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31. If a man walks at the rate of 5kmph, he misses a train by only 7min. However if he walks at the rate of 6 kmph he reaches the station 5 minutes before the arrival of the train.Find the distance covered by him to reach the station.
Answer: Option C
Explanation:Lets assume the required distance = x km.
Difference in the times taken at two speeds=12mins=1/5 hr.
Therefore (x/5-x/6)=1/5 or (6x-5x) = 6 or x = 6km.
So required distance = 6 km
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32. Yana and Gupta leave points x and y towards y and x respectively simultaneously and travel in the same route. After meeting each other on the way, Yana takes 4 hours to reach her destination, while Gupta takes 9 hours to reach his destination. If the speed of Yana is 48 km/hr, what is the speed of Gupta?
Answer: Option D
Explanation:Let the speed of Yana be v1 and Gupta be v2 and time taken by them be t1 and t2 respectively.
There is a shortcut for such questions as v1^2/v2^2= t2/t1
Putting the respective values we get v2= 32km/hr
1.6 km =1 miles
So 32/1.6= 20mph.
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33. Walking 5/6 of its usual speed, a train is 10min late. Find the usual time to cover the journey?
Answer: Option D
Explanation:New speed = 5/6 of usual speed
New time = 6/5 of usual time
Therefore, (6/5 of usual time) - usual time = 10min
Therefore Usual time = 50min
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34. Vivek travelled 1200km by air which formed 2/5 of his trip.One third of the whole trip , he travelled by car and the rest of the journey he performed by train. The distance travelled by tain was:
Answer: Option D
Explanation:Let the total trip be x km.
Then 2x/5=1200
x=1200*5/2=3000km
Distance travelled by car =1/3*3000=1000km
Journey by train =[3000-(1200+1000)]=800km.
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35. A girl rode her bicycle from home to school, a distance of 15 miles, at an average speed of 15 miles per hour. She returned home from school by walking at an average speed of 5 miles per hour. What was her average speed for the round trip if she took the same route in both directions?
Answer: Option C
Explanation:Avg speed=(total disance)/(total time)
Total distance= 15+15 = 30 miles.
Total time= 1+3=4 hrs.
So average Spedd = 30/4 = 7.5 miles/hr.
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36. Car A leaves city C at 5pm and is driven at a speed of 40 kmph. Two hours later another car leaves city C and is driven in same direction as car A. In how much time car B be 9 km ahead of car A if the speed of car B is 60 kmph.
Answer: Option D
Explanation:Given, car A speed = 40 kmph,
car B speed 60 kmph
Relative speed=(60-40)=20 kmph(as both are in same direction)
when car B started A has already covered distance = 40*2=80m
Time required for both car to be at same position = 80/20 = 4 hr
Additional distance = 9 km
Time required to cover this distance = (9/20)*60=27 min
Therefore, required time = 4hrs 27 min
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37. J is faster than P. J and P each walk 24 km. Sum of the speeds of J and P is 7 kmph. Sum of time taken by them is 14 hours. Then speed of J is equal to
Answer: Option C
Explanation:As per the question: J > P ---(1)
and J + P = 7 ----- (2)
So, the below option can satisfy above both eq. (1) & (2)
(6, 1), (5, 2), (4, 3)
sum of time taken by them = 14 hrs. ---- (3)
To satisfy the 3rd condition
If J = 4 and P = 3.
Times taken by them = 24/4 + 24/3 = 14.
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38. A person travels in a car with uniform speed. He observes the milestone,which has 2 digits. After one hour he observes another milestone with same digits reversed. After another hour he observes another milestone with same 2 digits separated by 0. Find the speed of the car?
Answer: Option D
Explanation:Let 10'splace digit is x and unit's place digit y
First milestone : 10x+y
Second milestone : 10y+x
Third milestone: 100x+y
Since the speed is uniform so
Distance covered in first Hr = Distance covered in Second Hr
(10y+x)-(10x+y) = (100x+y)-(10y+x)
By solving,
y=6x
but since x and y are digits so only possible combination is x=1 and y=6,
So average speed = 45 Kmph
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39. In a covering a certain distance, the speeds of A and B are in the ratio of 3:4. A takes 30 minutes more than B to reach the destination. The time taken by A to reach the destination is?
Answer: Option D
Explanation:As per the question:
Ratio of speeds = 3:4
Ratio of times taken = 4:3
Suppose A takes 4x hrs and B takes 3x hrs to reach the destination.
Then, 4x - 3x = 30/60 => x = 1/2
Time taken by A = 4x hrs = 4 * 1/2 = 2 hrs.
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40. A ship went on a voyage after 180 miles a plane started with 10 times speed that of the ship. Find the distance when they meet from starting point.
Answer: Option A
Explanation:Lets assume u is the speed of ship and 10u is of plane.
So, relative speed = 9u
Distance covered = (time to meet * speed of plane)
=180*10u/9u =200 miles
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Tags: Infosys