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OPERATION.

586. Questions in which the required numbers, or their parts, or their multiples, are increased or diminished by some given number which is no known part or multiple of the required number, or when a power or root of the required number is either directly or indirectly contained in the result given in the question, may be solved by two assumptions.

NOTE. – When the answer is obtained by means of a single assumption, the process is called Position; and when obtained by means of two assumptions, it is called DOUBLE Position.

Analysis by Position affords often a very compendious method of working questions, whose solution otherwise, except by algebra, would be lengthy and difficult. It is even found very useful in shortening some of the processes of algebra.

587. When the answer may be obtained by means of a single assumption or supposition.

Ex. 1. A schoolmaster, being asked how many scholars he had, replied, that if he had as many more as he now has, and half as many more, he should have 200. Of how many scholars did his school consist?

Ans. 80 scholars.

By having as many more, and Assumed number, 60 half as many more, he would have As many more,

60

had 24 times the original number; Half as many more, 3 0

therefore, the required number

must be as many as 200 contains 150 times 21, or 80. The same result

may be obtained, as in the opera150 : 200 :: 60:80, Ans. tion, thus : We assume the number

of scholars to be 60; if to 60 as many more, and half as many more, are added, the sum is 150. As this result has the same ratio to the result in the question as the supposed number has to the number required, we find the answer by a proportion.

RULE. - Assume any convenient number, and proceed with it according to the nature of the question. Then, if the result be either too much or too little, as the result found is to the result given, so will be the number assumed to the number required. 2. A person, after spending and of his money, had $ 60 what had he at first?

Ans. $ 144, 3. What number is that, which, being increased by 1, }, and 1 of itself, equals 125 ?

4. A's age is double that of B, and B’s is triple that of C, and the sum of all their ages is 140. What is each person's

Ans. A's 84, B's 42, C's 14 years.

left;

age ?

5. A person lent a sum of money at 6 per cent., and at the end of 10 years received the amount, $ 560. What was the sum lent?

Ans. $ 350. 588. When two or more assumptions or suppositions are required in finding the answer.

Ex. 1. A lady purchased a piece of silk at 80 cents per yard, and lining for it at 30 cents per yard ; the silk and lining contained 15 yards, and the price of the whole was $ 7. How many yards were there of each ?

Ans. 5 yards of silk ; 10 yards of lining.

OPERATION

Assume 6 yards of silk, $ 4.80 Assume 4 yards of silk, $ 3.20
Lining would be 9yd., 2.70 Lining would be 1lyd., 3.30
Their sum,
$ 7.50 Their sum,

$ 6.50 Sum in the question, 7.00 Sum in the question, 7.00 First error, + $ 0.50 Second error,

$ 0.50 .50 + .50 : 6 — 4 :: .50 :1; 6 yards — 1 yard = 5 yards;

15 yards -- 5 yards = 10 yards. Since the silk and lining contain 15 yards, cost $ 7.00, the average price per yard is 46f cents; and 80 cents - 463 cents

33} cents; 463 cents

- 30 cents = 16cents; and the quantity of lining will therefore be to that of the silk as 33} is to 163, or as to 2 is to 1. Hence, if the given number of yards, 15, he divided into 3 parts, two of those parts, or 10 yards, will be for the lining, and the other part, or 5 yards, will be for the silk.

The same result is obtained as in the operation, thus: We assume the quantity of silk to be 6 yards; then the lining would be 9 yards. We find the cost of each of these at the prices given, and, adding, have as the sum of their costs $ 7.50. This is a result too large by $0.50, when compared with the sum in question. By assuming the quantity of the silk to be 4 yards, and proceeding in a like manner, we obtain for the cost of the silk and the lining $ 6.50, a sum too small by $0.50. The first error, arising from a result too large, is marked with the sign plus (+), and the second error, arising from a result too small, is marked with the sign minus (-). Then, since ne of the esul is too large, and the other too small, we then say, as the sum of the errors, or .50 .t .50, is to the difference of the assumptions, or 6 · 4, so is .50 the less error, to 1, the correction; thich, being added to 4 (yards), the sum, 5, expresses the number of yards of silk; and consequently that of the lining must be 10 yards.

RULE. — Assume two different numbers, perform on them separately the operations indicated in the question, and note the ERRORS of the re

sults. Then, as the difference of the errors, if the results be both too great or both too small, or as the sum of the errors, if one result be too great and the other too small, is to the difference of the assumed numbers, so is either error to the correction to be applied to the number which produced that error.

Note 1. - The rule usually given fails in an important class of questions; but the rule here given, if not the simplest in the resolution of some questions, has the advantage of being applicable in every case.

Note 2. – In relation to all questions which in algebra would be resolved by equations of the first degree, the differences between the true and the assumed numbers are proportional to the differences between the result given in the question and the results arising from the assumed numbers. But the principle does not hold exactly in relation to other questions; hence, when applied to them, the above rule, or any other of the kind that can be given, will only produce approximations to the true results. In which case the assumed numbers should be taken as nearly true as possible. Then, to approximate more nearly to the required number, assume for a second operation the number found by the first, and that one of the first two assumptions which was nearer the true answer, or any other number that may appear to be still nearer to it. In this way, by repeating the operation as often as may be necessary, the true results may be approximated to any assigned degree of accuracy. This process is sometimes applied with advantage in extracting the higher roots, when approximate results, differing but slightly from entire correctness, will answer.

2. A and B invested equal sums in trade; A gained a sum equal to 1 of his stock, and B lost $ 225 ; then A's money was double that of B’s. What did each invest? Ans. $ 600.

3. A person, being asked the age of each of his sons, replied, that his eldest son was 4 years older than the second, his second 4 years older than the third, his third 4 years older than the fourth, or youngest, and his youngest half the age of the oldest. What was the

age

of each of his sons ?

Ans. 12, 16, 20, and 24 years. 4. A gentleman has two horses, and a saddle worth $ 50. Now if the saddle be put on the first horse, it will make his value double that of the second horse; but if it be put on the second, it will make his value triple that of the first. What was the value of each horse? Ans. The first, $ 30; second, $ 40.

5. A gentleman was asked the time of day, and replied, that

of the time past from noon was equal to 383 of the time to midnight. What was the time?

Ans. 12 minutes past 3. 6. A and B have the same income. A saves ik of his, but B, by spending $ 100 per annum more than A, at the end of 10 years finds himself $ 600 in debt. What was their income?

Ans. $ 480.

7. A gentleman hired a laborer for 90 days on these conditions: that for every day he wrought he should receive 60 cents, and for every day he was absent he should forfeit 80 cents. At the expiration of the term he received $ 33. How many days did he work, and how many days was he idle?

Ans. He labored 75 days, and was idle 15 days. 8. There is a fish whose head weighs 15 pounds, his tail weighs as much as his head and t of his body, and his body weighs as much as his head and tail. What is the weight of the fish?

Ans. 721b. 9. If 12 oxen eat 3 acres of grass in 4 weeks, and 21 oxen eat 10 acres in 9 weeks, how many oxen would it require to eat 24 acres in 18 weeks, the grass growing uniformly?

Ans. 36 oxen. 10. What number exceeds three times its square root by 11? (Note 2.)

Ans. 26.4201648.

SCALES OF NOTATION.

589. The scale of any system of notation is the law of relation existing between its units of different orders.

599.° The radix of any scale is the number of units it takes of one order to make a unit of the next higher. Thus, 10 is the radix of the decimal or denary system, 2 of the binary, 3 of the ternary, 4 of the quaternary, 5 of the quinary, 6 of the senary, 7 of the septenary, 8 of the octary, 9 of the nonary, 11 of the undenary or undecimal, 12 of the duodenary or duodecimal, 20 of the vigesimal, 30 of the trigesimal, 60 of the sexagesimal, and 100 of the centesimal.

591.° In writing any number in a uniform scale, as many distinct characters or symbols are required as there are units in the radix of the given system. Thus, in the decimal or denary scale 10 characters are required, in the binary scale 2 characters, in the duodenary or duodecimal 12 characters, and so on. In the binary scale use is made of the characters 1 and 0, in

OPERATION.

6)58 1
6)9 4

the ternary, 1, 2, and 0, &c., the cipher being one of the characters in each scale. In the duodenary scale, eleven characters being required beside the cipher, the first nine may be supplied by the nine digits, the tenth by t, the eleventh by e, and the twelfth by 0.

592.° To change any number expressed in the decimal scale to any other required scale of notation.

Ex. 1. Express the common number 75432, in the senary and duodenary scales.

Ans. 1341120 and 377e0.

By dividing the given 6)75432 12)75432

number by 6, it is distribut

ed into 12572 classes, each 6)125720 12)6286 0

containing 6, with O remain6)2095 2

12)523 10, or t der By the second division 6)349 1 12)437

by 6, these classes are dis

tributed into 2095 classes, 3 7 each containing 6 times 6,

or the second power of 6,

with a remainder of 2 of the 1 3

former class, each containing

6. By the third division the classes last found are distributed into 349 classes, each containing 6 of the latter, which were each the second power of 6, and therefore these are the third power of 6, with a remainder 1 time the second power of 6 In like manner, the next quotient expresses 58 times the fourth power of 6, with a remainder i time the third power of 6; the next quotient expresses 9 times the fifth power of 6, with a remainder 4 times the fourth power of 6; and the last quotient expresses 1 time the sixth power of 6, with a remainder 3 times the fifth power of 6. Hence, the given number is found to be equal to 1 x 66 + 3 X 67+ 4 X 64 +1 X 63 +1 X 62 + 2 X 6 + 0, or according to the senary system of notation 1341120.

By proceeding in like manner, we find the given number to be equal to 3 x 124 + 7 x 123 + 7 x 122 + 10° X 12 + 0, or, according to the duodenary scale, 377e0.

RULE. Divide the given number by the radix of the required scale repeatedly, till the quotient is less than the radix ; then the last quotient, with the several remainders in the retrograde order annexed, placing cijhers where there is no remainder, will be the the given number expressed in the required scale. 2. Change 37 from the decimal to the binary scale.

Ans. 100101. 3. Reduce 1000000 in the decimal scale to the ternary and also to the nonary.

Ans. 1212210202001, and 1783661.

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