Solve | 6×4×9 7×5×2 |
Set B to 6. Set 7 to T. Move T to 4. | ||||
(B now indicates answer to | 6×4 7 | ) | ||
Set 5 to T. Move T to 9. | ||||
(B now indicates answer to | 6×4×9 7×5 | ). | ||
Set 2 to T. Move T to 1. Read answer at B: 3.086. |
I. | Solve 12 : 7 :: 16 : x ? | ||
Set B to 12. Set 7 to T. Move B to 16. Read answer at T.
12 : 7 :: 16 : 9.333. | |||
II. | Solve 18 : 4 :: x : 53 ? | ||
Set B to 18. Set 4 to T. Move T to 53. Read answer at B. 18 : 4 :: 238.5 : 53. | |||
III. | Divide 8975 in the proportions 83 : 79 : 33 : 19. | ||
Set B to 8975. Set sum of required proportion, viz. 214, to T. Move T in succession to 83, 79, 33, 19, and read the corresponding proportions at B, viz. 3481, 3313, 1384 and 797. (On Model L this calculation necessitates "closing in" the cylinder. See page 9.) |
II. | Decimalise 3/32, 7/32, 15/32, 29/23. |
Set B to 32. Set 1 to T (32 and 1 being the constants in this series). Move B in succession to 3, 7, 15, 29 and read the corresponding answers at T, viz. .09375, .21875, .46875, .9062. |
I. | If 54 articles cost
£39.225, what is the price of 15? Set B to 39.225. Set 54 to T. (The cost of any number of articles at this price can now be obtained by moving T to the number required.) Move T to 15. Read answer at B: 15 articles cost £10.895. | ||||
II. | Find interest on £675 at 6 1/2% p.a. for 29 days. | ||||
(£675 × | 6.5 100 |
× | 29 365 |
) | |
Set B to 675. Set 1 to T. Move T to 6.5. Set 365 to T. Move T to 29. Read answer at B: £3.486. |
To close cylinder in. | Move T to bottom 1. Set top 1 to T. | |
To open cylinder out. | Move T to top 1. Set bottom 1 to T. |
1.0083.1 ×
3√63
× 4000 6 × 5√260000 × 421.82 | = .2495. |
1.025 × 3.98 × 4000
6 × 12.11 × 900.1 |
(a) | Set B to 1. Set 000 to T. Move B to 1055 (100+5 1/2%). Read log. .0232 at T. | |
(b) | Set B to 232. Set 1 (beginning of upper cylinder scale) to T. Move T to 14. Read 325 at B. | |
(c) | Set B to 1. Set 000 to T. Move T to .325. Set 1 (beginning of upper cylinder scale) to T. Move T to 25. Read 528.5 at B. (Answer: £528 10s.) |
0.0027×0.00031 =
0.000000837; approx. 0.003×0.0003 = 0.0000009. | ||
0.48×0.056 =
0.02688; approx. 0.5×0.05 = 0.025. | ||
11×305×29×49 = 4761000; approx. 10×300×30×50 = 4500000. | ||
577÷799 = 0.7225; approx. 6÷8 = 0.75. | ||
18×19×8 1.7×0.002×950 | = 847.8; | |
approx. | 20×20×10 2×0.002×1000 | = 1000. |
5430000, | 674, | 81.2, | 7.82, | 0.45, | 0.0421, | 0.00675 | ||
have | +7, | +3, | +2, | +1, | +0, | −1, | −2 | places |
To find the number of places (p) in the product (P=X×Y). Let X have m places, and Y have n places. | |||
RULE I. | p=m+n or p=m+n−1. | ||
(a) | When the result is
below the original setting p=m+n. | ||
Example: 3×4 (m=1; n=1). | |||
Set B to 3. Set 1 to T. Move T to 4. (This is below setting.) p=m+n=2. Answer = 12. | |||
(b) | When the result is above the setting, the product has m+n−1 places. | ||
Example: 3×3 (m=1; n=1) | |||
Set B to 3. Set 1 to T. Move T to 3. (This is above setting.) p=m+n−1 = 1. Answer = 9. | |||
(See note 4 below.) |
To find the number of places (q) in the quotient (Q = X/Y). | |||
RULE. II. | q=m−n or q=m−n+1 as follows: | ||
(c) | When the result is above the setting, the quotient has m−n places. | ||
Example: 3÷4 (m=1; n=1). | |||
Set B to 3. Set 4 to T. Move T to 1. (This is above setting.) q=m−n=0. Answer = 0.75. | |||
(d) | When the result is below the setting, the quotient has m−n+1 places. | ||
Example: 5÷4 (m=1; n=1). | |||
Set B to 5. Set 4 to T. Move T to 1. Read answer at B. (This is below setting.) Therefore q = 1−1+1 = +1. Answer = 1.25. | |||
(See note 5 below.) |
RULE III. | Two methods may be used in working out complex problems involving both multiplication and division. They are: | |||
(1) | Taking numerator and denominator alternately. | |||
(2) | Taking all the numerators first and then dividing consecutively by the denominators. | |||
Of these two methods, only the latter can be used if the position of the decimal point is required. If the other is used, the decimal point must be found by inspection. | ||||
First multiply consecutively the series of factors in the numerator and then divide consecutively by the factors of the denominator. | ||||
Take the algebraic sum of the places in the factors of the denominator from the algebraic sum of the places in the factors of the numerator, and to this result add the algebraic sum of the results obtained from the application of Rules I and II to the several steps of the problem. | ||||
Example: | ||||
432×32.4×0.0217×0.
98 0.00000621×412000×0.175×4.71 | = 141.14... | |||
Number of places in factors of— | ||||
Numerator = 3+2+(−1)+0 | = +4 | |||
Denominator = −5+6+0+1 | = +2
| |||
Difference | = +2 | |||
Results of various steps in calculation = -1 + 1 + 1 | = +1 | |||
Number of places in answer | = +3 | |||
Answer = 141.14. |
|
Andries de Man 10/29/1997