196 Answers to Queries. 
der is - Turn this fraction upside down, thus, = and multiply 
by 4 ths of 28.) the * ths of 28 is 16; and a multiplied by 16 
is 154. 154 feet is the length of the longer portion of the rod. 
Add 28 feet, and 182 feet, the length of the entire rod, is the 
answer required. Maree 
407, 409.—Arithmetical Problems.—The first query is not one 
that can be answered definitely. It is not possible to find, from 
the given data, how many of each coin there zwerve on any particu- 
lar occasion, but only how many there mzght have been, because 
the problem, as it stands, admits not of merely oze solution, but 
of an exormously large number of solutions. ‘Taking the numbers 
of all but three of the coins to be any numbers within certain 
limits that we please, it is generally possible to fit in the rest 
(often in several ways) so as to make up the required amount with 
the given number of coins. The following number of coins, for 
example, will satisfy all the conditions of the problem :— 
IQ, £00, 10, 10, 10, TO, 10,.10, 73% 1360. 
Bi; As ses. sly.) Ug kee 1)! oe oe 
But we can, for example, replace 2 sixpences and 3 pence by 5 
threepenny bits, or vice versa, any number of times without alter- 
ing the number of coins or their amount. Similarly, we can 
replace 2s. 6d. and 6d. by a two-shilling piece anda shilling. This 
leaves us no wiser than we were at the outset as to what actually 
was the number of each coin. 
The second question is one of the large class of arithmetical 
questions known par excellence as “problems.” No fixed rule 
can be given for solving these, but the method of solution must 
depend on the question. To proceed :—Call the longer and 
shorter pieces A and B (for the sake of distinction, and not to 
introduce algebra). When 3/11 of the piece B is added to A, 
8/11 is left. Of this remainder we are to further add 1/14 to A; 
this leaves 13/14—Ze, 13/14 X 8/11, or 52/77 of the original 
length of B not added to A. Since the piece A, with the two 
added lengths, is 3/7 of the whole, therefore the remaining piece 
which we have seen is 52/77 of the original length of B forms the 
remaining 4/7 of the whole. Therefore, the original length of B 
is 77/52 x 4/7 of the whole—z.e, 11/13 of the whole. The 
shorter length A will make up the remaining 2/13 of the whole 
length. The whole length is, therefore, 13/2 of the length of A, 
and since A is 28 feet long the whole length is 28 x 13/2 feet, 
or 182 feet. G. H. Bryan. 
409.—Another Rule Wanted.—I do not know of any special 
rule for these problems in arithmetic. ‘They are best worked by 
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