 | George Roberts Perkins - Arithmetic - 1849 - 356 pages
...per cent. Problem IV. Given the principal, the rate per cent., and the interest, to find the time. 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
...years to gain $ 36 as $ 18 is contained times in $ 36. Thus, $ 36 -i- $ 18 = 2 years for the answer. RULE. — Divide the given interest by the interest of the given principal for 1 year, and the quotient is the time. EXAMPLES FOR PRACTICE. 2. If the interest of $ 140 at 6 per cent,... | |
 | George Roberts Perkins - Arithmetic - 1850 - 364 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 I year, at the given rale per cent. EXAMPLES. 1. In what time will $37-13. at 4i per cent., give $0-75... | |
 | George Roberts Perkins - Arithmetic - 1850 - 356 pages
...per cent. Problem IV. Given the principal, the rate per cent., and the interest, to find the time. 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,... | |
 | Charles Guilford Burnham - 1850 - 350 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 - 1851 - 356 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 rale per cent. EXAMPLES. 1. In what time will $37-13. at 41 per cent., give $0-7518825... | |
 | Benjamin Greenleaf - 1851 - 332 pages
...many years to gain $ 36 as $ 18 is contained times in $ 36. Thus, $36 -=-$18 = 2 years for the answer. RULE. — Divide the given interest by the interest of the given principal for 1 year, and the quotient is the time. EXAMPLES FOR PRACTICE. 2. If the interest of $ 140 at 6 per cent.... | |
 | John Fair Stoddard - Arithmetic - 1852 - 320 pages
...9 days. PROBLEM. 4. — 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, at 1 per cent., for the given time. REMARK. — This rule is deduced from the fact, that the interest... | |
 | Benjamin Greenleaf - 1854 - 342 pages
...years to gain $ 36 as $ 18 is contained times in $ 36. Thus, $ 36 -H $ 18 = 2 years for the answer. RULE. — Divide the given interest by the interest of the given principal for 1 year, and the quotient is the time. , EXAMPLES FOR PRACTICE. 2. If the interest of $ 140 at 6 per... | |
 | Benjamin Greenleaf - Arithmetic - 1857 - 336 pages
...many years to gain $36 as $18 is contained times in $36. Thus, $36 -T- $18 = 2 years for the answer. RULE. — Divide the given interest by the interest of the given principal for 1 year, and the quotient will be the time. EXAMPLES FOR PRACTICE. 2. If the interest of $140 at 6 per... | |
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RULE. Divide the given interest by the interest of the given principal, for the given time, at 1 per cent. EXAMPLES. 1. 
