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7. To find the interest of 355/ 158 for 4 years, at 4 per cent.

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per annum.

Ans. 56/ 188 43d.

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10. To find the insurance on 205 15s, for of a year, at 4 per cent. per annum. Ans. 2/ 18 13d.

11. To find the interest of 319/ 6d, for 5 years, at 33 per cent. per annum. Ans. 68/ 158 94d. 12. To find the insurance on 1077, for 117 days, at 43 per cent. per annum. Ans. 1/ 128 7d. 13. To find the interest of 17 5s, for 117 days, at 43 per cent. per annum. Ans. 58 3d. 14. To find the insurance on 712/6s, for 8 months, at 7 per cent. per annum. Ans. 351 128 31d.

Note. The Rules for Simple Interest, serve also to calculate Insurances, or the Purchase of Stocks, or any thing else that is rated at so much per cent.

See also more on the subject of Interest, with the algebraical expression and investigation of the rules at the end of the Algebra, next following.

COMPOUND INTEREST.

COMPOUND INTEREST, called also Interest upon Interest, is that which arises from the principal and interest, taken together, as it becomes due, at the end of each stated time of payment. Though it be not lawful to lend money at Compound Interest, yet in purchasing annuities, pensions, or leases in reversion, it is usual to allow Compound Interest to the purchaser for his ready money.

RULES.-1. Find the amount of the given principal, for the time of the first payment, by Simple Interest. Then consider this amount as a new principal for the second payment, whose amount calculate as before. And so on through all the payments to the last, always accounting the last amount as a new principal for the next payment. The reason of which is evident from the definition of Compound Interest. Or else,

2. Find the amount of 1 pound for the time of the first payment, and raise or involve it to the power whose index is denoted by the number of payments. Then that power multiplied by the given principal, will produce the whole

amount,

RROBLEM VI.

To find the Number of Combinations of any Given Number of Things, by taking any Given Number at a time; in which there are several Things of one Sort, several of another, &c.

RULE.

FIND, by trial, the number of different forms which the things to be taken at a time will admit of, and the number of combinations there are in each.

Add all the combinations, thus found together, and the sum will be the number required.

EXAMPLES.

1. Let the things proposed be a aa bbc; it is required to find the number of combinations made of every 3 of these quantities?

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2. Let a a a b b b c c be proposed; it is required to find the number of combinations of these quantities, taken 4 at a time? Ans. 10.

3. How many combinations are there in a aa a b b c cde, taking 8 at a time? Ans. 13.

4. How many combinations are there in a aa a a b b b b b ccccddddee e efffg, taking 10 at a time? Ans. 2819.

PROBLEM. VII.

To find the Compositions of any Number, in an equal Number of Sets, the Things themselves being all different.

RULE.

MULTIPLY the number of things in every set continually together, and the product will be the answer required.

* Demonstr. Suppose there are only two sets; then, it is plain, that every quantity of the one set being combined with every quantity of the other, will make all the compositions, of two things in these two sets:

and

EXAMPLES.

1. Suppose there are four companies, in each of which there are 9 men; it is required to find how many ways 9 men may be chosen, one out of each company?

9

9

81

9

729

6561 the Answer.

Or, 9 x 9 x 9 x9= 6561 the Answer.

2. Suppose there are 4 companies; in one of which there are 6 men, in another 8, and in each of the other two 9; what are the choices, by a composition of 4 men, one out of each company?

Ans. 3888.

3. How many changes are there in throwing 5 dice?

Ans. 7776.

and the number of these compositions is evidently the product of the number of quantities in one set by that in the other.

Again, suppose there are three sets; then the composition of two, in any two of the sets, being combined with every quantity of the third, will make all the compositions of three in the three sets. That is, the compositions of two in any two of the sets, being multiplied by the number of quantities in the remaining set, will produce the compositions of three in the three sets; which is evidently the continual product of all the three numbers in the three sets.

And the same manner of reasoning will hold, let the number of sets be what it will. Q. E. D.

The doctrine of permutations, combinations, &c. is of very extensive use in different parts of the Mathematics; particularly in the calculation of annuities and chances. The subject might have been pursued to a much greater length; but what is here done, will be found sufficient for most of the purposes to which things of this nature are applicable.

PRACTICAL

PRACTICAL QUESTIONS IN ARITHMETIC.

QUEST. 1. The swiftest velocity of a cannon-ball, is about 2000 feet in a second of time. Then in what time, at that rate, would such a ball be in moving from the earth to the sun, admitting the distance to be 100 millions of miles, and the year to contain 365 days 6 hours.

Ans. 8. 4808

13149 years. QUEST. 2. What is the ratio of the velocity of light to that of a cannon-ball, which issues from the gun with a velocity of 1500 feet per second; light passing from the sun to the earth in 7 minutes? Ans. the ratio of 7822223 to 1.

QUEST. 3. The slow or parade-step being 70 paces per minute, at 28 inches each pace, it is required to determine at what rate per hour that movement is? Ans. 1 miles.

QUEST. 4. The quick-time or step, in marching, being 2 paces per second, or 120 per minute, at 28 inches each; then at what rate per hour does a troop march on a route, and how long will they be in arriving at a garrison 20 miles distant, allowing a halt of one hour by the way to refresh?

the rate is 3 miles an hour.

Ans. {and the time 74 hr. or 7 h 174 min.

QUEST. 5. A wall was to be built 700 yards long in 29 days. Now, after 12 men had been employed on it for 11 days, it was found that they had completed only 220 yards of the wall. It is required then to determine how many men must be added to the former, that the whole number of them may just finish the wall in the time proposed, at the same rate of working? Ans. 4 men to be added.

QUEST. 6 To determine how far 500 millions of gui neas will reach, when laid down in a straight line touching ⚫ one another; supposing each guinea to be an inch in diameter, as it is very nearly. Ans. 7891 miles, 728 yds, 2ft, 8 in.

QUEST. 7. Two persons, A and B, being on opposite sides of a wood, which is 536 yards about, they begin to go round it, both the same way, at the same instant of time; A goes at the rate of 11 yards per minute, and в 34 yards in 3 minutes; the question is, how many times will the wood be gone round before the quicker overtake the slower?

Ans. 17 times.
QUEST.

ALLIGATION ALTERNATE.

ALLIGATION ALTERNATE is the method of finding what quantity of any number of simples, whose rates are given, will compose a mixture of a given rate. So that it is the reverse of Alligation Medial, and may be proved by it.

RULE 1*.

1. SET the rates of the simples in a column under each other.-2. Connect, or link with a continued line, the rate of each simple, which is less than that of the compound, with one, or any number, of those that are greater than the compound; and each greater rate with one or any number of the less.-3. Write the difference between the mixture rate, and that of each of the simples, opposite the rate with which they are linked.-4. Then if only one difference stand against any rate, it will be the quantity belonging to that rate; but if there be several, their sum will be the quantity.

The examples may be proved by the rule for Alligation. Medial.

Demonst. By connecting the less rate to the greater, and placing the difference between them and the rate alternately, the quantities resulting are such, that there is precisely as much gained by one quantity as is lost by the other, and therefore the gain and loss upon the whole is equal, and is exactly the proposed rate and the same will be true of any other two simples managed according to the Rule. In like manner, whatever the number of simples may be, and with how many soever every one is linked, since it is always a less with a greater than the mean price, there will be an equal balance of loss and gain between every two, and consequently an equal balance on the whole. E. D.

It is obvious, from this Rule, that questions of this sort admit of a great variety of answers; for, having found one answer, we may find as many more as we please, by only multiplying or dividing each of the quantities found, by 2, or 3, or 4, &c: the reason of which is evident; for, if two quantities, of two simples, make a balance of loss and gain, with respect to the mean price, so must also the double or treble, the or part, or any other ratio of these quantities, and so on ad infinitum.

These kinds of questions are called by algebraists indeterminate or unlimited problems; and by an analytical process, theorems may be raised that will give all the possible answers.

EXAMPLES.

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