Question

A man has few cows and hens and the heads are 48 there feet are 140 find the hens and cow ​

Answer 1

Let the number of hens be x and the number of cows be y.

Then, x + y = 48 .... (i)

and 2x + 4y = 140 ⇒ x + 2y = 70 .... (ii)

Solving (i) and (ii) we get: x = 26, y = 22.

∴ The required answer = 26.

Answer 2

$\bold{\underline{Question:}}$

A man has few cows and hens. The no. of heads and feet is 48 and 140 respectively. Find the no. cows and hens. ​

$\underline{\bold{Solution:}}$

$\bold{\textsf{Let the no. of cows and hens be x and y respectively.}}$

From above question ,

Total no. of feet = 140.

$\bold{\boxed{\textsf{Note : Cows have 4 feet and hens have 2 feet.}}}$

• No. of feet of cows = $\bold{4*x=4x.}$
• No. of feet of hens = $\bold{2*y=2y.}$

So

$\bold{4x+2y=140.......(i)}$

Now ,

Total no. of heads = 48.

$\bold{\boxed{\textsf{Note : Both have 1 head.}}}$

So

$\bold{x+y=48........(ii)}$

$\bold{\underline{\textsf{Multiplying by 2 both side we get ,}}}$

$\implies{\bold{2x+2y=96......(iii)}}$

$\underline{\bold{\textsf{Now subtract equ.(iii)from equ.(i) we get,}}}$

$\implies{\bold{4x+2y-(2x+2y)=140-96}}$

$\implies{\bold{2x = 44}}$

$\implies{\bold{x=\frac{44}{2} = 22. }}$

$\bold{\underline{\textsf{Putting the value of x in equ. (ii)}}}$

$\implies{\bold{22+y=48}}$

$\implies{\bold{y=48-22=26.}}$

$\bold{\boxed{\textsf{No. of cows = 22 and hens = 26.}}}$