SECTION V. 1. What would be the half-yearly dividend from an investment of £3300 in the 3 per cents., made when the stock was standing at 91? 2. What sum of money must be invested in the 5 per cent. stock at 83, to enable the possessor to realize an income of £64 3s. 4d. per calendar month ? 1. For every £91 invested, the yearly dividend would be 35s.; the half-yearly dividend, £1 12s. 6d. Therefore, as £91 : £3302 :: {" LI 12s. 6d. 5 (a) of 3300 8) 16500 20621 or £1 d.} : the Answer. (b) £3300+ § of £3300 = £3300 + 2062 -£5362. (c) £53621÷91 = £58 18s. 6d. Answer. 2. (a) £83 realises an income of £5 10s. per annum. (b) £64 3s. 4d. (month's income) x 12 = £770 per Therefore, the problem stands thus : annum. As £5 10s. £770, or as 11 Half-sovereigns : 1540 × 83 II = 140 × 83=11620... Answer =£11620. SECTION VI. 1. What would be the expense of painting (at 3s. 2d. a yard) the walls of a room 27 feet long, 17 feet broad, and 11 feet high, the dimensions of 4 windows being 7 feet by 4 feet each? 2. What length of carpet, of a yard wide, would be required to cover the above room; and what would be the cost at 5s. 3d. per yard? I. Two walls, each 27 × 11=27 × 11 × 2 ft. = 27 × 23 ft. =621 sq. ft. Two walls, each 172 × 11 ft. = 171 × 11 × 2 ft. =1723 ft. = 402 sq. ft. = 7 × 4 × 4 ft. = 120 sq. ft. 621 +402120 sq. ft. = 903 100 = Four windows, each 7 x 4 ft. Therefore area to be painted sq. ft. = 100 sq. yds. sq. yds. at 3s. 2d. per yd. £ s. d. Top line × 3 as shillings, 301s. = price at 3s. = 15 I 2 2d.= 0 16 83 £15 17 10 Ans. 2. Floor of room = 27 × 17 ft. = 472 ft., or 52. Therefore No. of square yds. of carpet required is 52. But the carpet is only yd. wide, therefore more will be required. 52 3) 210 70 yards. Answer I. And the cost at 5s. 3d. per yd., will equal 5 × 70s. 70s. 350 17 36015. = £180 6 Answer II. SECTION VII. 1. Divide £10 Ios. between A and B, so that of A's portion shall equal of B's. 2. If 3 of a yard of French merino cost £1 14 41, what would of of a yard cost? = 1.—(a.) of 1's of B's. That is, of A's of B's. That is, the larger is to the smaller share as 4 to 3; and A's is the larger. Therefore the whole sum must be expressed by a number which will allow 4 parts to be allotted for the larger share and 3 for the smaller. This number is 7. (b.) Then as 7: 4 :: 210s.: the larger share. and as 73: 210s. the smaller. (c) 4x 210 And 7 3 x 210 S. S. = × 30s. = = 120S. = £6, the larger (A's) share. 3 x 30s. = 90s. = £4 10s. the smaller (B's) share. 2. (a.)—3—= 1 = 33; and † of }=}. Then if 33 (ninths) cost £1 14 4, what will 1 (ninth) cost? I (b.) Working. £ s. d. 33) I 14 4 I o Answer. SECTION VIIL 1. Prove the rule for the multiplication and division of decimals. 2. Find the value of 7.92 ÷ 3.84 and 1003'53÷ 1250 and of .8 x 016 x 0032. 1. (a.)-Rule. Multiply the numbers as integers. Mark off in the product as many decimal places as are found in all 'he factors put together. This Here, treating the numbers as integers, we increase the multiplicand 10 times and the multiplier 100 times. makes the product 10 x 100 (or 1,000) times too great, an error which we correct by marking off three figures with a decimal point, so as to reduce the local value of every figure 1,000 times. (b.) Rule.-Add ciphers to the number that has fewer decimal places, until the number of decimal places in both is equal; then omit the decimal points and treat the numbers as integers. Illustration in Proof. 284 256÷6.3. 6300) 284256 (45'12 being thus reduced to a common denominator, it is only necessary to divide the numerator of the one by the numerator of the other, as in division of vulgar fractions where a common denominator lecimal ve sum in the Price 3s. 6d. post free, THE INDUCTIVE ALGEBRA. By A. L. SPARKES, B.A., AUTHOR OF ‘ALGEBRAIC FACTORS SIMPLIFIED,' ETC. ETC. 802824 This work will prove specially useful to Pupil Teachers, and Pupils preparing for the College of Preceptors and the University Local Examinations. The greatest care has been taken to make every point thoroughly clear and simple, and to accustom Pupils to work neatly from the beginning of their course. Every chapter is supplemented by a large number of well-selected and graduated exercises for practice; and the work carefully studied, even without a Teacher, will enable the Student to pass any ordinary examination in Elementary Algebra. W. STEWART & CO., THE HOLBORN VIADUCT STEPS, LONDON, E.C. 384 *00004096 100 |