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Infosys Sample Distance Calculation Questions

Question 1
A train starts from station P towards Q with certain speed. Due to a problem, after crossing 50km, the train slows down to 2/3 rd of its actual speed and it reaches Q 50 minutes later than the planned time. Suppose the technical problem had happened after crossing 60 km and the train would had reached 40 minutes late. What is the actual(original) speed of the train and what is the distance between P and Q ?
a)30km/hr, 100km b)20km/hr,80km c)40km/hr,150km d)50km/hr,150km
Answer : a)30km/hr, 100km
Let the distance between P and Q be X and the speed initially be V.
Note that the train travels 50km with speed V and the remaining distance(X-50)km with speed 2/3 of V.
According to the above condition, with the formula " distance/speed = time", we can have
Time Taken For First 50 Km + Time Taken For Remaining (X-50) Km = Planned Time + Extra Time Due To Problem
[50/V]+[(X-50)/(2V/3)] = [X/V]+5/6 (here 50 minutes = 5/6 hours)
[100+3X-150-2X]/2V = 5/6
3X-5V = 150 ...eqn1
Suppose the technical problem had happened after crossing 60 km and the train would had reached 40 minutes late.
Using the same logic we used to arrive at eqn1, we get,
[60/V]+[(X-60)/(2V/3)] = [X/V]+2/3 (here 2/3 is for 40 minutes in hour.)
[120+3X-180-2X]/2V = 2/3
3X-4V = 180 ....eqn2
Solving eqn1 & eqn2, we get
X=100 & V=30
Hence the distance between P and Q is 100km and the actual speed of the train is 30km/hour.
Question 2
A train starts from A towards B with some velocity. Due to an engine problem, after travelling 3/8 of its journey, it slows to 3/5 of its actual velocity. The train reaches B 1 hour later than the actual planned time. If the engine had failed after travelling 80km and if it would had slowed down to 4/5th of its initial velocity for another 80km and covered remaining distance with 1/2 of its initial velocity, the train would had reached the destination one and half hours late. What is the distance between A and B in meters?
a)10000 b)48000 c)24000 d)52000
Answer : b)48000
Let the distance between A and B be X and the speed initially be V.
The train travels 3X/8km with speed V and the remaining distance(X - 3X/8)km with speed 3/5 of V. Ultimately the train was late by 1 hour.
According to the above condition with the formula " distance/speed = time", we can have
[(3X/8)/V]+[(X-(3X/8))/(3V/5)] = [X/V]+1
[3X/8V] + 5(8X-3X)/24V = [X/V]+1
9X+25X-24X / 24V = 1
10X-24V = 0 ..........eqn1
According to the question, if the train travelled 80km with speed V, another 80km with 4/5 th of V and the remaining distance(X-160)km with speed 1/2 of V then
[80/V]+[80/(4V/5)]+[(X-160)/(1V/2)] = [X/V]+3/2
80/V + 100/V + (2X-360)/V = X/V + 3/2
X-180 / V = 3/2
2X-3V = 360 .........eqn2
solving eqn1 and eqn2
we have, X=480 and V=200
Thus the distance between A and B is 480km and the speed of the train is 200km/hour.
Hence 480km = 480000meters is the answer.
Question 3
A man rides a bike with speed V for a distance X km. After completing 1/2 of his journey he slows down to 1/2 of his initial speed and completes his ride half an hour late than planned time. What will be the ratio of the distance to the speed?
a)1:2 b)2:1 c)1:1 d)2:3
Answer : c)1:1
Let the distance be X and the speed initially be V.
Using same logic as we used for previous two questions, we get,
X/2V + [X-(X/2)]/(V/2) = X/V + 1/2
X/2V + [4X-2X]/2V = X/V + 1/2
X/2V = 1/2
X = V or X/V = 1
Thus 1:1 is the required ratio.


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A.    cube
B.    sphere
C.    pyramid
D.    circle

D (because the circle is the only two-dimensional figure)

Question 2. Consider a language which uses the following set of characters:
Small set: { a b c }
Large set: { A B C }
Punctuation set: { x y }
This language must follow the following rules:
  1.    A punctuation character must end all series.
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Does the following series follow all the rules of the language defined above?
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The person in No.1 is:
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    C.   Unmarried, with at least one daughter
    D.   Unmarried, with at least one son
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