6. How many yards of drugget 3} yds. wide will cover a room 50 yds. by 35? yds. yds. yds. 35 Ans. : : 2 7. The papering of a room 13} ft. high, and 30 yds. round, cost £2:0:6 what will be the cost of papering a room 10 ft. high, and 20 yds. round? 27 2 £. 8. d. 600 Ans. : : 10 8. If io of an estate be worth £1500, what will of the same be worth? 9 £. 1500 Ans. 9 10 £=1 X 20 X 10£. 3 9. In what time can a man reap a field working 103 hours a day, if he occupies 3% days in reaping it, when he works 124 hours a day? hrs. days. 5 As 104 12 3 Ans. 2 : : 3 5 H 51 23 3 6 32 days 149 256 10. How many yards may be bought for £12: 12:0 if 74 yards cost 19s. 4fd.? £12: 12:0 = £12.6 74 yds. = 7.75 yds. 198. : 41 = £.96875. As £.96875 £12.6 7.75 yds. Ans. 12.6 : : In questions of Compound Proportion, several quantities of different kinds are involved, which are all connected with another quantity in such a manner, that they each vary directly or inversely as this other, when all the rest are supposed to remain unaltered. Certain values of the first-mentioned quantities, and the corresponding value of the last-mentioned, are given, and also other values of the first, from which it is required to find the corresponding value of the last. This is effected by the Rule of Compound Proportion, so called, because several proportions are compounded to form a Simple Proportion for the solution of the question. Rule-1. Put that quantity for the 3rd term of a Proportion, which is of the same kind with the answer required. 2. Take any pair of quantities of the same kind, and state them as the 1st and 2nd terms of the Proportion, precisely as though the answer depended entirely on them. 3. Take any other pair of quantities of the same kind, and state them as the 1st and 2nd terms of another Proportion, as though the answer depended entirely on them. 4. Proceed in the same manner with every pair of quantities. 5. Reduce the 1st and 2nd terms of each Proportion to the same name. 6. Multiply together all the 1st terms, and all the 2nd terms; make these respectively the 1st and 2nd terms of another Proportion, and put for the 3rd term the same as before. 7. Solve this Proportion as in Simple Proportion, the result will be the answer required. EXAMPLES. 1. If a man travel 126 miles in 5 days, walking 10 hours a day, in how many days would he travel 756 miles, walking 9 hours a day? 9 hours 10 hours 5 days 5 days Ans. 9 60 5 days dns. 5 : : } : or : : : : or : : 20 2. If £879 : 3:4 gain £17:11: 8 in 5 months, what sum will gain £20 in 10 months ? 1 Omos. bmos. £879: 3:4 £20 £8792 £879} Ans. £. -£ £500. : 3. If 20 men, working 12 hours a day, earn £54 : 7:6 in 15 days, how many days of 9 hours each must 27 men work to earn £163 : 2:6? } : Ans. 27 men 20 men 15 days 1305 20 x 12 x 15 days : 8 15 days : days 27 400 4 days = 44~ days. 9 : Ans. Def. 1. Interest is money, paid for the use of other money lent for a fixed time, at a given rate for every £100 for one year, called the rate per cent. Def. 2. The sum lent is called the Principal. Def. 3. When the Interest, being paid at fixed periods, is calculated only on the Principal, it is called Simple Interest. Def. 4. When the Interest, being left unpaid, is added to the Principal at fixed periods, and with it bears interest for the future, it is called Compound Interest. Def. 5. The sum of Principal and Interest is the Amount. Def. 6. Discount is an abatement made from a debt in consideration of its being paid before it is legally due. Def. 7. The present worth of a sum of money, due at a certain time, is the difference between the Principal and Discount, or is the sum, which, put out to interest for the given time, would amount to the given sum. A.- Simple Interest. I. TO FIND THE INTEREST DUE ON A GIVEN PRINCIPAL AT A GIVEN RATE PER CENT. FOR A GIVEN TIME. Rule 1. If the time be an exact number of years, multiply the Principal by the rate per cent. and the product by the number of years; divide the result by 100; the quotient will be the interest required. Rule 2. If the time be a number of years and months, reduce the time to months, and calculate interest as for an equal number of years; divide the result by 12, the quotient is the interest required. Rule 3. If the time be a number of years and days, calculate the interest for 1 year, multiply this by the number of days, and divide by 360 ; subtract from the result its 1-72nd part; the remainder will be the interest for the number of days nearly. If a further correction be required, add 1--72nd part of the former one. If the calculation be conducted by decimals, (which in most instances is desirable and sufficiently accurate,) multiply the interest for one year by 1–5th of the number of days, and divide by 73, or by 72, applying the same correction as before. Having thus found the interest for the days, add that for the years. Rule 4. To calculate the interest upon partial payments or an account current, multiply each sum, which lies at interest, by the number of days; add the products ; multiply the sum by twice the rate per cent. ; divide the product by 73000. Note. The division by 73 may be shortly effected thus :-Rule. Take 1–100th of the dividend, 1–3rd this result, 1-10th of this third, and 1-10th of this tenth. Add these results, and subtract from the sum .001 for every ten. To divide by 73000, take 1–100000th of the dividend, and proceed in the same way as before. EXAMPLES. £. s. d. 4 = Rate per cent. 2523 0 0 = Interest for 1 year at £400 per cent. 6 100)151,38 00 = Interest for 6 years... 20 7.60 s. 12 7,20 d. Interest for 6 years at £4 per cent. = £151 : 7:7}. Ans. 2. Find the interest on £2200 : 10:6 for 4 years 9 months at 5) per cent. £. s. d. 4 yrs. 9 mo. = 57mo. 2500 10 6 = Principal (P) 5= Rate per cent. 11.01d. £653 5 2.9175=Int. for 57 months 3. Find the interest on £742 : 13 : 4 for 175 days at 54 per cent. £. d. 54= Rate per cent. 20 19.80 s. 12 9.60 d. H 2 |