 | J. Brookes - Arithmetic - 1776 - 364 pages
...forbearance of any fum of money, for fome determined fpace of time. RULE I. When the Intereft is required for a Year, multiply the Principal by the rate per Cent, and divide that Product by loo, the Quotient is the Intercft required. What's the Intereft of 1 79/. for a Year,... | |
 | Michael Walsh - Arithmetic - 1807 - 290 pages
...is the sum per cent.-agreed on. Amount, is the principal and interest added together. GENERAL RULE. Multiply the principal by the rate per cent, and divide the product by 100, and the quotient is the answer for one year.. EXAMPLES. 1 . What is the interest of ¿£.496 for one... | |
 | Elijah H. Hendrick - Arithmetic - 1810 - 220 pages
...Simple and Compound.!, SIMPLE INTEREST is that which arifes from the principal only. General Rule. I. Multiply the principal by the rate per cent, and divide the product by loo, (or, which is the fame, cut-off two right hand figures in the pounds, which mint be reduced to... | |
 | Arithmetic - 1811 - 210 pages
...CASE*1!. When the time is one year, and the rate per cent, is pounds or dollars only. * • RULE.» Multiply the principal by the rate per cent, and divide the product by 100. PROOF. , , By the Single Rule of Three. EXAMPLES. 1. What is the interest of 525 L. for 1 year, at... | |
 | Nathan Daboll - Arithmetic - 1815 - 250 pages
...principal and interest added together. CASE I. To find the interest of any given sum for one year. RULE. Multiply the principal by the rate per cent, and divide the product by 100 ; the quotient will be the answer. EXAMPLE!. 1. What is the interest of 39Z. lls. 8id. for one year at 61. per cent,... | |
 | John Lyman Newell - Arithmetic - 1822 - 222 pages
...When the interest of any sum is required for one year, how do you proceed ? A. In finding the interest for a year, multiply the principal by the rate per cent., and remove the separatrix from its natural place two figures towards the left hand, then all those figures... | |
 | Nathan Daboll - Arithmetic - 1823 - 262 pages
...principal and interest added together. CASE I. To find the interest of any given sum for one yeai» RULE. Multiply the principal by the rate per cent. and divide the product by 100 ; the quotient will be the answer. EXAMPLES. 1. What is the interest of S9/. Us. 8itI. for one year, at 61. per cent.... | |
 | Stephen Pike - Arithmetic - 1824 - 212 pages
...1. To find the interest when the time is one year, and the rate per cent. is pounds or dollars only. Multiply the principal by the rate per cent. and divide the product by 100: the quotient will be the interest for 1 year. PROOF. By the Single Rule of Three. EXAMPLES. 1. What is the interest... | |
 | Nathan Daboll - Arithmetic - 1825 - 248 pages
...principal and interest addeâ together. CASE i. Vb find {he interest of any given sum for one ysar. RULE. Multiply the principal by the rate per cent. and divide the product by 100 ; the quotient will be the answer. EXAMPLES. 1. What is the interest of 391. 11s. SJd. for one year. at 61. per cent,... | |
 | Nicolas Pike, Dudley Leavitt - Arithmetic - 1826 - 214 pages
...insurance of his property, and is generally at so much per cent. RULE. — Multiply the sum to be insured by the rate per cent. and divide the product by 100 ; — the quotient will be the premium or answer. EXAMPLE. What will be the annual premium for insuring a house against... | |
| |
RULE.* Multiply the principal by the rate per cent, and divide the product by 100: the quotient will be the interest for 1 year. 
