of 120 lb. worth 16 cents per lb. ? Ans. 36 lb. at 8 cents; 12 lb. at 12 cents; 24 lb. at 18 cents; 48 lb. at 22 cents.

4. A gold beater has gold 15, 17, 18, and 22 carats fine, of which he wishes to make a compound of 40 oz. 20 carats fine. How much of each kind must he take? Ans. 25 oz.

22 carats fine; and 5 oz. of 15, 17, and 18 carats fine.

5. How much water of no value, and how much wine at 90 cents per gallon, must be taken to make 100 gallons, worth 60 cents per gallon? Ans. 33 gallons of water, and 63

of wine.

gallons

QUESTIONS.-What is Alligation? What is Case 1st? What is the rule? What is Case 2d? What is the rule? What is the note? What is Case 3d? What is the rule? What is Case 4th? What is the rule?

POSITION.

Position is a rule by which answers are obtained 'to such questions as cannot be solved by the common direct rules, by assuming any convenient number or numbers, and then working according to the nature of the question.

SINGLE POSITION.

When the question can be solved by the assumption of a single number, the operation is called Single Position.

The following sum will serve for an illustration.

Ex. 1. A teacher being asked how many scholars he had, replied, "If I had once, one half, one third, and one fourth as many more as I now have, I should have 185." How many had he? We will suppose the number to be 24. As he first doubles his number, 24 must be doubled. To this amount, one half his original number, 12, must also be added. He then increases his number by one third of his original

number, viz. 8, and also by one fourth, viz. 6. Whole amount, 74.

Now it is evident that we have not supposed the right number, otherwise the amount would have been 185, as given in the sum. We have, however, increased the number we supposed, viz. 24, by the same or similar additions, as the teacher did the true number of his scholars; consequently, 74, the number we obtained, must have the same ratio to 24, the number assumed, as 185 has to the real number of scholars in the school. Therefore, 74: 24:: 185: the number required; viz. 60. Proof, 60+60+30+20+15=185.

We have then the following rule:

RULE.-Take any convenient number and proceed with it according to the conditions of the question, and observe the result; then say, as the number thus obtained is to the given number, so is the assumed number to the true one. Or the numbers may be canceled by arranging the terms as directed in Simple Proportion.

Ex. 2. A man being asked how much money he had, replied, that,,, of his money added, made $57. How much money had he? Ans. $60.

3. What number is that, which, being multiplied by 9 and divided by 4, the quotient will be 27?

Ans. 12.

4. A man borrowed a sum of money on interest, which in 10 years amounted to $1800 at 6 per cent.; what was the sum? Ans. $1125.

5. Two boys were playing at marbles. Says one to the other, 3, 4, and of my marbles added together make 45, and if you can now tell how many I have, you may have them. How many had he? Ans. 60.

6. A boy wishing to try the skill of his companions in figures, said he had a pile of apples, of which, if he gave to A., to B., and to C., there would remain 28 for D.; and requested them to tell him how many there were in all. What was the number? Ans. 112.

7. A person being asked his age, said that if of the years he had lived were multiplied by 7, and 3 of them added to the product, the sum would be 292. How old was he? Ans. 60 years.

8. A. saves of his income, but B. who has the same income, spends twice as fast as A., and thereby contracts a debt What is their income? Ans. $360. B., and C's ages, is 132 years.

of $120 annually.

9. The sum of A., is 1 the age of A; and C's What are their respective ages? and C's, 72 years.

B's age

age is twice as great as B's. Ans. A's age is 24; B's, 36;

QUESTIONS.-What is Position? What is Single Position? What is the rule? How may the operations be canceled ?

DOUBLE POSITION.

By Double Position, we solve such sums as require two suppositions.

In this rule, the numbers supposed to be the true ones bear no certain or definite proportion to the required answers.

RULE.-Assume any two convenient numbers and proceed with each according to the conditions of the question, and compare the result of each with the sum or result given in the question, and find their differences. Call each difference an error.

Multiply the first assumed number by the last error; and the last assumed number by the first error.

If both errors are too great or too small, divide the difference of these products by the difference of the errors, and the quotient will be the number sought. But if one of the errors be too large, and the other too small, divide the sum of the products by the sum of the errors.

Note. The errors are said to be too large or too small, when by operating on each supposed number according to the nature of the question, the number obtained is greater or less than the corresponding number in the sum.

Ex, 1. Three men found a purse of money containing $80,

which they agree to divide in such a manner, that A. shall have $5 more than B, and that B. should have $10 more than C. What was each man's share of the money?

$85, a sum of money greater

than that found; therefore, $85-$80-$5, 2d error.

If now the above operations be compared with the rule and the note following, it will be seen that the first error is too small, and the last one too large; therefore, 15, number first supposed × 5, the last error=75; and 20, the number last supposed 10, the first error=200; and 200+75 275, the sum of the products; and 10+5=15, the sum of errors. Therefore, 275-15-$18.333+, C's share; and $18.333+$10= $28.333+, B's share; and $28.333+$5=$33.333, A's share.

2. Four individuals having $100 to divide among themselves, agree that B. should have $4 more than A.; C., $8 more than B.; and D. twice as much as C. What was each man's share?

$80

and 100

70=30, 1st error. Hence, 100-80-20, 2d error.

Here both errors are too small, therefore, 6×20=120; and 8×30=240; then, 240-120-120, the difference of the products; and 30—2010, the difference of errors.

There

fore, 120-10=12, A's share; 12+4=16, B's share; and 16 +8=24, C's share; and 24+24=48, D's share.

Proof, 12

3. Three men hired a piece of wall built, for which they paid $500. Of this, A. paid a certain part; B. paid $10 more than A., and C. paid as much as A. and B. both. What did each man pay? Ans. A. paid $120; B. $130; and C. $250.

Sums like the preceding are solved with ease by analysis. Since we have the sum they all paid, we know that C. paid $250, because he has paid as much as the other two, that is, one half of the whole. Therefore, A. and B. together paid $250. But B. paid $10 more than A, hence, 250—10—240, twice the number of dollars A. paid, and 240÷2=120, A's share; then, 120+10=130, B's share; and 120+130=250, C's share..

A.

4. Two persons lay out equal sums of money in trade. gains 120 £,, and B. loses 80 £. A's money was then treble B's. With what sum did they commence? Ans. 180 £.

5. A farmer hired a laborer 40 days, on condition that he should receive 20 cents for every day he wrought, and forfeit 10 cents every day he was idle. At the expiration of the 40 days he received $5. How many days did he work, and how many was he idle? Ans. He wrought 30 days, and was idle 10 days. 6. What is the length of a fish whose head is 10 inches long, his tail as long as his head and half the length of his body, and his body as long as his head and tail both? Ans. 80 inches.

7. Two persons, A. and B., have the same income. A saves of his, but B., by spending $150 per annum more than A., at the end of 8 years finds himself $400 in debt. What was their income, and how much did each spend annually? Ans. Income, $400. A spends $300, and B. $450.

8. A man bequeathed his property to his three sons, on the following conditions; viz. to A., one half, wanting $50; to B., one third; and to C., the remainder, which was $10 less than B's share. How much did each son receive, and what was the whole estate? Ans. A. received $130; B. $120; and C. $110. The whole estate was $360.

9. A farmer bought a certain number of oxen, cows, and calves; for which he paid 130 £. For every ox he paid 7 £.; for every cow, 5 £.; and for every calf, 1 £. 10s. There were two cows for every ox, and three calves for every How many were there of each kind?

COW.

10 cows, and 30 calves.

10. A person after spending $10 more than

Ans. 5 oxen,

of his annual