Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

EXAMPLES.

1. A person after spending and of his money, has yet remaining 601; what had he at first ? Suppose he had at first 1201.

Proof. Now of 120 is 40 5

of 144 is 48 of it is 30

of 144 is 36

their sum is 70 which taken from 120

their sum

84 taken from 144

leaves 50 Then, 50 : 120 :: 60 : 144, the Answer.

leaves

60 as per question.

2. What number is that, which being multipied by 7, and the product divided by 6, the quotient may be 21? Ans. 18.

3. What number is that, which being increased by 1, }, and of itself, the sum shall be 75?

Ans. 36. 4. A general, after sending out a foraging and of his men,

had yet remaining 1000: what number had he in command ?

Ans. 6000. 5. A gentleman distributed 52 pençe among a number of poor people, consisting of men, women, and children ; to each man he

gave 6d, to each woman 4d, and to each child 2d: moreover there were twice as many women as men, and thrice as many children as women. How many were there of each?

Ans. 2 men, 4 women, and 12 children. 6. One being asked his age, said, if } of the years I have lived, be multiplied by 7, and of them be added to the product, the sum will be 219. What was his age ?

Ans. 45 years.

DOUBLE

DOUBLE POSITION.

DOUBLE POSITION is the method of resolving certain questions by means of two suppositions of false numbers.

To the Double Rule of Position belong such questions as have their results not proportional to their positions : such are those, in which the numbers sought, or their parts, or their multiples, are increased or diminished by some given absolute number, which is no known part of the number sought.

RULE 1*. Take or assume any two convenient numbers, and proceed with each of them separately, according to the conditions of the question, as in Single Position; and find how much each result is different from the result mentioned in the question, calling these differences the errors, noting also whether the results are too great or too little.

BS.

* Demonstr. The Rule is founded on this supposition, namely, that the first error is to the second, as the difference between the true and first supposed number, is to the difference between the true and second supposed number; when that is not the case, the exact answer to the question cannot be found by this Rule.-That the Rule is true, according to that supposition, may be thus proved.

Let a and b be the two suppositions, and A and B their results, produced by similar operation ; also r and s their errors, or the differences between the results A and B from the true result N; and let x denote the number sought, answering to the true result n of the question. Then is N - A = r, and N

And, according to the supposition on which the Rule is founded, r:s :: x-a : x-b; hence, by multiplying extremes and means, rx - rb = sx - sa ; then, by transposition, rx

sr = rb

sa; and, by division, rb

= the number sought, which is the rule when the results are both too little.

If the results be both too great, so that A and B are both greater than N then N

- ,

- s, or r and s are both negative; hence -1:-S:: X a : x - b, but - -:-s :: + 7 : + 8, therefore r :s :: x-Q; x b; and the rest will be exactiy as in the former case.

But if one result A only be too little, and the other e too great, or one error r positive, and the other s negative, then the theoren

rbt sa becomes 2 =

which is the Rule in this case, or when

rts the errors are unlike,

Then

[ocr errors]
[ocr errors]

A

and n

[ocr errors]

Then multiply each of the said errors by the contrary supposition, namely, the first position by the second error, and the second position by the first error. Then,

If the errors are alike, divide the difference of the products by the difference of the errors, and the quotient will be the

answer.

But if the errors are unlike, divide the sum of the products by the sum of the errors, for the answer.

Note, The errors are said to be alike, when they are either both too great or both too little; and unlike, when one is too great and the other too little.

EXAMPLES.

1. What number is that, which being multiplied by 6, the product increased by 18, and the sum divided by 9, the quotient shall be 20? Suppose the two numbers 18 and 30. Then, First Position.

Second Position. Proof. 18 Suppose

30

27 6 mult.

6

6

[blocks in formation]

FIND, by trial, two numbers, as near the true number as convenient, and work with them as in the question ; marking the errors which arise from each of them.

Multiply the difference of the two numbers assumed, or found by trial, by one of the errors, and divide the product by the difference of the errors, when they are alike, but by their sum when they are unlike.

Add the quotient, last found, to the number belonging to the said error, when that number is too little, but subtract

it when too great, and the result will give the true quantity sought *

EXAMPLES

1. So, the foregoing example, worked by this 2d rule, will be as follows : 30 positions 18;

their dif. 12
- 2 errors + 6;

least error 2
sum of errors 8 ) 24 ( 3 subtr.

from the position 30
leaves the answer 27

Ex. 2. A son asking his father how old he was, received this answer : Your

age

is

now one-third of mine; but 5 years ago, your age was only one-fourth of mine. What then are their two ages ?

Ans. 15 and 4-5. 3. A workman was hired for 20 days, at 3s per day, for every day he worked; but with this condition, that for every day he played, he should forfeit 1s. Now it so happened, that upon the whole he had 21 4s to receive. How many of the days did he work?

Ans. 16. 4. A and B began to play together with equal sums of money : À first won 20 guineas, but afterwards lost back of what he then had; after which, B had 4 times as much as Ai What sum did each begin with ? Ans. 100 guineas.

5. Two persons, A and B, have both the same income, A saves of his; but B, by spending 50l per annum more than A, at the end of 4 years finds himself 1001 in debt. What does each receive and spend per annum? Ans. They receive 125l per annum; also a spends 1001,

and B spends 150l per annum,

* For since, by the supposition, r: 8:: x - a : x-h, there, fore by division, 2-s:8;: baixb, which is the 2d Rule,

PRACTICAL

PRACTICAL QUESTIONS IN ARITHMETIC.

[ocr errors]

QUEST. 1. The swiftest velocity of a cannon-ball, is about 2000 feet in a second of time. Then in what time, at that rate, would such a ball be in moving from the eartlz to the sun, admitting the distance to be 100 millions of miles, and the year to contain 365 days 6 hours ?

Ans. 8490 years. QUEST. 2. What is the ratio of the velocity of light to that of a cannon-ball, which issues from the gun with a velocity of 1500 feet per second; light passing from the sun to the earth in 7 minutes ? Ans. the ratio of 782222 to 1.

Quest. 3. The slow or parade-step being 70 paces per minute, at 28 inches each pace, it is required to determine at what rate per hour that movement is? Ans. 1.1 miles.

QUEST. 4. The quick-time or step, in marching, being 2 paces per second, or 120 per minute, at 28 inches each; then at what rate per hour does a troop march on a route, and how long will they be in arriving at a garrison 20 miles distant, allowing a halt of one hour by the way to refresh?

Ans. { and the time 72 hr, or 7h 17 min. QUEST. 5. A wall was to be built 700 yards long in 29 days. Now, after 12 men had been employed on it for 11 days, it was found that they had completed only 220 yards of the wall. It is required then to determine how many men must be added to the former, that the whole number of them may just finish the wall in the time proposed, at the same rate of working

Ans. 4 men to be added. QUEST. 6. To determine how far 500 millions of guineas will reach, when laid down in a straight line touching one another; supposing each guinea to be an inch in diameter, as it is very nearly. Ans. 7891 miles, 728 yds, 2 ft, 8 in.

QUEST. 7. Two persons, A and B, being on opposite sides of a wood, which is 536 yards about, they begin to go round it, both the same way, at the same instant of time; A goes at the rate of 11 yards per minute, and B 34 yards in 3 minutes; the question is, how many times will the wood be gone round before the quicker overtake the slower ?

Ans. 17 times.

QUEST.

« ΠροηγούμενηΣυνέχεια »