When, in order to find a fourth proportional, several circumstances require to be considered, it is called Compound Proportion. If 14 horses eat 56 bushels of oats in 16 days ; how many bushels will be required for 20 horses for 24 days? Practical arithmetic for the use of adults - Page 103by P. Fletcher - 1848 - 120 pagesFull view - About this book
| Francis Walkingame - 1833 - 204 pages
...Reduction of Vulgar 1' radiums, Case 6. PROOF. By two operations of the Single Rule of Three. (!)• If 11 horses eat 56 bushels of oats in 16 days, how many bushels will serve 20 horses 24 days ?t • See ulso Supplemental Questions, Nos. 6 and 7. t By Rule ! Ï .*••••... | |
| Francis Walkingame - 1835 - 270 pages
...the quotient of these will be the answer, in the same denomination as the third term. EXAMPLES. (1) If 14 horses eat 56 bushels of oats in 16 days, how many bushels will be sufficient for 20 horses 24 days? S Dividend x xM = ,0 x 3 x 4 J?' An,. Divisor W x *6 2 (2) If 8 men... | |
| Luther Ainsworth - Arithmetic - 1837 - 298 pages
...after the rate of 874 cents the square yard ; to what did it amount at that rate 1 Ans. $108,28 cts. 3. If 14 horses eat 56 bushels of oats in 16 days, how many bushels will be sufficient for 20 horses 24 days ? Ans. 120 bushels, 4. If 112 acres of grass can be mowed in 14 days,... | |
| Richard W. Green - Arithmetic - 1840 - 300 pages
...will last 7 men 14 days, how much will last 3 men 21 days ? Ans. Srrri = SJLFJl = Ans. 36 loaves. 2. If 14 horses eat 56 bushels of oats in 16 days, how many bushels will be sufficient to keep 20 horses 24 clays? Ans. 120 bushels. 3. If 40s. will pay 8 men for 5 days' work,... | |
| Commissioners of National Education in Ireland - 1842 - 160 pages
...fourth proportional, several circumstances require to be considered, it is called Compound Proportion. If 14 horses eat 56 bushels of oats in 16 days ; how maiij bushels will be required for 20 horses for 24 days ? frttffA. RULE wiTHExAMPLE. — Write horses... | |
| Alfred Crowquill - Arithmetic - 1843 - 156 pages
...for a divisor, and the other three for the dividend, and the quotient will be the answer. EXAMPLES. If 14 horses eat 56 bushels of oats in 16 days; how many bushels will be sufficient for 20 horses for 24 days ? By two single rules. .* Or, in one stating, worked thus : hor.... | |
| Ireland commissioners of nat. educ - 1850 - 160 pages
...fourth proportional, several circumstances require to be considered, it is called Compound Proportion. If 14 horses eat 56 bushels of oats in 16 days ; how...bushels will be required for 20 horses for 24 days? bush. RULE wiTHExAMPLE. — Write horses 14 : 20 : : 56 down for the third term that days 16 : 24 number... | |
| Commissioners of National Education in Ireland - Arithmetic - 1850 - 162 pages
...to be considered, it is called Compound Proportion. If 14 horses eat 56 bushels of oats in 16 <lays; how many bushels will be required for 20 horses for 24 days ? l«ih. RULE wrrnExAMPLE — Write horses 14 : 20 : : 56 down for the third term that days 16 : 24... | |
| John Radford Young - Arithmetic - 1852 - 230 pages
...fraction - : which, J 11x18 by striking out common factors, becomes simply 5x2 or 10. Exercites. 1. If 14 horses eat 56 bushels of oats in 16 days, how many horses will 120 bushels keep for 24 days? 2. If a person walking 12 hours a day travel 250 miles in... | |
| J L. Ellenberger - 1854 - 338 pages
...weighed 14 ounces when wheat was 4s. per bushel, what must it weigh when wheat is 6s. 9d. per bushel ? 9. If 14 horses eat 56 bushels of oats in 16 days, how many bushels will serve 20 horses 24 days ? 10. What must be the breadth of a piece of ground which is 14$- yards long,... | |
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