Mr. Kumar drives to work at an average speed of 48 km/h. The time taken to cover the first 60% of the distance is 10 minutes more than

Mr. Kumar drives to work at an average speed of 48 km/h. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining 40%. What is the total distance to his workplace

Question:
Mr. Kumar drives to work at an average speed of 48 km/h. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining 40%. What is the total distance to his workplace?

Solution:

1. Let the total distance to work be $D$ kilometers.
   The first part of the journey is 60% of the distance, so:

   $$\text{Distance for the first part}=0.6D$$The second part of the journey is 40% of the distance:

   $$\text{Distance for the second part}=0.4D$$

2. Time taken for each part of the journey:
   Using the formula Time=DistanceSpeed\text{Time} = \frac{\text{Distance}}{\text{Speed}}Time=SpeedDistance​, the time for each part is:
   • Time for the first part:

     T1=0.6D48T_1 = \frac{0.6D}{48}T1​=480.6D​

   • Time for the second part:

     T2=0.4D48T_2 = \frac{0.4D}{48}T2​=480.4D​

3. Given condition:
   The time for the first part is 10 minutes more than the time for the second part. Convert 10 minutes to hours:

   10 minutes=1060=16 hours10 \, \text{minutes} = \frac{10}{60} = \frac{1}{6} \, \text{hours}10minutes=6010​=61​hoursSo:

   T1=T2+16T_1 = T_2 + \frac{1}{6}T1​=T2​+61​

4. Substitute the values of T1T_1T1​ and T2T_2T2​:

   0.6D48=0.4D48+16\frac{0.6D}{48} = \frac{0.4D}{48} + \frac{1}{6}480.6D​=480.4D​+61​

5. Simplify the equation:
   Multiply through by 48 to eliminate the denominator:

   $$0.6D=0.4D+8$$Subtract $0.4D$ from both sides:

   $$0.2D=8$$

6. Solve for $D$:

   D=80.2=40 kmD = \frac{8}{0.2} = 40 \, \text{km}D=0.28​=40km

Answer:
The total distance to Mr. Kumar’s workplace is 40 kilometers.