ABCDEL

10

17

307418

24

123m

31

35

42

49

56

63

must have 10, C 17, D 24, E 31, and F 38, as in the Margent: But (by the tenure of the Problem) the fhare of F fhould be but juft double to that of A, but it is much more; wherefore fubftract A's fhare 3, from F's fhare, 38, and the remainder will be 35; which is the number of Pounds that A must have; and then B must have 7 more (viz 42.) and C 49, D 56, E63, and F 70, which is just double to what A had; which answers the Question as to that part: Then, these shares added together, do make 315 Pound, and that was the number of Pounds to be divided.

F 38 70

35 315

There were Five Bags of Money, and in the biggest Bag there was as many more Pounds as there was in the leaft, and in the middle Bag there was half as much as there was in the biggest and leaft together; and each Bag from the leaft had in it 30 Pound more than the other: What number of Pounds was in each Bag?

L

GHKLM

4

34

120

150 64 180

210

ET the Five Bags be represented by these Five Letters, GHKL and M, and suppose there was in the leaft Bag G 4 Pound, then must there be in the Bag H 34, and in K 64, and in L 94, and in M 124. But the Bag M holds much more than double that of G, wherefore fubftract G 4, from M124, and there will remain 120, and so much Money must be in the leaft Bag G, as in the Margent: Then (because every Bag exceeds other by 30 Pound), there must be in H 150 Pound, in K 180 Pound, in L 210 Pound, and in M 240 : And then Mis double to G, for 120 doubled is 240, equal to M, and G and M together is equal to 360, the half whereof is 180, and fo much is there in the middle Bag K; and the Money in all the Bags is equal to 900 Pounds: And fo is the Question answered in all particulars.

94 M 124

120

If 136 Pound be to be divided into two parts between A and B, so that B fball have 25 Pound more than A, what is each share?

F

"Rom 136 (the Sum) fubftra&t 25 (the excess that B must have more than A ), and the remainder will be 111, the half whereof is 55 l. 10s. for A; and that added to 25 the excefs, it makes 80/. 10 s. for B; for 55. 10 s. and 80 %. 10 s. is equal to the Sum 136 1.

PROBL.

If 560 Pound be to be divided between A and B, in fuch proportion as 2 Pound bath to7 Pound, what must each have?

TH

HE Sum of the two Proportional Terms 2 and 5 make 7; wherefore by the Golden Rule,

As 7 is to 560, fo is 2 the leffer Proportional to 160 the leffer Share; and fo is 5 the greater Proportional to 400 the greater Share; which together make 560 Pound, the Sum to be divided.

There was a May-Pole which confifted of three pieces of Timber, of which the firft (or lowermoft) was 13 foot long, the third (or uppermost piece) was as long as the lowermoft piece, and half the middle piece; and the middle piece was as long as the uppermost and lowermoft together: How high was this May Pole, and how long each Piece?

M

Ultiply the length of the firft piece 13, by 3 and by 4, fo fhall the two Products be 39 and 52, for 13 by 3, is 39 for the length of the uppermoft piece; and 13 by 4, is 52 for the length of the middle piece, which 52 is equal to 13 the lower piece, and 39 the upper piece added together; and the height of the May-Pole was therefore 104 Foot.

PROBL. VIII.

A Grandfather living in the house with his Son, and his Son's Wife, and their Three Sons, and difcourfing concerning their Ages, the Father faid, I am as old as my Wife and Second Son, and four years over. The Wife faid that he was as old as all her three Sons, and three years over.

The Eldeft Son faid that he was as old as his two younger Brothers,and 3 years over. And the Second Son faid he was four times the age of his younger brother, and one year over. And when all their Ages were added together, they proved to be the juft Age of the Grandfather, who was then 125 years old. What was the age of the Father, Mother, and all the three Sons?

OR the refolving of this and the like Questions, fuppofe the young

FOR eft Son to be One year old, then the fecond must be 4 years, and I

over, and the eldeft as old as both the other, that is, 5 years, and 3 over. Alfo the Mother must be as old as all the three Sons, which is 10 years, and 3 over: And confequently by the Tenor of the Queftion, the Father being the Age of his Wife and Second Son, muft be 14 years old, and

4 over:

Which fet down in this manner,
1 Year.

The Youngest Son was old
The Second Son was then
The Eldeft Son was then

The Mother then must be
The Father must then be

All which added together make

But according to the tenor of the Question, all their Ages together fhould be equal to the Age of the Grandfather, which was 125 years: Wherefore, to find the Age of the youngest Son, you must work by the Rule of Proportion thus: Saying,

The Sum here is 113. 968 years, to which the 11 years over being added in their due places, as in the fecond part of the Example they are, the Sum will be 124. 968 years, equal to the Age of the Grandfather; and the Question refolved in every particular.

PROBL. IX.

If 222 Crowns be to be divided among three Perfons A B and C, in fuch proportion that C shall have 9 times as many as B, and 6 times as many as Awanting Ten. And B (ball have 7 times as many as A, and 25 over: How many Crowns must A have, and alfo B and C ?

Sup

Uppose A to have only one Crown, then by the Question, B must have 7 Crowns, and 25 over; and C must have 9 times as many as B, that is 63, and 6 times as many as A, that is 6, in all 69 wanting 10; which fet in this manner.

A 1 Crown.

B7 Crowns, and more 25.

C 69 Crowns wanting 10.

The Sum 77 Crowns, more 25, wanting 10.

Substract the 10 wanting from the 25 more, and then there will remain 15 over; Subftract 15 from 222, and there will remain 207, for a dividend, and the Sum 77 must be 207 by 77, and the Quotient will be 2

the divifor; wherefore divide (or in Decimals 2. 688) for the fhare of A, and then A having 2. 688, B must have 7 times as many, that is 18. many, that is 18. 816, and 25 over; and C muft have nine times as many as B, which is 169.344, and fix times as many as A, which is 16. 128, which

together

together makes 185.472, wanting to; all which being added together as above, do make 206. 976, to which the 25 over being added, and the 10 wanting fubftracted in their due places, the Sum will be 221. 976, that is, 222 Crowns, and anfwers the Queftion.

There are four Weights A B C D, all which together do weigh 324 Pound, of which B weighs more than Aby 9 Pound, Cweighs three times as much as A, and four times as much as B; and D weighs 6 times A, 3 times B, and once and a half C; their respective weights are demanded.

L'

ET the weight A be fuppofed to weigh only one Pound, then B muft weigh Pound and 9 over; and C muft weigh 3 times A which is 3, and 4 times as much as B which is 4, which together make 7 for C; and then D must weigh 6 times A, 3 times B, once and a half C, and 4 times B, which together make 19; all which added together do make 28 more by 9, which should be equal to 324; Subftract the 9 which is over, from 324, and the remainder will be 315 for a Dividend, and the Sum 28 is the Divisor : Divide 315 by 28 (or decimally 315. 5. by 28. (5) the Quotient will be 11. 07 for the

A I.
B

I more 9.

C 7. D 19.

I

Sum 281⁄2 more 9. weight of A. Then

Three Merchants build a Ship which coft 6000 Pound,of which A paid part,

B part, and Cpart; What did each pay?

M

i;, and they

Ultiply the Denominators of the several Fractions
make 120; divide this 120 by each feveral Denominator, as by

4, by 5 and by 6, and the Quotients will be 30, 24 and 20,

.

There are three Weights ABC, of which, the leaft Weight A, weighs 30 Pound, and the other two together do weigh feven times as much as A, and they are in proportion one to the other as 4 is to 5.

HE Weight A weighing 30 Pound, the other two Weights together weighing 7 times A which is 30, muft weigh 210 Pounds; and the Proportional Numbers 4 and 5 being added together do make 9. Then by the Rule of Proportion:

As 9, the Sum of the Proportional terms

Is to 210 the Sum of the two unknown Weights;
So is 5, the Greater Proportional Number,
To the Weight of the Greater Weight.

Wherefore, Multiply 210 by 5, the Product will be 1050, which divided by 9, gives in the Quotient 116 for the Weight of the bigger Weight; and that fubftracted from 210, the remainder is 93, for the leffer Weight, and fo the three Weights are

Two Travellers fet out from two Towns which are 140 Miles diftant, upon one and the fame day; One travels 8 miles a day, the other 6: In how many days Travel will they meet each other?

Suppose

one D (or Day) for the time that they fhall meet, in which time one would have travelled 8, and the other but 6 miles of the Journey, which together make 14 miles; whereas they should have been 140 miles: Wherefore, divide 140 by 14, the Quotient will be 10, and in fo many days will they meet: For 10 times 8 is 80 Miles; and 10 times 6 is 60 Miles; which together make 140 Miles the whole Journey.

PROBL