Question

A farmer has some buffalos and peacocks. One day he observed that his animals witch are normal, have 78 eyes and 110 legs. How many animals of each type does he have ?

Answer 1

Answer: there were 16 buffalos and 23 peacocks.

Explanation:

Let x represent the number of Buffaloes.

Let y represent the number of peacocks.

A Buffalo has 2 eyes while a peacock has 2 eyes. If the Buffalo and the peacock has 78 eyes in total, then

2x + 2y = 78

Dividing both sides of the equation by 2, it becomes

x + y = 39

x = 39 - y

A Buffalo has 4 legs while a peacock has 2 legs. If the Buffalo and the peacock has 110 in total, then

4x + 2y = 110- - - - - - - - - - - - 1

Substituting x = 39 - y into equation 1, it becomes

4(39 - y) + 2y = 110

156 - 4y + 2y = 110

- 4y + 2y = 110 - 156

- 2y = - 46

y = - 46/- 2

y = 23

x = 39 - y = 39 - 23

x = 16

Answer 2

23 and 16

Explanation:

Answers

Let x be the number of peacocks and y be the number of buffalows.

Each Peacock has 2 eyes and 2 legs.

Each buffalow has 2 eyes and 4 legs.

Total no of eyes = 78

So 2x + 2y = 78......(1)

Total no of legs = 110

So 2x + 4y = 110......(2)

Substracting (1) and (2) we get,

2x + 4y = 110.

-2x + 2y = 78

2y = 32

y = 32/2

y = 16

Put y in equation 1

2x + 2y = 78

2x + 2*16 =78

2x + 32 = 78

2x = 78 - 32

2x = 46

x = 46/2

x = 23

x = 23 and y = 16