Question:

Shruti travels a distance of 224 km in four parts for a total travel time of 3 hours. Her speeds in these four parts follow an arithmetic progression, and the corresponding time taken to cover these four parts follow another arithmetic progression. If she travels at a speed of 960 meters per minute for 30 minutes to cover the first part, then the distance, in meters, she travels in the fourth part is:

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When both speeds and times form arithmetic progressions, express each term in terms of the first term and common difference, then use: \[ \text{Total distance} = \sum (\text{speed} \times \text{time}) \] to solve for the unknown common difference.
Updated On: Dec 5, 2025
  • \(72000\)
  • \(80000\)
  • \(86400\)
  • \(90000\)
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The Correct Option is C

Solution and Explanation

Step 1: Convert units and find first-part speed. First part: - Speed: \(960\) m/min - Time: \(30\) min Convert speed to km/h: \[ v_1 = 960 \,\text{m/min} = 0.96 \,\text{km/min} = 0.96 \times 60 = 57.6 \,\text{km/h}. \] Time of first part: \[ t_1 = 30 \text{ min} = 0.5 \text{ h}. \] Total distance: \[ D = 224 \text{ km},\quad \text{Total time } T = 3 \text{ h}. \]
Step 2: Times form an AP. Let the four times be \(t_1, t_2, t_3, t_4\) in AP with common difference \(d_t\). \[ t_1 = 0.5,\quad t_1 + t_2 + t_3 + t_4 = 3. \] Sum of 4-term AP: \[ \frac{4}{2}\bigl(2t_1 + 3d_t\bigr) = 3 \Rightarrow 2(1 + 3d_t) = 3 \Rightarrow 1 + 3d_t = 1.5 \Rightarrow 3d_t = 0.5 \Rightarrow d_t = \frac{1}{6}. \] Thus: \[ t_2 = t_1 + d_t = \frac{3}{6} + \frac{1}{6} = \frac{4}{6}, \] \[ t_3 = t_1 + 2d_t = \frac{5}{6}, \] \[ t_4 = t_1 + 3d_t = \frac{6}{6} = 1 \text{ h}. \]
Step 3: Speeds form an AP. Let speeds be \(v_1, v_2, v_3, v_4\) in AP with common difference \(d_v\): \[ v_1 = 57.6,\quad v_2 = 57.6 + d_v,\quad v_3 = 57.6 + 2d_v,\quad v_4 = 57.6 + 3d_v. \] Total distance: \[ D = v_1 t_1 + v_2 t_2 + v_3 t_3 + v_4 t_4 = 224. \] Substitute: \[ 57.6\left(\frac{3}{6}\right) + (57.6 + d_v)\left(\frac{4}{6}\right) + (57.6 + 2d_v)\left(\frac{5}{6}\right) + (57.6 + 3d_v)(1) = 224. \] Multiply both sides by 6: \[ 57.6(3) + 4(57.6 + d_v) + 5(57.6 + 2d_v) + 6(57.6 + 3d_v) = 1344. \] Separate constants and \(d_v\): Constant part: \[ 57.6(3 + 4 + 5 + 6) = 57.6 \times 18 = 1036.8. \] Coefficient of \(d_v\): \[ 4 + 10 + 18 = 32. \] So: \[ 1036.8 + 32d_v = 1344 \Rightarrow 32d_v = 1344 - 1036.8 = 307.2 \Rightarrow d_v = \frac{307.2}{32} = 9.6 \,\text{km/h}. \]
Step 4: Speed and distance in the fourth part. Fourth-part speed: \[ v_4 = 57.6 + 3 \times 9.6 = 57.6 + 28.8 = 86.4 \,\text{km/h}. \] Fourth-part time: \[ t_4 = 1 \text{ h}. \] Distance in fourth part: \[ d_4 = v_4 \cdot t_4 = 86.4 \text{ km}. \] Convert to meters: \[ 86.4 \times 1000 = 86400 \text{ m}. \] Thus, the distance she travels in the fourth part is \[ \boxed{86400}. \]
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