The speed of a car is $X$ km/hr and the speed of a bus is twice the speed of the car. The car starts from point A at 10 am and reaches point B at 5 pm. The bus starts from point A to reach point B and covers $(672 + 2X)$ km in 3 hours. If the speed of the car decreases by 25%, then how much more time will the car take to reach point B compared to its original time? (Note: The car halts two times, each halt lasting 5 minutes.)

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**Solution:**

1. **Identify given data:**
   - Speed of car = $X$ km/hr
   - Speed of bus = $2X$ km/hr
   - Car starts at 10 am and reaches at 5 pm, so total time taken by car = $7$ hours
   - Bus covers $(672 + 2X)$ km in 3 hours

2. **Calculate distance between A and B:**
   - Distance covered by bus in 3 hours = $(672 + 2X)$ km
   - Speed of bus = $2X$ km/hr
   - Distance = Speed × Time = $2X 	imes 3 = 6X$ km

3. **Equate distances:**
   $$6X = 672 + 2X$$

4. **Solve for $X$:**
   $$6X - 2X = 672$$
   $$4X = 672$$
   $$X = \frac{672}{4} = 168 	ext{ km/hr}$$

5. **Calculate distance between A and B:**
   $$	ext{Distance} = 6X = 6 	imes 168 = 1008 	ext{ km}$$

6. **Calculate original speed and time of car:**
   - Speed of car = $X = 168$ km/hr
   - Time taken by car = 7 hours (given)
   - Check distance with car: $168 	imes 7 = 1176$ km (This contradicts distance 1008 km)

Since the car took 7 hours, distance should be $X 	imes 7 = 7X$ km.

But from bus calculation, distance is $6X$ km.

This suggests a discrepancy. Let's re-examine the problem statement carefully.

The problem states:
   - Car starts at 10 am and reaches at 5 pm → 7 hours
   - Bus covers $(672 + 2X)$ km in 3 hours
   - Speed of bus = $2X$

So distance between A and B is the same for both vehicles.

Therefore, distance = $7X$ (car) = $3 	imes 2X = 6X$ (bus)

Equate:
   $$7X = 672 + 2X$$
   $$7X - 2X = 672$$
   $$5X = 672$$
   $$X = \frac{672}{5} = 134.4 	ext{ km/hr}$$

7. **Calculate distance:**
   $$	ext{Distance} = 7X = 7 	imes 134.4 = 940.8 	ext{ km}$$

8. **New speed of car after 25% decrease:**
   $$	ext{New speed} = X - 0.25X = 0.75X = 0.75 	imes 134.4 = 100.8 	ext{ km/hr}$$

9. **Calculate new time taken by car (excluding halts):**
   $$	ext{Time} = \frac{	ext{Distance}}{	ext{Speed}} = \frac{940.8}{100.8} = 9.3333 	ext{ hours} = 9 	ext{ hours } 20 	ext{ minutes}$$

10. **Calculate original time taken by car (excluding halts):**
    - Car halts two times, each for 5 minutes → total halt time = 10 minutes
    - Original total time = 7 hours = 420 minutes
    - Time excluding halts = 420 - 10 = 410 minutes = 6 hours 50 minutes = 6.8333 hours

11. **Calculate additional time taken by car:**
    - New time excluding halts = 9.3333 hours
    - Original time excluding halts = 6.8333 hours
    - Additional time excluding halts = 9.3333 - 6.8333 = 2.5 hours = 2 hours 30 minutes

12. **Add halt time to additional time:**
    - Since halts are fixed at 10 minutes, the additional time due to speed decrease is only the difference in running time.
    - Total additional time including halts = 2 hours 30 minutes + 10 minutes = 2 hours 40 minutes

**Answer:** 2 hours 40 minutes

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