Abbildungen der Seite
PDF
EPUB
[ocr errors]

100

100

Question 29. Several Merchants enter into Partnerthip, every
one put into the Stock 65 times as many Pounds as there were
Partners; with that Stock they traded and gained as many Pounds
per 100 l. as there were Partners. Now if 101. To s. be added to,
and fubftracted from, their Gain, the Product of that Sum and
Difference will be 649 il. 6s. 3 d.
Quare, How many Merchants there were, Egc.
Lecl i

a= the Number of Merchants.
I x 55 2165 a = every one's Sum put into Stock.
2 xal 3165 A a Šthe whole Stock

da
And 4 100: 4:: 65 aa:

65 a a a

by cho Queftion.

100
65 a a a
Viz. 5

the whole Gain.

65 a a a
5 +10,5
6

+ 10,5
100

65 a a a
5 10,51 7

16,5
8
4225 aaaaaa

-110,25=6491,3125, by the Quer: 8 x 10000 9 4225 a6 - 1102500 = 64913125

9 + 10 4225 46 = 66015625 10 • 4225 I 26 =

=

4225 Il wwo 12 a=v 15625 =s the Number of Merchand. 12 55 131652 = 325 the Number of Pounds each put in. Question 30. Three Merchants join Stocks together; the firft Man's Stock was less than the second Man's by 131. the second and third Man's Stock was 175 l. in trading they gain 481. more than their whole Stock was, the first Man's proportional Part of the Gain was 78, What was each Man's Stock and Part of the Gain?

Let b, d, jy reprefent each Man's Srock.

ilafety as the whole Stock.

25+48= the whole Gain.
Ana
a{i

+

・モリー and to 17548

6 and 2

6x7

[ocr errors]

10000

66015625 315625

[ocr errors]

6:7 Then

[ocr errors][ocr errors]

401,3175}by the Question

[ocr errors]

6 and 21 71s + 48 = 223 +2..

But 81175 +a: 223 +a::a:78 per Question.

8:19 aa + 223a =78 a + 13650
9-78 a 10 aa * 145 a = 13650
10 Conaa + 145 a + 5256,25 = 18906,25

11 w21212 + 72,5 =V 18906,25 = 137,5 12 - 72,5 13 a = 137,5 – 72,5 = 65

3, 14 e = a +13=78
4 – 14 15 y = 97

Then 1665: 78::78:931. 12 5. Se's Gain.
Again 17 65:-78::97:1161. 8 s. y's Gain.
Proof

18 1761.8s. +931. 125. +781.=2881. the Gain.

19/65 +78 +97=240. the whole Stock. 18 - 19/201288—240=48 the Gain more than the Stock.

[ocr errors]

Question 31. A Father at his Death left his three Sons his Money in this manner; to the eldeft he gave half of it, wanting 44 Pounds; to the second be gave one third of it, and 14 Pounds more; to the youngest he gave the Remainder, which was less than the Share of the second Son, by 82 Pounds : What was each Son's Share?

[ocr errors]

98

Let a, e, y be the three Shares, and z = the whole Sum.

ilate+y=%

2a=iz- 44.
Then

by the Question:
34z + 14
4 y=42+14-82)

z 2+3+41 52 fet y = +

3

2 I and 5 62

-98

3 6x31 737=22+

32

294 7 * 2 86z=4% +32-588

8 + 912=588, the whole Sum that was left. 2 and 9 10 =

101=--44 = 250, the eldeft Son's Share. 3 and 9 ore="! to 14 = 210, the second Son's Share. 4 and 9112ly = y + 14 - 82=128, the youngest, &c.

[ocr errors]

+

[ocr errors]

2

Question 32. A Man playing at Hazard or Dice, won the first Throw juft so much Money as he had in his Pocket, the

second

second Throw he won the Square Root of what he then had, and five Shillings more; the third Throw he won the Square of all he then had; after which his whole Sum was uz l. 16s. What Money had he when he began to play?

Suppose | 11 a= his first Sum. Then
I X2 22 a = his Sum after the first Throw.

And 35+ V 2a = the Winnings at the ad Throw. 2+3 424 +5 + 2a= the Sum after the 2d Throw, 4 Q2

54 a 2 + 22 a +25+4 av 20: +10V 2a= the

Winnings at the 3d Throw: and therefore 4+5) olfaa +242 +30+400 za tri za=2256 Shil. But to avoid these Surd Quantities, let us, instead of supposing the first Sum, make a second Trial, viz.

Let 1, 2a a = the first Sum.
Ix2 24 a a = the Sum after the first Throw.
Then

3 2 2 + 5 = the Sum won at the ad Throw.
2 +31 44 99 +2a+5=his Sum after the ad Throw.

516a+ +16a3-+4422 +200 + 25 = the Win

nings at the 3d Throw; and therefore 4 +5 6116a+ +16a} +48aa +222 +30=2256 Shil.

a

4 Q2

2

1 X 2

[ocr errors]

Yet again, to avoid these high Equations, let us make a third Supposition ; thus,

Let - the first Sum.

2a a = the Sum after the first Throw.
Then 39 +5 = the Winnings at the 2d Throw.
2 + 3

4 a atat 5 = the Sum after the 2d Throw. Subfti. Sle=aata + 5.

Glee = the Winnings at the 3d Throw. Then
5 + 6 zlete=2256 Shillings by the Question.
7 CO

82ete+0,25 = 2256,25
8 w gle+0,5 = V 2256,25 = 47,5
9-0,5 104 = 47
5 and 10 ulaatat 547

5 12 aa ta = 42
12, C |13|aata + 0,25 = 42,25

I w211414 +0,5 =V 42,25 = = 6,5
14-0,5 15 a=6
15 Q2|16|aa

The Shillings he had in his
18
2

Pocket when he began to play.

II

36

36.

16217

[ocr errors]

: Note, Note, In resolving of the last Question, I have made three different Suppositions for the Thing fought, purely as an Instance, to shew the young Learner how well he ought to consider the Nature of the Question, when he first states it, and make Choice of representing the Things fought, so as to avoid running it into Surds, if possible, viz, as in the first Suppofition of a = the first Sum, &c. Not but that such Equations may be solved, as shall be thew'd in the next Chapter. However, it is moft like an Artift to perform Things of this Nature the nearest and easieft Way they can be done.

Question 33. Suppose there were two equal Circles, whose Peripheries (viz. Circumferences) are divided into 44310 equal Parts; and that those Circles were to placed upon one Axis, as to move the contrary IVay to each other ; and suppose one of them to move but one of these equal Parts the first Day, two Parts the second Day, three Parts the third Day, and so on in Arithmetical Progression, viz. 1, 2, 3, 4, 5, &c. and the other to move every Day the Cube of those Parts, 1, 8, 27, 64, 125, &c. of the same Parts; How many Parts, and how many Days must each Circle move, before the same two Points meet that were together when they began to move ?

In order to give a ready Solution to this Question (or any other in this kind) it will be convenient to premife this Lemma.

LE M M A.

The Sum of any Series of Cubes whose Roots are in Arithmetick Progression (the first Term, and common Difference being Unity or 1) is equal to the Square of the Sum of all those Roots. : As in these

Terms in Arith. &c. their Cubes.

[blocks in formation]

Let la= the Sum of all the Parts the 1 Circle moves.

Then 2 aa= the Sum of all the Parts the 2d moves Consequen. 3ja a ta= 44310 by the Queft.

(per Lem. 2 Col4laatat 0,25 = 44310,25

.

4 w 51a +0,5 = V 44310,25 = 210,5

5—0,5 a=210{ the Number of Parts the firt Circle

6 @

44100

Circle moves. Next to find the Number of Days they moved ; there is given the firft Term =1, the common Difference = 1, and the Sum of all the Terms = 210, thence to find the last Term, which in this Cafe is the fame with the Number of all the Terms.

Let a = 1 the first Term, cal the common Difference, and s=210 the Sum of all the Terms, to find y = the last Term; as per Sect. 1. Chap. 6. Then yy tey=zstaa- ae by the 16 Step, Page 186. that is, yy ty= 210 x2 = 420 &c. Hence y=20 the Number of Days required.

I shall now proceed to give an Example or two of the Method used in arguing about unlimited Questions; viz. such Questions which admit of various Answers, such as those in Alligation Alternate promised in Page 117.

In order to shorten that Work, it will be convenient for the Learner to know the two Signs of Comparison, 7 and 5: The Sign 7 is of 6cater than ; as b za fignifies that b is greater than a. The Sign is or Letter than ; as bsd signifies that b is lesser than d, &c.

Ε Χ Α Μ Ρ Ι Ε 1.

Question 34.

A Tobacconist bath three Sorts of Tobacco, viz. one of 2 s. 8 d. the Pound, another of 20 d. the Pound, and a third Sort of 16 d. the Pound; of these he would make a Mixture to contain 56 Pound, that may be fold for 22 d. the Pound: How much of each Sort may he take?

Let a = the Quantity of that worth 32 Pence the Pound, e = that of 20 Pence the Pound, and y = that of 16 Pence the Pound;

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