Monday

January 16, 2017
Posted by **Jack** on Wednesday, September 9, 2009 at 10:53pm.

- Math -
**radhika**, Monday, January 9, 2017 at 10:46pmWe 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