ch1.4.3-solution-of-questions

Ch 1.4.3 — Of the Solution of Questions relating to the preceding Chapter

Summary: Twenty-one fully worked word problems illustrate the equation-setup method: partitioning sums, inheritance with chained conditions, arithmetic progressions, trade and pricing, the identical equation, and the elegant “tenth-part” inheritance problem.

Sources: chapter-1.4.3

Last updated: 2026-05-03

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Partitioning a sum (§585–586)

Q1–Q2: Divide a number into two parts whose difference is known.

Let the greater part be $x$; then the smaller is $(a-x)$. The condition $x-(a-x)=b$ gives

$x=\frac{a+b}{2},a-x=\frac{a-b}{2}.$

The greater part equals half the sum plus half the difference; the lesser equals half the sum minus half the difference. This theorem reappears in ch1.4.4-resolution-two-or-more-equations as the two-unknown version.

Inheritance with chained offsets (§586–590)

The standard setup: express every share in terms of the smallest (or the total), sum all shares, equate to the known total.

Q3 (§586): Three sons share £1600; eldest has £200 more than second, second £100 more than youngest. Let youngest’s share = $x$: then second = $x+100$, eldest = $x+300$. Sum: $3x+400=1600$, so $x=400$.

Q4 (§587): Four sons, more complex chaining ($\text{eldest}=2\cdot \text{second}-100$, etc.). Reducing to one variable in the youngest: $41x-3700=8600\Rightarrow x=300$.

Q6–Q7 (§589–590): Shares expressed as fractions of the unknown total $x$:

$\frac{1}{2}x+\frac{1}{3}x+\frac{1}{4}x-2400=x\Rightarrow \frac{1}{12}x=2400\Rightarrow x=28800.$

This requires collecting $x$ on both sides before solving.

Arithmetic-progression problems (§595–596)

Q12 (§595): Divide 48 into 9 parts forming an arithmetical progression with common difference $\frac{1}{2}$. The first term is $x$, the ninth is $x+4$. Applying the sum formula $S=\frac{n(a+z)}{2}$:

$9x+18=48\Rightarrow x=3\frac{1}{3}.$

Q13 (§596): Find a progression with first term 5, last term 10, sum 60. Two-step solution:

1. Let the number of terms be $x$. Use $S=\frac{(5+10)x}{2}=60\Rightarrow x=8$.
2. Let the common difference be $z$. The eighth term is $5+7z=10\Rightarrow z=\frac{5}{7}$.

Links to arithmetical-progressions and ch1.3.3-arithmetical-progressions.

The identical equation (§597)

Q14: A sequence of operations (double, subtract 1, double, subtract 2, divide by 4) on $x$ is claimed to yield $x-1$. Tracing through:

$x\stackrel{\times 2}{\to }2x\stackrel{-1}{\to }2x-1\stackrel{\times 2}{\to }4x-2\stackrel{-2}{\to }4x-4\stackrel{÷4}{\to }x-1.$

The “equation” $x-1=x-1$ is trivially true for every $x$. Euler calls this an identical equation and notes it shows $x$ is indeterminate — any value satisfies it.

Trade and proportional problems (§598–603)

Q15 (§598): Cloth bought at 7 crowns per 5 ells, sold at 11 crowns per 7 ells, gain = 100 crowns. Using the Rule of Three for both cost and revenue:

$\text{cost}=\frac{7x}{5},\text{revenue}=\frac{11x}{7}.$

$\frac{11}{7}x=\frac{7}{5}x+100\Rightarrow \frac{6}{35}x=100\Rightarrow x=583\frac{1}{3}\text{ ells}.$

Q19 (§602): General banker-change problem. $a$ pieces of kind A = 1 crown; $b$ pieces of kind B = 1 crown; customer wants $c$ pieces total worth exactly 1 crown. Give $x$ of kind A and $(c-x)$ of kind B:

$\frac{x}{a}+\frac{c-x}{b}=1\Rightarrow x=\frac{a(b-c)}{b-a},c-x=\frac{b(c-a)}{b-a}.$

The tenth-part inheritance (§604)

Q21: The most intricate problem in the chapter. A father’s estate is divided so that:

• 1st child takes £100 plus one-tenth of the remainder;
• 2nd child takes £200 plus one-tenth of the new remainder;
• etc., each child receiving an equal share $x$.

Let total fortune = $z$. Computing the difference between consecutive shares and setting it to zero gives $x=900$. Substituting back yields $z=8100$, and $z/x=9$ children. This anticipates the multi-variable techniques of ch1.4.4-resolution-two-or-more-equations.

Practice questions (pp. 187–189)

25 problems with answers, including:

No.	Problem	Answer
9	Trader doubles savings for 3 years, stock doubles	£1480
14	Cistern filled in 12 min by two pipes, 20 min by one	30 min (other pipe alone)
16	Privateer at 10 mph overtakes ship at 8 mph, 18 miles ahead	Ship runs 72 miles
20	Three-digit number with AP digits, quotient by digit-sum = 48, subtract 198 inverts digits	432
25	Hare 50 leaps ahead; greyhound’s 2 leaps = hare’s 3; hare takes 4 leaps per 3 of greyhound	300 leaps

Related pages

• ch1.4.1-solution-of-problems-in-general
• ch1.4.2-resolution-simple-equations
• ch1.4.4-resolution-two-or-more-equations
• equations
• arithmetical-progressions
• geometrical-proportion
• compound-interest