Elementary ArithmeticHarper & Brothers, 1878 |
From inside the book
Results 1-5 of 23
Page 40
... gallons are 110,040 gallons - 90,621 gallons ? 56. At the battle of Bunker Hill the Americans lost 449 men , and the British 1054. How much did the British loss exceed the American ? SECOND METHOD . 37. Ex . From 7,623 subtract 4,856 ...
... gallons are 110,040 gallons - 90,621 gallons ? 56. At the battle of Bunker Hill the Americans lost 449 men , and the British 1054. How much did the British loss exceed the American ? SECOND METHOD . 37. Ex . From 7,623 subtract 4,856 ...
Page 51
... gallons in 6 hogsheads , each containing 124 gallons ? 744 . 22. A market gardener bought 5 acres of land , at $ 635 per How much did the land cost him ? acre . 23. Nine cars of a freight train are loaded with flour , and each car ...
... gallons in 6 hogsheads , each containing 124 gallons ? 744 . 22. A market gardener bought 5 acres of land , at $ 635 per How much did the land cost him ? acre . 23. Nine cars of a freight train are loaded with flour , and each car ...
Page 52
... gallons are 8 times 27,645 gallons ? 33. How many pounds are 5 times 32,051 pounds ? 34. What is the product of 6 times 1,026,348 ? 6,158,088 . CASE II . The Multiplier any number of Tens , 52 INTEGERS .
... gallons are 8 times 27,645 gallons ? 33. How many pounds are 5 times 32,051 pounds ? 34. What is the product of 6 times 1,026,348 ? 6,158,088 . CASE II . The Multiplier any number of Tens , 52 INTEGERS .
Page 54
... gallons , how many gallons of kerosene in the cargo ? 301,000 . 48. A wholesale grocer bought 500 chests of tea , each con- taining 56 pounds . How many pounds did all the chests con- tain ? 28,000 . 49. In one hour there are 60 minutes ...
... gallons , how many gallons of kerosene in the cargo ? 301,000 . 48. A wholesale grocer bought 500 chests of tea , each con- taining 56 pounds . How many pounds did all the chests con- tain ? 28,000 . 49. In one hour there are 60 minutes ...
Page 59
... gallons . How many gallons of oil in the cargo ? 68,640 . 88. How much would the yearly pay of an army of 650,000 men amcunt to , at $ 192 per man ? $ 124,800,000 . 89. If the factors are 9,654 and 21,800 , what is the pro- duct ? 90 ...
... gallons . How many gallons of oil in the cargo ? 68,640 . 88. How much would the yearly pay of an army of 650,000 men amcunt to , at $ 192 per man ? $ 124,800,000 . 89. If the factors are 9,654 and 21,800 , what is the pro- duct ? 90 ...
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Common terms and phrases
12 rods acres amount annex apples blocks bushels cents ciphers Commence Compound Numbers contained cords of wood cost Count by 9's cubic inches dealer decimal figures decimal point Divide Division divisor dollars dreds Dry Measure equal EXPLANATION EXPLANATION.-Since Express by figures farm farmer fence foot fourth gallons Hence higher denomination hundred hundred-thousandths hundredths improper fraction inches long inches wide integer interest Lake Itasca loads lowest terms Manual measure millionths minuend mixed number month multiplicand Multiply ounces paid partial dividend pile pint place the decimal pounds PROBLEMS pupils quarts quires quotient railroad Read Reduce remainder result right-hand figure rods long SECOND SOLUTION sell sheets similar fractions sold square miles subtract subtrahend ten-millionths ten-thousandths tenths third thousand thousandths tons week weighing Write in words write the numbers written yards
Popular passages
Page 177 - Multiply the numerators together for a new numerator, and the denominators together for a new denominator.
Page 78 - Cut off the ciphers from the divisor, and the same number of figures from the right of the dividend.
Page 138 - A pile of wood 8 feet long, 4 feet wide, and 4 feet high, contains 1 cord; and a cord foot is 1 foot in length of such a pile.
Page 55 - Zeros at the right of the partial products, since they do not affect the result of the addition, should be omitted as in (2). Care must be taken, however, to put the right-hand figure of each partial product directly under the figure of the multiplier used to obtain it.
Page 32 - It shows that the number after it is to be subtracted from the number before it.
Page 78 - III. Multiply the divisor by this quotient figure, subtract the product from the partial dividend, and to the remainder annex the next figure of the dividend.
Page 64 - It shows that the number before it is to be divided by the number after it. The expression 6 -H 2 = 3 is read, 6 divided by 2 is equal to 3.
Page 183 - This quotient may be obtained by multiplying the numerator of the dividend by the denominator of the divisor for the numerator of the quotient, and the denominator of the dividend by the numerator of the divisor for the denominator of the quotient Hence, the Rule.
Page 41 - RULE. — Place the less number under the greater, so that units of the same order shall stand in the same column. Commencing at the right hand, subtract each figure of the subtrahend from the figure above it.
Page 79 - Then divide the remaining figures of the dividend by the remaining figures of the divisor.