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x 60 = (x+20) x 60 = 60 x + 1200. Ans. I 2 60 x + 1200 = 6000; x=80; x + 40 = 120.

Ans. 80 yd. and 120 yd. 9. The number of square feet in the walls of a room 164 ft. long, 148 ft. wide, and 13} ft. high, may be obtained by adding the bases of the four sides, — 161 +144 +161 +144= 624,- and multiplying this by their common height, 131. Dividing by 9 gives the number of square yards. The operation of finding the cost may be indicated as follows:

10€ x 125 x 40 _ 25000 ¢ = $9.2545. Ans. $9.26.

9X2 X3 27

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10. (20 x 171) -- 21 gives the number of feet of carpet. Dividing this result by 3 gives the number of yards.

11. 414 lb. X (154 x 19 x ). Cancel.

12. The “development” will be a modification of the one given in problem 20, Arithmetic, Art. 818. In drawing the development, the pupil should be required to approximate the proper proportions, and to place the faces in the proper order. It is not necessary to have the top and bottom faces in the positions shown in Art. 818.

The surface of the four vertical faces should be obtained in one operation, as in 9 ; also the surface of the two horizontal faces :

[2x (34+24) x 1}]+[2 x (34 x 24)]. 13. (1351 - 12%) ft. 14. [(128 x 152 x 105) = 2150.4] bu. 15. [(77 x 45 x 54) -- 231] gal.

16. 40 acres = (160 x 40) sq. rd. = 6400 sq. rd. The dimensions are 80 rd. by 80 rd., making 320 rd. of fence necessary. There will be 640 posts, at 15$ each; and 5 times 640 rails, or 3200 rails, at 10$ each.

17. There will be 16 fields, 4 rows of 4 fields each. Five parallel fences, each a mile long, and five other parallel fences of the same length, and perpendicular to the first, will be required.

19. 1728 cu. in. of water weigh 1000 oz.; 1 cu. in. weighs (1000 = 16) lb. = 1728; 231 cu. in. weigh [(1000 x 231) = (16 x 1728)] lb.

21. Number of square feet = (320+ 210 + 320 + 210) x 6. Divide by 9 to obtain square yards.

22. Area of outer rectangle in square feet = 332 x 222; of inner rectangle = (320 x 210) sq. ft. Divide the difference by 9 to obtain square yards.

23. Area of outside plot = (320 x 210) sq. ft.; of inner plot = (308 x 198) sq. ft.

972. 1. Each yard measured with the short yardstick contains yd.; the true length = 25 yd. x

Or, each so-called yard is zo yd. short, and 25 yd. are the yd. short; and the piece contains 25 yd. - j yd. = 2431 yd. Ans.

4. 32 boys'=20 men. If 15 men do the work in 12 da., 20 men do it in 12 da. x .

20. Change 6 lb. 14 oz. and 23 lb. 12 oz. to ounces, or to pounds and fractions.

976. 3. A and B can mow 1 of the field in 1 da., all three can mow of the field in 1 da. O mows (1 - 1) of the field in 1 da., or at of it; in 5 da. he does x 5, or of the work, for which he should receive 4 of $25. 4.

5 bu. @ 80¢ = 400%
5 bu. @ 60$ = 300%.

x bu. @ 30¢= 30 x¢.
: (x+10) bu. @ 50$ = (30 x + 700)¢,

50 x + 500 = 30 x + 700,

20 x = 200,
x= 10.

Ans. 10 bu. of oats.

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7. Total cost = (65¢ x 128) +80°; quantity sold=128 gal. – 16 gal. Selling price per gallon = of [(65$ x 128) + 80%] :(128 - 16).

It is advisable to accustom children to understand that a gain of } of cost makes the selling price of cost.

10. On sofas sold for $1125 there was a loss of }, making the selling price of cost. On the remaining sofas, the selling price, $1125, represents of cost.

4 fifths of cost = $1125, selling price of 25 sofas.
1 fifth of cost = $281ļ, loss on first lot.
6 fifths of cost = $1125, selling price of remaining sofas.

1 fifth of cost = $1871 = gain on second lot.
Loss on the transaction, $281.25 – $187.50.
The following is an algebraic solution without fractions :
Let

x = loss on first lot; then

5x = cost of first lot.
5 x — x=4x = selling price = 1125.

x = 2814 = loss in dollars.

x=gain on second lot;

5x= cost of second lot.
5x+x=6x = selling price=1125.
x = 1871 = gain in dollars.

etc.

Let

11. Let x = cost per egg.

18 x = cost of 18 eggs.
18x = selling price per egg.

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He sold it at 3f of asking price; i.e., 5 of

399 x 400

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20. Length of field in rods =(1600 + 146) +- 18.

21. Let 10x represent cost, then loss = x; and selling price = 9x = $117; x=13; cost = $130. A gain of 10% would be $13, making the price at which he should have sold it to gain 10% = $130+ $13 = $143. Ans.

22. Wife receives anti; son, of , or 2 ; daughter, $ 5000.

+ 5000 ==.

977. The ten problems of this section will call for no special treatment. An occasional problem of this kind has already been given, although, perhaps, with smaller numbers. After the pupils understand in 1, that the joint capital, $ 700, is the basis upon which the profits are distributed, they will have no difficulty in understanding that B is entitled to 40% of $182, and that C is entitled to 40% of $182. Cancellation should be employed. See Art. 1121, No. 5.

There is no need in this connection of discussing the subject of business partnerships that are continued for a year or longer. The division of profits in these cases is the subject of a special agreement, and is rarely made solely on the basis of the amounts invested. A yearly gain of $4000 made in regular business by two partners, one of whom invested $1000, and the other $3000, might be divided in various equitable ways. The partner investing the larger sum, might first take out $120 as interest on his excess of capital; and the remaining $3880 might be equally divided, giving one of them $1940, and the other $ 2060. Another arrangement might permit each partner to withdraw a fixed sum for services, say $1000, leaving $2000 to be divided on the basis of 1 to 3. This would make the shares ($1000 +$500) and ($1000+$1500), or $ 1500 and $2500.

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10. Cases of this kind are found only in the books.

979. The cost of the first item is given. The second item contains 451 sq. ft. ; the third contains (4 x 42 x 73) sq. ft., or 1204 sq. ft.; the fourth contains (3 x 43 x 74) sq. ft., or 989 sq. ft.; the total of the three items being 2644 sq. ft. The cost @ 10d. is

£110 – 3-4
First item, 24 – 7-8

£134 – 11
Less 27% (o), 3 – 7–34+

Ans. £ 131 – 3 –89 The fraction of a penny in the discount is , which is nearly 1 farthing, written d.

980. German currency being a decimal one, the bill is computed in the ordinary way. The duty is found by reducing the marks to dollars by multiplying by $.238, and taking 35% of the result.

981. Too much stress cannot be laid upon the importance of requiring children to estimate the probable answer to every “written " problem before placing a figure on paper. The mere drill on the “approximations" found in the text-book is of com

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