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Or, the problem may be solved as follows (Arithmetic, Art.

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5. The date not being given, the number of days is taken as 120+3. The proceeds for 120 da. = $ 490.

8. The interest on $635 for 205 da. at 5% = $18.08.

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XVII

NOTES ON CHAPTER FOURTEEN

1115. 1. The total interest on the given sums of money is equal to the interest of $3000 for 1 mo. As the total sum is $1000, the interest of $3000 for 1 mo. equals the interest on $1000 for 3 mo.

1116. 3. Since the time is required, the products may be expressed in years (or months or days), and their total divided by the total of the multipliers.

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5. [(15 da. × 200) + (30 da. × 300) + (45 da. × 400)] ÷ (200 +300+400).

6. [(1 mo. x 210) + (2 mo. x 210) + (3 mo. x 210) + (4 mo. X 210)]÷840.

Since the sum due at each period is the same, the 210 may be omitted; (1 mo. + 2 mo. + 3 mo. + 4 mo.) ÷ 4.

7. [(2 mo. x 320) + (4 mo. × 160) + (5 mo. × 240)+(6 mo. × 240)] ÷ 960.

Or, (2 mo. x) + (4 mo. × ) + (5 mo. x ) + (6 mo. × 1) 41⁄2 mo. Ans.

8. (2 mo. ×)+(3 mo. × 1)+(4 mo. × ¿) + (12 mo. × 18) 71% mo. = 7 mo. 26 da. Ans.

9. (0 mo. x)+(3 mo. × 1) + (6 mo. × 1) + (9 mo. × 1) = 4 mo. Ans.

10. [(0 da. × 300) + (30 da. × 800) +(60 da. × 1000)]÷(300 +800+1000) = 40 da. July 1 + 40 da. = Aug. 10. Ans.

1117. 11. The total amount received ÷ number of bushels sold average price.

The arrangement of the work may follow that given in Art. 1116, Problem 3.

15 x 12 180

==

12. The first puts in the equivalent of 180 cows for a week; the second, 120 cows for a week; the third, 180 cows for a week; a total of 480 cows for a week. The first should pay 188 of $84;

48

the second, 120 of $84; the

480

third, 188 of $84.

Proportion is commonly employed in working examples of

20 × 6=120
18 x 10 = 180

480: 180::$84: x

480 120 $84: y

480 120 $84: z

this kind. As the whole number for a week (480) is to the first man's number for a week (180) so is the whole rent ($84) to the first man's share (x). The second proportion is used to ascertain the second man's share (y); and the third, to ascertain the third man's share (z).

B

2000 × 2 = 4000

1000 × 1 = 1000 5000

13. A furnishes $2000 for 2 yr. and $1000 for 1 yr., which is the equivalent of $5000 for a year. furnishes the equivalent of $6000 for a year. The total is $11000 for a year, and

the profits of $1100

3000 × 2 =

=

6000

11000: 5000: $1100: A 11000 6000: $1100: B

are distributed in the ratios of 500000 and 1000, as indicated by

11000

the proportions here given. A receives $500 of the profits and his capital of $3000, or $3500 in all. B receives $3600.

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that is, x bu. @ 16. A does

=

2

in 1 da.

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60¢ + 80 bu. @ 50¢ = (x + 80) bu. @ 52¢.

in 1 da.; B does in 1 da.; both do 2% +3% They finish the work, therefore, in 12 da., and receive $60. If A does in 1 da., in 12 da. he does, and should receive 12 of $60 = $36. B should receive $24. 17. C's capital of $4000 for yr. may be considered as $2000 for a year. This, with $4000 furnished by A and B, makes the capital $6000. A takes 1508, or of the profits; etc. See 13.

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6000'

180 tons : A's share.

180 tons : B's share.

600: 315 :: 180 tons : C's share.

19. Let x= number of quarts of water, then 40

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ber of quarts of milk. 40 quarts of the adulterated article at 5¢ per quart=200¢. (40-x) quarts of milk at 6 per quart= (240 – 6x)¢. x quarts of water cost nothing.

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The can contains 6 qt. water and 331 qt. milk.

1120.

1. $999 rent for 1 yr. 10. mo. 6 da., or 22 mo.

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per

=

5
= $999 ÷ 221 = $999 × 1; for 12 mo. =

111

cent of 55 oz. is 121 oz.?

of 55 = 121; 55 x = 12100; x= = 220.

The

= $999

3. The field contains 1600 sq. rd.; each side measures 40 rd., or 220 yd.; etc.

4. A solves 20 per hour; B solves 15 x, or 16, per hour; both solve 36 per hour. To solve 100 will require (100÷364) hours.

6. Six faces, each containing (15 × 15) sq. in.

7. Cost 90. See Arithmetic, Art. 924, 7. Selling price =90 × 1.431.

8. March 4, 1861, +4 yr. 1 mo. 11 da.

Apr. 15, 1865,

56 yr. 2 mo. 3 da.

=

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Apr. 15, 1865;

Feb. 12, 1809. Ans.

9. The selling price, $4.50 90% of cost; the latter is, therefore, $5 per barrel. Loss per barrel, 50; on 50 bbl., $25. A profit of 6% 30 per barrel; the gain on 100 bbl. = $30. Net gain = $5. Ans.

=

10. 60% of 66% =40%. If 40% of a number = 810, the number = 810 x 100 = 2025.

NOTE.-Observe the difference between this example and 8 of Art. 1101.

1121. 1. Having used of = box. If 14 box = $17, a $9.52. Ans.

=

box, the remainder of box box = $17 × 14 = 68¢ × 14 =

=

3. The 40-ft. ladder forms the hypotenuse of two rightangled triangles; CE and DE, Arithmetic, Art. 1250, Problem 9. CA, one perpendicular, measures 21 ft.; DB, the other, measures 33 ft. AB is the width of the street.

AE=√402-212; EB = √402-332; AE+ EB = AB. 4. 16 oz.: 12 oz. :: $28 : x.

5. A partnership is the association of two or more persons for the transaction of business on joint account. One advantage of a partnership is the employment of a larger capital, with a smaller percentage of expenses than would be the case were each member to establish a separate business. The firm obtains the combined business experience of its several members, and can

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