Final answer:
To obtain 400 kgs of steel with 20% purity, one must take 150 kgs from the 17% steel rod and 250 kgs from the 25% steel rod, by solving the system of linear equations derived from the conditions given.
Step-by-step explanation:
The question you're asking relates to a mixture problem in which two different purities of steel must be combined to achieve a new mixture with a specific purity. We are looking for a method to obtain 400 kgs of steel with 20% purity, by mixing steel from a rod with 17% purity and another rod with 25% purity.
To solve this, we'll let x be the amount of the 17% steel rod and y be the amount of the 25% steel rod. We want the total weight to be 400 kgs, so x + y = 400. We also want the total amount of pure steel (which is the weight multiplied by the purity percentage) to be 20% of 400 kgs, so 0.17x + 0.25y = 0.20 × 400.
Now we solve the system of equations:
0.17x + 0.25y = 80 (1)
x + y = 400 (2)
If we multiply equation (2) by 0.17, we get 0.17x + 0.17y = 68 (3). Subtracting equation (3) from equation (1), we have 0.25y - 0.17y = 80 - 68, which gives us 0.08y = 12. Solving for y gives us y = 150 kgs. Substituting y = 150 into equation (2) gives us x = 400 - 150, which is x = 250 kgs.
Therefore, the answer is c) Take 150 kgs from the 17% rod and 250 kgs from the 25% rod.