ALGEBRA. [KTAOTLM nc pHOORissioir, Era 



2. A merchant *old ome brandy for 39, and gained 

 M much per cent. M the brandy cost him. What wa 

 the original prioa t 



Let x - tho cost price, then tho gain was JT per 

 100, 



. . 100 : x : : x : * , tho number of pound* gained. 



But tho number of pound* gained wa* Jo39 * 

 g* 



* .'. x + lOOx - 3900 .-. z +100x + 50 J 



- 3900 + 2500 - 0400. 

 .. x + 60 -+80 .'. * -80 or 130. 



.-. tlio cost price wa* 30 ; the negative value of z, 

 though satisfying the algebraical condition*, being ex- 

 cluded by tho nature of the question. 



3. The sum of throe numbers in arithmetical progres- 

 sion is 9, and tho sum of thoir cubes 153. What aro the 

 numbers ? 



Let x y, 35, x + y represent tho numbers ; then 

 -i_ x 4- v = 3j; = 9 . . x = 3 



3 ~ ar3 - 4 .-. y = + 2, .-. tho numbers aro either 



1, 3, 6 ; or 5, 3, 1 ; being tho same set, whether the 

 common difference y be taken positively or negatively. 



4. A person travels a distance of 198 miles as follows : 

 namely, 30 miles the first day, 28 the second, 20 tho 

 third, and so on : in how many days will he have finished 

 tin- journey? 



Here wo have an arithmetical progression, in which 

 a = 30, d = 2, and S = 193, to find n, the number of 

 terms or days. 



By the formula, p. 478, S=g-n J2a + (n l)dj> 

 that is, 



- 2 n JOO (n 1)2} =31n n a .-. n 31n 



198 



- 198. 

 t .-. 2.131 



V (31* 792) 

 31_+ 13 _ 

 "2 " 



.n = 



22 or 9. 



It would seem from this result, that ho might arrive 

 at hia journey's end cither in 9 days, or in 22 days. But 

 you must always remember, in solving particular questions 

 by algebra, that the equation which embodies the conditions 

 of the question is not necessarily restricted to those con- 

 ditions. The solution will furnish aM tho values which can 

 satisfy tho equation; whereas, from tho limited natureof the 

 question, some of these may be inadmissible. You per- 

 ceive, in a moment, that tho question solved above is of 

 wiiler meaning than that proposed : it includes tho pro- 

 posed and something more ; for it is this namely. The 

 sum of an arithmetical series is 198, tho first term is 30, 

 and tho common difference 2: find tho number of terms. 

 The answer is perfectly general ; for the series consists 

 of either tho nine term* 30+28 + 20+24 + 22+20 

 + 18 + 1C + 14 198, or of the twenty-two terms fur- 

 iu-!ii-d by extending tho series to the thirteen additional 

 terms + 12 + 10 + 8 + + 4 + 2 +02 40 

 8 10 12 ; for you see that these additional terms 

 mount to ; and therefore that the sum of nine terms, 

 or of ttcenty-two, equally amounts to 193. It is plain, 

 therefore, that the algebra would have been imperfect if 

 n 22 had not been contained in the result, though the 

 limitation of the question excludes it : this limitation 

 ntly U, that each day's travel addt to the distance 

 from the starting-point : tho traveller is presumed not to 

 BO back, to that a lubtractive day's journey is for- 



6. The sum of three number* in geometrical progrcs- 



Ito* Quadratic EqutUom ; mti, p. 471. 



* JM., 1U1* 11. ; anlt, p. 474. 



ion i* 13 ; and the pro 1 net of the mean and sum of the 

 extreme* i* 30 ; required the number*. 



Lot the numbers be represented by -, x, and ry ; tho 



condition that they are to form a geometrical progres- 

 sion being thus secured. 

 Then the other two conditions are as follow namely, 



- + * 4- *y - 13 

 and (* + zy) x-30 

 From the first of these equations 



5 + *y=13-* ..... (A). 



From the second I- zy ,.-. 13 x = ' r 



.-. 13z 3? = 30, or x 1 IS* = 30, 

 Consequently, by Rule II. of quadratic equation*, 

 page 474, 



2x 13 = V ( 120 + 13 I )= J 49 - + 7, 



13 + 7 

 .-. x = y = 10, or 3. 



Therefore, substituting 3 for x in equation (A), we 

 have 9 - + 3y = 13 3 = 10 .-. 3 + 3y' = lOy, or 3y* 



lOy = 3 ; 

 j Gt/ 10 



( 30 + 100)= V<54= + 8. 

 10 + 8 1 



y = -~ = 3 > or 3 



Hence, putting 3 for x, and 3 for y, in -, z, xy, the 



numbers are 1, 3, 9 ; or putting 3 for x, and g for y, 



the numbers are 9, 3 ; 1 ; forming the same geometrical 

 progression written in reverse order. If, instead of 

 x = 3, the other value z = 10 had been taken, the cor- 

 responding values of y would have been imaginary, or 

 impossible : thus, putting 10 for z in (A), the equation 

 becomes 



- + 



= 13 10 = 3 .-. 10 + 10y s 

 or 10y s 3y = 10, 



.-. 20y 3 = V ( 400 + 9) = V 391, 

 3+ V :i!U 



These two values of ;/ are imaginary; BO that tho 

 three numbers, 1, 3, 9, determined above, aro the only 

 real numbers that fulfil tho conditions of the question. 



L Divide 49 into two parts such, that tho greater 

 increased by 6 may be to the loss diminished by 11 as 

 9 to 2. 



2. Two hundred stones are placed in a straight line at 

 intervals of 2 feet, the first stono being 20 yards in 

 advance of a basket : suppose a person starting from the 

 basket collects tho stones, and returns them ono by one 

 to tho basket. How much ground docs he go over ( 



3. The sum of four numbers in geometrical progres- 

 sion is equal to 1 added to the common ratio ; and -? T is 

 tho first term : required tho numbers. 



4. From two towns 165 milos apart, A and B set out 

 to meet each other. A travels 1 mile the first day, 2 

 the second, 3 tho third, and so on: B travels 20 milrt 

 the first day, 18 the second, 1C the third, and so on : in 

 how many days will they meet 1 



6. The sum of tho first and second of four numbers 

 in geometrical progression is 15, and tho sum of tho 

 third and fourth GO : what are the numbers I 



6. Tho sum of three numbers in geometrical progres- 

 sion is 35 ; and the mean term is to the difference of tho 

 extremes as 2 to 3 : what are tho numbers ? 



I Sec Quadratic Equations, Rule 11. ; anlr, p. 471. 



