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 1 
 
 2 
 
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 4 
 
 5 
 
 6 
 

 ^"^ laEBEAic foemulj: 
 
 V 
 
 
 1 
 
 SHOWING THE METHOD OF 
 
 I 
 
 M DEDUCING THE MOST IiMPORTANT RULES 
 
 Zu„Uct 33084 
 
 Irit^'iuctit aiib lllciisunitioii : 
 
 WITH EXAMPLES 
 
 1 L L U S T n A T 1 N G T II E I U USE A X D A P P L I C A T I O X, 
 
 BY JOHN SANGSTEE, 
 
 TKACnSR IX THE -NORMA!. AXD MODEL SCHOOL?, XOROXTO. 
 
 TORONTO : 
 
 K D I r O A T I O N A I. D E P O S I T O R 
 
 Dfp.irtmrnt of ^ublir Engtrurtion for JHpacr ifanatij. 
 
 1853. 
 
 r 
 
 (1 
 
 cl 
 
 
 § 
 
 m 
 
 fj 
 
 
 :>-'■) 
 
 c 
 
 

 BIBLIOTHEQUB DE LA 
 VILLE DE MONTREAL 
 
 COLLECTION 
 GAGNON 
 
 
 30 O^-/ 
 
 29 
 
 I'orme iriK(l-|-'20 
 
 r 
 
 
 :\ 
 
 i 
 
f\ 
 
 ALGEBRAIC FOEMUIi: 
 
 SHOWINa THK METHOD 07 
 
 DEDUCING THE MOST IMPORTANT RULES 
 
 OF 
 
 
 WITH EXAMPLES 
 
 ILLUSTBATINQ THEIR USB AND APPLICATION. 
 
 BY JOHN SANOSTEB, 
 
 TKACBKR IN TUB NORMAL AND MODEL SCHOOLS, T050RT0. 
 
 TORONTO: 
 
 EDUCATIONAL DEPOSITORY, 
 
 2?epai;tintnt of ^ubU( Instructton for Slpp f anaOa. 
 
 1853. 
 
 Price. *l\ii. each. 
 
ALGEBRAIC FORMULiE. 
 
 I. RATIOS AND PROPORTION. 
 
 Let a, hf c, and d be any four quantities such that a : b H c I 
 i\ then a : 6 = ^ ; r; : d = ^ ; and •; a\h-=zc \d\ T=.-\ad 
 
 . 11. INTEREST 
 
 LetP= Principal, /= Interest, A=: Amount, i = Time in 
 5 ears, and ? = Interest o^ one pound for one year. IzzPrt 
 
 (2); .-. P=^^(3);r = i^(4)ia„d<=l(5)i4 = P+/= 
 P + Prt = P (l+rt) (6); ;. F = -|-- (^)■,r=:^ (8) , 
 
 and « = -^ — (9). From (9) the time in which any sum 
 
 « .„ . r,. 1 . , nS-S n-'\ 
 
 as o will amount to wtimes S is represented by f ::: ^ 
 
 S/ 
 
 n-l 
 
 (10); :. n=:«/--f.lCU)iandr=--- (12) 
 
III. DISCOUNT. 
 
 T> = Discount J other symbols as before 
 
 -D = ^ ~P= from (6) and (7) Ki+If) ±_, ^ j^£ 
 
 •Present Worth or P= -— ru^ 
 
 IV. COMPOUND INTEREST. 
 
 Since £1 at the end of 1st year amounts to 1 -f- r. 
 
 1 : 1 + r :: P :i'(14-r) = AmountofPattheendofthel8t 
 year. 
 
 1 : I + r :: P (1 + r) : P (1 4- r)2 == « 2nd « 
 
 J : 1 + »• :: P (1 + 0'^ : i' (i + r)^ == « Spj « 
 
 And so on, therefore at the end of the tth year A^r( 1 +r)t (1 SV 
 
 (18). From (18) th3 time in which any sum, as S, will 
 amount to n times S at Compound Interest is represented 
 
 by t =. —""-t^- (m 
 ^ log.(l+r)^^^> 
 
 V. arithm:etical progression. 
 
 Let a = first term, / = last term, d = common difference, n 
 « number of terras, and S = sum of the series. 
 
 S=a-}-{a + d)+ (a. + 2d)+(a + Sd) + (/-^A 
 
 + /, Reversing the eeries. 
 
 ^*=/ + ('-'') + ('-2'0 + (/-3<;)+ ... (a +<i)+„ 
 
 2S = (a + + (a + + (a + 0+(« + 0...(a + /) 
 to « terms = «_(o-|-;). ^ ■ / v n y 
 
 tl 
 
 n 
 
5 
 
 At 
 
 Wz-— 1_(23). Alsof=.rt+ (7t- l)ri(24). 
 
 •^ a = / _ (?i - 1) J (25); ^== i^(26); n = 1 +i^ 
 (27). Substitute (24-) in (20); and, 
 S = ||2a+(;^-l)c^j(2S);;.a = |-(^)^(29)., 
 
 tf = —^7 — ^ (30); and n = ^^^irr^ 
 
 n (n — 1) ^ ^ '^ ^ 
 
 2d 
 
 n 
 
 (31). Substitute (25) in 28 anil, 5'=<2/— (« — l)rfS. (32)} 
 
 . , = H^-Z^ (33); ^^£ + (1^1)^ (34), 
 w(n — 1)^'^ n 2 ^. 
 
 <f + 2 /-|-v^(2/4-rf)'^"~ir"8 <f S (35). Substitute (27) ia 
 
 n 
 
 a 
 
 _d +\^(2l + d)'^—SdS 
 
 (38); Z= 
 
 v/Oia-^)'^+«^^'- -^ 
 
 (39). 
 
 n-2 + ^^n~l 
 
 VI. GEOMETRICAL PROGRESSION". 
 
 Let r = common ratio, other symbols as before. 
 
 S = a-{- ar -^ ar2 ■{■ ar^ -\-ar^ -^ ar 
 
 (A). Multiplying by r. 
 
 Sr == ar -^ ar^ -\- ar^ -{- ar^ -\- ar -{-ar ~\-<if 
 
 (U). Subtracting (A) from (B). 
 
 Sr—S=ar^—a,ov,S(r—l)=.a(r^--l) .*. 5=^^!!^) 
 
 (4.0) ; a = s( ^;i^j) (4^0 5 Also from {J) l^ar^'"^ (42) j 
 
 ^ /io\ 11 'o?. / — log. a .... , 
 
 ,. a == ^43) ; « == 1 -j ^ — —j-:r — ('^^^ > «*>** 
 
 n— 1 
 
 log. r 
 
 , .f«. <■> «... * 
 
6 
 
 1 
 
 ^ ^{^aY ("^^^ * multiplying (42) by r and substituting 
 it in (40.) 
 
 S='!"'l(^6)i«=^/-S(r-l)(47);/ = S-i^ 
 (48) J '"= ^^37 C^9)« When w is infinite, and r a pc^jfr frac- 
 
 72 
 
 tion, ar in (lO) becomes = o. Hence for an Infinite Scries 
 5= ^ (50) J a = S (1 - r) (51) ; and r= -^(52). 
 
 VII. ANNUITIES AT SIMPLE INTEREST. 
 
 Let ^=a single payment of the Annuity, ilf= Amount, 
 e = number of payments, and r = Interest of one pound for one 
 period. Then when the annuity is forborne any number of pay- 
 ments, the last payment being received at the time it falls due, 
 r= A ; last but one = 7l -j- Ar, last but two = A-\- 2Ar, last 
 but three = A -\- 3Ar, 1st = ^ + (^ — ]) ^r ; hence M=.A 
 + (A + Ar) -\-{A + 2Ar) + (A + 3^/) + (A + 
 
 (t — 1) Ar). Whence from (28), M= AtU-^- ^Lz}jl\^^^>^^ 
 
 a 2M ,_,, 2(M^M) 
 
 ^8r _ H- (2 - r)'-^ - (2 - r) 
 
 2 r 
 
 (56"! ; Let v = present value of 
 an annuity to continue an; number of payments; from (6)1; (1 -f- rt) 
 ^M = M (1 + ^-^) ; hence . = ^i^Jll 
 /-J^VsTVr- 2(^£-,;) 2v(^;-+l) 
 
 (59). ^ ;^ 
 
 as 
 
 (1 
 
ubdtituting 
 
 r 
 
 oper frac- 
 
 Ue Series 
 
 Amount, 
 1(1 for one 
 ?r of pay- 
 alls due, 
 lAr, last 
 
 ,...(^ + 
 ^'■)(53), 
 
 and ^ = 
 
 value of 
 
 [1 + rt) 
 — l)r 
 
 VIII* ANNUITIES AT COMPOUND INTEREST. 
 
 Symbols fame as before. Then last payment ueing rccrvcd 
 as before = ^, last but one = .^ (1 + »") Ja»' but two = ^ 
 (l-{-ry jlast but three = ^ ( I + ry ; and so on, hence 1st payment 
 
 •^ (1 + 0' + '^ (^ + '*) ' a geometrical progrescon, 
 
 whence by (40) JW = — ii-i-^ ^ (60;j Jl =^ ^^_^^.y_^ 
 
 (6,);a„d.='2i:(^ld^r^:^(g2).From(,5).(l+r) 
 
 = ^=:l((l±iO!.-:il);. .^dULt^L-^ <63)., ^ 
 
 ^1(1+1)1 (6,)^ ,„a , ^ log../?-log.(.^-.r ) ^^ 
 
 (1+r)'— 1^ ^ log. (1+r) 
 
 find the present value of an annuity which is to commence after 
 s years and continue for t years j from (63) v A^ i' -t * years 
 
 /. for t years to commence after s years, v = — -I . ^ i y \g" — 
 
 -) (66). 
 
 When an annuity lasts for ever, as in the case of landed pro- 
 perty, ,, } , - in (63) = — = 0; hence for a perpetual annuity 
 *^ •' (l + O* 
 
 V = — (67) ', J = vr (68) ; r = — (69). The present value 
 of a freehold estate to a person to whom it will revert after 5 
 years is found from (66) and is represented by Vs= rCl+ rT 
 (70). 
 
 .■,|||,--*-.«».A....*».iS» ir 
 
8 
 
 IX. AREAS OP SURFACES. 
 
 Let^rr area,s = side, J = diagonal, then 
 
 • Square.-A = s^ (71) ; s = y'J (72) 5 ^ = f (73) ; d^ 
 
 y^A (74). 
 
 Rectangle.— hex h = base, and p = perpendicular, A = bp 
 (75) J 6 =. 4 (76); p=j (77). ^^^^ ^ ^ ^^^^^-^___ 
 (78). Parallelogram.— A = lyp (79). 
 
 TWa;z5/..-J = -^ (80) J ^ =-^ (81) ; ^ = M (82). 
 , Let.a, ^ c, be the three sides of any triangle and let s^^ + ^ + g> 
 -A = /,. (s--a)_(s~6)(5 — c)(83); when the triangle is equa- 
 IateraM==^!^(8i). 
 
 Quadrulatero l in a circle or \y\\o^e o ppcs<ite angles = two right 
 
 haffthr^ l-'^K^r '^ (.- ^)^i^c)-(J=rrf) (85), where .L 
 half the gum of the four sides. 
 
 /Je'gi./^r P(p/y^o«.^Let s== side, n == number of sides, p^ 
 
 apothein or perpendicular from the centre. A= ^-^ 
 
 ^ (86). 
 
 Gircle.—Ui c = circumferance, r = radius, ^= diameter, 
 fc-3.Hl6. c = 2r^(87);r = 2-^(88); ^^ H.^ (S^)\ 
 
 CT 
 
 = "s-C^O) ; 4 = ,nc 2 (!>]) ; where »> = -0796. 
 Sector.-A^rLi^^). ,vhere / = length of circular arc 
 
 '^^ ■360^(^'^) 5 ^^ei'e n = number of degrees in the arc. 
 
 Circular Annvhs-A = m {c^d){c-c') (94); where (f 
 
 e= circumference of inner circle ; and Jl = M^_+J2J^z£} 
 
 " 4 " 
 
 (65) J where «?'= the diameter of the inner circle. 
 
 * 
 
 '•Vir-'*" 
 
 .#•'■ 
 
9 
 
 r3); d= 
 
 '(dTIT) 
 
 -(82). 
 
 f 6_+£, 
 
 2 
 1 is equa- 
 
 wo right 
 ere s = 
 
 les, p ss 
 36). 
 
 iameter, 
 " (89) i 
 
 liar arc 
 
 arc. 
 
 lere cf 
 
 Ellipse. — Let C= circumference, i= transverse axis, <7=s 
 conjugate axis,a = abscis3a,o = ordinate, r? = tlistance of abscissa 
 
 from centre. C =^L(L±i) (96) ; .^ = ^(97); o = ^ 
 -/(^^■^(SS); a=^±rtf, and 6/= -^^Z I (^+o) (^--o) | 
 
 en f c 
 
 ot 
 
 Parabola. — Let p = parameter, a, and a' = abcifsas, oand(/ 
 = ordinates, 6 = base,= double ordinate, /= length of parabolic 
 
 curve, p =^ (102) ; d = v(^)(^03) ; a' =fl /^j ' (104.) ; 
 
 ^= .M_(i05)jand/ = 2v(o'-^ + i^\(106^. 
 
 Hyperbola. — Symbols same as in ellipse* 
 =i^(T+T)"i (107) J a= ^ + ^; an(U = i ^/f! + oA 
 
 and A = '\^ca\ 3>/ 7^> (7 ^ + n «) + -^ V^^~ (^^^'^ 
 
 .75^ "" 
 
 Regular solids. — Let s = surface, v= volume, e= one edge 
 
 Tetraedron.— V = t.^{\ 12); and s ^e'-^ -/3"(1 13). J/<?iB-. 
 
 12 
 aedron or cube. — F=e' ( 1 1 ^)) ^=6 e'^ ( 1 1 5) . Octacdron, — 7= 
 
 j!l^(116);s = 2eV3~(ll'^)- 
 o 
 
 X)^;r/cc«f^/ym.— F-= 5e5 X 1.53262 (118) ; s = 15e2 X 
 
 1.376385(119). 
 
 Jcosac(i/o;i.— V = ^T" X 2.6 1803 (120) ; and s = 5e^^ yT 
 
 .^^^M- 
 
 !..ilMiV«4fb„'4|k »^JM«**4i» -*'.«•» 
 
 "i. .-' 
 
10 
 
 Cylinder or pHsni—S = i)h + 2 J. (122) ; where p = pei- 
 imeterof base, h = perpendicular height, and jj = area of base. 
 
 rijrami(lorcone.—S= -~—\- A (123); where / = length 
 of slant side. 
 
 Frus'.um of Pyramid or Cone. S= P+P' l-{-A-{- A' (124'); 
 
 where V and A' represent the perimeter and area of smaller base. 
 
 Sphere. S = ^^r'^ (125); S = cd (126). 
 
 Sphericcd Segment. S = 2+r ;-/i (127) ; where h = height of 
 segment. 
 
 X. VOLUME OF SOLIDS. 
 
 r= volume, /j= height, then, 
 
 Pnsm or Cylinder. V = Ah (128) ; A = ~ (129). 
 
 h 
 
 Pyramid or Cone. V= ~ (130); A = -^ (131). 
 
 h 
 
 h 
 
 Frustum ofPijramid or Cone. V = -^{A + A' + »/'JfJ,) 
 
 (132), where A and A> represent the areas of the ends of the 
 Frustum. • 
 
 Sphere. F= l^^ll (133); r =^(g.)(134). 
 
 Spherical Segment. F= —■ {3d — 2h)h^ (135) ; F = ^ 
 
 (3?-2 -|_ /^c) /^ ^i3g^^ ^Yhere r = radius of base of segment, and 
 d= diameter of !?phere. 
 
 Spherical Zone. F= {r'^ + r'^ + ^ ^2) -^*(137) ; where 
 r and r' = the radii of the ends ; for the middle zone, 
 
 t 
 
 '^iUtr'm- uJl*.«* ■*<••*♦ •• • -v. n 
 
11 
 
 of base. 
 = length 
 
 ^' (124); 
 
 Ilerbase. 
 
 y _ (,72_ 1 7,2) i^ (138); or F= (cZ'^ + f /i^)-^ (139) j 
 where <Z and ^' = the diameters oflhe sphere and of the segment. 
 
 Spheraid. V = ^ (14-0), where P =^ Polar and E = Eiiua- 
 lorial axis.* 
 
 height of 
 
 EXAMPLES 
 
 I of the 
 
 
 ft 
 
 6 
 ;nt, and 
 
 where 
 zone, 
 
 Note.— The numbers inclosed within parentheses, JmcZcs serving to num- 
 ber the questions, indicate the Formula to be used. 
 
 (1) Find the fourth proportional — 
 
 to 9,36 and 17. Ans. d=68: 
 
 Also to 11, 165 and 14. Ans. d=210. 
 
 (2) What is the Interest of £537 12s. 6d. lor 5 years at 4 per 
 
 cent^ Ans. I=£107 10s. 6d. 
 
 (3) What principal will produce £117 Is. lO^d. in 7 years at 6 
 percent? Ans. P=je278 16s. 
 
 (4) At what rate per cent will £672 16s. 4d. produce £518 Is. 
 
 44d. in 7 years'? Ans. r=ll per cent. 
 
 (5) In what time will £2000 produce £280 at 3,^ per cent? 
 
 Ans. t=4 years. 
 
 (6) What is the amount of £219 16s. for 3 years at 6 per cent? 
 
 Ans. A=£259 7s. 3.id. 
 
 * A spheroid is generated by the revolution of an ellipse about one of its 
 axes; the fixed axis :^ called the Polar, the revolving one the Equatorial. 
 In an oblate spheroiu the transverse axis revolves : in a prolate spheroid the 
 conjugate. , 
 
 I 
 
 •%.«'■ 
 
pi 
 
 12 
 
 (7) Whnt principal will amount to ^£387 16s. 3]d. in 5] years at 
 3| per cent? Ans. P=je319 Os. 6d. 
 
 (8) At what rate per cent will £537 12s. 6d. amount to £645 3s. 
 in five years? Ans. r=4 years. 
 
 (9) Jn what time will £926 12s. amount to £1 130 9s. O^d. at 4 
 percent? Ans. t=5^ years. 
 
 (10) In what time will £130 double itself at 7 per cent? Ans. 
 t= 14^ years. 
 
 In what time will any sum of money quadruple itself at 6 
 per cent ? Ans. t=50 years. 
 
 (11) To how many times itself will any sum of money amount in 
 25 years at 8 per cent ? Ans. n=3. 
 
 (12) At what rate per cent will any sum of money amount to 6 
 times itself in 20 years ? Ans. r=2.i . 
 
 (13) What is the discount on a bill of £550 10s. payable in nine 
 months at 5 per cent? Ans. D=£19 17s. ll^d. 
 
 (U) What is the present worth of a bill for £l00*payable in 2 
 years at 5 per cent. Ans. r=£90 18s. 2 ^ d. 
 
 (15) What is the amount and compound interest of £142 for 8 
 years at 3 per cent half-yearly ? Ans. A=£227 17s 
 4^d. and I=£85 17s. 4id. 
 
 (16) What sum will rmount to £432 in 8 years at 3 per cent 
 Compound Interest. Ans. P=£341 Os. 4d. 
 
 (17) At what rate per cent will £500 amount to £629 17^ 1 1 ' d 
 m 3 years allowing- Compound Interest? Ans.r=8 per c'ent.* 
 
 (18; In what time will £1500 arise from £12 19 12s. 9,|. at 3 per 
 cent allowing Compound Inlerest ? Ans. t=7 years. 
 
 (19) In what time wiill any sum of money double itself at 5 per 
 
 cent Compound Interest. Ans. 1=14^ years. 
 
 (20) Find the sum of the series of which 6 is the first term 796 
 the last term, and 80 the number of terms. Ans. S=32080. 
 
 (21) Find the first term of the series whose sum=444, last term 
 =70 and number nf fprmc io a_^ „ < 
 
 ''•■*'*s: ^ m't'.Jit^ -m. 
 
 ,^^*r- 
 
 «fr ,|io*.»..«».i^,^jV)».- 
 
 ■^^5**«W»I-»«*4»M«<«».«W.».-*fcW»»».,MI».:«» . 
 
 '-'(':C» 
 
13 
 
 
 (22) 
 (23) 
 
 (24) 
 
 (25) 
 
 (26) 
 (27) 
 
 (28) 
 
 (29) 
 
 n,796 
 
 I (30) 
 
 52080. 
 
 1 
 
 t term 
 
 1 (31) 
 
 A person travels 300 miles in 10 days, his first day*s jouiney 
 was one mile, what was his last? Ans. 1=59 miles. 
 
 A boy won 33700 marbles, his winnings were in arithmetical 
 progression — on the first day he won 3 and on the last 671 ; 
 how many days did he play] Ans. n=100 days. 
 
 and (28) A body falling through space, falls through 16 feet 
 the first second, throuo;h 48 feet the next second, 80 feet 
 the next and so on, what will be its entire descent during 
 21 seconds and through how many feet will it fall the last 
 second ] Ans. 1==656 feet, and ;S=7056 feet. 
 
 A person purchasing an estate agrees to pay for it in 29 pay- 
 ments, each payment being J625 more than the preceding 
 one ; what must the first payment be that the last may 
 =£897] Ans. n=£l97. 
 
 Given the number of terms=:7j last term =27 ; first terra. 
 =3 ; to find common difference. Ans. d=4<. 
 
 A drover bought a flock of sheep, paying 8s. for the first, 8s* 
 6d. for the second, 9s. for the third, and j£2 4s. for the last, 
 how many sheep were there in the flock ] Ans. n:=73 sheep. 
 
 Find the sums of the following series : 
 
 1_}_5 f 9+13 -f &c. tol8 terms. Ans; S=630. 
 2-i-2i+2'^-|-34-&c. to 16 terms. Ans. S=56. 
 f-i-14-lf+&c. to 15 terms. Ans. S=40f . 
 and of— 9— 7—5— 3— &c. to 20 terms. Ans. S=200. 
 
 A person worked 47 weeks upon condition that his wages 
 should be increased 2s. every week, at the end of the time 
 he received £110 9s. as the amount oi his wages, what wag 
 the wajjes of the first week] Ans. a=l shillinof. 
 
 Find the common difference of the series of which the first 
 term=2,sum=8675,and number oftcrms=50. Ans. d=7. 
 
 Given the sum of the serips=34850, first term=2, and com- 
 mon difference=:7 : to find the number of terms, Ans, n — IQO 
 
14 
 
 (32) The last term of a series=157, common difference=3, anil 
 number of terms=51 ; find the sum of the series. Ans. 3 
 
 :=41S2. 
 
 (33) Given the last term of a series=97, number of terms=ll, 
 sum of the series=:2489, to find the common difference. 
 Ans. d==20|. 
 
 (34) A person spends ^d. more on the 2nd January than on the 
 first ; ^d. more on the 3rd than on the 2nd, and so on ; at 
 the end of the year he finds that he has spent «£155 2s. 
 6d., what was his outlay on the 31st December. Ans. 1 
 = 16s. Id. 
 
 (35) Find the number of terms in the series of which 18000 is the 
 sum, 10 the common difference, and 595 the last term. 
 Ans. n=60. 
 
 (36) Find the sum of the odd numbers from 1 to 99 inclusive. 
 
 Ans. 8=2500 : Also of the even numbers from 2 to 100 
 inclusive. Ans. S=2550. 
 
 (37) Given the sum of a series=2625, first term=5, and last term 
 
 =245, to find the common difference. Ans. d=12. 
 
 (38) Find the first term of the series of which 'he sum=2288,last 
 
 term=95, and common difference=2. Ans. a=9. 
 
 (39) What is the last term of the series of which the first term 
 
 = — 5, the sum=196, and the common difference=ll. Ans. 
 1=61. 
 
 (40) Required the sum of the series 1,2, 4, 8, 16, &c. to 10 terms. 
 Ans. S=1023. Required the sum of these.ies 2, 6, 18, 
 34, &c. to 8 terms. Ans. S=6560. 
 
 (41) What is the first term of th ? geometrical series of which the 
 sum=682,the number of terms=5, and the common ratio 
 =4 1 Ans. a=2. 
 
 (42) Required the last term of a series of which the first term=3, 
 the number of terms= 10, and the common ratio=8. Ans, 
 1=402653184, 
 
15 
 
 (43) What is the first term of a series when thelasttcrm=1024000, 
 
 the common ratio=4', and number of terms=6 1 Ans. a=s 
 
 1000. 
 (44-1 Given the first term=2, the last term=512, and the common 
 
 ratio =4, what is the number of terms 1 Ans. n=5. 
 (45) What is the common ratio of a series of which the first term 
 
 =«=64, last term=jy3^3^, and number of terms=12 ? Ans. 
 
 r=j 
 
 (46) The extremes of a series are 12 and 175692 and the com- 
 
 mon rations 11 ; what is the sum? Ans. S=l 93260. 
 
 (47) Find the first term of a series of whii;h tl<e common ralio=3, 
 
 Ihe last term=4374, and the sum=6560. Ans. a==2. 
 
 (48) Given the sum of the series=4095, common ratio=2, first 
 
 term=l, to find the last term. Ans. 1=2048. 
 
 (49) Gi^en the sum of the series=1023, the last term=512,and 
 
 the first term=l, to find the common ratio. Ans. r=2. 
 
 (50) Find the sum of an Infinite series of which the first term=-j7j 
 and the ratio=Y'ij. Ans. 8=^. 
 
 Find the value of -463 ad infinitum. Ans. S=J|J. 
 
 (51) Given the sum of an Infinite series=2, and the common ratio 
 
 =^ to find the first term. Ans. a=l. 
 
 (52) Find the common ratio of an Infinite series of which the first 
 
 term is 17 and the sum 18. Ans. r=-y'jj. 
 
 (53) What is the amount of an annuity of £436 forborne 12 years 
 
 at 3,^ per cent, simple interest ] Ans. M=je6239 3s. 2^d. 
 
 (54) What annuity will amount to £385 ; if forborne 5 years at 5 
 
 per cent, simple interest? Ans. A=£70. 
 
 (55) At what rate per cent, will an annuity of £356 amount to 
 £3972 19s. 2^d.; if forborne 9 years, allowing simple 
 interest? Ans. r=6 per cent. 
 
 (56) In what time will an annuity of £37 amount to £508 15s. 
 
 at 5 per cent, simple interest ? Ans. t=ll years. 
 
1« 
 
 (57) How much ready money should be paid for a pension of £30 
 to continue for 5 years at 4^ per cent, simple interest ? 
 Ans. v=jei33 9s. 4^d. 
 
 (58) Tfi;i33 9s. 4^d. ready money be paid for an annuity of £30 
 to continue 5 years, what is the rate per cent, simple inter- 
 est? Ans r=4^ per cent. 
 
 (59) A person sells 6 years of his pension, which is payable quar- 
 
 terly with simple interest at 5 per cent, per annum, for £263 
 18s^ lOd.j what was the annual value of his pension? 
 
 (60) What will an annuity of £33 17s. 9d. amount to, in U years 
 at 4 per cent, compound interest? Ans. M=£619 17s. 4d. 
 
 (61) What annuity, accumulating at3| percent, compound inter- 
 
 est, will amount to £600, in 40 years ? Ans A=^6 13s 
 lid. 
 
 (62) In how many years will an annuity of £8 per annum, amount 
 
 to £187 6s. 3|d. at 3 per cent, compound interest ? Ans. 
 t=18 years. 
 
 (63) What is the present value of an annuity of £154 for 19 
 years, at 5 per cent, compound interest ? Ans. v— £1861 
 2s. 7id. 
 
 (64) What annuity to continue 12 years at £4 5s. per cent, com- 
 
 pound interest may be purchased for £231 5a. 2Ad. ready 
 money ? Ans. A=£25. 
 
 (65) For how many years may an annuity of £22 be purchased for 
 
 £308 12s. lOd., allowing compound interest at 4 per cent? 
 Ans. t=2l years. 
 (Q6) Required the present value of a deferred annuity of £80 to 
 be entered upon at the expiration of 12 years, and then to 
 be contmued for 7 years, at 4 per cent, compound interest ? 
 Ans. V— £299 18s. Id. 
 (67) What is the present value of an estate whose rental is $1500 
 allowmg 5 per cent, compound interest? Ans. v=30000 
 dollars, or 20 years' purchase. 
 
17 
 
 (68) What is tbe annaal rental of a freehold estate purchased for 
 JC3000 when the rate of interest is at 4 per cenf 1 Ana. 
 A=jei20. 
 
 (69) If a perpetuity of JC563 can be purchaseil for j£11260 ready 
 money, what is the rate of interest allowed ? Ans. r=:5 per 
 cent. 
 
 (70) A freehold estate producing JC75 per annum, is mortgaged 
 
 for the period of 14 years ; what is its present value reck- 
 oning compound interest, at 4 percent? Ans. vmjBldSX 
 15s. 4d. 
 
 MENSURATION. 
 
 (71) How many acres are there in a squa leld whose side is 
 809 links t Ans. A=6 acres, 2 roods, 7^ poles. 
 
 (72) What is the side of a square whose area is 3025 yards t 
 
 Ans. s=55 yards. 
 
 (73) What is the area of a square room whose diagonal is 31 
 
 feet? Ans. A=480i feet. 
 
 (74) Required the diagonal of a square table whose area is 16 
 
 square feetl Ans. d=:5.65685 feet. 
 
 (75) Find the number of square inches in a sheet of paper whose 
 length is 11 inches and breadth 8^ inches. Ans. A=93j. 
 inches. 
 
 (76) If the area of a rectangle, whose perpendicular is 11 yards, 
 be 2112 yards, how long is its base ? Ans. b=192 yards. 
 
 (77) The area of a rectangular pond is 43750 square yards, one 
 
 side is 350 yards, what is the length of the other? Ans. 
 P=125 yards. ' 
 
 (78) Find the area of a rectangle whose base is 21 and diagonal 
 
 35 yards. Ans. A=588 yards. 
 
 (79) Find the area of a parallelogram whose base=90, and per- 
 
 pendicular heightrr 12 ^ feet. Ans. A= 1125 feet. 
 
 (80) Reqcired the area of a triangle whose base=8l feet, and 
 
 altitude 46 feet. Ans. A=1863 feet. 
 
 B 
 
It 
 
 (81) What is the base of a triangle whose area=:2560, and per- 
 pendicular height 40 feet? Ans, b=128 feet. 
 
 (82) Required the altitude of a triangle whose area:=l 17.5625 
 square yards, and base=4<9 feet 6 inches. Ans. P=:42 
 feet, 9 inches. 
 
 (83) Find the area of a triangular field, whose sides are 1200, 1800 
 and 2400 links. Ans. A=:10 acres, 1 rood, 33 pole's. 
 
 (84) Find the area of an equilateral flower bed whose side=25 
 yards. Ann. A=i270.625 yards. 
 
 (85) The four sides of a trapezium inscribed in a circle are 75, 40 
 60 and 55 links, what is its area 1 Ans. A=3146.427 links 
 
 j(86) Find the area of a park in the form of a regular octagoo, 
 whose side=:12 chains, and apothem 14.485 chains. Ans. 
 < . . A=69 acres, 2 roods, 4.6 poles. 
 
 (87) What is the circumference of a circle whose diameter is 44 
 ^ feet T Ans. c=138.23 feet. 
 
 (88) Required the diameter of a circle whose circumference is 
 78.54 yards. Ans. d=2r=25 yards. 
 
 (89X What is the area of a circle whose diameter=80 feet ? Ans 
 Az=5026.56 feet. 
 
 (90) Required the area of a circular garden whose diameter is 200 
 yards and circumference 628.32 yards. Ans. A=31416 
 square yards. 
 
 (91) Find the area of a circle whose circumference==200 feet. 
 Ans. A=3]84 square feet. 
 
 (92) The radius of a circle is 50 feet, frhat will be area of the 
 
 sector whose circular arc is 30 feet in length t Ans A 
 
 750 feet. 
 
 (93) Find the area of a sector, the circular arc of which contains 
 
 40*, the diameter being 60 feet. Ans. A=314.16 feet. 
 
 (94) Find the area of a circular annulus, the circumferences of the 
 
 containing circles being 90 and 60. Ans. A=358.2. 
 
 m 
 
 it**: ■ 
 
19 
 
 (95) The diameters of two concentric circles are 50 and 30 f«#t, 
 
 what is the area of the included annulus 1 Ans. A=125d. 
 64 feet. 
 
 (96) Required the circumference of an ellipse whose diameters are 
 
 600 and 400. Ans. C==l 570.8. 
 
 (97) What is the area of an ellipse whose diameters are 5 and 10 f 
 
 Ans. A=39.27. 
 f 98) The axes are 30 and 10, and one abscissa 24 ; fio4 the 
 
 ordinate. Ans. o=4. 
 (99) The axes are 70 and 50, and an ordinate 20 ; what are the ' 
 
 abscissas I Ans. a=56 and 14. 
 (100) The conjugate axis is 10, the smaller abscissa 6, and the 
 
 ordinate 4 ; required the transverse axis. Ans. t=30. 
 C\01) The transverse axis is 280, an ordinate 80, and one absciss^ 
 56 ; what is the conjugate axis? Ans. c=200. 
 
 (102) If an ordinate of a parabola is 20, and its abscissa 36, what 
 
 is the parameter? Ans. p=ll.l. 
 
 (103) Two abscissas are 9 and 16, and the ordinate of the former 
 
 is 6, find that of the latter. Ans. o'=8. 
 
 (104) Given the two ordinates 6 and 8, and the abscissa of the 
 
 former=9 ; to find that of the latter. Ans. a'=16. 
 
 (105) Find the area of a parabola whose base or double ordinate 
 
 is 15, and height or abscissa 22. Ans. A=220. 
 
 (106) Kequired the length of the parabolic curve whose abscissa 
 
 is 6, and ordinate 12. Ans. 1=27.71. 
 
 (107) The transverse axis of an hyperbola is lb, the conjugate 
 
 axis 9, the smaller abscissa 5 ; required the ordinate. Ans.^ 
 0=6. 
 
 (108) The transverse and conjugate axes are 60 and 45, and onie 
 
 ordinate is 30 ; what are the abscissas t Ans. 8=67 J 
 and 7^. 
 
 (109) The transverse axis is 60, an ordinate 24, the smalAer 
 
 abscissa 20 ; what is the coBjugate axis 1 Ans. c=36. 
 
20 
 
 (110) The conjugate axis is 45, the less abscissa 30, and tie 
 
 ordinate 30 ; required the transverse axis. Ans. t=r90. 
 
 (111) What is the area of an hyperbola whose transverse and 
 
 conjugate axes=15 and 9, and the less abscisia=51 Ans. 
 
 A=37.919. 
 (^J?) and (113) Find the solidity and surface of a tetrahedron 
 
 whose edge=8. Ans. v=60.3 : and S=l 10.85. 
 (114") and (115) Find the volume and surface of a cube or 
 
 hexahedron whose edge is 11. Ans. v=:1331 : and SS=726. 
 (UC) and (117) Find the volume and surface of an octahedroi* 
 
 whose edge is IG. \ns. v=4'71.4: and S=346.4«. 
 (I IS) and (119) Find the volume and siuface of a dodecahedron 
 
 whose edge is 4. Ans. v=490.4.4 : and S=r330.33. 
 f 1?J0) and (121) Find the volume and surface of an icosahedroni 
 
 whose edge is 6. Ans. v=4.71.24r5 : and S=311.76. 
 (12|X) What is the surface of a right cylinder whose length ('.i> 
 ■ " =20 and circumferrnce (p)=:t) "? Ans. 8=125.73. 
 
 (123) What is the surface of a regular pentagonal pyramid, each 
 - side of its base being Ig feet, and its slant side 10 feet? 
 
 Ans. 8=46.4456. 
 
 (124) Find the surface of a frustum of a right cone, its length 
 
 being 31, and the circumferences of its two ends 62.832,, 
 and 37.6992. Ans. 8=1985.49. 
 
 (125) What is the surface of a sphere whose diameter=800 
 
 inches'? Ans. 8=2010624 inches. 
 
 (126) Find the surface of a globe whose diameter=rl2 and! 
 
 circumference 37.6992. Ans. 8=452.39. 
 
 (127) Find the surface, of a spherical segment whose height is 
 
 2, the diameter of the sphere be'rg 10. Ans. 8=62.832. 
 
 (128) What is the volume of a prism whose length is 18 feet, 
 its base being a regular hexagon each side of which is 16 
 inches? Ans. v=83. 138 feet. 
 
 (it9i) If the volume of a triangular prism is 'i.656,and its length 
 lOi J what is the area of the base 1 Ans. A=-7.29. 
 
 (130) 
 (131) 
 (132) 
 
 (133) 
 (134) 
 (135) 
 
 (136j 
 (137; 
 
 fi38; 
 
 (139 
 (140 
 
 I' 
 

 81 
 
 (130) What is the solidity of a cone whose altitude is 12 feet, 
 
 the di;imeter of iU base being 10 feet 1 Ans. v=3 14^16. 
 
 (131) Find the area of the base of a cone whose volume is 282.74» 
 
 and altitude 30. Ans. A='28.274. 
 
 (132) Required the volume of a frustrum of a square pyramid, 
 
 the side of the greater base being 16, of the lesser 10; 
 and its length 18. Ans. v=:37152. 
 
 (133) What is the solidity of a sphere whose diameter is 30 1 
 
 Ans. v=14.l37.2. . 
 
 (134) What is the diameter of a sphere whose volume is equal to 
 C5449.85 feet. Ans. d=n2r=50 feet. 
 
 (135) What is the solidity of a segment of a sphere, the height 
 
 of the segment being 2, the diameter of the sphere 10 1 Ans* 
 
 V=::54.4544. 
 
 (136) What is the volume of a spherical segment, whose height 
 . is 10, and the diameter of its base 20 1 Ans. v=2094.4. 
 
 (137) Find the volume of a spherical zone, the diameters of its 
 
 ends being 10 and 12, and its height 2. Ans. v=195.915^. 
 
 (138) Reqvired the solidity of the middle zone of a sphere, its 
 
 height being 32 feet, and the diameter of the sphere 40. 
 
 Ans. v=3163:^.8. .,,.,. 
 
 (139; Find the volume of the middle zone of a sphere, its height 
 
 being 8, and end diameters 6. Ans. v=494.278. 
 (140) Find the solidity of an oblate spheriod whose axes are 20 
 
 and 12. Ans. v=2513.28. 
 What is the volume of a prolate spheriod, its polar axis. 
 
 heino- 7^ and equitorial axis 5 1 Ans. v=91.63.