Answer :
40 cattle can feed for 4 days
if there were 10 more cattle, number of cattle = 40+10 = 50
let number of days they can feed = x
(note: since number of days the cattle can feed is inversely proportional to the number of cattle,
no. of cattle × no. of days = constant
⇒ 40×4 = 50×x
⇒ 50x = 160
⇒ x = 160/50 = 3.2 days
His food would last for 3.2 days if he had 10 more cattle.
if there were 10 more cattle, number of cattle = 40+10 = 50
let number of days they can feed = x
(note: since number of days the cattle can feed is inversely proportional to the number of cattle,
no. of cattle × no. of days = constant
⇒ 40×4 = 50×x
⇒ 50x = 160
⇒ x = 160/50 = 3.2 days
His food would last for 3.2 days if he had 10 more cattle.
Answer:
3.2 days
Step-By-Step Explanation:
If the number of animals increases, then it will take fewer days to last.
Then the two quantities are in inverse proportions.
Let the required number of days be p.
[tex] \because x_1y_1 = x_2y_2 [/tex]
Where,
[tex] x_1 = 40, \: y_1 = 4, \: x_2 = 50 [/tex] and [tex] y_2 = p [/tex] (let)
=> 40 × 4 = 50 × p
=> 160 = 50p
=> p = 160/50
=> p = 3.2
Hence the required number of days = 3.2