In a Farm there are 20 animals .buffalo gives 4 liters of milk. cow gives half liter of milk.goat quater liters of milk. per day they give 20 liters of milk. how many animals are there? buffalos? cows? goats?
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Answer:
Assume number of respective animals are x,y,z.
Assume number of respective animals are x,y,z.x+y+z=20---(1) as the total number of animal has to be 20
Assume number of respective animals are x,y,z.x+y+z=20---(1) as the total number of animal has to be 20Amt of milk will be 4x+(1/2)y+(1/4)z=20---(2)
Assume number of respective animals are x,y,z.x+y+z=20---(1) as the total number of animal has to be 20Amt of milk will be 4x+(1/2)y+(1/4)z=20---(2)Solving equation 1 and 2 we get
Assume number of respective animals are x,y,z.x+y+z=20---(1) as the total number of animal has to be 20Amt of milk will be 4x+(1/2)y+(1/4)z=20---(2)Solving equation 1 and 2 we get15x+y=60 ----(3)
Assume number of respective animals are x,y,z.x+y+z=20---(1) as the total number of animal has to be 20Amt of milk will be 4x+(1/2)y+(1/4)z=20---(2)Solving equation 1 and 2 we get15x+y=60 ----(3)Since buffalo gives 4litre and total milk is 20, x < 5
Assume number of respective animals are x,y,z.x+y+z=20---(1) as the total number of animal has to be 20Amt of milk will be 4x+(1/2)y+(1/4)z=20---(2)Solving equation 1 and 2 we get15x+y=60 ----(3)Since buffalo gives 4litre and total milk is 20, x < 5But from eq 3, x cannot be more than 4;
Assume number of respective animals are x,y,z.x+y+z=20---(1) as the total number of animal has to be 20Amt of milk will be 4x+(1/2)y+(1/4)z=20---(2)Solving equation 1 and 2 we get15x+y=60 ----(3)Since buffalo gives 4litre and total milk is 20, x < 5But from eq 3, x cannot be more than 4;Further if x=1 or 2; y>20... Not possible, since total animal is 20
Assume number of respective animals are x,y,z.x+y+z=20---(1) as the total number of animal has to be 20Amt of milk will be 4x+(1/2)y+(1/4)z=20---(2)Solving equation 1 and 2 we get15x+y=60 ----(3)Since buffalo gives 4litre and total milk is 20, x < 5But from eq 3, x cannot be more than 4;Further if x=1 or 2; y>20... Not possible, since total animal is 20Thus, x=3, y=15, z=2
Assume number of respective animals are x,y,z.x+y+z=20---(1) as the total number of animal has to be 20Amt of milk will be 4x+(1/2)y+(1/4)z=20---(2)Solving equation 1 and 2 we get15x+y=60 ----(3)Since buffalo gives 4litre and total milk is 20, x < 5But from eq 3, x cannot be more than 4;Further if x=1 or 2; y>20... Not possible, since total animal is 20Thus, x=3, y=15, z=2Ans: 3 buffalos, 15cows, 2goats.
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