Direction :"You are presented with two quantities, labeled as Quantity I and Quantity II. Your task is to solve both quantities and determine the correct relationship between Quantity I and Quantity II. After solving, choose the appropriate option that describes the relationship correctly." Quantity-I: The time required for tap 'P' to fill a water tank is 4 hours more than the time taken by tap 'P' and tap 'Q' together to complete the same task. If tap 'P' takes 'p' hours to fill the tank independently, deduce the value of '25p,' given that tap 'Q' can independently fill the tank in 15 hours. Quantity-II: Determine the number of distinct ways to partition a group of 7 individuals into three subgroups. One of these subgroups must consist of 3 members, while the remaining subgroups should each have 2 members."

### Concept & Pipes and Combinatorics For Quantity I, we use the combined work formula for pipes and cisterns. If two pipes take $a$ and $b$ hours individually, together they take $\frac{ab}{a+b}$ hours. For Quantity II, we use the formula for partitioning $n$ distinct objects into groups of specific sizes, adjusting for groups of identical sizes to avoid overcounting. $$ \text{Group Division} = \frac{n!}{a! \cdot b! \cdot c! \cdot k!} $$ (Where $k$ is the factorial of the number of groups having the same size). ### Step-by-Step Solution **Solving Quantity I:** 1. Let time taken by tap P alone be $p$ hours. 2. Time taken by tap Q alone = 15 hours. 3. Time taken by P and Q together = $\frac{15p}{15 + p}$ hours. 4. According to the question, P takes 4 hours more than P and Q together: $p = \frac{15p}{15 + p} + 4$ $p - 4 = \frac{15p}{15 + p}$ $(p - 4)(15 + p) = 15p$ $15p + p^2 - 60 - 4p = 15p$ $p^2 - 4p - 60 = 0$ 5. Factoring the quadratic equation: $p^2 - 10p + 6p - 60 = 0$ $(p - 10)(p + 6) = 0$ Since time cannot be negative, $p = 10$. 6. The value required is $25p = 25 \times 10 = 250$. **Quantity I = 250** **Solving Quantity II:** 1. We need to partition 7 individuals into subgroups of 3, 2, and 2. 2. Using the partition formula for identical group sizes (two groups of size 2): $\text{Number of ways} = \frac{7!}{3! \cdot 2! \cdot 2! \cdot 2!} = \frac{5040}{6 \cdot 2 \cdot 2 \cdot 2}$ $\text{Number of ways} = \frac{5040}{48} = 105$. **Quantity II = 105** **Comparison:** Comparing both quantities, $250 > 105$, which means Quantity I > Quantity II. ### Exam Strategy & Shortcut For Quantity II, remembering standard factorials up to $7!$ (5040) greatly speeds up permutation/combination calculations. In Quantity I, directly testing simple values for $p$ (like 10, since 10 and 15 make 6 hours together, and $10 = 6 + 4$) can bypass solving the full quadratic equation. ### Common Pitfall A common error in Quantity II is forgetting to divide by $2!$ for the two identical subgroups of size 2, leading to an overcounted result of 210, which might alter comparisons in closer problems. ### Final Answer Therefore, the correct answer is **Quantity-I > Quantity-II**.