Schillings and fruit...show all work to be voted best answer?

Herr Vitami buys 3 kinds of fruit for 40 schillings, 10 schillings, and 1 schilling each. He pays 259 schillings for 100 pieces of fruit. How many of the cheapest kind of fruit did he buy?

Schillings and fruit...show all work to be voted best answer?
Let's visualise these by size, Water Melons cost 40 schillings, apples cost 10 schillings and cherries cost 1 schilling each.



Suppose our Austrian fruit and veg merchant buys 100 cherries at 1 schilling each, he spends 100 schillings. For the bigger fruit he puts a cherry back on the stall and takes a water melon instead for 39 further schillings, or ditto an apple for 9 further schillings.



So we ask "is 39 a factor of 159 (259 - 100)?" and "is 9 a factor of 159?" and the answer is No as 159 is prime. So we have to rule out solutions that involve buying no water melons and solutions that involved buying no apples.



We notice that 4 water melons is the maximum as 4 x 39 = 156 and more than that would burst the budget.



1 water melon would mean 120 schillings for apples but 120 (modulo 9) = 3



3 water melons would mean 42 schillings for apples but 42 (modulo 9) = 6



2 water melons would mean 81 schillings for apples and 81 (modulo 9) = 0 (it is 9 squared)



So clearly the answer is 2 water melons, 9 apples and 89 cherries remain having swopped 11 pf them for larger items of fruit.
Reply:89 (the number of cheapest fruit must end in a 9)



259-89=170



2x40=80



170-80=90



89 fruit at 1 schilling

9 fruit at 10 schilling

2 fruit at 40 schilling



gives you 100 fruit for 259 schilling.
Reply:40x+10y+z=259

x+y+z=100 -

There is no method to find a sole for two equations whit 3 variables.

The solution is

x=2

y=9

z=89
Reply:Let 40 sch fruit s no. is x, 10 sch. is y and 1 sch. is y. Thus ,

X+Y+Z=100

40X+10Y+Z=259

NOW SINCE WE VE ONLY 2 EQNS BUT 3 VARIABLES SO WE SOLVE IT BY HIT AND TRIAL. ACC. TO ME Z=89 THAT IS THE CHEAPEST FRUIT.AND X=2 AND Y=9.

THUS, 2+9+89=100

AND 2X40+9X10+89=259
Reply:minimum of 9
Reply:sushantdhamija

is right



Let 40 sch fruit s no. is x, 10 sch. is y and 1 sch. is y. Thus ,

X+Y+Z=100

40X+10Y+Z=259

NOW SINCE WE VE ONLY 2 EQNS BUT 3 VARIABLES SO WE SOLVE IT BY HIT AND TRIAL. ACC. TO ME Z=89 THAT IS THE CHEAPEST FRUIT.AND X=2 AND Y=9.

THUS, 2+9+89=100

AND 2X40+9X10+89=259