(6.) If a freehold estate which commences 4 years hence, be sold for 61977. 6s. 53d., allowing the purchaser 4 per cent. for his money, what ought the yearly rent to be?

Note. The preceding rules and examples include all kinds of annuities which do not depend upon chance.

SIMPLE INTEREST BY DECIMALS.

Put p the principal, or sum put to interest.

=

r the ratio, being the rate per cent, divided by 100.
t= the time, or years, the money is at interest.
i= the interest for the time t.

the amount.

Then, at 2 per cent. r = 025 | At 4 per cent. r = '04

3

3

1 day
1 week

1 month

1 quarter

1 half

And

r = '03 r = .035

35 of a year =

r=045

4
5

r= ⚫05, &c.

Decimals.

⚫002739726, &c.

33 of a year

019178082, &c.

of a year

⚫083'

Hence the decimal parts of a year, for any number of days, weeks, months, &c. may be readily found.

Proposition 1. Given the principal, time, and rate per cent., to find the interest or the amount.

Rule. Multiply the principal, time, and ratio, together, the last product will be the interest; to which add the principal to find the amount.

Theorem, ptri, and ptr+pa. When p, t, and r, are given. Prop. 2. Given the amount, (or the interest,) time, and rate, to find the principal.

Rule. Multiply the time by the ratio, and add au unit to the product; by this sum divide the amount, and the quotient will be the principal.-Or, divide the interest by the product of the time and ratio, and the quotieut will be the principal.

Prop. 3. Given the principal, time, and amount, (or the interest,) to find the rate per cent.

Rule. Divide the difference between the amount and the principal (viz. the interest) by the product of the principal and time, and the quotient will be the ratio, which multiply by 100 to obtain the rate per cent.

pt

=

pt

=r, when p, t, and a, (or i,) are given.

Prop. 4. Given the principal, rate, and amount, (or interest,) to find the time.

Rule. Divide the difference between the amount and the principal (viz. the interest) by the product of the principal and ratio, and the quotient will be the time.

Theo.

a-p i

pr pr

t. When p, r, and a, (or i,) are given.

Examples to Proposition 1.

(1.) What will 5671. 10s. amount to in 9 years, at 4 per cent. per annum ?

£567.5= p
9= t

(2.) What is the amount of 2351. at simple interest, for 3 years, at 5 per cent. per annum?

(3.) What is the interest of 5501. for 5 years at 31⁄2 per cent. per annum?

(4.) What is the interest of 7157. 15s. for 240 days, at 5 per cent. per annum?

(5.) What will 510l. amount to in 5 years 120 days, at 5 per cent. per annum?

Examples to Prop. 2

(6.) What principal, in 9 years, will amount to 7717. 16s. at 4 per cent. per annum?

9t •04=r

•36=tr

1.

Now a 7711. 16s.£771.8, dividend.
Hence 771.8-1.36=£567.5=5671. 10s. answer.

1.56tr+1 divisor.

(7.) What principal, put to interest for 9 years, will gain 2041. 6s. interest, at 4 per cent. per annum?

9=t

⚫04

•36=tr

Hence 204-3-36=£567·5—5671. 10s. answer.

(8.) What principal in 3 years, will amount to 2791. 1s. 3d., at 5 per cent. per annum?

(9.) What principal put to interest for 5 years, will amount to 810l. 16s. 6d.}, at 3 per cent. per annum? (10.) What principal, put to interest for 240 days, at 5 per cent. per annum, will gain 237. 10s. 71⁄2d.?

(11.) What principal put to interest for 65 days, at 5 per cent. per annum, will gain 37. 38. 7d. interest?

Examples to Prop. 3.

(12.) At what rate per cent, will 5677. 10s. amount to 7717. 168. in 9 years time?

£771 16 = £771.8=a
567 10= 567.5=P

204·3a-pi, dividend.

567.5x95107.5=pt, divisor.

Then 204-3-5107.504. Heuce the rate is 4 per cent.

(13.) At what rate per cent. will 2357. amount to 2791. 1s. 3d. in 34 years?

(14.) At what rate per cent, will 7157. 15s. amount to 9431. 17s. 10 d. in 7 years?

(15.) At what rate per cent, will 3577. 10s. gain 37, 3s. 72d.95 in 65 days?

(16.) At what rate per cent. per annum will 5107. amount to 6791. 8s. 44d. in 5 years and 120 days?

t

Examples to Prop. 4.

(17.) In what time will 5677. 10s. amount to 7717. 16s. at 4 per cent. per annum?

Then 204-3-22·7 = 9 years, the time required.

(18.) In what time will 7007. 10s. amount to 8107. 168. 62d., at 3 per cent. per annum?

(19.) In what time will 7157. 15s. amount to 9437. 17s. 102d., at 44 per cent. per annum?

(20.) In what time will 7157. 15s. gain 237. 10s. 74d. at 5 per cent. per annum?

(21.) In what time will 510l. amount to 6791. 8s. 41d.73 at 5 per cent. per annum?

EQUATION OF PAYMENTS AT SIMPLE INTEREST, BY DECIMALS,

ON MALCOLM'S PRINCIPLES.

Proposition. Having two debts, due at different times, to find the equated time for paying the whole at once, without loss either to the debtor or creditor.

Rule 1. Divide the sum of the debts by twice the first payment, multiplied by the ratio; to the quotient add half the time between the two payments, and call the sum the first number found.

2. Multiply the second payment by the time between the two payments, and divide the product by the first payment multiplied by the ratio; call the quotient the second number found.

3. From the square of the first-found number subtract the second, and extract the square-root of the difference. -The first-found number, diminished by this root, will give the equated time, reckoning from the time the first payment is due.

Note. The preceding rule is the same as Mr. Malcolm's, though expressed in a different, and it is apprehended, more intelligible, manner. This rule is built upon a supposition, That the sum of the interests of the debts, due before the equated time, from the time they become due to that time, ought to be equal to the sum of the discounts of the debts due after the equated time from that time to the time they become due. According to this supposition, the rule given above is universally true for two payments. But, when three or more pay ments are to be equated for, Mr. Malcolm's directions for finding an equated time for the two that are first payable, then their sum and a third, &c. is not strictly true, according to the supposition on which his rule is founded; nor would it be an easy matter to give general rules or theorems for all the possible cases, on account of the variation of the debts, and the difficulty of finding between which of the payments the equated time would fall. Besides, in long and tedious operations, mistakes are frequently made; and the answer, when obtained, admitting it to be true, differs a mere trifle from the answer found by the old rule: hence, a rule, founded upon simple interest and Mr. Malcolm's principles, may, I think, with propriety, be considered as an useless curiosity.

Examples.

(1.) A person has now due to him 320l. and at the end of 5 years 967. more will be due from the same debtor. Now both parties have agreed that the whole shall be paid at once, viz. at that time when the interest of the 3207. shall be equal to the discount of the 967. both being calculated at 5 per cent. per annum. The time of payment is required.

1st. 320+96=4161. sum of the debts.

320×2×05=32, the product of twice the first payment by the

ratio.

416-52-13, quotient. Then 13+3=15-5, the first number found. 2dly. 96×5-320 × ⋅05=30, the second number found.

3dly.1552-30=√/210•25=14•5, and 15′5—14·5—1 year, the time which must elapse (after the first payment is due) before the whole ought to be paid together according to the conditions of the question.

(2.) There is 100l. payable 1 year hence, and 105/ payable 3 years hence; what is the equated time, allowing simple interest at 5 per cent. per annum?

(3.) At Michaelmas 1815, I lent 3207., and at Michaelmas 1820, 2021. will be due to me from the same person. Now, on what day, and in what year, may I receive both the debts together, viz. 5227., reckoning interest at 5 per cent. per annum?