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Question no.9:
Farmer Brown had 25 cows and chickens. They had a total of 72 legs. How many cows and how many chickens did Farmer Brown have?
Solution:
Let the number of cows be "x" and the number of chickens be "y".
According to question,
x + y = 25 4x + 2y = 72 -->(ii)
=> 4(x + y) = 25 × 4
=> 4x + 4y = 100 -->(i)
By Elimination method,
Equation (i) - (ii)
(4x + 4y) - (4x + 2y) = 100 - 72
=> 4x + 4y - 4x - 2y = 28
=> 2y = 28
=> y = 28/2
=> y = 14
Putting the value of "y" in Equation (ii)
4x + 2y = 72
=> 4x + 2 × 14 = 72
=> 4x + 28 = 72
=> 4x = 72 - 28
=> 4x = 44
=> x = 44/4
=> x = 11
Therefore,
Total number of cows = 11
Total number of chickens = 14
Note: In the second equation of this question I've taken 4 and 2.
The reason behind it is simple. Since, I've assumed the number of cows is "x" and a cow has four legs that why infront of "x" I've put 4.
And a chicken has 2 legs and I've assumed it as "y" therefore infront of "y" I've put 2.
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Question no.10:
Farmer Brown had 28 cows and chickens. They had a total of 96 legs. How many cows and how many chickens did Farmer Brown have?
Solution:
Let the number of cows be "x" and the number of chickens be "y".
According to question,
x + y = 28 4x + 2y = 96 -->(ii)
=> 4(x + y) = 28 × 4
=> 4x + 4y = 112 -->(i)
By Elimination method,
Equation (i) - (ii)
(4x + 4y) - (4x + 2y) = 112 - 96
=> 4x + 4y - 4x - 2y = 16
=> 2y = 16
=> y = 16/2
=> y = 8
Putting the value of "y" in Equation (ii)
4x + 2y = 96
=> 4x + 2 × 8 = 96
=> 4x + 16 = 96
=> 4x = 96 - 16
=> 4x = 80
=> x = 80/4
=> x = 20
Therefore,
Total number of cows = 20
Total number of chickens = 8
Note: In the second equation of this question I've taken 4 and 2.
The reason behind it is simple. Since, I've assumed the number of cows is "x" and a cow has four legs that why infront of "x" I've put 4.
And a chicken has 2 legs and I've assumed it as "y" therefore infront of "y" I've put 2.