Time And Distance - MCQ 4
20 Questions MCQ Test Quantitative Techniques for CLAT | Time And Distance - MCQ 4
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A car after travelling 100 km from point A meets with an accident and then proceeds at 3/4 of its original speed and arrives at the point B 90 minutes late. If the car meets the accident 60 km further on, it would have reached 15 minutes sooner. Find the original speed of the train?
let distance between A and B be D km and real speed of the car be S km/hr
First time car takes 90 minutes more and second time car takes 75 minutes more than scheduled time.
So, T1 – T2 = 15/60 = 100
15/60 = [100/s + (D -100)/(3s/4)] – [160/s + (D – 160)/(3s/4)] Solve this D will be cancelled and S comes out to be 80km/hr
A boy goes to school from his home at a speed of 4 km/hr and returns at a speed of 6 km/hr. If in total he took 5 hours in going and coming back, the distance between his house and school?
5 = d/4 + d/6 (d is the distance between school and home)
A police officer is 110 meter behind a thief. The officer runs 21 meters and thief runs 14 meters in a minute. In what time the officer will catch the thief ?
let the thief covers x more distance before get caught, so officer have to cover 110+x in the same time in which thief covers x distance ,
x/14 = (110 +x)/21, x = 220 so time = 330/21 = 110/7 = 15.5/7 minutes
Two trains P and Q are separated by 220 km on a straight line. One train starts at 8 am from one station A towards B at 40 km/hr and another train starts from B towards A at 9 am at 60 km/hr. At what time will both train will meet?
In one hour first train will cover 40 km, so distance between them remains only 180. Now
x/40 = (180 –x)/60, we get x = 72, so time = 72/40 = 1 hour 48 minutes
so both will meet at 10:48 am
Two girls begin together to write out a book containing 900 lines. The first girl starts with the first line, writing at a speed of 150 lines per hour and the second girl starts writing from the last line at the rate of 200 lines per hour. At which line will they meet?
Solve this question as analogous to distance method,
At some x page they will meet, so
x/150 = (900 – x)/200, we get x = 385.7 = 386th line (approx.)
Two women start walking towards each other at same time from two points separated by 84 km. First women walks at 5km/hr and second walks at 2 km/hr in the first hour, 2.5 km/hr in second hour, 3 km/hr in third hour and so on. Find the approximate time after which both women will meet?
Let after T time both women meet each other, so distance travelled by first women is 5T and second women forms an AP, so
distance = (T/2)*[4 + (T -1)1/2] so, 5T + (T/2)*[4 + (T -1)1/2] = 84, solve for T, u will get T = 9 hrs
take √2073 = 45 (approx.)
Two places are A and B are 200 kms from each other. A train leaves from A for B at the same time another train leaves B for A. The two trains meet at the end of 8 hours. If the train travelling from A to B travels 10km/hr faster than the other. Find the speed of the faster train?
speed of trains = s + 10 and s km/hr. At some P distance from A both train will meet. So,
8 = P/(s +10) = (200 – P)/s
Solve both equation, we get s = 7.5 so faster train speed 17.5 km/hr
Two bullets are fired at an interval of 15 minutes but a car approaching the place of firing hears the second bullet after 14 minutes. Speed of sound is 330 meters per second. Find the speed of the car?
distance travels by sound in one minute is equal to the distance travels by car in 14 minutes i.e. 60*330 = s*14*60
s = (330/14)*(18/5) = 85km/hr (approx.)
A truck driving on a highway passed a man walking at the rate of 12km/hr in the same direction. He could see the truck for 2 minutes and up to 500 meters. Find the speed of the truck?
500 = (s – 12*5/18)*2*60
s = 27km/hr
Excluding stoppages the speed of bus is 60 km/hr and including stoppages it average speed becomes 48km/hr. For how much time the bus stops.
Due to stoppages bus covers 12 km less. To cover 12km with 60km/hr speed it takes 12min
A man walks at certain place and rides back in 10 hours. He could ride both ways in 8 hours. The time taken by him to walk both ways?
W + R = 10
and 2R = 8, R = 4
so, W = 6 so, walking both sides, he will take = 12 hours
Two trains start from same place at same time at right angles to each other. Their speeds are 36km/hr and 48km/hr respectively. After 30 seconds the distance between them will be
Using pythagarous theorem,
distance travelled by first train = 36*5/18*30 = 300m
distance travelled by second train = 48*5/18*30 = 400m
so distance between them =√(90000 + 160000) = √250000 = 500m
A train covers a distance between two stations P and Q in 30 minutes. If the speed of the train is reduced by 10km/hr, then the same distance is covered in 45 minutes. The distance between P and Q
D = S*1/2 and D = (S-10)*3/4
solve both equation. u will get D = 15km
An inspector is 228 meter behind the thief. The inspector runs 42 meters and the thief runs 30 meters in a minute. In what time will the inspector catch the thief?
inspector s 228 meter behind the thief and now after some x distance he will catch the thief. So,
x/30 = (228 + x)/42, we will get x = 570m
so time taken by inspector to catch the thief = (228+570)/42 = 19 minutes
Two trains 210 meters and 180 meters are running on parallel track at the speed of 72km/hr and 45km/hr respectively. The time taken by them to cross each other, if they are running in opposite direction?
210 + 180 = (72 + 45)*5/18*T, T = 12 seconds
A car travels from P to Q at a constant speed. If its speed were increased by 20km/hr, it would have taken two hour lesser to cover the distance. It would have taken further 30 minutes lesser if the speed was further increased by 10 km/hr. The distance between the two cities
Let the distance between two cities = D,
D = x*t (x = usual speed and t is the actual time)
D = (X + 20)*(T -2) and D = (X +30)*(T – 2.5)
Solve the above equation, u will get T = 5 and X = 30, so distance = 30*5 = 150 km
A girl goes to her school from her house at a speed of 6km/hr and returns at a speed of 4 km/hr. If she takes 10 hours in going and coming back, the distance between her school and house is
Let distance be D
10 = D/4 + D/6
A bus leaves the stop 30 minutes before the scheduled time. The driver decreases its speed by 30km/hr. At the next bus stop 180 km away, the bus reached on time. Find the original speed of the bus?
Distance = 180km, actual speed = x and actual time = t
180 = x*t
180 = (x – 30)*(t +1/2)
Solve both equation, we will get x = 120km/hr
A walks with a speed of 6 km/hr and after 5 hr of his start, B starts running towards A at a speed of 8 km/hr. At what distance from start will B catch A.
In 5hrs, A will cover 30 km. Now, at some distance ‘x’. So A will cover X distance and B will cover 30 + X.
x/6 = (30+x)/8
x = 90. So distance from start after which B will catch A = 90+30 =120km
Without any stoppages a person travels a certain distance at an average speed of 40 km/hr and with stoppages he covers the same distance at an average of 20 km/hr. How many minutes per hour does he stop?
In one hour the distance covered at actual speed = 40km and with stoppages it covers only 20 km
so to travel 20 km at original speed i.e. 20 = 40*t, so t = 1/2 hour = 30
minutes
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