Hence it appears, that as the time elapsed is 54 years, the period of the lease unexpired must necessarily be 45; and this solution agrees with the conditions of the problem.

PROBLEM VI.

It is proposed to divide the number 50 into two such parts, that the sum of three-fourths of the one, and five-sixths of the other may be equal to 40.

x.

Let 50 a, and 40 = b; and if one of the parts of a be denoted by x, the other must necessarily be a By the conditions of the problem we shall then have the following equation:

One of the parts of 50 then is 20, and the other 30; which answers the conditions of the problem; for 15, the three-fourths of 20, added to 25, the five-sixths of 30, is just equal to 40.

PROBLEM VII.

It is proposed to divide 100 into two such parts, that if a third of the one be taken from a fourth of the other, the remainder shall be 11.

Let 100 a, and 11 = b; also let one of the parts

be expressed by x, and the other by ax.

a-x

Then

x

3

will denote the third of the one part, and the fourth

4

of the other; and by the conditions of the problem we shall have the following equation:

The two parts of 100 then are 24 and 76; for if 8, the third of 24, be taken from 19, the fourth of 76, the remainder will be 11.

PROBLEM VIII.

Two persons sat down to play, one of whom had 72 guineas, and the other only 52; after a certain number of games they separated, the former carrying with him three times as many guineas as the other. How much did he win?

Let a represent the 72 guineas of the former, b the 52 guineas of the latter, and the loss of the second player.

The money of the first player when they give over play will therefore be a + x, and that of the other b—-x; but as a +x, by the question, is three times as great r, we shall have:

as b

As it here appears that the loss of the second player was 21 guineas, leaving him only 31, the first must have carried off 93 guineas, which answers the conditions of the problem.

G

PROBLEM IX.

The minute hand of a clock being at 12, and the hour hand at 1, at what point between 1 and 2 will they both be in conjunction?

If r represent the space, between the hours of I and 2 passed over by the hour hand before it is overtaken by the minute hand, and a the interval between 12 and 1; as the space passed over by the minute hand will be twelve times as great as that passed over by the hour hand, a +x will be equal to 12r; and we shall have the following equation:

From which we may conclude, that the minute hand will overtake the hour hand after the latter has passed over part of the space between the hours of 1 and 2.

PROBLEM X,

If two bodies move towards each other with unequal velocities, the ratio of which is known, as well as the distance between the bodies, to determine the point at which they will meet,

Let the velocities be as 12 to 1, and let a represent the distance between the bodies, and r that part of it passed over by the body having the least velocity, when they meet.

The space then passed over by the body which has the greatest velocity, will be a r, and we shall have the following proportion:

The solution of this problem is general; and consequently applicable to all cases where the distance of the bodies and the ratio of the velocities are known.

PROBLEM XI.

To divide 90 into two parts, which shall be to each other in the same ratio as 2 to 3.

Let 90 be represented by a, the least of the two parts by x, and the other by a x. We shall then have the following proportion:

Consequently, the least of the numbers will be 36, and the other 54; and indeed 36 + 54 = 90, and 36: 54:23.

PROBLEM XII.

Application of analysis to the solution of the 11th problem of Divining Arithmetic, in which it is proposed to tell the number of spots on all the bottom cards of several heaps arranged on a table.

It is here supposed, that a complete pack of 52 cards is employed; and that as many cards are placed over

the first of each heap as are necessary to make the sum of the spots and cards together to amount to 12.

Let a represent 52, the whole number of cards, and b that of the remaining cards. The number of cards in all the heaps will then be a b. If the number of spots to be guessed be expressed by r, and the sum of all these spots and the cards over them, as they are known by c; we shall have the following equation:

That is to say, if four heaps which are equivalent to a be deducted, r will be equal to the sum of the remaining cards, and the number of the spots and cards which are in the other heaps. The truth of this operation may be easily proved.

PROBLEM XIII.

What number is that, the of of which is equal to 1?

Proof of 2 are ; and 3 of 3 = = 1.

PROBLEM XIV.

What number is that of of which+of of it, is equal to 11?

Let x, as before, be the required number.

Therefore 11r 11 x 12, and x = 12.