1. With a uniform speed a car covers the distance in 8 hours.Had the speed been increased by 4km/hr, The same distance could have been covered in 7 1/2 hrs.What is the distance covered?

A. 420 km B. 480 km C. 640 km D. None of these

Answer Answer: Option B Explanation: Let the distance be x km then
x/7 1/2 - x/8 =4 = 2x/15 - x/8 = 4 = x = 480km

A. 25 m/sec B. 1500 m/min C. 90 km/hr D. None of these

Answer Answer: Option D Explanation: 25 m/sec = (25 * 18/5) km/hr = 90 km/hr.
And 25 m/sec = (25 * 60) m/min = 1500 m/min.
So, all the three speeds are equal.

3. Walking 6/7th of his usual speed, a man is 12 minute too late. The usual time taken by him to cover that distance is?

A. 1 hr B. 1 hr 12 mins C. 1 hr 15 mins D. 1 hr 20 mins

Answer Answer: Option B Explanation: New speed = 6/7 of usual speed
New time = 7/6 of usual time
(7/6 of usual time) - (usual time) = 1/5 hrs
1/6 of usual time = 1/5 hrs
usual time = 6/5 hrs = 1hr 12mins

4. Two men starting from the same place walk at the rate of 5 km/hr and 5.5 km/hr respectively. What time will they take to be 8.5 km apart, if they walk in the same direction?

A. 4 hrs 15 min B. 8 hrs 30 min C. 16 hrs D. 17 hrs

Answer Answer: Option D Explanation: To be 0.5 km apart, they take 1 hour.
To be 8.5 km apart, they take (1/0.5*8.5) hrs = 17 hrs.

5. The speed of a car increases by 2 km after everyone hour. If the distance traveled in the first one hour was 35 km, what was the total distance traveled in the first one hour was 35 km, what was the total distance traveled in 12 hours?

A. 456 km B. 482 km C. 552 km D. None of these

Answer Answer: Option C Explanation: Total distance traveled in 12 hours = (35+37+39+..... up to 12 terms).
This is an A.P. with first term, a = 35, number of terms, n=12, common difference, d=2.
Therefore required distance = 12/2 (2*35+(12-1)*2) = 6(70+22) =552 km.