Question

hi ,friends
how r u ?
I think u all r fine ;
plz solve this question
in a animal husbandry there r 3 types of animals. cow, buffalo, goat. cow gives 2 liter, buffalo gives 5 litre ,goat gives 250 ml milk. no. of animals r 20 and gives milk 40 litres. then no. of each animals? explain also. ​

Answer 1

Answer:

Let no. of buffalos, cows and goats = b,c and g respectively.

• •

• the total number of animals is 20

b+c+g= 20 —-> (i)

)The amount of milk given by 1 buffalo is 5 litres. therefore, amount of milk given by b buffalos = 5b litres...

Similarly,

Similarly,amount of milk given by c cows = 2c litres

2c litresamount of milk given by g goats = 250/1000= 0.25g litres

g litresTotal amount of milk = 40 litres.

So,

5b + 2c + 0.25g = 40 —-> (ii)

)Multiplying (ii) by 4 we get,

) by 4 we get,20b + 8c + g = 160 —->(iii)

)Subtracting (i) from (iii) we get,

) we get,19b + 7c = 140 —-> (iv)

)Now, b,c and g is non negative integers as they represent the number of animals.

negative integers as they represent the number of animals.As we have been provided with only 2 equations having 3 unknowns, we have to do calculated guesswork at this stage, since we cannot use any other equation which could be solved with (iv) simultaneously.

) simultaneously.(iv) can also be written as

) can also be written as(140–19b)/7 = c

0–19b)/7 = c=> (140/7) - (19b/7) = c

0/7) - (19b/7) = c=> 20 - (19b/7) = c —->(v)

)For c be a non negative integer, there are only 2 possible values of b which are 0 and 7.

)For c be a non negative integer, there are only 2 possible values of b which are 0 and 7.This is because c can be an integer when (19b/7) is an integer. Thus b can only be a multiple of 7. The only values of b that satisfy this constraint and the equation (v) are 0 and 7.

Thus,b=0 or b=7..

Bro, put these values into eq to find answers..