Posted by Jack on Wednesday, September 9, 2009 at 10:53pm.
We use the assumption method that we need at least 1 of each animal..
Cow @ $50
Sheep @ $3
Pig @ $2
Chicks @ 10cents
We can easily observe that only 1 cow can be purchased to enable purchase of other animals.
Now we are left with 99 more animals to buy in $50
Easily observable that we need to buy Chicks in multiples of 10
So
10 20 30 40 50 60 70 80 90
Let us draw a chart now
10 chicks $1 $49 (1sheep-15 sheep)(2pigs-23pigs) even if we count max animals in this assumption
(10chicks + 1sheep + 23pigs =34 but we need 99..)
So we have to increase the number of chicks..atleast by 6-7 times more
Lets assume 80 chicks..
80 chicks will cost us $8
We will remain with ($100-$50-$8=42)
Total animals 80chicks+1cow = 81
Remaining amount $42 and animals 19
Lets try some combinations
42-6($3*2) =36
36/$2= 18
Hence 2 sheep and 18 pigs is wrong since it equals 20 and we need only 19..
Guess we are pretty close
Lets try
42-12($3*4)= $30
30/$2= 15
BANG ON.. 4 SHEEP AND 15 PIGS = 19
HENCE ANSWER IS
1 COW $50
4 SHEEP $12
15 PIGS $30
80 CHICKS $8