Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

John Fraser, a teacher of navigation in Newport, Rhode-Island, and another person under the signature Loxodromicus, have in the public newspapers lately made various attempts to persuade the people of these states, "that a ship steering a due east or west course "does not sail on a parallel of latitude, but on "tain spiral line which would ultimately carry her "to the equator" They might just as well have said that a due east or west course will carry a ship to the

moon.

a cer

ARTICLE IV.

NEW QUESTIONS, to be answered in the next number. I. QUESTION, by William Scott, A. M. New-York.

Ir is required to establish a general rule for finding in the shortest manner possible, how much New-York currency is equal to any given sum of sterling money, suppose 657l. 188, 9d.

II. QUEST. by W. Scott, A. M.

It is required to establish a general rule for finding in the shortest manner possible, how much sterling money is equal to any given sum of New-York currency, suppose 976/. 10s. 8d.

III. QUEST. by John Boyd, of New-York.

My age in years is a number consisting of two digits, of this number is a mean proportional between these two digits, and two years hence, my age will be a third proportional to the same two digits, directly as they stand in my present age; How long will it be before I am sixty years old?

IV. QUEST. by R. Tagart, New-York.

In baking a hemispherical loaf of bread of 10 inches radius, the crust was every where of an equal thickness, and the solidity of the crust was equal to half the solid content of the whole loaf; What were the dimensions of the interior soft part?

V. QUEST. by Diarius Yankee, Bunker's-Hill.

A wretch, who spurn'd the virtuous path below,
To heaven above by other means would go :
He by mechanic art a ladder rais'd,

Its height the vulgar and the learn'd amaz'd.
Thus Jacob did, said he, and why not I
Scale heaven as well as he? at least I'll try.
He mounts his ladder, rapidly ascends,
And bids adieu to all terrestrial friends.---
Now wrapp'd in clouds unpierc'd by human eyes---
And now environ'd with the azure skies

But now a ball red-hot he lets fall down,
Which fires a cannon at the base, whose sound
Ascends his ear, in thrice the length of time
The ball was falling from this height sublime.
Ye sages tell if he to heaven has got,

Or how far soaring from this earthly spot ?

VI. QUEST. by Walter Folger, jun. Nantucket.

A gentleman has bought a rectangular piece of land whose perimeter is to be 100 rods; and he is to pay 1 dollar for each rod in the length of the land, and 3 dollars for each rod in the breadth. It is required to determine the length and breadth so that the quantity of land may be had at the cheapest rate possible.

VII. PRIZE QUEST. by George Baron, New-York.

In surveying a field in the form of an oblique angled rectilineal triangle, the first or least side measured 24 chains, the second or next greater side measured 37.44 chains; on the third or greatest side grew two large hemlock trees whose distance asunder was 16.80 chains; a straight line drawn from the obtuse angle to the tree nearest the first side, was perpendicular to the second side, and another straight line drawn from the same angle to the other tree, was perpendicular to the first side. Required the third side, the distance of each tree from the obtuse angle, and the area of the field.

ARTICLE V.

ANSWERS to the Questions proposed in Article IV.

I. QUEST. answered by Wm. Green, New-York.

SINCE 9 sterling sixpences are equal to a dollar, or 16 sixpences New-York currency; it is evident that 9 is to 16 as any sum of sterling money is to its equivalent in New-York currency.

Hence 9 16: 6571. 188. 9d.: the sum required, and 18 16 :: 13157. 178. 6d. : the sum required (by art. 1. § 17. Math. Corr.). Here of the first term subtracted from the first, leaves a remainder equal to the second; and consequently of the third term subtracted from the third will leave a remainder equal to the fourth (by art. 1. § 9. Math. Corr.). The general rule in question is therefore as follows: Multiply any sum of sterling money by 2, and of the product, subtracted from the product, leaves the NewYork currency required. Hence,

1. 8. d.

657 18 9 sterling.

2

9)1315 17 6

146 4 2

Answer 1169 13 4 New-York currency.

II. QUEST. answered by Samuel Moor, N. York.

From the reasons advanced in the preceding solu tion, we have 16:9: 976l. 10s. 8d. the sum required, and 16: 18 :: 976/. 10s. 8d.: twice the sum required (by art. 1. cor. to § 17. Math. Corr.). But hereof the first term added to the first, is equal to the second; and of consequence of the third term, No. 2.

C

added to the third will be equal to the fourth (by art. 1. § 8. Math. Corr.). The following is therefore the general rule sought, viz. The eighth part of any sum of New-York currency, added to that sum, is equal to twice the sterling money required. Hence,

[merged small][ocr errors][merged small][merged small][merged small]

III. QUEST. answered by William Lenhart, York-Town, Pennsylvania.

Let x = the digit in the ten's place, and xy2 = the digit in the unit's, then it is evident from the ques, tion that y is greater than 1 but not greater than 3, and that 10x+xy2 = the age of the proposer. By the second condition of the quest. 7xy = 10x + xy2, or Tyy2 10, and each side of this equation sub

=

49
4

49

tracted from leaves - 7y + y2 = the square

=

4

roots of which (since y is not greater than 3) give -y, whence y 2. The two digits originally expressed by x and xy2, will now become x and 4x and the age of the proposer 10x + xy2 will be = 14x. Again by the third condition of the quest. x: 4x: 4x: 16x = 14x+2, and hence x = 1; xy2 = 4, and 14 years is the age of our young mathematical friend, who, in 46 years will attain the age of 60.

The same answered by R. Tagart, N. York.

Put the proposer's age, y = the digit in the ten's place and x → 10y will be the digit in the place of units. Then by the second condition of the quest. x = 7 X- 10y Xy, or x2 49xy 490y2.

Further, by the third condition of the quest. y : x

10y :: x 10y:

x02

X2

20xу100y2

y

= x + 2, or

21xy 100y2+2y, and this equated with the former value of x2 gives 49xy

- 100y2 + 2y, whence x =

490y* = 21xy

195y + 1

which by

" 14 But since per quest.

the quest. is a whole number.
196y
y, and of consequence
14
1964 195y +1 y 1

therefore

14

=

14.

are also whole numbers;

=

14

must either be

[ocr errors]

p

a whole number or 0, which being assumed gives y 14h1. Here it is evident that the digit y has but one possible value, that must be =

[merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][ocr errors][merged small][merged small]

IV. QUEST. answered by Robert Adrain, York-Town,

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small]
« ΠροηγούμενηΣυνέχεια »