Question

A man has some hens and cows if no. of heads be 48 and the no. of feet equals 140 then no. of hens will be mca

Answer 1

let the hens are x and cows are y.
we know that both have one head,and cow has 4 legs but hen has two legs only.
x
$x + y = 48 \\ 2x + 4y = 140 \\ multiply \: eq1 \: by \: 2 \: and \: subtract \: from \: 2 \\ 2x + 2y = 96 \\ 2x + 4y = 140 \\ - \: \: \: - \: \: \: \: \: \: \: \: - \\ - 2y = - 44 \\ y = \frac{44}{2 } \\ y = 22 \\ x + y = 48 \\ x + 22 = 48 \\ x = 48 - 22 \\ = 26$

so number of hens he purchased were 26 and number of cows he purchased were 22.