112 ARITHMETIC. 



EXAMPLES. 



I. Let It be proposed to find the value of 140Z. 8dv/t; 

 of gold, at 3I. 19s. I id. an ounce. 



2,0)115,05. 9d. z4o<\' 

 Ans. 57I. I OS. 9d. 2Toq' 



Explanation. The three terms being stated by the 

 general rule, as above, the second term is reduced 

 to pence, and the third to penny-weights, these being 

 their lowest denominations, as directed in the first note. 

 The first term is also reduced to dwts. that it may agree 

 with the third, by the same note. The second term is 

 then multipHed by the third, and the product divided by 

 the first, according to the general rule, when the answer 

 comes out 13809 pence, and 12 remaining ; which 

 remainder being reduced to farthings, and these divided by 

 the same divisor 20, by the second note, the quotient is 2 

 farthings, 8 remaining. Lastly, the pence are divided 

 by 1 2, to reduce th,em to shillings, and these again by 20 for 

 pounds; when the final sum comes out 57I. los. pd. 2q. 

 for the answer. 



2. How 



