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:

• A. \(72000\)
• B. \(80000\)
• C. \(86400\)
• D. \(90000\)

Hint:

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.

Answer 1

The Correct Option is C

Step 1: Convert units and determine the initial speed. During the first part of the journey, the speed is \(960\) m/min. Converting this into km/h, \[ 960 \text{ m/min} = 0.96 \text{ km/min} = 0.96 \times 60 = 57.6 \text{ km/h}. \] The time taken for the first part is \[ 30 \text{ min} = 0.5 \text{ h}. \] The total distance covered is \(224\) km and the total time taken is \(3\) h. Step 2: Determine the time intervals using AP. Let the times taken in the four parts be \(t_1, t_2, t_3, t_4\), which are in arithmetic progression with common difference \(d_t\). Given, \[ t_1 = 0.5 \text{ h}, \quad t_1 + t_2 + t_3 + t_4 = 3. \] Using the sum formula for a 4-term AP, \[ \frac{4}{2}(2t_1 + 3d_t) = 3. \] Substituting \(t_1 = 0.5\), \[ 2(1 + 3d_t) = 3 \Rightarrow 1 + 3d_t = 1.5 \Rightarrow d_t = \frac{1}{6}. \] Thus, the time intervals are \[ t_2 = \frac{4}{6}, \quad t_3 = \frac{5}{6}, \quad t_4 = 1 \text{ h}. \] Step 3: Find the speeds using AP. Let the corresponding speeds be \(v_1, v_2, v_3, v_4\) in arithmetic progression with common difference \(d_v\). \[ v_1 = 57.6, \quad v_2 = 57.6 + d_v, \quad v_3 = 57.6 + 2d_v, \quad v_4 = 57.6 + 3d_v. \] The total distance is given by \[ v_1 t_1 + v_2 t_2 + v_3 t_3 + v_4 t_4 = 224. \] Substituting the values, \[ 57.6\left(\frac{3}{6}\right) + (57.6 + d_v)\left(\frac{4}{6}\right) + (57.6 + 2d_v)\left(\frac{5}{6}\right) + (57.6 + 3d_v)(1) = 224. \] Multiplying throughout by 6, \[ 57.6(3) + 4(57.6 + d_v) + 5(57.6 + 2d_v) + 6(57.6 + 3d_v) = 1344. \] Separating constants and variables, \[ 57.6(18) + (4 + 10 + 18)d_v = 1344, \] \[ 1036.8 + 32d_v = 1344. \] Solving, \[ 32d_v = 307.2 \Rightarrow d_v = 9.6 \text{ km/h}. \] Step 4: Calculate the distance in the fourth part. The speed during the fourth part is \[ v_4 = 57.6 + 3(9.6) = 86.4 \text{ km/h}. \] Since the time taken in the fourth part is \(1\) h, the distance covered is \[ d_4 = 86.4 \times 1 = 86.4 \text{ km}. \] Converting into meters, \[ 86.4 \times 1000 = 86400 \text{ m}. \] Hence, the distance travelled in the fourth part is \(86400\) m.

Answer 2

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 \[ \boxed{86400}. \]