9. A and B can, together, do a piece of work in 8 days; A and C can, together, do it in 9 days; B and C can, together, do it in 10 days. How many days would it require for each to perform the work alone?

Solution.

Since A and B can do the work in 8 days, they can, in 1 day, do part of it; for a similar reason, A and C can do part of it in 1 day; B and C can do part of it in 1 day. Adding these fractional parts together, and observing that each individual has been included twice, we shall get, by dividing the sum by 2, the following fraction, (++7%)÷2=4}}, which is the fractional part of the work which they all together would perform in 1 day.

1

9

7209

We have already seen that the part which B and C can perform in 1 day is; ·· 424-7%=7, is the fractional part which A could perform in 1 day. Hence, the time in which A could alone perform the work is 720-14

49

days.

Again, the fractional part which A and C together could perform, is ; .. —, is the fractional part which B could perform in 1 day; hence, the time in which B could alone perform the work, is 2=173} days. The fractional part which A and B together could perform, is; •• !?!—1=7%, is the fractional part which C could perform in 1 day; hence, the time in which he could alone perform the work is 720=23,7 days.

10. A and B have each the same income. A contracts an annual debt, amounting to of it; B lives on of it; and, at the end of ten years, B lends to A enough to pay off his debts, and has $160 left. What is the income?

Solution.

- 35

Since B lives on of his income, he must save of it. A's debt for 1 year being of the income, B will have left, after paying A's debt, of his income. And, since this would in 10 years amount to $160, of his income must equal $16. Hence, the income was 35 of $16-$280.

11. A merchant supported himself 3 years for $50 a year; at the end of each year, he added to that part of his stock which was not thus expended, a sum equal to of this part. At the end of the third year, his original stock was doubled. What was the stock?

Solution.

After supporting himself the first year, he will have his original stock, -$50; this, increased by its third part, will become of his original stock, — of $50; living upon another $50, he will have left of his original stock, of $50-$50. This must again be increased by its third part, giving 1 of his original stock, of $50-4 of $50.

Again, living upon $50, he will have left of his original stock, of $50- of $50-$50; increasing this once more by its third part, we get of his original stock, of $50-1 of $50- of $50. This, by

6 4

the question, is equal to twice his original stock, or to of his original stock.

Hence,

19 of his original stock must equal 4 of $50+ of $50+ of $50-14 of $50=740° dollars; .. his stock was 149÷19=$740.

12. Fourteen oxen have, in 3 weeks, eaten all the

grass which grew on 2 acres of land, in such a manner, that they not only ate all the grass which at first was there, but also that which grew during the time they were grazing. In like manner have 16 oxen, in 4 weeks, eaten all the grass upon 3 acres of land. How many oxen can, in this way, graze for 5 weeks upon 6 acres of land?

of 3 acres, with its growth for 3 weeks, will keep 7 oxen 9 weeks, or 63 oxen for 1 week, which result corresponds with (4) in the adjoining table.

Again, by the second condition of the question, if the grass of 3 acres, with its growth for 4 weeks, keep 16 oxen for 4 weeks, then will the grass of 3 acres, with its growth for 4 weeks, keep 64 oxen for 1 week, which corresponds with (6).

By carefully comparing (4) and (6), we see that the growth of 3 acres for 1 week is sufficient to keep 1 ox 1 week; consequently, the growth of 6 acres for 1 week will keep 2 oxen for 1 week; or, which is the same thing, the growth of 6 acres for 5 weeks will keep 2 oxen for 5 weeks.

By (4) we have seen that the grass of 3 acres, with its growth for 3 wecks, will keep 63 oxen for 1 week; but the growth of 3 acres for 3 weeks, will keep 3 oxen

1 week; consequently, the grass alone of 3 acres will keep 60 oxen 1 week, and the grass of 6 acres will keep 120 oxen for 1 week, or it will keep 24 oxen for 5 weeks. Hence, finally, the grass of 6 acres, with its growth for 5 weeks, will keep, during 5 weeks, 2+24=26 oxen.

13. A person expends just £100 for live stock, consisting of geese, sheep, and cows; for each goose he paid 1 s., for each sheep £1, and for each cow £5. How many did he purchase of each kind, so as to have just 100 in all?

Solution.

Since the average price of the animals was £1=20s., the price of a goose was 19 s. below the average, and the price of a cow was 80 s. above the average.

Hence, were he to purchase the geese and cows only in the ratio of 80 geese to 19 cows, their whole cost would be just as many pounds sterling as there were animals; thus, by purchasing 80 geese and 19 cows, he would have 99 animals, together worth £99. Now, by adding 1 sheep, worth £1, he will have 100 animals, together worth £100. So that he bought 80 geese, 1 sheep, and 19 cows.

14. A and B leave Utica for Albany at the same time that C leaves Albany for Utica. If A goes 8 miles each hour, B 12 miles, and C 9, when will C be equally distant between A and B, if the distance between Utica and Albany is 95 miles?

Solution.

The average velocity of A and B is 10 miles an hour; hence, if D, a fourth person, leave Utica at the same time, going 10 miles each hour, he will always be equally

distant between A and B, and the time sought will be when he is met by C. Now D and C together travel 19 miles each hour; consequently, they will meet in 9519-5 hours, at which time C will be equally distant from A and B.

15 A, B, C, D, and E, play together on this condition that he who loses shall give to all the rest as much as they already have. First, A loses, then B, then C, then D, and at last also E. All lose in turn, and yet, at the end of the fifth game, they have all the same sum, viz., each $32. How much had each before they began to play?

Solution.

The solution of this question is the most readily effected by a reverse process; that is, by beginning with the last game, and playing them all in a reverse order, as follows: first, take from A, B, C, and D, half they have, and add it to E's money; second, take from A, B, C, and E, half what they now have, and add it to D's; third, take from A, B, D, and E, half what they now have, and add it to C's; fourth, take half of A's, C's, D's, and E's, and add it to B's; lastly, take half of B's, C's, D's, and E's, and add it to A's.

These successive operations may be exhibited as follows:

D.

E.

$32 $32 $32 $32 $32, end of 5th game,

96, end of 4th game,

48, end of 3d game,

24, end of 2d game,

12, end of 1st game, 6, before playing.