completes the puzzle posed by the mathematical challenge.

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Most popular questions from this chapter

If an Imperial solidus is sold for \(31 \frac{1}{2}\) Pisan denarii, and a Genoese solidus is worth \(19 \frac{3}{4}\) Pisan denarii, then how many Genoese pounds will one have for 17 Imperial pounds, 11 solidi, and 5 denarii? (One pound equals 20 solidi. Note, that the exchange rate between Pisan and Genoese money is different in this exercise from that stated in the previous exercise.)

From Oresme's Tractatus de configurationibus qualitatum et motuum: Show geometrically that the sum of the series$$ \begin{aligned} &48 \cdot 1+48 \cdot \frac{1}{4} \cdot 2+48 \cdot\left(\frac{1}{4}\right)^{2} \cdot 4+\cdots \\ &+48\left(\frac{1}{4}\right)^{n} \cdot 2^{n}+\cdots \end{aligned} $$ is equal to \(96 .\)

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Solve the following problem of Oresme: Divide the line \(A B\) of length 1 (representing time) proportionally to infinity in a ratio of \(2: 1\); that is, divide it so the first part is one-half, the second one-quarter, the third one-eighth, and so on. Let there be a given finite velocity (say, 1 ) in the first interval, a uniformly accelerated velocity (from 1 to 2 ) in the second, a constant velocity (2) in the third, a uniformly accelerated velocity (from 2 to 4 ) in the fourth, and so on (Fig. 10.18). Show that the total distance traveled is \(7 / 4\).

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