Direction: In the following question, two quantities I and II are given. Compare quantity I and quantity II on its basis. (Only quantity is to be considered) Train A crosses train B in 13 seconds while running in the opposite direction and train C crosses train A in 100 seconds while running in the same direction. Average of speeds of trains A and B is 15 m/s and the speed of train C is 4 m/s more than that of train A. Train A can cross a 140 m long platform in 20 seconds and the speed of train B is 14 m/s. Quantity I: If speed of train C were 95% of its original speed, then in what time train C would cross train B while running in the same direction? Quantity II: If a series is formed, whose 1st term is the length of train A, 2nd term is (1st term - 2²), 3rd term is (2nd term - 3²), 4th term is (3rd term - 5²) and 5th term is (4th term - 7²), then what is the 5th term of the series?

### Concept & Relative Speed Dynamics This problem merges relative speed kinematics with logical sequence generation. Remember that when crossing a platform, $\text{Distance} = \text{Train Length} + \text{Platform Length}$. When trains cross each other, $\text{Distance} = \text{Sum of their Lengths}$. Relative speed is additive in opposite directions and subtractive in the same direction. ### Step-by-Step Solution * **Determine Speeds:** * Average speed of A and B is $15 \text{ m/s} \Rightarrow \frac{v_A + v_B}{2} = 15 \Rightarrow v_A + v_B = 30 \text{ m/s}$. * Given $v_B = 14 \text{ m/s}$, so $v_A = 30 - 14 = 16 \text{ m/s}$. * Speed of Train C is $4 \text{ m/s}$ more than Train A $\Rightarrow v_C = 16 + 4 = 20 \text{ m/s}$. * **Determine Lengths:** * Train A crosses a $140 \text{ m}$ platform in $20 \text{ s}$: $\frac{L_A + 140}{16} = 20 \Rightarrow L_A + 140 = 320 \Rightarrow L_A = 180 \text{ m}$. * Train A crosses Train B in $13 \text{ s}$ (opposite direction): $\frac{L_A + L_B}{v_A + v_B} = 13 \Rightarrow \frac{180 + L_B}{30} = 13 \Rightarrow 180 + L_B = 390 \Rightarrow L_B = 210 \text{ m}$. * Train C crosses Train A in $100 \text{ s}$ (same direction): $\frac{L_C + L_A}{v_C - v_A} = 100 \Rightarrow \frac{L_C + 180}{20 - 16} = 100 \Rightarrow L_C + 180 = 400 \Rightarrow L_C = 220 \text{ m}$. * **Calculate Quantity I:** * New speed of Train C $= 95\%$ of $20 = 19 \text{ m/s}$. * Time for C to cross B in same direction: $\text{Time} = \frac{L_C + L_B}{v_C - v_B} = \frac{220 + 210}{19 - 14} = \frac{430}{5} = 86 \text{ seconds}$. * **Calculate Quantity II:** * $1^{\text{st}} \text{ term} = L_A = 180$. * $2^{\text{nd}} \text{ term} = 180 - 2^2 = 176$. * $3^{\text{rd}} \text{ term} = 176 - 3^2 = 176 - 9 = 167$. * $4^{\text{th}} \text{ term} = 167 - 5^2 = 167 - 25 = 142$. (Subtracting squares of consecutive prime numbers). * $5^{\text{th}} \text{ term} = 142 - 7^2 = 142 - 49 = 93$. * **Compare:** Quantity I ($86$) < Quantity II ($93$). ### Exam Strategy & Shortcut Organize your scratchpad systematically: first dedicate a section to speeds ($v_A, v_B, v_C$), then to lengths ($L_A, L_B, L_C$). This prevents confusion when multiple interacting variables exist. For series problems embedded in algebra, don't compute until you are sure of the subtraction logic (here, consecutive prime squares). ### Common Pitfall A major pitfall is missing the fact that the series subtracts the squares of **prime numbers** ($2, 3, 5, 7$), not odd numbers. If a student subtracts $1^2, 3^2, 5^2$, they will derive an incorrect sequence. ### Final Answer Therefore, the correct answer is **Quantity I < Quantity II**.