SOLUTION.

Since A and B can do the work in 8 days, they can, in one day, do part of it; for a similar reason, A and C can do f part of it in one day; B and C can do part of it in one 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: (+} +1%) ÷ 2=121, which is the fractional part of the work, which they all together would perform in one day.

121

4

720

We have already seen, that the part which B and C can perform in one day is ;.. part which A could perform in one day. which A could alone perform the work is

is the fractional Hence, the time in 30=142 days.

49

Again, the fractional part, which A and C together could perform, is;.. 421-4, is the fractional part which B could perform in one day; hence, the time in which B could alone perform the work is 12-1723 days. The fractional part which A and B together could perform, is }; ... - is the fractional part which C could perform in one day; hence, the time in which he could alone perform the work is

23

days.

10. A and B have each the same income.

121

726

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.

Since B lives on of his income, he must save of it. A's debt for one year being of the income, B will have left after paying A's debt, 1—4—3 of his income. And since this would in ten years amount to $160, of his income must equal $160. 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 one-third 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 of his original stock-16 of $50-4 of $50.

Again, living upon $50, he will have left of his original stock-of $50-4 of $50-$50; increasing this once more by its third part, we get 4 of his original stock-64 of $50of $50-4 of $50. This, by the question, is equal to twice his original stock, or to 4 of his original stock.

Hence, 4-1 of his original stock must equal of $50+ of $50+ of 50 dollars=148 of $50-7400 dollars; ... his stock was 749019-$740.

27

12. Fourteen oxen have in three 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 four weeks, eaten up all the grass upon 3. acres of land. How many oxen can, in this way, graze for 5 weeks, upon 6 acres of land?

SOLUTION.

By the first condition of the question, we see that the growth of 2 acres for 3 weeks is 2 × 3-6 times the growth of one acre

for one week; .. the quantity which 14 oxen ate in 3 weeks is 2 acres, together with 6 times the growth of onè acre for one week. Hence, one ox in one week would eat of of 2 acres, together with of of 6 times the growth of one acre for one week.

Again, by the second condition of the question, we see that the growth of 3 acres for 4 weeks is 3x4-12 times the growth of 1 acre for 1 week ; ... the quantity which 16 oxen in 4 weeks ate, is 3 acres, together with 12 times the growth of 1 acre for 1 week. Hence, one ox, in one week, would eat of of 3 acres, together with of of 12 times the growth of one acre for one week.

Now, by the nature of the question, an ox, in the former case, must eat the same as one in the latter; therefore,

of 2 acres, together with of acre for one week must equal

of

of 6 times the growth of one of 1 of 3 acres, together with

of of 12 times the growth of one acre for one week, or which is the same thing, of an acre + of the growth of one acre for one week is equal to of an acre+ of the growth of one acre for one week.

64

Hence, -=134 of an acre must equal-of the growth of one acre for one week: therefore, the growth of one acre for one week is 13442% of an acre.

Now, by the third condition of the question, there are 6 acres, which increase by the growth of grass for 5 weeks; and since this increase is of an acre, for one acre for one week, it follows that 6 acres in 5 weeks will increase 6×5×1 of an acre; so that there will be 6 acres to be eaten in 5 weeks.

Now, we have already seen, by the first condition of the question, that one ox in one week would eat of of 2 acres

+of of 6 times the growth of one acre for one week, or which is the same thing, one ox in one week would eat of an acre of an acre of an acre.

=

Consequently, one ox would in 5 weeks eat 5 ×

of an acre. Therefore, the number of oxen necessary to eat the 6 acres in 5 weeks, must be ÷1=26 oxen, the answer.

13. 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:

14. A father left to his three sons, whose ages are 8, 10, and 13 years, $10000, to be so divided that the respective parts being placed out at 5 per cent., compound interest, should amount to equal sums when they became 21 years of age. What are the parts?

SOLUTION.

By the question, their respective shares would be at interest 13, 11, and 8 years.

We find, by table under Art. 70, the present worth of $1 for 13, 11, and 8 years respectively, at 5 per cent., compound interest, to be $0.530321, $0.584679, and $0.676839. Now it is obvious that the parts must be to each other in the same ratio as the numbers 530321, 584679, and 676839; the sum of these numbers is 1791839. Hence, the parts are as follows:

The first one's part is 53032, of $10000-$2959.646.

1791839

The second one's part is 584679, of $10000-$3263.011.

1791839

The third one's part is 973 of $10000=$3777.343.

1791839

15. Find what each of the four persons, A, B, C, and D, are worth, by knowing,

First, that A's money, together with of B's, C's, and D's, is equal to $137.

Secondly, that B's money, together with of A's, C's, and D's, is equal to $137.

Thirdly, that C's money, together with of A's, B's, and D's, is equal to $137.

Fourthly, that D's money, together with of A's, B's, and C's, is equal to $137.