A FARMER SPENDS $1000 AND BUYS 100 ANIMALS. THE PIGS COSTS $100 EACH, THE GOATS $30 AND THE CHICKS $5. THE FARMER BUYS AT LEAST ONE OF EACH ANIMAL, WHAT DOES HE BUY?
100 p + 30 g + 5 c = 1,000
p+g+c = 100 so c = 100 - p - g
100 p +30 g +5(100-p-g) = 1,000
95 p +25 g = 500
we know p+g < 100 or we have <1 c
p and g must be integers, no half goats allowed
well 95 times what is a multiple of 25
95 = 5*19
so 95*5 = 19*5*5 =475:)
try 5 pigs
95 p + 25 g = 500
using p = 5
475 + 25 g = 500
g = 1
so
c = 100 - p - g
c = 100 -5 -1 = 94
so
5 pigs, 1 goat, 94 chickens
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check
5*100 + 1*30 + 96*5 = 1,000 whew!
p+g+c = 100
100p+30g+5c = 1000
A little checking will show that he bought
94 chickens
1 goat
5 pigs
Let's assume the farmer buys x pigs, y goats, and z chicks. Based on the given information, we can create the following equations:
x + y + z = 100 (Equation 1)
100x + 30y + 5z = 1000 (Equation 2)
Let's solve these equations step-by-step to find the values of x, y, and z.
Step 1: Solve Equation 1 for x in terms of y and z:
x = 100 - y - z
Step 2: Substitute the value of x from Step 1 into Equation 2:
100(100 - y - z) + 30y + 5z = 1000
10,000 - 100y - 100z + 30y + 5z = 1000
-70y - 95z = -9,000 (divide through by -5)
Step 3: Simplify Equation 3:
14y + 19z = 1,800 (Equation 3)
Now we have two equations:
14y + 19z = 1,800 (Equation 3)
y + z = 100 (Equation 1)
Step 4: Solve Equation 5 for y in terms of z:
y = 100 - z
Step 5: Substitute the value of y from Step 4 into Equation 3:
14(100 - z) + 19z = 1,800
1,400 - 14z + 19z = 1,800
5z = 400
z = 80
Step 6: Substitute the value of z from Step 5 into Equation 4 to find y:
y + 80 = 100
y = 20
Step 7: Substitute the values of y and z from Step 6 into Equation 1 to find x:
x + 20 + 80 = 100
x = 0
Based on these calculations, the farmer buys 0 pigs, 20 goats, and 80 chicks.
To solve this problem, we can start by figuring out the number of each animal the farmer buys. Let's assume the farmer buys x pigs, y goats, and z chicks.
According to the information provided, the pigs cost $100 each, so the cost of x pigs is 100x dollars.
Similarly, the cost of y goats would be 30y dollars, and the cost of z chicks would be 5z dollars.
Since the farmer spent a total of $1000, we can write the equation:
100x + 30y + 5z = 1000
Now, we need to find the values of x, y, and z that satisfy this equation while also meeting the given conditions. Let's analyze the conditions in the problem:
1. The farmer buys at least one of each animal. This means that x, y, and z should be greater than or equal to 1.
2. The total number of animals bought is 100. Therefore, x + y + z = 100.
Now we can solve these equations to find the values of x, y, and z.
One approach to solving this system of equations is by trial and error or guess-and-check. Start with a reasonable value for one variable and then work on the other variables to satisfy both equations.
For example, assuming x = 1, we can solve the remaining equations:
100(1) + 30y + 5z = 1000 (substituting x=1 into the first equation)
y + z = 99 (we subtract 1 from both sides of x + y + z = 100)
From here, we can choose values for y and z that satisfy the equation y + z = 99. Since the farmer must buy at least one of each animal, it's reasonable to assume that y and z should be greater than 0.
Some possible values could be:
- y = 98, z = 1
- y = 1, z = 98
- y = 49, z = 50
- etc.
By substituting these possible values back into the equation 100(1) + 30y + 5z = 1000, we can see if the equation holds true. If it does, then we have found a valid combination of animals the farmer can buy.
Repeat this process by assuming different values for x (greater than or equal to 1) and trying different values for y and z that satisfy the remaining equation. Eventually, you will find a combination of values for x, y, and z that satisfy both equations and make the total cost equal to $1000.
To summarize, the farmer can buy different combinations of pigs, goats, and chicks to satisfy the conditions. There are multiple correct answers to this problem, and finding them requires solving a system of equations through trial and error.