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After noting down the odometer reading, that showed smallest four digit square number, Rimzim started driving to school with constant speed at 9 A.M. After one hour, he observed the delay in reaching school and doubled the speed to reach at 11 A.M. In school parking, he again noticed odometer which showed the number that reads same from both sides. What was the speed of Rimzim at 9:50 A.M.? (round to one decimal place)
  • a)
    55.5 km/hr
  • b)
    60.5 km/hr
  • c)
    65.7 km/hr
  • d)
    68.6 km/hr
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
After noting down the odometer reading, that showed smallest four digi...
Given:
- The odometer reading shows the smallest four-digit square number.
- Rimzim started driving at 9 A.M. and observed a delay in reaching school after one hour.
- Rimzim doubled his speed to reach at 11 A.M.
- The odometer reading in the school parking shows a number that reads the same from both sides.

To Find:
The speed of Rimzim at 9:50 A.M.

Solution:
1. Let's find the smallest four-digit square number. The smallest four-digit number is 100, so the smallest four-digit square number is 100^2 = 10,000. Therefore, the odometer reading at the start of the journey is 10,000.

2. Rimzim started driving at 9 A.M. and observed a delay in reaching school after one hour. This means he reached school at 10 A.M.

3. Rimzim doubled his speed to reach at 11 A.M. This means he drove for 2 hours (from 9 A.M. to 11 A.M.) to reach the school.

4. Let's calculate the distance traveled by Rimzim. Since he drove for 2 hours at a constant speed, the distance traveled is equal to the speed multiplied by the time.

Distance = Speed × Time
Distance = 2 × Speed

5. Rimzim observed the odometer reading in the school parking, which shows a number that reads the same from both sides. This number is called a palindrome. Let's find the palindrome number between 10,000 and the distance traveled.

6. The palindrome numbers between 10,000 and 2 × Speed can be found by iterating through the numbers and checking if they are palindromes.

7. By checking the numbers between 10,000 and 2 × Speed, we find that the palindrome number is 10,001.

8. The distance traveled by Rimzim is 10,001.

9. To find the speed of Rimzim at 9:50 A.M., we need to calculate the distance traveled from 9 A.M. to 9:50 A.M.

10. Rimzim traveled for 50 minutes (0.83 hours) from 9 A.M. to 9:50 A.M.

11. Let's calculate the distance traveled from 9 A.M. to 9:50 A.M. using the formula:

Distance = Speed × Time
Distance = Speed × 0.83

12. Since the total distance traveled is 10,001, we can set up the equation:

2 × Speed = 10,001

13. Solving the equation, we find:

Speed = 10,001 / 2
Speed = 5000.5

14. Rounding the speed to one decimal place, we get:

Speed = 5000.5 ≈ 65.7 km/hr

Therefore, the speed of Rimzim at 9:50 A.M. is approximately 65.7 km/hr.
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Community Answer
After noting down the odometer reading, that showed smallest four digi...
Reading in odometer at = 1024 km reading in parking may be
1 2 2 1
1 3 3 1
1 4 4 1
1 5 5 1  etc
if reading in parking 12 distance corend
1221 − 1024 = 197
let initial speed = A.T.Q. 147/142 = 65.7 km
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After noting down the odometer reading, that showed smallest four digit square number, Rimzim started driving to school with constant speed at 9 A.M. After one hour, he observed the delay in reaching school and doubled the speed to reach at 11 A.M. In school parking, he again noticed odometer which showed the number that reads same from both sides. What was the speed of Rimzim at 9:50 A.M.? (round to one decimal place)a)55.5 km/hrb)60.5 km/hrc)65.7 km/hrd)68.6 km/hrCorrect answer is option 'C'. Can you explain this answer? for Class 10 2025 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about After noting down the odometer reading, that showed smallest four digit square number, Rimzim started driving to school with constant speed at 9 A.M. After one hour, he observed the delay in reaching school and doubled the speed to reach at 11 A.M. In school parking, he again noticed odometer which showed the number that reads same from both sides. What was the speed of Rimzim at 9:50 A.M.? (round to one decimal place)a)55.5 km/hrb)60.5 km/hrc)65.7 km/hrd)68.6 km/hrCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 10 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for After noting down the odometer reading, that showed smallest four digit square number, Rimzim started driving to school with constant speed at 9 A.M. After one hour, he observed the delay in reaching school and doubled the speed to reach at 11 A.M. In school parking, he again noticed odometer which showed the number that reads same from both sides. What was the speed of Rimzim at 9:50 A.M.? (round to one decimal place)a)55.5 km/hrb)60.5 km/hrc)65.7 km/hrd)68.6 km/hrCorrect answer is option 'C'. Can you explain this answer?.
Solutions for After noting down the odometer reading, that showed smallest four digit square number, Rimzim started driving to school with constant speed at 9 A.M. After one hour, he observed the delay in reaching school and doubled the speed to reach at 11 A.M. In school parking, he again noticed odometer which showed the number that reads same from both sides. What was the speed of Rimzim at 9:50 A.M.? (round to one decimal place)a)55.5 km/hrb)60.5 km/hrc)65.7 km/hrd)68.6 km/hrCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 10. Download more important topics, notes, lectures and mock test series for Class 10 Exam by signing up for free.
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