Hogshead, 1=1=3=6=48 | 108=12=6—3—2=1 Butt.

NOTE.--A gallon, beer or ale measure, is 282 cubic inches.

IX. PROBLEMS,

WITH RULES, AND QUESTIONS TO ILLUSTRATE THE RULES.

DEFINITION.

A Problem is a proposition or a question requiring something to be done; either to investigate some truth or property, or to perform some operation.

PROB. I.-The sum of two numbers, and the difference of their squares given, to find those numbers.

RULE 1.-Divide the difference of their squares by the sum of the numbers, and the quotient will be their difference.

2. Subtract the difference from the sum, and half the remainder will be the smaller number. Then add the difference to the smaller number and you have the larger number.

EXAMPLES.

1. A and B played at marbles, having 14 apiece at first; but after playing several games, B having lost some of his, would not play any longer, and it was found that the difference of the squares of what each then had was 336. How many did B lose?

Thus, 14+14)336(12 diff.; 14 half sum, and 12÷2=6, half diff.

Then 14+6=20, A retired with. And 14-6-8 B had left; then, 14-8-6 marbles that B lost.

4509

2. What are the two fractions whose sum is 3, and whose difference is ? Questions of this nature are solved by the latter clause of the preceding Rule. Thus 8-9-287, then 2874-887, less fraction, and 383+4=487, greater fraction. PROB. II. The difference of two numbers, and the difference of their squares given, to find those numbers.

RULE.-Divide the difference of their squares by the difference of their numbers, and the quotient will be their sum. You then have their sum and difference to proceed by PROB. I.

EXAMPLE.

Said Ai to Lorenzo, Father gave me $12 more than he gave Charles, and the difference of the squares of our separate parcels is 288. How much did he give each. ?

Thus, 288-12-24, the sum; then 24-12÷÷2-6; then 12+6=$18, Ai's share; 12-6-$6, Charles had given him.

PROB. III.-The product of three factors, and two of those factors given, to find the third factor.

RULE. Divide the given product by the product of the two known factors, and the quotient is the factor required.

EXAMPLES.

1. What must be the length of a stick of hewn timber that is 10 inches wide, and 1 ft. 3 in. deep, in order to contain 1 ton?

First, consider 1 ton the product of 3 factors. Then, 40 cubic feet 1 ton=69120 in.; and 10 in. ×1 ft. 3 in.—150 in. Then, 69120-150-12-383 feet in length, Ans.

2. Suppose wood to be piled on a base 15 ft. 6 in. long, and 7 ft. 9 in. wide, what must be the height of the pile to contain 16 cords ?

First, 16 cords=2048 solid feet, the product of 3 factors; and 15 feet 6. in. X7ft. 9in. 1204ft., the product of the 2 given factors. Then 2048-1204-17,048907+ft. in height, the required factor.

PROB. IV. To find a divisor that will divide two or more numbers without a remainder.

RULE 1.-Divide the larger number by the smaller, and this divisor by the remainder, and thus continue dividing the last divisor by the last remainder till nothing remains, and the last divisor is the divisor sought, if only two numbers are giv

en :

2. But if more than two numbers are given, first find a divisor for any two of the numbers; then find a divisor to the found divisor and another of the given numbers, and thus proceed till all the given numbers are brought in, and the last divisor used is the divisor sought.

EXAMPLES.

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1. Suppose a hall to be 154 feet long and 55 feet wide, is the length of the longest pole that will exactly measure both the length and width of said hall?

Thus, 154-55-2,+44 rem.; then 55÷44-1, +11 rem.; again, 44÷÷11=4; consequently, a 11 foot pole will be the

answer.

2. A owns 720, B 336, and C 1736 rods of land. They agree to divide it into equal house lots, fixing on the greatest number of rods for a lot, that will allow each owner to lay out all his land. How many rods must there be in a lot, how many lots has each, and how many lots in all ?

PROB. V.-To find the dividend that will contain two or more numbers given, without a remainder, when either of the given numbers is used as a divisor.

RULE. Call the numbers the denominators of so many fractions, and a common denominator found thereto, in the manner prescribed for finding a common denominator to vugar fractions, is the dividend sought.

EXAMPLE.

3 63 42 32

Allowing 63 gallons to fill a hogshead, 42 a tierce, and 32 a barrel, what is the smallest quantity of molasses that can be first shipped in some number Operation. of full hogsheads, then discharged and re-shipped in some number of full tierces, and again discharged and reshipped in some number of full barrels.

7

63 21 16

3 9 3 16

gallons

42 ×3× 1 × 16—2016, Ans.

PROB.VI.-The sum of two numbers and their quotient given, to find those numbers.

RULE.-Add 1 to the quotient and divide the sum of the two numbers by this sum, which will give the less number;

subtract the less number from the sum and you will have the greater number.

EXAMPLE.

Divide 100 into two such parts, that if the greater be divided by the less, the quotient may be just 30.

Thus, 30+1)100(37, less part; then 100-37=963, greater number.

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

RULE.-Divide the difference of the two numbers by the quotient, less 1, and you will have the less number. Add the less number to the difference, and this sum is the greater number.

EXAMPLES.

1. A grey hound, in pursuit of a hare, run three times as fast as the hare; and when he overtook her, he had run 30 rods more than she.

How many rods did each run?

Thus, 30÷÷3—1=15 rods, the hare run; then 15+30=45 rods, the hound run.

2. A and B start at opposite points, to skate to the other's starting point distance, 8 miles. A, by having the advantage (hence B, the disadvantage) of a uniform wind, performs his task 2 times the quickest, and 48 minutes the soonest. Required, the time that each is skating, and the force of the wind per minute.

Thus, 48-21-1-32 minutes, A's time; then 32+48=1 hour and 20 minutes, B's time.

Then, 80+32-2-56 minutes, each, if it had been calm; and 56-32-24 minutes A was forwarded; and B was retarded the same.

Then, if 56:8::24:181029 ft., 24 m. 754 ft. per minute.

PROB. VIII. To multiply £ s. d. and gr. by £ s. d. and gr. RULE.-£X£=£; £ × s. —s.; £xd.=d.; £Xqr.=qr.; s. Xs. 20th of a s.; s.Xd.=20ths of a d.; s.Xqr.=20ths of a qr; d.Xd.=240ths of a d.; d.Xqr.=240ths of a qr.; and farthings multiplied by farthings, are 960ths of a farthing,

EXAMPLE.

Multiply £19 19s. 11d. 3 qr. by £19 19s. 11d. 3 qr.

Hence, the product of two denominations, having the same integer, takes the name of the least, and is of such value as the larger denomination implies. Thus, 6s. X7d. d.; here the product takes the name of pence, and is 20ths, because the larger denomination is shillings, 20s. being the integer. Also, 6 inches X8 inches=4 of a foot, or 4 inches,

etc.

PROB. IX.-To divide a larger denomination by a smaller denomination, when the divisor and dividend used in such a division have the same integer.

RULE.-Reduce the number to be divided to the same denomination as the divisor. Divide, and the quotient is the answer, in the same denomination as was the dividend, before it was reduced to the denomination of the divisor.

EXAMPLES.

1. Divide $1 by 4 cts.; or 1 by ,04:-Thus, $1=100 cts., then 100-4-$25, Ans.;for 1,00,04-25, Ans.

2. Divide £1 by 1s.-First, £1=20s., then 20÷1= £20,

Ans.