 | 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 on... | |
 | 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, 4 months. 3. In what... | |
 | 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... | |
 | 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. ? Answer, 4 months. 3. In... | |
 | 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 given principal, for the given time, at 1 per cent. EXAMPLES. 1. The interest of $100, for 9 months and 10 days, is $3.50.... | |
 | 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... | |
 | 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 cent. ?... | |
 | George Roberts Perkins - Arithmetic - 1849 - 346 pages
...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... | |
 | Benjamin Greenleaf - Arithmetic - 1849 - 336 pages
...per cent., $ 48 will be as many per cent. as $ 6 is contained times in $48. which gives 8 per cent. for the answer. RULE. — Divide the given interest by the interest of the given sum at 1 per cent. for the given time, and the quotient will be the rate per cent. required. QUESTIONS.... | |
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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. 
