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shtirl [24]
3 years ago
11

2. In a barnyard, there is an assortment of chickens and cows. Counting heads, one gets 80; counting legs, one gets 184. How man

y of each are there? Note: To receive credit for this problem you need to define the variables, write a system of equations and solve the system to answer the question.
Mathematics
1 answer:
Anvisha [2.4K]3 years ago
4 0

Answer:

68 chickens and 12 cows.

Step-by-step explanation:

Let x represents the number of chicken and y represents the number of cows in the barnyard,

Given,

Total heads = 80

⇒ x + y = 80 ------(1),

Also, total legs = 184,

Since, a chicken has two legs and cow has 4 legs,

⇒ 2x + 4y = 184 -----(2),

Equation (2) - 2 × equation (1),

We get,

4y - 2y = 184 - 160

2y = 24

y = 12

From equation (1),

x + 12 = 80 ⇒ x = 80 - 12 = 68

Hence, the number of chicken = 68,

And, the number of cows = 12

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