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5d. 24.

LESSON 2.

1817 at 3 qrs. Here as in lesson first we say 1817d. which sum, divided by 12 and by 20, will make £7 11s. 5d. as before ; then the half of this sum, and the half of that half, will be equal to the amount at 3q.: because 2 q. are equal to half a penny, and Iq. is equal to half of 29. OPERATION.

PROOF.
12 1817d.

29. of 1d. 1817 at 3q.
201 151s. 5d,
29. of 1d. £ 7 11 5d.

19. of 29. 908d. 2q.

454 1
19. of 29.1 3 158. 8d. 29.
1 17 10 1

12 1362d. 39.

• 20 113s. 68. 39. Answer, £5 13 6 3

Answer'£ 5 13s. 6d. 3q.

LESSON 3. 1691 at id. 29. In this lesson we say 1691 at 18. makes £ 84 11s. Then 1d. is jy of a shilling and 29. 1 of a penny; that is, the 1 of £ 84 11 and the half of that je will give the answer, viz. £ 10 11 4 2 OPERATION.

PROOF. • 2011691 at 1s.

1d. is of 1s.[1691 at 1d. 29. id. of 1s. £84 lls.

29. į of 1d. 140s. 11d. 29. 1 of id. 7 Os. 11d.

70s.
1
3 10s. 5d. 29.

• 20 211s. 4d. 29.
Answer, £ 10
11s. 4d. 29.

Answer, £ 10 11s. 4d. 29.

LESSON 4. 1881 at 25. 3d. 29. Say 1881 at 20s. = { 1881. Then 2s. is zs of a pound, 3d. ; of 2s. and 2q. of 3d. OPERATION.

PROOF. 2s. io of £1 £1881

3d. 1 of 18./1881 at 2s. 3d. 29.

2s. 3d. 1 of 25. 188 2s. 29. of 3d. 23 iOs. 3d.

3762s. 3 18s. 4d. 24.

29. of 3d. 470s. 3d.

785. 4d. 29. Answer, £215 10s. 7d. 29.

- 20 4310s. 7d. 29.

Answer, $ 215 10s. 78. 29. LESSON 5.

478 at 3s. 2d. 19. OPERATION.'

Proof, 2s. To of £1 £478

2d. 5 of 1s. 478 at 3s. 2d. 19.

3s. 1s. of 2s.

47, 16s. 2d. of is. 23 18s.

1434s. 3 198. 8d.

19. of 2d.j 79s. 8d. 9s. 11d. 24.

9s. 11d. 29

19. 5 of 2d.

Answer, £76 38. 7d. 29.

+ 2015238. 7d. 29. Answer, ' 76 38. 7. 29.

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LESSON 6 and 7. What will 568 pairs of mittens come to at 2s. 6d. a pair, and at 3s. 4d.

Ans. £71 at 2s. 6d. and 94 135. 4d. at 3s. 4d.

EXPLANATION. In the first place we consider 568 pairs at 20s. equal to £ 568. Then as 2s. 6d. is of a pound, we divide by 8, the quotient is the answer at 2s. 6d.

Secondly, we say 35. 4d. is of a pound, and divide by 6, the quotient will be the answer at 38. 4d.

OPERATIONS. 2s 6d. į of £11568 at 2s. 6d. 38. 4d. of £1|1568 at 3s. 4d. Answer, £71

Answer, £ 94 138 4d.

LESSON 8 and 9. 678 at 5s, and at 6s. Bd.

OPERATIONS. 55. of £1)678 at 5s.

6s. 8d. į of £1)678 at 6s. 8d. Answer, 169 10s.

Answer, 226

LESSON 10. 678 at 78. 6d. 24.

DIRECTIONS. Say 678 at 20s.<£ 678. Then take parts of that sum thus: 45.5 of a pound ; 28.=} of 48.; 18.= of 2s.; 6d.=1 of 1s. and 2q=1 of 6d. OPERATION.

PROOF. 48. } of 1£. £678 at 7s. 6d. 29.

6d. of 1s.[678

7 2s. į of 4s. 135 12s. 1s. 1 of 2s. 67 16s.

4746s. 6d. į of 1s. 33 18s.

29. le of 6d. 339s. 16 19s.

18./678

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28s. 3d, 1 8s. 3d.

· 2015113s. 3d. Answer, £255 138. 3d.

Answer, £255 13s. 3d.

LESSON 11. 459 at 15s. 8d. 39.

Here we will suppose that 459 cost £459. Then say, 10s.=} of a pound ; 58.5 of 10s.; 6d.= 1 of 58.; 2d.= of 6d. ; 29.= of 2d. and 19.= of 29. OPERATION.

PROOF. 10s. į of 1£. 1459 at 15s. 8d. 39.

6d. of 1s. 459
1

х 3
58. Ź of 10s. 229 10s.
6d. 3 of 58. |114 158.

1377 2d. of 6d. 11 98. 6d.

x 5 29. of 2d.

3 16s. 6d. 19. of 29. 0 198. id. 2q.

6885s. 9s. 6d. 39.

2d. 4 of 6d. 229s. 6d.

24. of 2d. 769. 6d. Answer, £360 198. 8d. 19.

19. 1 of 29. 198. 1d. 27.

29. 1 of 6d./

9s. 6d. 3q.

20172195. 8d. 19. Answer, € 360 195. 8d. 17.

LESSON 12. 642 at 37s. 6d. 29.

STATEMENT. Say 642 pounds; this will include the amount of the integers at 208. each: then for 175. 6d. 29. say 10s.=į of a pound; 58.=the half of 108.; 25. 6d.=the half of 5s. and 24.=és of 2s.6d. OPERATION.

PROOF. 10s. of 1£. £642

6d. I of 1s.1642

15.1642 58. of 10s. 321

x 378. 2s. 6d. I of 5s. 160 10s. 29. do of 2s. 60 80 58.

4494 1 6s. 9d.

1926

Answer, 1205 1s. 9d.

24. i's of 6d.) 3215

23754s.
321s.
26s. 9d.

20124101s. 9d.

Answer, £ 1205 ls. 9d.

LESSON 13. What cost 15 tons 9 cwt. 3 qrs. 25 lb. of iron, at £2 168. a hundred ?

Answer, £867 188. 6d.

DIRECTIONS. 1. Multiply the price of 1 cwt. by 20, the product will be the price of 1 ton.

2. Multiply the price of 1 ton by 15, the product will be the price of 15 tona.

3. Multiply the price of 1 cwt. by 9, the product will be the price of 9 cwt.

4. For the 3 qrs. 25 lb. say, 2 qrs.=1 of 1 cwt.; 1 qr.=) of 2 qrs.; 14 lb. = of 1 qr.; 7 lb.=1 of 14 lb.; and 4 lb.= part of 1 cwt.

OPERATION. 2 qrs. of 1 cwt.[£2 16s. price of 1 cwt. 1 4

£ 840 Os. price of 15 ton. € 2 16s. x 9 = 25 4 price of 9 cwt. 11 4

1 qr. of 2 qrs. 1 8. price of 2 qrs. 5

141b. of 1 qr. 14 price of 1 qr.

7lb. of 141b. 7 price of 141b. Price of 1 ton, 56 0

3 6d. price of 71b. 3 4lb.= as of 1 cwt. 2 Od. price of 41b.

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$ 8400 Proof in dollars and cents, at the rate of 8s. to the dollar. 2 yrs. of 1 cwt./87. cwt.18% price of 1 cwt.

cwt. 1 ton.

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$ 2169.8125 Answer, $ 2169.814 cents.

INTEREST.
How to find the time a Note or Bond has been on Interest.

LESSON 1. Suppose a note was given on interest April 2, 1813, and paid March 4, 1816; how many years, months, and days does it draw interest ?

Answer, 2 years 11 mo. and 2 days.

OPERATION.
March 4, 1816,

1816 years 2 mo. 4 days.
April 2, 1813,

1813 3 2

2

Answer,

11 2 Allow 30 days to be 1 month and 12 months 1 year.

OBSERVATIONS. 1. The Note was paid on the 4th of March 1816, which time we call 1816, years on New-year's day last before the payment, and the 4th of March we say is 2 months and 4 days after New-year's day.

2. So again for the time of signing the note, we say that the 2nd of April was 3 months and 2 days after New-year's day 1813. Subtract and carry at 30 and at 12.

APPLICATION. What is the interest of a Note of $ 450.75 for 2 years 11 mo. and 2 days at 7 per cent. per annum?

Answer, $ 92.20+

DIRECTIONS. First, find the interest for three years ; then take one twelfth of one year's interest from that sum, the difference will be 2 years 11 mo. interest. Secondly, take the 15th part of one month's interest for 2 days.

OPERATION. 45075 cents.

7 per cent.

3155.25 one year's interest.
3 years

the time.

9465.175 three years' interest.

262.93 one month's interest subtracted. 9202.182 two years' and 11 months'interest.

17. 528 two days' interest or 15th of 1 mo.

9220.1348 Answer, $ 92.20 cents, and .348 + thou

sandths of a cent.

LESSON 2.
If a note were given on interest, 5th May, 1812, and paid March 2,
1817, what time should be allowed for interest ?
Ans. 4 years, 9 mo. 27 days.

OPERATION.
March 2, 1817,

1817 years, 2 mo. 2 days.
May 5, 1812,

1812

4 5

27

4

9

APPLICATION. What is the amount of $16.50, for 4 years, 9 months and 27 days, at 7 per cent. per annum ?

Aps. 822.07+ OBSERVATIONS. 1. In this case obtain the interest for 4 years; then lake half the interest for 1 year, and the half of that half for 9 months.

2. For 27 days, divide the interest of 1 year by 12, the quotient will be the interest of 1 month, viz. 9.6250 cents; subtract to of a month's interest, that is 3 days, from that sum, the remainder will be 27 days' interest, viz. 8.6625 cents.

OPERATION.
1650 cents, principal. 12)115.50 cts. 1 year.
7 rate per cent.

109.6250 1 month. 6 mo. 4 a yr.)115]50 interest for 1 year.

.9625 3 days.

8.6625 27 days. 462100

interest for 4 years. 3 mo. of 6) 57 75 interest for 4 a year.

28.875 interest for of a year.
816625 interest for 27 days.

557|2875 interest for 4 years, 9 mo. 27 days.
1650 principal.

2207|2875 amount, or answer. COMPOUND INTEREST, CALCULATED BY LOGARITHMS.

RULE. Add the rate per cent. to 100; find the Logarithm of that sum, and subtract from it the Logarithm of 100; call the difference, Logratio ; multiply the Logratio by the number of years, and to that product add the Logarithm of the principal; the natural number of the last Logarithm will be the amount in Compound Interest.

EXAMPLE 1. What is the amount of $500 for three years, at 7 per cent. per annum, compound interest ?

Ans. $612.50
OPERATION.
Rate per cent. 7 + 100 = 107. Logarithm of 107 is 2.02938
Logarithm of 100 is

2.00000

Logratio or difference,
Number of years,

0.02938

3

x

Product,

0.08814

[Continued.

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