Arithmetic for High Schools: Containing the Elementary and the Higher Principles and ... by James B Dodd

Author:James B Dodd
Language: eng
Format: epub
Publisher: Pratt
Published: 1859-03-25T05:00:00+00:00

5lb. at 13 cts. ' IS'^ 5 + 2x7Z6. at 13 cts. 2lb. at 14 cts. 14-*^ 2Z6. at 14 cts.

We take any one rate which is less than the mean rate 10, and any one which is greatery and adjust the proportions for those two rates. We proceed in like manner with either of these two rates and another, or two others, until all are included ; and add together the proportional terms found for the same rate.

«

Different results will be obtained, according to the difierent ways of coupling the ingredients.

Find other Answers to the preceding question.

Ans. lib. at $ cents, Zlb. at 8 cents, lib. at 13 cents, 5lb. at

14 cents. Alb. at 5 cents, llh. at 8 cents, 2lh. at 13 cents, IJh, at

14 cents ; lU). at 5 cents, Alb. at 8 cents, 6lb. at 13 cents, lib. at

14 cents.

COMPOUND RATIO.

(163.) A Compound Eatio is the ratio of the prodttct of two or more antecedents to the product of their consequents.

Thus the compound ratio of 3 and 4 to 5 and 7 is the ratio of the product 3 x4 to the product 5 x 7,=^.

COMPOUND PROPORTION.

(154.) A Compound Proportion consists in an equahty between a compound and a simple ratio. Thus

2 6

; g > : : 5 : 10 is a Compound Proportion;

in which the compound ratio 2 x 6 : 3 X 8 is equal to the simple ratio 5 : 10.

Compound Proportion is applicable to the solution of questions which would require two or more simple proportions. .

RULE XXXVI.

(165.) For Solving Questions in Compound Proportion.

1. Take for the third term that which is of the same kind as the answer to the question.

2. Take the remaining terms in couples of the same kind, and place each couple as required for questions in simple proportion (146 ... 2).

3. Multiply the first terms together for a divisor, and the second and third together for a dividend; the quotient will be the ans'vv^er required.

4. Each antecedent and its consequent must be taken in the same order of units, and the third term reduced, when necessary, as in finding a fourth proportional.

E XAMPL £.

If a footman can go 150 miles in 5 days, by walking 12 Hours each day, in how many days may he go 275 miles, by walking 10 hours each day?

150m. : 275m. ) . . ^ j -• • »

-Q7 .gi f • • o aays ; ttmerequired,

"We take 5 days for the third term, because the answer will be the number of days in which he would go 276m. *

It would require a greater number of days to go 275m. than it would to go 150m. ; the greater of these two terms must therefore be taken for the second term (146 ... 2).

When he walks lOA. a day he will require a greater num-. ber of days than when he walks 12^. a day ; hence the greater of these two terms must be taken ibr the second term. The operation is

(275xl2x5)-^(150xl0) = 16500-^1500 = ll days.

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