1. Let 6498 men be so formed that the number in rank may be double the number in file.

6498÷2-3249, and 1/3249-57 in file; and

57×2-114 in rank, Ans:

2. A man wishes to plant 1875 trees in an orchard which is 3 times as long as it is broad, so that they may stand in rows in equal distances apart; how many rows must he make, and how many trees must he plant in each row ? Ans. 25 rows of 75 trees.

3. There is a certain lot of land containing 6 acres 2 roods and 18 rods in the form of a long square, the length of which is twice as much as the breadth; required the length and the breadth.

Ans. The width is 23 rods, and the length 46 rods. PROB. V.-The diameter of a circle being given, to make another circle which shall be proportionably greater or smaller than the given circle.

Note. The areas, or contents of circles, are in proportion to the squares of their diameters or circumferences.

RULE.

Square the given diameter; then multiply by the given proportion, if greater, (but divide if smaller,) and extract the square root of the product, (or quotient,) which will give the required diameter.

1. Suppose there is a certain garden whose diameter 9 rods, and it is required to lay out another which shan contain 3 times as much; what must be its diameter ?

/9x9x3=15,58+ rods, Ans. 2. There is a circular field whose area, or contents, is 64 rods, the diameter of which is 9 rods; required the diameter of another which shall contain just one-fourth as much. Ans. 4,5rds.

3. The quantity of water discharged through a certain pipe, which is 2,5 inches in diameter, will fill a certain cistern in one hour; what is the diameter of another pipe which will fill another cistern four times as large, in the same time? √/2,5×2,5×4=5 inches diameter, Ans. PROB. VI.-The sum of two numbers and their product given, to find those numbers.

RULE.

From the square of their sum subtract 4 times their product, and the square root of the remainder will be their difference; half the said difference added to the half sum will be the greater of the two numbers; and half the said difference subtracted from the half, will be the lesser number.

1. The sum of two numbers is 58, and their product is 792; what are those numbers?

58x58 3364 square of their sum. 792 × 4

3168=4 times their product.

√196=14, difference of the numbers.
29, the half sum.

7, the half difference.

36, the greater number.

22, the lesser number.

PROB. VII.-The difference and the product of two num bérs given, to find those numbers.

RULE.

Add the square of half the difference of the numbers to their product, and the square root of that amount will be half the sum of the two numbers. Then to the half sum add half the difference, gives the greater number; and from the half sum, subtract half the difference, gives the lesser number.

EXAMPLE.

1. The difference of two numbers is 9, and their product is 442; what are those numbers?

Difference of the numbers 9 Product 442

EXTRACTION OF THE CUBE ROOT.

We have seen that any number multiplied into itself produces a square, and that the square multiplied again by that number produces a cube; and likewise, that the number itself is the root of the given cube.

Hence, to extract the cube root of any given number is, to find a number which, being raised to its third power, that is, multiplied into its square, shall produce the given number.

A solid body having six equal sides, and each of the sides an exact square, is a cube; and since the length, breadth and thickness, are the same, it is evident that the length of one side of the given body is the cube root of that body; for, the length, multiplied by the breadth, multiplied by the thickness, will give the cubic contents, &c.

Thus, the cubic contents of a square block a foot long, a foot wide and a foot thick is 1X1X1=1 foot. The cubic feet contained in a block 2 feet long, 2 feet wide and 2 feet thick, is 2x2x2=8 cubic feet. Hence the cube root of 8 is 2, because 23, that is, 2×2×2=8.

RULE.

I. Separate the given number into periods of three figures each, by putting a point over the unit figure, and every third figure from the place of units, towards the left, and if there be decimals, point them from the unit's place towards the right in the same manner.

II. Find the greatest cube in the left hand period, and put its root in the quotient.

III. Subtract the cube thus found from the said period, and to the remainder bring down the next period, calling this the dividend.

IV. Multiply the square of the quotient by 300, calling it the divisor.

V. Seek how often the divisor may be had in the dividend, and place the result in the quotient, (root.)

VI. Multiply the divisor by this last quotient figure, and place the product under the dividend.

VII. Multiply the former quotient figure, or figures, by the square of the last quotient figure, and that product by 30, and write the product under the last; then place the cube of the last quotient figure under these two products, and call their amount the subtrahend.

VIII. Subtract the subtrahend from the dividend, and to the remainder bring down the next period for a new dividend, with which proceed as before; and so on till the whole is finished.

Note 1. If the subtrahend happens to be larger than the dividend, the last quotient figure must be made one less, and a new subtrahend found.

2. When it happens that the divisor is not contained in the dividend, we must put a cipher in the quotient, and bring down the next period for a new dividend; and multiply the square of the whole quotient by 300 for a new divisor,

3. When there is a remainder after bringing down all the periods, we may annex periods of ciphers, and continue the operation to decimals.

EXAMPLES.

1. What is the length of the side of a cubic block, which contains 12167 solid or cubic inches?

* We have seen that the square of any number contains just twice as many figures as the number itself, or at least, but one less than twice that number. So also the cube (being a number multiplied into its square) contains just 3 times

32 x 32 x 300=307200)1560125 dividend.

1536000

32 × 5 × 5 × 30= 5x5x5=

24000

125

1560125 subtrahend.
0000

as many places of figures as the number itself, or at least, but two figures less than 3 times that number. Hence, by pointing any number into periods of 3 figures each, as directed in the Rule, we may at once find how many figures the root will consist of.-Pointing the above exampie, we have two periods: hence we find that the root will consist of two figures, viz.: a figure of tens and a fig. ure of units.

We then seek for the greatest cube in the first or left hand period, 12 [thousands ;] this we find to be 8 [thousands,] the root of which is 2 [tens:] placing the 2 [tens] for the first figure in the root, and its cube, 8 [thousands,] under the 12 [thousands, and subtracting it therefrom, the remainder is 4 [thousands,] to which we bring down the next period, 167, making 4167 inches, which remain.

We have now disposed of 8000 inches in a cube, the length of each side of which being 2 [tens]=20 inches, and 20x20x20=8000. Now suppose we make a cubic block, and suppose each side of it to be 2 [tens]-20 inches, it will contain 8000 cubic inches. We must now enlarge this block by the addition of 4167 cubic inches, so that the block shall retain its cubic form; and in order to do this it is plain that we must make the addition on three sides of it. Now the square contents of each of these sides is 20 X 20-400, and 400 × 3, the number of sides on which the addition is to be made, gives 1200, the square contents in the given sides. (But we may obtain the square contents in these sides by neglecting the cipher in the 2 tens 20, and multiply the square of this quotient figure 2 by 300, and it will produce the same. Thus, 2X2X300-1200 as above; and thus the rule is formed.)

Now it is evident that the 4167 inches, which are to be added to this block divided by 1200, the square inches contained in the sides on which the additions are to be made, will show the thickness of the additions to be made on each of the three sides. Thus, 1200 is contained in 4167, 3 times, which shows that the thickness of the additions must be 3 inches. We place the 3 in the root, and multiply the square contents, 1200, by the thickness, 3 inches: that is, the last quotient figure; making 3600 cubic inches contained in these additions, which we place under the dividend. Now after these additions are made to the cube, there are 3 vacancies on the corners, each of which is 3 inches wide, and 3 inches thick, and 20 inches long, containing 3X3X20-180 cubic inches. This, multiplied by 3, gives the whole number of inches in the three vacancies, 540 cubic inches. But by the rule we neglect the cipher, and multiply the former quotient figure, 2 [tens,] by the square of the last, and that product by 30, which produces the same effect. Thus, 2X3X3X30=540, which we place under the former. Now if we