Elements of Algebra

228 ELEMENTS SECT. IV. $\frac{3x-3}{2}$, if he had remained three months in the partnership: and as the profits gained in the same time are in proportion to the principals, we have the following proportion, $x : 99 - x :: 100 - x : \frac{3x-3}{2}$ And the equality of the product of the extremes to that of the means, gives the equation, $\frac{3x^2-3x}{2} = 9900 - 199x + x^2$; then multiplying this by 2, we have $3x^2 - 3x = 19800 - 398x + 2x^2$; and subtracting $2x^2$, we obtain $x^2 - 3x = 19800 - 398x$. Adding $3x$, gives $x^2 = 19800 - 395x$; then by the rule, $x = -\frac{395}{2} + \sqrt{(\frac{156025}{4} + \frac{79200}{1})} = -\frac{395}{2} + \frac{485}{2} = \frac{90}{2} = 45$. The first partner therefore contributed 45l. and the other 55l. The first having gained 54l. in three months, would have gained in one month 18l.; and the second having gained 44l. in two months, would have gained 22l. in one month: now these profits agree; for if, with 45l., 18l. are gained in one month, 22l. will be gained in the same time with 55l. 654. Question 9. Two girls carry 100 eggs to market; the one had more than the other, and yet the sum which they both received for them was the same. The first says to the second, If I had had your eggs, I should have received 15 pence. The other answers, If I had had yours, I should have received $6\frac{2}{3}$ pence. How many eggs did each carry to market? Suppose the first had $x$ eggs; then the second must have had $100-x$. Since, therefore, the former would have sold $100-x$ eggs for 15 pence, we have the following proportion: $(100-x) : 15 :: x : \frac{15x}{100-x}$ Also, since the second would have sold $x$ eggs for $6\frac{2}{3}$ pence, we readily find how much she got for $100-x$ eggs, thus: As $x : (100-x) :: \frac{20}{3} : \frac{2000-20x}{3x}$. Now, both the girls received the same money; we have