 | Zadock Thompson - Arithmetic - 1832 - 186 pages
...the answer. Therefore, ... j \ I V. The principal, rate and interest being given, tofind the time. RULE.— Divide the given interest by the interest of the given principal for 1 year at the given rale, and the quotient will be the time m years and decimal parts. 2. If the mterest... | |
 | Zadock Thompson - Arithmetic - 1832 - 186 pages
...answer. Therefore, -• - IV. The principal, rate and interest being gtven, to fjidt!wtim$, ROLE. — Divide the given interest by the interest of the given principal for 1 year at the given rate, and the quotient will be the time in years and decimal parts. 2. If the interest... | |
 | Charles Guilford Burnham - Arithmetic - 1837 - 266 pages
...ratio of the interest for 1 year is to the given interest as 1 year is to the time required. Hence the RULE : DIVIDE the given interest by the interest of the given principal for 1 year, and the quotient will be the answer. 2. In what time will $240 gain $4,80, at 6 per cent? Answer,... | |
 | Calvin Tracy - Arithmetic - 1840 - 326 pages
...8 months 1 Ans. $360. Prob. 37. Given the principal, interest, and time, to find the rate per cent. RULE. — Divide the given interest by the interest of the given principal, at one per cent, for the given time. Ex. 1. A man having $4000 on interest, at the expiration of one... | |
 | George Roberts Perkins - Arithmetic - 1841 - 274 pages
...and the interest, to find the rate per cent. RULE. Divide the given interest by the interest oftte given principal, for the given time, at 1 per cent....The interest of $100, for 9 months and 10 days, is $3.50. What is the rate per cent. ? In this example we find the interest of $100, for 9 months and... | |
 | Charles Guilford Burnham - Arithmetic - 1841 - 324 pages
...of the interest for 1 year, is to the given interest, as 1 year is to the time required. Hence the RULE : Divide the given interest by the interest of the given principal for 1 year, and the quotient will be the answer. 2. In what time will $240 gain $4.80, at 6 per cent. ?... | |
 | George Roberts Perkins - Arithmetic - 1846 - 266 pages
...$4090.909. PROBLEM III. Given the principal, the time, and the interest, to find the rate per cent. RULE. Divide the given interest by the interest of...The interest of $100, for 9 months and 10 days, is $3.50. What is the rate per cent. ? In this example, we find the interest of $100, for 9 months and... | |
 | Zadock Thompson - Arithmetic - 1848 - 184 pages
...months (145), the answer. Therefore, IV. Tlte principal, rate and interest being given, tQjind the time. RULE. — Divide the given interest by the interest of the given principal for 1 year at the given rate, and the quotient will be the time m years and decimal ^,arts. 2. If the interest... | |
 | George Roberts Perkins - Arithmetic - 1849 - 346 pages
...as many times greater than the interest for one year, as the particular time is greater than 1 year. Hence, we have this RULE. Divide the given interest by the interest of the given principal, for 1 year, at the given rate per cent. EXAMPLES, 1. In what time will $37-13. at 4£ per cent., give $0-7518825... | |
 | Rufus Putnam - Arithmetic - 1849 - 276 pages
...it must be on interest as many years as there are times $21 in $87.50. Ans. 4 j yrs. = 4 yr. 2 mo. RULE. Divide the given interest by the interest of the given principal for 1 year. 4. A note of $340 amounted to $381.65 ; how long was it on interest, the rate being 6J per... | |
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RULE. — Divide the given interest by the interest of the given principal for 1 year, and the quotient is the time. 
