THE APPLICATION OP ALGEBRA 
84 
be expressed by 40 — a : moreover, since he was to re 
ceive 20 pence for every day he worked, the whole num 
ber of pence gained by working, will be 20 a; and for 
the like reason, the number of pence forfeited by play 
ing, or being absent, will be 8 x 40 — a, or 320 — 8a; 
which deducted from 20a, leaves 28a—320, for the sum 
total of what he had to receive : whence we have this 
equation, 28a—320 — 380; from which 28a — 380 -|- 
320 zz: 700, and consequently a zr — 25, equal 
to the number of days he worked ; therefore 40 — 25 
— 15, will be the number of days he played. 
PROBLEM XX. 
A farmer would mix two sorts of strain, viz. wheat, 
worth 4s. a bushel, with rye, worth 2s. 6(1. the bushel, so 
that the whole mixture may consist of 100 bushels, and be 
worth 3s. and 2d. the bushel: now it is required to find 
how many bushels of each sort must be taken to make up 
such a mixture. 
Let the number of bushels of wheat be put zz x, and 
the number of bushels of rye will be loo — a : but the 
number of bushels multiplied by the number of pence 
per bushel, is equal to the number of pence the whole 
is worth; therefore 48a is the whole value of the wheat, 
and 30 x too — a, or 3000 — зоа, that of the rye; 
and consequently, 48a + 3000 — 30a, the sum of these 
two, the whole value of the mixture: which, by the 
question, is equal to 100 x 38, or 3800 pence: hence 
we have 48a -f 3000 — 30a — 3S00; and therefore 
x — — 44$, the number of bushels of wheat: 
whence the number of bushels of rye will be loo —- 
44$ - 55$. 
PROBLEM XXI. 
A farmer sold, to one man, 30 bushels of wheat and 40 
of barley, and for the whole received 270 'shillings; and to 
another he sold 50 bushels of wheat and 30 of barley, at