Leonhard Euler · 1770 (1828 English ed.) · Internet Archive
Elements of Algebra
All 627 leaves · leaf 237
Transcription
200 ELEMENTS SECT. IV.
E. C. E.
As $7 : 11 :: x : \frac{11}{7}x$ crowns;
and this receipt ought to exceed the expenditure by 100 crowns. We have, therefore, this equation:
$\frac{11}{7}x = \frac{7}{5}x + 100$.
Subtracting $\frac{7}{5}x$, there remains $\frac{6}{35}x = 100$; therefore $6x = 3500$, and $x = 583\frac{1}{3}$.
There were, therefore, $583\frac{1}{3}$ ells bought for $816\frac{2}{3}$ crowns, and sold again for $916\frac{2}{3}$ crowns; by which means the profit was 100 crowns.
599. Question 16. A person buys 12 pieces of cloth for $140l.$ of which two are white, three are black, and seven are blue: also, a piece of the black cloth costs two pounds more than a piece of the white, and a piece of the blue cloth costs three pounds more than a piece of the black. Required the price of each kind.
Let the price of a white piece be $x$ pounds; then the two pieces of this kind will cost $2x$; also, a black piece costing $x+2$, the three pieces of this color will cost $3x+6$; and lastly, as a blue piece costs $x+5$, the seven blue pieces will cost $7x+35$: so that the twelve pieces amount in all to $12x+41$.
Now, the known price of these twelve pieces is 140 pounds; we have, therefore, $12x+41 = 140$, and $12x = 99$; wherefore $x = 8\frac{1}{4}$. So that
A piece of white cloth costs $8\frac{1}{4}l$.
A piece of black cloth costs $10\frac{1}{4}l$.
A piece of blue cloth costs $13\frac{1}{4}l$.
600. Question 17. A man having bought some nutmegs, says that three of them cost as much more than one penny, as four cost him more than two pence halfpenny. Required the price of the nutmegs?
Let $x$ be the excess of the price of three nutmegs above one penny, or four farthings. Now, if three nutmegs cost $x+4$ farthings, four will cost, by the condition of the question, $x+10$ farthings; but the price of three nutmegs gives that of four in another way, namely, by the Rule of Three. Thus,
$3 : x+4 :: 4 : \frac{4x+16}{3}$.
So that $\frac{4x+16}{3} = x+10$; or, $4x+16 = 3x+30$; therefore $x+16 = 30$, and $x=14$.
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