 | Nathan Daboll - Arithmetic - 1824 - 264 pages
...«ceived for them all I3O/, being paid for every ox 71. fix every cow 5/. and for every calf 1 1. 10s. there were twice as many cows as oxen, and three times as many catves as cows; how many were there of each sort? .ins. 5 oxen, 10 coit-S, and 50 wives. 9. A, B and... | |
 | Nathan Daboll - Arithmetic - 1825 - 248 pages
...being paid for every ox 71. for every cow 5l. and for every calf 1/. 10s. there were twice as many cews as oxen, and three times as many calves as cows ;...many were there of each sort ? Ans. 5 oxen, 10 cows, and S0 calves. 9. A, B and C, playing at cards, staked 324 crowns } but disputing about tricks, each... | |
 | Warren Colburn - Algebra - 1825 - 400 pages
...each, the cows at $18, and the calves at $4. There were three times as many cows as oxen, and twice as many calves as cows. How many were there of each sort ? Ans. 3 oxen, 9 cows, and 18 calves. 14. A merchant bought a quantity of flour for $132; for one half of... | |
 | Nicolas Pike, Dudley Leavitt - Arithmetic - 1826 - 214 pages
...paid at the rate of $24 per ox, $16 per cow, and $6 per calf: there were as many oxen as cows, and 4 times as many calves as cows : How many were there of each sort ? Ans. 5 oxen, 5 cows, and 20 calves. PERMUTATION. PERMUTATION is the method of finding how many different ways the... | |
 | Warren Colburn - Arithmetic - 1826 - 264 pages
...and some sheep, which he sold for $749 ; the oxen at $28, the cows at $17, and the sheep at $7.50. There were twice as many cows as oxen, and three times as many sheep as cows. How many were there of each sort ? 131. A man sold 16 bushels of rye, and 12 bushels... | |
 | Warren Colburn - Algebra - 1826 - 290 pages
...each, the cows at $18, and the calves at $4. There were three times as many cows as oxen, and twice as many calves as cows. How many were there of each sort ? 14. A merchant bought a quantity of flour for $132 ; for one half of it he gave $5 per barrel, and... | |
 | Nicolas Pike, Dudley Leavitt - Arithmetic - 1826 - 222 pages
...paid at the rate of $24 per ox, $16 per cow, and $6 per calf: there were as many oxen as cows, and 4 times as many calves as cows : How many were there of each sort ? Ans. 5 oxen, 5 cows, and 20 calves. PERMUTATION. PERMUTATION is the method of finding how many different ways the... | |
 | Nathan Daboll - Arithmetic - 1828 - 250 pages
...received for them all 130?. being paid for every ox 71. for every cow 51. and for every calf I1. 10s. there were twice as many cows as oxen, and three times...calves as cows; how many were there of each sort ? Ans. Boxen, 10 cows, and 30 calves. 9. A, B, and C, playing at cards, staked 324 crowns ; but disputing... | |
 | Nathan Daboll - Arithmetic - 1828 - 268 pages
...71. for every cow 5l. and for every calf \l. 10s. there were twice as many cows as oxen, and thr«e times as many calves as cows; how many were there of each sort ? , Ans. 3 oxen, 10 cows, and 3O calces. 9. A, Band C, playing at cards, staked 324 crowns; but disputing about... | |
 | Warren Colburn - Algebra - 1828 - 330 pages
...each, the cows at $18, and the calves at $4. There were three times as many cows as oxen, and twice as many calves as cows. How many were there of each sort ? -^ cr>*-' * " • 14. A merchant bought a quantity of (lour for $132 ; for one half of it he gave... | |
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I1. 10s. there were twice as many cows as oxen, and three times as many calves as cows; how many were there of each sort ? Ans. Boxen, 10 cows, and 30 calves. 9. A, B, and C, playing at cards, staked 324 crowns ; but disputing about tricks, each man took... 
