A car leaves point A at 3:30 pm and moves towards point B at a uniform speed. A train leaves point B

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16 Mar 2022 Time Distance & Speed

A car leaves point A at 3:30 pm and moves towards point B at a uniform speed. A train leaves point B at 4:03 pm and moves towards point A at a uniform speed which is 5/4 that of the car. They meet at 5:15 pm at a point which is 161 km away from A. What is the distance between A and B?

OPtion

1) 273 km
2) 237 km
3) 299 km
4) 256 km
5) 286 km
6) 268 km
7) 301 km
8) 271.4 km
9) 249.32 km
10) None of these

Solution

Till they meet, car covers 161 km in 1 hrs 45 min or 7/4 hrs.
.'. Speed of a car = 161/(7/4) = 92 km/hr
Speed of train = (5/4)*92 = 115 km/hr
Time taken by train till they meet = 1 hr 12 min = 6/5 hrs
.'. Distance covered by train = Speed*Time = 115*6/5 = 138 km.
Total distance between A and B = Sum of the distances covered by car and train till they meet = 161+138 = 299 km.

Correct Option: 3 Best Solution (2)

G.S.Kantharao   2 years AGO

The distance from A they met=161Km
Car travelled time =7/4hrs
Speed of the car =161*4/7=92Kmph
Now speed of the train =92*5/4=115Kmph
Train travelled time =1.2hrs
It travelled =138Km
Total distance from A to B =161+138=299Km
So my option is 3👈

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RAMADURAI NATESAN   2 years AGO

Let Speed of car be = S km/hr. Spd. of train = 1.25 S.
Car travels 161 km in 1.75 hrs. Car spd . = 92 km.
Train spd. = 115 km/hr.In 1.2hrs, train will cover 138 km
Distance from A to B 138 + 161 =299

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