A Key to Chase's Common School Arithmetic: With Explanations and Remarks Upon the Peculiar Features of the Work, and Operations of the More Difficult ExamplesA. Hutchinson, 1853 |
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Page 6
... deficient in the first principles of numbers , that their teach- ers compelled them , very much against their will , to spend several weeks on the introductory chapters of Chase's Arithmetic . The result was an understanding of numbers ...
... deficient in the first principles of numbers , that their teach- ers compelled them , very much against their will , to spend several weeks on the introductory chapters of Chase's Arithmetic . The result was an understanding of numbers ...
Page 64
... deficiency . As the assumed quan- tities cost too little , we must diminish the quan- tity whose price is below the price of the mixture , or increase the one above it . The difference between the price of the mixture , 7 , and the ...
... deficiency . As the assumed quan- tities cost too little , we must diminish the quan- tity whose price is below the price of the mixture , or increase the one above it . The difference between the price of the mixture , 7 , and the ...
Page 65
... Deficiency 14 ct . Take 1 less at 5 ct . , 1 at 5 ct and 1 at 8 ct .; 1 at 5 ct . , 5 at 8 ct .; 5 at 5 ct . , 7 at 8 ct .; 1 at 5 ct . , 3 at 8 ct . 76-80 . Assume 2 at .70 , and 3 at 1.00 . Excess .40 . Take 2 gall . less at 1.00 , 2 ...
... Deficiency 14 ct . Take 1 less at 5 ct . , 1 at 5 ct and 1 at 8 ct .; 1 at 5 ct . , 5 at 8 ct .; 5 at 5 ct . , 7 at 8 ct .; 1 at 5 ct . , 3 at 8 ct . 76-80 . Assume 2 at .70 , and 3 at 1.00 . Excess .40 . Take 2 gall . less at 1.00 , 2 ...
Page 66
... Deficiency of 1 ct . Add 1 oz . of 18 , 1 of 12 , 2 of 18 , and 1 of 20 ; or we may assume 1 oz . of 12 , 2 of 18 , and 3 of 20. Excess 6 ct . Add 11⁄2 of 12 , & c .; or assume 2 oz . each . Deficiency 2 ct . 2 oz . of 18 carats , & c ...
... Deficiency of 1 ct . Add 1 oz . of 18 , 1 of 12 , 2 of 18 , and 1 of 20 ; or we may assume 1 oz . of 12 , 2 of 18 , and 3 of 20. Excess 6 ct . Add 11⁄2 of 12 , & c .; or assume 2 oz . each . Deficiency 2 ct . 2 oz . of 18 carats , & c ...
Page 67
... Deficiency 5 ct . Add 11 gall . 32 ct . , gall . at .32 and 5 of water . 161 For 19 ct . , assume 10 gall . each . Excess 2.85 . Add 15 gall . water , 10 at .32 and 25 of water . = For 15 ct . , assume 20 gall . each . Excess 5.05 . Add ...
... Deficiency 5 ct . Add 11 gall . 32 ct . , gall . at .32 and 5 of water . 161 For 19 ct . , assume 10 gall . each . Excess 2.85 . Add 15 gall . water , 10 at .32 and 25 of water . = For 15 ct . , assume 20 gall . each . Excess 5.05 . Add ...
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A Key to Chase's Common School Arithmetic: With Explanations and Remarks ... Admiral Paschel Stone No preview available - 2017 |
A Key to Chase's Common School Arithmetic: With Explanations and Remarks ... Admiral Paschel Stone No preview available - 2017 |
A Key to Chase's Common School Arithmetic: With Explanations and Remarks ... Admiral Paschel Stone No preview available - 2016 |
Common terms and phrases
11 gall 11 spaces 20 gall 9 mo A.'s distance A.'s share Add 11 gall Amount due Amount of $1 Assume 1 lb assume 1 oz Assume 10 gall bank discount carats cent circ COMPLEX ANALYSIS complex fraction compound interest COMPOUND NUMBERS cords cost cube root Decillions Decimals Deficiency denominator diameter divide dividend Duodecillions DUODECIMALS Excess 10 ct feet frac gain Hence hour inches last root figure last term least common multiple length Mensuration miles mill MISCELLANEOUS EXAMPLES mixture Nonillions obtain Octillions Octodecillions operation of Example perform Powers Present worth principal pupil Quindecillions quotient ratio remainder Required the contents rods Septillions square SUBTRACTION Sum of products Table Take 1 less teacher text book tons Tredecillions trial divisor Vigintillions Whole Numbers worth of $1
Popular passages
Page 92 - RULE. — Multiply the length (in feet) by the width (in inches) and divide the product by 12 — the result will be the contents in square feet.
Page 24 - ... thirds, and we wish to divide it into 6ths : We have, therefore, simply to reduce thirds to sixths. 2 sixths make a third, for the unit is divided into twice as many parts, and therefore the parts are one-half as large. Hence the RULE. Divide the required denominator by the denominator of the given fraction, and multiply the quotient by the numerator. The product will be the required numerator. Art. 58. — To reduce a whole number to an equivalent fraction, having a given denominator. 1. Reduce...
Page 92 - But. in measuring timber, you may multiply the breadth in inches, ami the depth in inches, and that product by the length in feet, and divide the last product by 144, which will give the solid content in ftet, &c.
Page 91 - С in. long, 14ft. wide, and 10ft. high. The room contains 4 windows, each 3 ft. 6 in. by 5 ft. 8 in. ; 2 doors, each 6 ft. 4 in. by 2 ft.
Page 69 - The roots of fractions are obtained by extracting the root of the numerator, and of the denominator, separately.
Page 44 - The remainder will form a new principal, upon which interest is to be cast to the time of the next payment.