Practical Arithmetic, by Induction and Analysis |
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Page 9
... dollar , one pound . 2. Number is a term signifying one or more units ; as , one , five , seven cents , nine men . 3. Numbers are expressed by ten characters , called Figures ; as , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 0 . 4. Arithmetic ...
... dollar , one pound . 2. Number is a term signifying one or more units ; as , one , five , seven cents , nine men . 3. Numbers are expressed by ten characters , called Figures ; as , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 0 . 4. Arithmetic ...
Page 21
... dollars , and when used is placed before the figures . 3. There are 43 sheep in one pasture , 21 in another , and 14 in another : how many sheep in all ? Ans . 78 . REVIEW . - 20 . Why are units placed under units , and tens under tens ...
... dollars , and when used is placed before the figures . 3. There are 43 sheep in one pasture , 21 in another , and 14 in another : how many sheep in all ? Ans . 78 . REVIEW . - 20 . Why are units placed under units , and tens under tens ...
Page 28
... dollars left is 23 ; that is , 135-112 = 23 . 23 remainder . 2. A farmer having 245 sheep , sold 123 : how many had he left ? Ans . 122 . 3. A man bought a farm for $ 751 , and sold it for $ 875 : how much did he gain ? What is the ...
... dollars left is 23 ; that is , 135-112 = 23 . 23 remainder . 2. A farmer having 245 sheep , sold 123 : how many had he left ? Ans . 122 . 3. A man bought a farm for $ 751 , and sold it for $ 875 : how much did he gain ? What is the ...
Page 32
... dollars had I left ? Ans . $ 1365 . 22. A man has property worth $ 10104 , and owes debts to the amount of $ 7426 : when his debts are paid , how much will be left ? Ans . $ 2678 . 23. A man having $ 100000 , gave away $ 11 : how many ...
... dollars had I left ? Ans . $ 1365 . 22. A man has property worth $ 10104 , and owes debts to the amount of $ 7426 : when his debts are paid , how much will be left ? Ans . $ 2678 . 23. A man having $ 100000 , gave away $ 11 : how many ...
Page 36
... dollars by 2 yards , or 25 cents by 25 cents , is as absurd as to propose to multiply 8 apples by 2 potatoes . ART . 31. When the Multiplier does not exceed 12 . EXAMPLES . 1. How many yards of cloth are there in 3 pieces , each ...
... dollars by 2 yards , or 25 cents by 25 cents , is as absurd as to propose to multiply 8 apples by 2 potatoes . ART . 31. When the Multiplier does not exceed 12 . EXAMPLES . 1. How many yards of cloth are there in 3 pieces , each ...
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Common terms and phrases
acres amount annexed apples bank discount barrels bought bushels cancel ciphers cloth common fraction composite number Compound Numbers contained cost cube root cubic denominator denotes diameter difference dividend divisible dollars Dry Measure equal expressed feet figure find the interest gain gallons Give examples given number greatest common divisor Hence hundred hundredths improper fraction inches least common multiple lowest terms MEASURE meter mills mixed number multiplicand multiply NOTE number of terms OPERATION payment pecks pints pounds prime factors principal proper fraction proportion quarts quotient rate per cent ratio Ray's Test Examples Reduce remainder Rule selling side simple fraction Simple Numbers sold solid contents SOLUTION square root subtract tens tenths third thousand thousandths U. S. Money units weight whole number write yards
Popular passages
Page 158 - Reduce compound fractions to simple ones, and mixt numbers to improper fractions ; then multiply the numerators together for a new numerator, and the denominators for. a new denominator.
Page 179 - To multiply a decimal by 10, 100, 1000, &c., remove the decimal point as many places to the right as there are ciphers in the multiplier ; and if there be not places enough in the number, annex ciphers.
Page 63 - TABLE. 10 Mills (m.) = 1 Cent . . ct. 10 Cents = 1 Dime . . d. 10 Dimes = 1 Dollar . $. 10 Dollars = 1 Eagle . E.
Page 137 - To reduce a mixed number to an improper fraction, — RULE : Multiply the whole number by the denominator of the fraction, to the product add the numerator, and write the result over the denominator.
Page 150 - We have seen that multiplying by a whole number is taking the multiplicand as many times as there are units in the multiplier.
Page 222 - Compute the interest to the time of the first payment ; if that be one year or more from the time the interest commenced, add it to the principal, and deduct the payment from the sum total. If there be after payments made, compute the interest on the balance due to the next payment, and then deduct the payment as above; and, in like manner, from one payment to another, till all the payments are absorbed ; provided the time between one payment and another be one year or more.
Page 219 - The rule for casting interest, when partial payments have been made, is to apply the payment, in the first place, to the discharge of the interest then due. If the payment exceeds the interest, the surplus goes towards discharging the principal, and the subsequent interest is to be computed on the balance of principal remaining due.
Page 52 - III. — 1. Cut off the ciphers at the right of the divisor, and as many figures from the right of the dividend. 2. Divide the remaining figures in the dividend by the remaining figures in the divisor.
Page 282 - Hence, when the first term, the common difference, and the number of terms, are given, to find the last term...
Page 219 - If the payment be less than the interest, the surplus of interest must not be taken to augment the principal; but interest continues on the former principal until the period when the payments, taken together, exceed the interest due, and then the surplus is to be applied...