Arithmetic, Part 1 · Section 36
The Rule of Falſe Poſition.
THis Rule ſerves to reſolve queſtions, which are not preſently fit for the Golden Rule; and therefore in ſtead of the true Number which is ſought: Suppoſe any number great or ſmall, and make trial of it, whether it reſolve the queſtion without any errour; if ſo, it is the true Number: if not note what errour is at the end of the worke, and whether it be too much or too little: if too much, mark it thus + but if too little, thus −
Then ſuppoſe again another Number, (it imports not whether it be neerer or further off) and try as before, and mark that errour alſo;
And, then
THE RULE.
Multiply the firſt Poſition by the ſecond Errour, and the ſecond Poſition by the firſt errour, and (if the errours be both + or both −) Subtract the leſſer product from the greater, and keep the remaine for a Dividend, and the difference of the errours for the Diviſor: the quotient of that diviſion is the true Number required.
But if the errours had been one +, the other −, then the ſumme of the products added together had been the Dividend: and the ſumme of the errours, the Diviſor; the reſt of the work the ſame as before.
Queſtion 1.
A man is to drive 48 young Turkies 40 miles: and for every Turkie which comes alive to the end of the Journey, he is to receive 3 d. but for every one which dyes by the way, he is to pay 6 d. at the end, he received 72 d. How many dyed by the way?
Let the firſt ſuppoſition be
| That by the way there dyed | 20 | |
| For them he was to pay 120 d. and for 28 which lived, he was to receive 84 d. ſo he paid more then he received | 36 d. | |
| and ſhould have got clear | 72 d. | Adde 108 |
| Wherefore the firſt errour is | −108 |
| Let the ſecond ſuppoſition be | 10 |
| For theſe he paid 60 d. and for the reſt he received 114 d. the difference is 54, and ſhould be 72: ſo the ſecond errour is | −18 |
| Now 20 multiplyed by 18, produceth | 360 |
| and 10 by 108 produceth | 1080 |
and the difference is 720 for the Dividend, likewiſe the difference of the errours is 90 for the Diviſor, and the quotient is 8, which is the true Number of thoſe he loſt by the way. As may be proved by trial.
Queſtion 3.
If it was required to make up a pound Sterling of ſhillings and groats only: and ſo as the number of groats, may be to the Number of ſhillings, as 7 to 1, how many ſhillings muſt there be?
| Firſt, ſuppoſe the ſhillings | 4 |
| then the groats muſt be equal to 16 s. that is 48 groats; but the ſhillings taken 7 times are 28, to which 48 ſhould be equal, but is | +20 |
| Secondly, ſuppoſe the ſhillings | 2 |
| then the groats (making 18 s.) are 25, | |
| which ſhould be equal to 7 times 2, but is | +40 |
| Multiply 4 by 40 product is 160, then | |
| Multiply 2 by 20, the product is 40, which taken from 160, reſts for the Dividend | 120 |
| And the difference of errours is | 20 |
| Laſtly, 120 divided by 29, quotient is | 6 |
| The number of ſhillings therefore is | 6 |
| And the number of groats is | 42 |
But as 7 to 1, ſo is 6 times 7 which is 42 to 6 times 1, which is 6: ſo the work is done.
Queſtion 3.
If there be 4 ſeveral weights, A. B. C. D. of which D is 24 ounces, and C is double to B, and triple to A, and D with twice A is double to C. and quadruple to B: How much doth every one of theſe weights weigh?
| Firſt, ſuppoſe A to be then D with twice A is 24, and 16, that is 40, of which … being the halfe is …0, and B 10. | 8 |
| Now thrice A is 24, to which C ſhould be equal, but is | −4 |
| Secondly, let A be ſuppoſed | 4 |
| then D more, twice A is 32, and C 16 and B is 8, but thrice A is 12, to which 16 ſhould be equal, but is | +4 |
Then 8 multiplied by 4, gives 32, and 4 by 4, produceth 16: both theſe products gives 48 for the Dividend: and the ſumme of the errours (becauſe the firſt is −, the other +) gives 8, for the Diviſor; and the quotient will be 6, to which A is equal, and twice A more D is 36, of which C, (being halfe is 18, and B is 9, and thrice A is equal to C, namely, 18, and all right.
Whereas the firſt errour is equal here to the ſecond, it followes that the Poſitions were equally falſe: and therefore their difference which is 4, being parted into two equal parts, 2 and 2, if 2 be taken from 8, the remaine is the true number 6, or if … be added to 4, (which was the ſecond poſition) the ſumme will be alſo 6.
And further, whenſoever the erronrs be one +, the other −, though they be not equal; yet then if the difference between the poſitions be parted into two parts, which are in proportion one to another, as the two errours are one to another reſpectively: then if the firſt part be taken from the firſt poſition (if that be the greater) or adde to it (if it be the leſſe, the ſame number required is thereby had.
As, let the laſt queſtion be reſumed,
| And let the firſt poſition for A be | 15 |
| Then the firſt errour will be | −18 |
| Then let the ſecond poſition be | 3 |
| And ſo the ſecond errour will be | +6 |
| And the difference of poſitions is | 12 |
which divided into two parts 9 and 3, which have that proportion one to another as have the errours 18 and 6, then if the firſt part 9, be taken from the firſt poſition 15, there remains the true Number 6. Or elſe if the ſecond part 3, be added to the ſecond poſition 3: thereby alſo is made the true Number 6.
The way of parting 12, (or any other) into two parts proportional with the errours is eaſily done by the Golden Rule, thus:
As the ſumme of the Errours 24,
is to the difference of poſitions 12;
So is the greater errour 18
to the greater part required, namely, 9.
Many other queſtions are in other bookes exemplifyed and wrought by this Rule; but ſeeing I intend not to write a great book; and alſo becauſe ſome of thoſe queſtions may be eaſily reſolved without this Rule, I will adde no more: ondy mention one of thoſe queſtions.
I there be a Ciſterne with 4 Cocks, which holls 8 barrels of water, and the firſt cock will runit all out in 6 houres, the ſecond in 4, the third in 3, and the laſt in 2 houres: in what time ſhal all of them runne it out?
| If the firſt in 6 houres run | 8 |
| the ſecond in the ſame time would …un | 12 |
| the third | 16 |
| the laſt | 24 |
| In all | 60 |
Then ſay, If 60 require 6… what 8?
The anſwer 48/60, that is 4/5 o… an hour; in which time all the 4 Cocks together would run out all the 8 Barrels of water.