Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση
[ocr errors]

SOLUTION.

Since £6 was added to g of A's money, and subtracted from g of B's money, the sum of A's and B's money, after this ad. dition and subtraction, is the same as it would have been, had no such addition and subtraction been made. Therefore of A's and B’s money is, by the question, £72.

Again, by the question, of A’s, increased by £6, is equal to

of g of B’s, diminished by £6; .. of A's, increased by £6, is to z of B's, diminished by £6, as 7 to 8. But we have already seen, that of A's, increased by £6, added to g‘of B's, diminished by £6, is £72. Hence, if we divide £72 into two parts, which are to each other as 7 to 8, these parts will beg of A's, increased by £6, and g of B’s, diminished by £6. The parts of £72 pounds, which are to each other as 7 to 8, are is of £72=£33, and is of £72=£38. Therefore of A's, increased by £6, is equal to £33}, consequently, s of A's is £333 - £6= £277; and the whole of A's money is of £273=£413. And g of B's, diminished by £6, is equal to £38. Therefore, of B’s is £38 + £6=£44}; and the whole of B’s money is of £444=£66%.

17. A purse of $2350 is to be divided among three persons, A, B, and C. A's share is to be to B's, as 6 to 1), and C is to have $300 more than A and B together. What is each one's share ? :

SOLUTION. Since C is to have $300 more than A and B together, it follows that A and B must have half of what is left, after sub. tracting $300: hence, A and B together have $1275; this divided into two parts, which are to each other as 6 to 11, gives A's=ily of 1275=$450 ; B's=1 of 1275=$825; C's is evidently $1575. ^

of what is de 1975; this

18. Two persons, A and B, purchase in company, 100 acres of land for $1000, of which the southern portion is of rather the bost quality. In the division of it, A says to B, let me have of the southern portion, and I will pay for my part $11'per acre more than you pay for yours. How much land must each have, and at what price per acre ?

Note.-This, and like questions, can be solved by the following

RULE.

Divide-half the whole cost by the whole number of acres, and to the square of the quotient add the square of half the difference of the prices per acre; then extract the square root of the sum, and to this root add the quotient of half the whole cost, divided by the whole number of acres. This last sum, increased by half the difference of the prices per acre, will give the price per acre of the best land; and diminished by the same, will give the price per acre of the poorest.

Applying the above rule, we find the quotient of half the whole cost, divided by the whole number of acres, to be 5, which squared, gives 25 ; this increased by the square of half the difference of the price per acre, becomes 25+12,206 = 32,7%, whose square root is 1 ; this root added to 5, gives 30 =106. Therefore, the price per acre, of A's land, is 1036 +18=$105. The price of B's land is 1036 -18=$93.

$500-$101=47 1 acres, for A's portion.$500--$94 = 521acres, for B’s portion. . 19. A boy divided his apples among his four companions, in the following manner: To the first he gave half an apple more than half his whole number; to the second, he likewiso gave half an apple more than half the number which he then had; in the same manner he divided with the third and fourth companion, giving to each half an apple more than half the number which was left after giving to the preceding one.

After having divided with the fourth companion, he had but one apple left. How many had he at first ?

SOLUTION.

This question is most easily solved by beginning with the last companion, and reversing each successive operation.

Since he had but one apple left, after making the last division, he must have had 3 after the third division, because 3 diminished by } more than its half, leaves 1; for a similar reason, he must have had 7 after the second division. In this way we may retrace the process by multiplying the number by 2, and adding 1 to the product, for each successive step. In this way we find that after the first division he must have had 2x7+1=15. And before he divided with the first, he must have had 2x15+1=31 apples.

Note. From the above method of solving this question, we see that it would not be difficult to extend the solution to the case of any num. ber of divisions. Indeed, we see that the number required is always one less than 2, raised to a power whose index is one more than the number of divisions.

Thus, had there been 10 companions to divide with, after the above method, he must have had 211-1=2047 apples at first.

This method of division is effected without dividing an individual apple.

20. A hare is 50 leaps before a greyhound, and takes 4 leaps to the greyhound 3, but two leaps of the hound are equal to 3 of the hare's. How many leaps must the greyhound take before he catches the hare?

SOLUTION. Since 2 leaps of the greyhound equals 3 leaps of the hare, it follows that 6 leaps of the greyhound equals 9 leaps of the hare.

But while the greyhound takes 6 leaps, the hare takes 8 leaps; therefore, while the hare takes 8 leaps, the greyhound gains on her 1 leap. Hence, to gain 50 leaps she must take 50 X 8=400 leaps ; but while the hare takes 400 leaps, the greyhound would take 300 leaps, since the number of leaps taken by them were as 4 to 3.

21. From a cask of wine a tenth part is drawn out, and then the cask is filled with water; after which, a tenth part of the mixture is drawn out; again, the cask is filled, and again a tenth part of the mixture is drawn out. Now suppose the fluids rix uniformly at each time the cask is replenished, what fractional part of pure wine will remain after the process of drawing out, and replenishing has been repeated ten times ?

SOLUTION.

[ocr errors]

wine, or is

Since to of the wine is drawn out at the first drawing, there must remain . And after the cask is filled with water, the of the whole being drawn out, there will remain % of the mix. ture; but % of the mixture, we have already seen, is pure wine; therefore after the second drawing there will remain 4 of 4 of pure wine, or 2. By similar reasoning, we see that after the third drawing there will remain % of % of % of pure wine, or 103

From this, we see that the part of pure wine remaining is expressed by the ratio 1, raised to a power whose exponent is the number of times the cask has been drawn from. Hence, in the present question, the fractional part of pure wine is 910

=134867:44:1=0.3486784401, which is nearly 35 per cent.

22. Suppose, from an acorn, there shoots up a single stalk at the end of the year ; that at the end of each year there. after, this stalk puts forth as many new branches as it is years old : also suppose all the branches to follow the same law, that is, to produce as many new branches as they are years old. How many branches will this oak tree consist of at the end of 20 years ?

[ocr errors]
« ΠροηγούμενηΣυνέχεια »