15. A farmer turned 152 beasts into a meadow; to every horse there were 3 cows, and to every cow, 5 sheep; how many of each sort were there? Suppose there were x horses, then would there be 3x cows, and (3xx5=) 15x sheep; whence by the problem (x+3x+15x=) 19x=152: therefore x=8 horses, 3 x=24 cows, and 15x=120 sheep. 16. A man has six sons, whose successive ages differ by 4 years, and the eldest is thrice as old as the youngest; required their several ages? Let the age of the youngest be x. Then will that of the second youngest be x+4. Whence by the problem, x+20=3x, or 2 x=20, and x=10 years =the age of the youngest; also x+4= (10+4=) 14=the age of the second, x+8=18=that of the third, x+12=22=that of the fourth, x+16=26=that of the fifth, and x+20=30=the age of the eldest. 17. Two butchers bought a calf for 40 shillings, of which the part paid by A, was to the part paid by B, as 3 to 5; what sum did each pay? Let x=A's part, then 3: 5 :: * : =B's part; whence by 5x 3 Let 3x A's part, then will 5x=B's, and their sum 8x= 40, whence x=5; therefore 3x=15s.=sum paid by A, and 5x= 25s.=sum paid by B, the very same as before. 18. A person rented a house on a lease of 21 years, and agreed to do the repairs when 2 93 of the part of the lease elaps of the part to come; how long will he have 9 been in possession when the repairs are begun ? also Let x=the time elapsed, then will 21-x=the time to come; 2x 3 168-8x 9 blem, 2x 3 8 two thirds of the time elapsed; and (x21-x=) eight ninths of the time to come; whence by the pro 168-8 x 9 ; this cleared of fractions, is 18x=504 24 x, whence 42x=504, and x=12 years, the time required. 19. Two country girls went to a fair, A laid out as much above 4 shillings, as B did under 6; and the sum spent by A was to that spent by B as 7 to 8; how much did each lay out? Let 4+x be the sum A spent, then will 6-x be the sum B spent; then by the problem 4+x: 6-x:: 7 : 8: whence by multiplying extremes and means, (4+x×8=6¬x×7, or) 32+8x= 42-7x; therefore 15x=10, and x= 10 2 of a shilling) 8 =sum spent by A, and 6—x= (6s.— 20. The sum of the ages of a man and his wife is 55 years, and his age exceeds her's by 7 years; required the age of each? Let x=the man's age, then 55-x=the woman's ; subtract the latter from the former, and 2x-55-7 by the problem; whence 2x=62, and x=31 years=the man's age, therefore 55-x = (55—31=) 24 years=the wife's age. Or thus, Let x=the wife's age, then 55-x=the man's; subtract the former from the latter, and 55-2x=7 by the problem; whence 2x=(55—7=) 48, and x=24=the woman's age; also 55—x= (55—24=) 31=the man's age, as before. Or thus, Let x=the man's age, then x-7=the woman's; add both equations together, and 2x-7=55 by the problem; whence x= 31, and x-7=24, as before. Or thus, Let x=the woman's age, then x+7=the man's; add both equations together, and 2x+7=55 by the problem; whence. x=24, and x+7=31, as before. Or thus, Let x=the man's age, y=the woman's; then by the problem x+y=55, and x—y==7; this may be solved by each of the three methods in Art. 90, 91, and 92, whence x will be found=31, and y=24, as before. 21. What two numbers are those, the sum of which is 140; and if four times the less be subtracted from three times the greater, the remainder is 70. Let x=the greater, then 140-x=the less; 3 times the greater=3x, and 4 times the less (4 × 140—x=) 560—4 x ; therefore by the problem (3 x−560—4x=) 7x−560=70; whence 7x=630, and x=90=the greater, also 140—x= (140—90=} 50 the less. 22. A labourer received his week's wages, amounting to twenty shillings, in half crowns and sixpences, and there were twenty pieces in all; how many of each did he receive ? Let x=the number of half crowns, then 20—r=the number of sixpences, also 5x=the number of sixpences in x half crowns, and 40 the number in 20 shillings; therefore (5x+20-x=) 4x+20=40 by the problem; whence 4x=20, and x=5=the number of half crowns, and 20—x= (20—5=)15=the number of sixpences. 23. A paid B 20 guineas, and then B had twice as much money as A had left; but if B had paid A 20 guineas, A would have had thrice as much as B had left; what sum did each possess at first ? Let x=A's sum at first, y=B's, then x-20=As remaining sum, and y+20=B's sum after the payment was made; also x+ 20=A's sum, and y—20 B's remainder, had B paid A 20 guineas; whence by the problem y+20= (2 × x−20=) 2 x−40, and x+ 20= (3 × y—20=) 3 y—60; from the former y=2x—60, and x+80 3 from the latter y=- ; whence 2x-60= x+80 or 6x-180 =x+SO, or 5x=260, therefore x=52 guineas; also y=2x−60 = (104—60=) 44 guineas. 24. It is twelve o'clock, and the hour and minute hands of my watch are exactly together; at what o'clock will they be next together, and how often does the minute hand pass the hour hand in twelve hours? Let x=the space the hour hand has passed (or its distance from 12) when they come together: now it is evident that the minute hand must go once round the dial besides the space x, in order to overtake the hour hand; whence 1+x=the space through which the minute hand moves in the same time. Likewise the minute hand moves 12 times as fast as the hour hand, wherefore 12 : 1 :: 1+x: x, by the problem; whence 12 x=1+x, or 11 x= 5 =5 minutes past one o'clock; also 12÷ =12x 12 11 11 12 number of times the minute hand passes the other in 12 hours. 25. Two beggars went to an alehouse to share their booty; after dividing it equally, A spent 5d. and B 8d. they then tossed up, and A won 20d. of B, after which A's cash was double of B's; what sum had each at first? Let x=the share of each. Then x -5= A's sum x-8=B's sum } after spending. And (x−5+20=) x+15=A's sum (x-8-20=) X- -28=B's sum } after tossing up. Therefore x+15= (2 × x−28=) 2x-56 by the problem; whence x=71 pence=5s. 11d.=the share of each. 26. What fraction is that, which if 1 be added to its numerator, the value of the fraction will be; but if 1 be added to its denominator, the value will be ? Let x=the numerator, y=the denominator, then - y the y+1 get y=15, and consequently x= (+1=) 4, whence the fraction required. 27. A Gentleman left 561. between two persons, whose shares 2 3 were respectively as to ; what sum did each receive? 7 8 Let x=the least share, then 56—1=the greater, and by the problem x: 56−x :: 2 3 3x 112-21 ; therefore 7 896 x=(=241.2162=) 241. 4s. 3d4. and 56—x=(56—24.2162 37 31.7838=) 31l. 15s. 8d. 28. Some fishermen having caught a shark, and cut it into three pieces, the tail weighed 60 pounds; the head weighed as much as the tail and the body; and the body weighed as much as the head and tail together; required the weight of the shark, and of each of its parts ? I Let x=the weight of the body, then 60+ =weight of the I 4 head; and (60+ +60=) —+120=x, by the problem; whence 4 1=160 pounds=the weight of the body; 60+=(60+40=) 100 pounds the weight of the head; and (160+100+60=) 320 pounds the weight of the shark. 29. A hare is 50 of her own leaps before a greyhound, and takes 4 leaps to the greyhound's 3; but 2 of the greyhound's leaps are as much as 3 of the hare's: how many leaps must each take before the hare is caught ? Let x the greyhound's leaps, y=the hare's after the dog's starting, then y+50=the whole of the hare's leaps. But 3:4 4x ::x:y by the problem, whence 3y=4x, and y== " 3 also y +50 4x 4x +50; whence by the problem 2:3::x:+50, therefore 3 3 8x 4 x +100 3 x; whence x=300=the greyhound's leaps, and + 3 3 50=450=the hare's leaps. 30. A market-woman bought a number of eggs at 2 a penny, and as many at 3 a penny, and sold them all at 5 for twopence, whereby she lost fourpence; how many eggs had she in all? Let x=the number of each sort, then 2x=her whole stock, also =the value of those at 2 a penny, and the value of |