Quantitative Aptitude
SPEED TIME AND DISTANCE MCQs
Time And Distance, Time & Distance, Speed Time & Distance
Total Questions : 1223
| Page 6 of 123 pages
Answer: Option A. -> 10 min
 - Due to stoppages, it covers 9 km less. Time taken to cover 9 km = (9/54 x 60) min = 10 min
 - Due to stoppages, it covers 9 km less. Time taken to cover 9 km = (9/54 x 60) min = 10 min
Answer: Option C. -> 4 1/2 hrs
 - Distance covered in first 2 hours = (70 x 2) km = 140 km
Distance covered in next 2 hours = (80 x 2) km = 160 km
Remaining distance = 345 – (140 + 160) = 45 km.
Speed in the fifth hour = 90 km/hr
Time taken to cover 45 km = 45 90 hr = 1 2 hr
Total time taken = 2 + 2 + 1 2 = 4 1 2 hrs
 - Distance covered in first 2 hours = (70 x 2) km = 140 km
Distance covered in next 2 hours = (80 x 2) km = 160 km
Remaining distance = 345 – (140 + 160) = 45 km.
Speed in the fifth hour = 90 km/hr
Time taken to cover 45 km = 45 90 hr = 1 2 hr
Total time taken = 2 + 2 + 1 2 = 4 1 2 hrs
Answer: Option C. -> 48 kmph
 - Speed on return trip = 150% of 40 = 60 kmph
Average speed = 2 x 40 x 60 40 + 60 km/hr = 4800 100 km/hr = 48 km/hr.
 - Speed on return trip = 150% of 40 = 60 kmph
Average speed = 2 x 40 x 60 40 + 60 km/hr = 4800 100 km/hr = 48 km/hr.
Answer: Option C. -> 480 km
 - Suppose they meet x hours after 14.30 hrs
Then, 60x = 80 (x – 2) or x = 8
Required distance = (60 x 8) km = 480 km
 - Suppose they meet x hours after 14.30 hrs
Then, 60x = 80 (x – 2) or x = 8
Required distance = (60 x 8) km = 480 km
Answer: Option C. -> 60 kmph
 - Number of gaps between 21 telephone posts = 20 Distance traveled in 1 minute = (50 x 20) m = 1000 m = 1 km Speed = 60 km/hr
 - Number of gaps between 21 telephone posts = 20 Distance traveled in 1 minute = (50 x 20) m = 1000 m = 1 km Speed = 60 km/hr
Answer: Option B. -> 11 : 9
 - In the same time, they cover 110 km and 90 km respectively. Ratio of their speeds = 110 : 90 = 11 : 9
 - In the same time, they cover 110 km and 90 km respectively. Ratio of their speeds = 110 : 90 = 11 : 9
Answer: Option C. -> 7
 - Relative speed = (2 + 3) = 5 rounds per hour So, they cross each other 5 times in an hour and 2 times in half an hour Hence, they cross each other 7 times before 9.30 a.m.
 - Relative speed = (2 + 3) = 5 rounds per hour So, they cross each other 5 times in an hour and 2 times in half an hour Hence, they cross each other 7 times before 9.30 a.m.
Answer: Option C. -> 11 am
 - Suppose they meet x hrs after 8 a.m.
Then, (Distance moved by first in x hrs) + [Distance moved by second in (x-1) hrs] = 330
60x + 75(x – 1) = 330
x = 3
So, they meet at (8 + 3), i.e. 11 a.m
 - Suppose they meet x hrs after 8 a.m.
Then, (Distance moved by first in x hrs) + [Distance moved by second in (x-1) hrs] = 330
60x + 75(x – 1) = 330
x = 3
So, they meet at (8 + 3), i.e. 11 a.m
Answer: Option C. -> 552 kms
 - Total distance travelled in 12 hours = (35 + 37 + 39 + ...... upto 12 terms) This is an A.P. with first term,
a = 35, number of terms, n = 12, common difference, d =2.
Required distance =12/2 (2 x 35 + (12 - 1) x 2) = 6(70 + 22) = 552 kms
 - Total distance travelled in 12 hours = (35 + 37 + 39 + ...... upto 12 terms) This is an A.P. with first term,
a = 35, number of terms, n = 12, common difference, d =2.
Required distance =12/2 (2 x 35 + (12 - 1) x 2) = 6(70 + 22) = 552 kms
Answer: Option D. -> 5 km/h