In a mixture of three varieties of oil, the ratio of their weights is 4 : 5 : 8. If 5 kg of the first variety, 10 kg of the second variety, and some quantity of the third variety are added, the ratio of the three varieties becomes 5 : 7 : 9 in the final mixture. What quantity of the third variety of oil was added in kilograms?

Introduction / Context:
This question is based on ratios and mixtures involving three components. We are given an initial ratio of three varieties of oil and then information about additional quantities added to each variety which results in a new ratio. Such problems are common in ratio and proportion sections and test your ability to convert ratios into algebraic expressions for actual quantities.

Given Data / Assumptions:

Initially, the ratio of the three varieties of oil is 4 : 5 : 8 by weight.
Let the initial amounts be 4k, 5k and 8k kilograms respectively.
Additional quantities added: 5 kg of the first variety, 10 kg of the second, and x kg of the third variety.
After addition, the new ratio of the three varieties is 5 : 7 : 9.
We must find x, the quantity of the third variety added, in kilograms.

Concept / Approach:
We convert the ratios into algebraic expressions using a common multiplier k. Then we express the new quantities after addition in terms of k and x. Since we are given that the final ratio is 5 : 7 : 9, the ratios of corresponding new quantities must be equal. This yields equations that we can solve to find k and x. Typically we equate two ratio pairs at a time to obtain a solvable system of equations.

Step-by-Step Solution:
Step 1: Represent initial amounts using a common multiplier. Initial amounts: first = 4k, second = 5k, third = 8k kilograms. Step 2: Write expressions for new amounts after addition. New amount of first variety = 4k + 5. New amount of second variety = 5k + 10. New amount of third variety = 8k + x. Step 3: Use the new ratio 5 : 7 : 9. We have (4k + 5) : (5k + 10) : (8k + x) = 5 : 7 : 9. First equate the ratio of first two types. (4k + 5) / (5k + 10) = 5 / 7. Step 4: Solve this equation for k. Cross multiply: 7 * (4k + 5) = 5 * (5k + 10). This gives 28k + 35 = 25k + 50. Rearrange: 28k - 25k = 50 - 35, so 3k = 15 and k = 5. Step 5: Use k to find x from the third ratio. Now the new amounts become: first = 4 * 5 + 5 = 25 kg, second = 5 * 5 + 10 = 35 kg, third = 8 * 5 + x = 40 + x. The final ratio is 5 : 7 : 9, so third : second = 9 : 7. Therefore, (40 + x) / 35 = 9 / 7. Cross multiply: 7 * (40 + x) = 9 * 35. This gives 280 + 7x = 315. So, 7x = 315 - 280 = 35, hence x = 35 / 7 = 5 kg.

Verification / Alternative Check:
With k = 5 and x = 5, the final amounts are 25, 35 and 45 kg. The ratio 25 : 35 : 45 simplifies by dividing each term by 5, giving 5 : 7 : 9, which matches the given final ratio exactly. This confirms that 5 kg of the third variety must have been added.

Why Other Options Are Wrong:
Option 15 kg would make the final third amount 55 kg, giving the ratio 25 : 35 : 55, which does not simplify to 5 : 7 : 9.
Option 25 kg would yield 65 kg for the third variety, again not in the correct ratio with the first two types.
Option 35 kg would give 75 kg as the third amount, resulting in 25 : 35 : 75, which cannot be simplified to 5 : 7 : 9.

Common Pitfalls:
Students sometimes forget to use a common multiplier k or mix up the order of the terms in the ratio. Another frequent error is to incorrectly set up the fraction equations or make algebraic mistakes while cross multiplying. It is also important to check the final ratio with the found values to ensure that no arithmetic slip occurred during calculations.

Final Answer:
The quantity of the third variety of oil added is 5 kg.