15. B's gain of $1400 is of total gain; of total = $200; A's gain, of total = $1000. 1167. 4. Let x cost per barrel. 75% of 500 x x .02} = 80.85. 5. A does work, or, and in 1 da.; B does B does in 1 da. of it, or 3, leaving A does 12 of the to be done by C require 4 da. in 4 da. To do the whole work C would 6. A ditch 20 yd. × 18 in. × 4 ft. is dug in (3 × 10) hours by A ditch 30 yd. x 27 in. × 5 ft. is dug in (9 × 15) hours 72 men. by? men. 7. £2400 income is produced at 3% by bonds whose face value is £80000. Their cost £80000 × .94% = £75500 = = $367433, 12% of which $44092.00. 8. x 40x 100 = $4.86 × (2400.03) × 948 × .12. 30+30% of 30= 39. 10. [(15+ 10 + 15 + 10) × 93] + (15 × 10) = number of and ceiling = 637 sq. ft. = 70% sq. yd. The cost 21d. x 425 1487d.; etc. 1168. 1. + = 1,5 = 1.41666+; ++$/ 148 1.41429- = Clearing of fractions, 37 650 x; x = 3. Ans. 49 = = == 32 of 2 of 8 of 7×17××1. x= smaller number; x+= larger. 32 x+x+18=1; 126x+126x+49=113; 252 x = 11364; x = 64 2842 = 1; x+18= }} +18=126 +128=126 +++ 14. Ans. 18 and 4. = 3. 16s. 4 d. 1963d. Ans. = £(§ of 1961) ÷ 240. mile = 1000 meters; 1 mi. = 1000 m. ÷ § = 1600 m. 17 mi. = 1600 m. × 17 = 27200 m.; 6 furlongs=200 m. × 6 = 1200 m.; 824 yd. 1600 m. 165 = 1760 2 27200 m. +1200 m. + 75 m. = 27 yd. 2 ft. 9 in. 1005 in.; 17 yd. 1 ft. 11 in. = 635 in. (635 × 1) sq. in. cost $25.40; 1 sq. in. = $2540 ÷ 635; (1005 × 7) sq. in. = ($25.40 ÷ 635) × 1005 × 7 = $35.171⁄2. A's rate is of B's; B's time is of A's. To run the whole distance, A needs 34 min. ++7=36 min. If he runs 2 miles in 16 min., in 1 min. he runs 2 mi. runs (2 min. 16) × 36 = mi. × 5 × 365 mi. Ans. 164, and in 36 min. he 6. For the information of the teacher, the following method We learn in algebra that the sum of two numbers (x + y) multiplied by their difference (xy) gives the difference of their squares (x2 — y2). If the sum of √2 and 1 be multiplied by their difference (√2 −1), we obtain the difference of their squares (2-1). Multiplying the numerator also by √2-1, we retain the equality and obtain a divisor that has no decimals. See Art. 1169, 3. It is not expected that this method should be given to the pupils. 7. For 42 da., 50 men were at work. To do the same work, 30 men would have required 70 da., or (70+40) da. to do the whole work. 110 da. - 84 da. = number of days the contractor would have been behindhand. = = 8. The number of square feet in the wall (233 +15+23 +15) × 11=9281 sq. ft. The two windows contain (19 × 5) sq. ft. 95 sq. ft.; the fireplace contains (4 × 6) sq. ft. 27 sq. ft.; the door contains (731) sq. ft. = 261 sq. ft.; a total of 95 sq. ft.+27 sq. ft.+261 sq. ft. 1481 sq. ft. There remain = to be papered 9281 sq. ft. 1481 sq. ft. 780 sq. ft. 780 sq. yd. A roll of paper contains 12 × 28 sq. yd.; the number of 10. The "present worth" of $365 due in 30 da. = $365÷ 1.005 = $363.18+ the cost of the horse in cash. The " ent worth" of the selling price = pres $435 1.02 $426.47+. = = Gain $426.47-$363.18 $63.29, which is 17.43% of the cost. If the seller has the note for $435 discounted at a bank, he will receive in cash $435-$8.70 $426.30. If he uses this money to buy the note he has given, he should pay, at bank rates, $365 $1.82 $363.174. The profit would be $426.30 = $363.17 $63.12, which is 17.38% of the cost. 11. £57 ls. 8d. = 13700d.; £2 11s. 4 d. = 13. A man that does only of a day's work, does 14 da. less work in 84 da. than the average. The contractor therefore loses, in 84 da. on three men, 14 da. + 12 da. + 9 da. = 35 da. He gains on two others 10 da. +8 1618 da. = da. 18% da. The net loss The extra 17 men have to do the equivalent of 1618 days' work; each has, therefore, to do 1618 days' work÷ 178 of a day's work, or less than the average. 14. Making no allowance for waste, etc., two strips, each 260 ft. long, will be needed for two sides; two strips, each (93 — 3 — 3) ft., or 87 ft. long, will be needed for the other two, or 520 ft.+ 174 ft. 231 running yards, 1 yd. wide, making 231 sq. yd. Cost at 90 = $208.20. 694 ft. = The surface to be carpeted = (2605) ft. by (935) ft. = 85 yd. × 291 yd. Cost $2.09 × 85 × 291 38 = $4936.80. Total, $4936.80 +$208.20 = $5145. Ans. 15. Since the meeting-place is twice as far from A as from B, the first man goes twice as fast as the other; the latter, there fore, walks 2 mi. per hour. If x is the distance between A and х B, the first will require hr., and the second hr. 5 21 1; x=5. Ans. 5 mi. 16. Let x= number of miles between A and C. Then x 15 distance between B and C. = first train to run from B to C; = time required for second train to run from A to C. As the latter train leaves 3 hr. later and arrives one-half hour later, the running time of the first 1169. 2. The width of the road = (60 ÷ 161) rd.; its area [104 x (60161) 160] acres. = (104 × 60 ÷ 161) sq. rd. = [104 × (60 ÷ 161) = The cost of fencing = $x 104 × 5 = $286. |