You have R200 to buy a cow costing R10.00 sheep costing R2.00 a chicken costing R0.10,no change must be leftover from R200 total number of animals is to be 100

Question

You have R200 to buy a cow costing R10.00 sheep costing R2.00 a chicken costing R0.10,no change must be leftover from R200 total number of animals is to be 100

Answer 1

To solve this problem, we can use a technique called "trial and error." We will start by assuming an initial number of cows, sheep, and chickens, and then check if the cost and the total number of animals are satisfied. If they are not, we will adjust our assumptions and try again until we find a solution that meets the requirements.

Let's assume we have a certain number of cows, sheep, and chickens, and their respective quantities are represented by the variables C, S, and H.

Since the total number of animals should be 100, we can use the equation: C + S + H = 100.

Now, we need to consider the cost of these animals. The cost of cows is R10.00 each, so the total cost of cows will be 10C. Similarly, the total cost of sheep will be 2S, and the total cost of chickens will be 0.10H.

According to the problem, the total cost of all the animals should be R200. Therefore, we can set up the equation: 10C + 2S + 0.10H = 200.

Now, we can start with a reasonable assumption and try to solve for C, S, and H using trial and error.

Let's assume we have 10 cows (C = 10). Substituting this assumption into the first equation, we have: 10 + S + H = 100, which simplifies to S + H = 90.

Now, let's substitute these values into the second equation: 10(10) + 2S + 0.10H = 200, which simplifies to 100 + 2S + 0.10H = 200, or 2S + 0.10H = 100.

We can solve this equation for either S or H. Let's solve it for S by isolating S: 2S = 100 - 0.10H, or S = (100 - 0.10H)/2.

Now, we can substitute this expression for S into the first equation: (100 - 0.10H)/2 + H = 90. We can simplify this equation further and solve for H.

Multiply the equation by 2 to get rid of the fraction: 100 - 0.10H + 2H = 180.

Combining like terms, we have: 1.90H = 80.

Divide both sides of the equation by 1.90: H = 80/1.90 ≈ 42.11.

However, since the number of animals must be a whole number, we need to adjust our assumption for H. Let's assume H = 42.

Now, we can substitute this value for H into the first equation: 10 + S + 42 = 100, or S = 100 - 52 = 48.

Since C = 10, S = 48, and H = 42 satisfy both of our equations, we have found a valid solution.

Thus, you can buy 10 cows, 48 sheep, and 42 chickens, costing a total of R200, with no change remaining, and have a total of 100 animals.