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Q1 Option A
Distance covered by 1st man at S km/hr = X km
Time taken by him = t hours.
Therefore, Speed = S = X/t km/hr.
Distance covered by 2nd man at R km/hr = X/2 km
Time taken by him = 2t hours.
Therefore, Speed = R = (X/2)/2t km/hr = X/4t km/hr.
Required ratio = S:R = X/t : X/4t = 1 : 1/4 = 4:1.

Q2. Option D
Initial speed of the cyclist = 7km/hr.
Distance covered in 1st 3 hours = 7 x 3 = 21 km.
After increasing, 1km/hr for every 3 hours period,
Distance covered in 2nd 3 hours period = 8 x 3 = 24 km.
Distance covered in 3rd 3 hours period = 9 x 3 = 27 km.
Distance covered in 4th 3 hours period = 10 x 3 = 30 km.
Total distance covered = 21 + 24 + 27 + 30 = 102 km.
Remaining km to cover = 113 – 102 = 11 km.
Speed in 5th 3 hours period = 11 km/hr.
Time to cover 13 km at 11km/hr = 11/11 hours = 1 hour.
Now, the total time taken by him for 113 km = (3 + 3 + 3 + 3 + 1) = 13 hours.

Q3. Option B
Speed of the rider = 60km/hr.
Distance covered in 1st 2 hours = 60 km.
He increased his speed in every 2 hours by 3 km/hr.
Distance covered in every 2 hours will be, 60, 63, 66,… upto 12 terms.(for 24 hours).
The above series is an A.P series;
Sum of first n terms = (n/2)(2a+(n-1)d)
Here, a = 60, d = 3 and n = 12.
Sum of first 12 terms = (12/2)(2(60)+(11)3) = 6(120 + 33) = 6(153) = 918.
Hence, he covers 918 km in 24 hours.

Q4. Option C
Initial speed of the car = 50km/hr
Due to engine problem, speed is reduced to 10km for every 2 hours(i.e., 5 km per hour).
Speed of the car at 11 am = (50 – 5) = 45km/hr
Time to cover 10 km at 45 km/hr = distance/speed = 10/45 hours. = 2/9 hours
= 2/9 x 60 minutes = 40/3 minutes = 13 minutes + 1/3 minutes
= 13 minutes + 1/3 x 60 seconds
= 13 minutes and 20 seconds.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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