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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

Directions for question number 1-5: Study the following table carefully and answer the questions that follow:
RAILWAY TIMETABLE : NEW DELHI – BHUBANESWAR RAJDHANI EXPRESS

The longest run for the train between the two successive halts is

Correct

Solution: With the run is of 440 km, the longest run is between New Delhi-Kanpur Central.

Incorrect

Solution: With the run is of 440 km, the longest run is between New Delhi-Kanpur Central.

Question 2 of 10

2. Question

1 points

Category: Quantitative Aptitude

Directions for question number 1-5: Study the following table carefully and answer the questions that follow:
RAILWAY TIMETABLE : NEW DELHI – BHUBANESWAR RAJDHANI EXPRESS

The average speed that the train maintained between two successive stations was the highest between

Correct

Solution: The average speed of 93.22 kmph is the highest between NewDelhi – Kanpur Central.

Incorrect

Solution: The average speed of 93.22 kmph is the highest between NewDelhi – Kanpur Central.

Question 3 of 10

3. Question

1 points

Category: Quantitative Aptitude

Directions for question number 1-5: Study the following table carefully and answer the questions that follow:
RAILWAY TIMETABLE : NEW DELHI – BHUBANESWAR RAJDHANI EXPRESS

The average speed that the train maintained between New Delhi and Bhubaneswar was nearly equal to

Correct

Solution: The average speed that the train maintained between New Delhi and
Bhubaneswar = 1800 km/24 hrs and 25 min ~ 73 kmph.

Incorrect

Solution: The average speed that the train maintained between New Delhi and
Bhubaneswar = 1800 km/24 hrs and 25 min ~ 73 kmph.

Question 4 of 10

4. Question

1 points

Category: Quantitative Aptitude

Directions for question number 1-5: Study the following table carefully and answer the questions that follow:
RAILWAY TIMETABLE : NEW DELHI – BHUBANESWAR RAJDHANI EXPRESS

If we consider a journey that begins in New Delhi and ends in Bhubaneswar, the train has the longest halt at

Correct

Solution: The train has the longest halt of 15 minutes at Kharagpur JN.

Incorrect

Solution: The train has the longest halt of 15 minutes at Kharagpur JN.

Question 5 of 10

5. Question

1 points

Category: Quantitative Aptitude

Directions for question number 1-5: Study the following table carefully and answer the questions that follow:
RAILWAY TIMETABLE : NEW DELHI – BHUBANESWAR RAJDHANI EXPRESS

The train begins its return journey from Bhubaneswar to New Delhi Seventeen hours after it has arrived at Bhubaneswar. If the train left New Delhi on Tuesday on what day will it have returned to New Delhi? (Assume that on the return journey that train maintains the same average speed as on the onward journey).

Correct

Solution: It takes 24 hours and 25 minutes to reach Bhubaneswar, i.e it reaches on Wednesday t 17.30, then after 17 hours it left i.e at 12.30 on Thursday and will take another 24 hour and 25 min and reach New Delhi on Friday.

Incorrect

Solution: It takes 24 hours and 25 minutes to reach Bhubaneswar, i.e it reaches on Wednesday t 17.30, then after 17 hours it left i.e at 12.30 on Thursday and will take another 24 hour and 25 min and reach New Delhi on Friday.

Question 6 of 10

6. Question

1 points

Category: Quantitative Aptitude

Directions For questions 6-8: There takes a race between A and B under different conditions. Answer the following questions accordingly.

A runs 5/3 times as fast as B. If A gives B a start of 80 m, how far must the winning post be so that A and B might reach it at the same time?

Correct

Solution: Speed of A : Speed of B = 5/3 : 1 = 5 : 3 i.e., In a race of 5 m, A runs 5 m and B runs 3 m i.e., A gains 2 m over B in a race of 5 m, therefore A gains 80 m over B in a race of 5/2×80 = 200 m. Hence, winning point should be 200 m away from the starting point.

Incorrect

Solution: Speed of A : Speed of B = 5/3 : 1 = 5 : 3 i.e., In a race of 5 m, A runs 5 m and B runs 3 m i.e., A gains 2 m over B in a race of 5 m, therefore A gains 80 m over B in a race of 5/2×80 = 200 m. Hence, winning point should be 200 m away from the starting point.

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

Directions For questions 6-8: There takes a race between A and B under different conditions. Answer the following questions accordingly.

In a 100 metres race. A runs at a speed of 2 metres per seconds. If A gives B a start of 4 metres and still beats him by 10 seconds, find the speed of B.

Correct

Solution: Speed of A = 2 m/s, Time taken by A to run 100 m distance/speed=100/2 = 50 seconds. A gives B a start of 4 metres and still A beats him by 10 seconds =>B runs (100-4)=96 m in (50+10)=60 seconds, Speed of B = distance/time = 96/60 = 1.6 m/s

Incorrect

Solution: Speed of A = 2 m/s, Time taken by A to run 100 m distance/speed=100/2 = 50 seconds. A gives B a start of 4 metres and still A beats him by 10 seconds =>B runs (100-4)=96 m in (50+10)=60 seconds, Speed of B = distance/time = 96/60 = 1.6 m/s

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

Directions For questions 6-8: There takes a race between A and B under different conditions. Answer the following questions accordingly.

A and B run a km race. If A gives B a start of 50 m, A wins by 14 sec and if A gives B a start of 22 sec, B wins by 20 m. The time taken by A to run a km is?

Correct

Solution: Let time taken by A to run 1000 m = a and time taken by B to run 1000 m = b. If A gives B a start of 50 m, A wins by 14 sec, distance travelled by A = 1000 and distance travelled by B = (1000-50)=950. Time taken by B to run 950 m – Time taken by A to run 1000 m = 14 sec.
(950/1000 )b−a=14 therefore (95/100)b−a=14—(Equation 1)
If A gives B a start of 22 sec, B wins by 20 m i.e., A starts 22 seconds after B starts from the same starting point. i.e., here B runs for b seconds where A runs for (b-22) seconds => Distance Covered by B in b seconds – Distance Covered by A in (b-22) seconds = 20 =>1000−1000/a×(b−22) =20 therefore 1000/a×(b−22) =980., hence 1000b−22000 =980a, 100b−2200 =98a therefore 50b−1100 =49a therefore 50b−49a=1100—(Equation 2)
(Equation 1)×49 ⇒95×49/100*b−49a =14×49 ⇒46.55b−49a=686—(Equation 3) Now, (Equation 2) – (Equation 3) ⇒3.45b =414 ⇒b=414/3.45=120 seconds. Substituting this value of b in Equation 1⇒95×120/100 −a= 14⇒114−a=14 ⇒a=114−14=100
i.e., The time taken by A to run a km = 100 seconds.

Incorrect

Solution: Let time taken by A to run 1000 m = a and time taken by B to run 1000 m = b. If A gives B a start of 50 m, A wins by 14 sec, distance travelled by A = 1000 and distance travelled by B = (1000-50)=950. Time taken by B to run 950 m – Time taken by A to run 1000 m = 14 sec.
(950/1000 )b−a=14 therefore (95/100)b−a=14—(Equation 1)
If A gives B a start of 22 sec, B wins by 20 m i.e., A starts 22 seconds after B starts from the same starting point. i.e., here B runs for b seconds where A runs for (b-22) seconds => Distance Covered by B in b seconds – Distance Covered by A in (b-22) seconds = 20 =>1000−1000/a×(b−22) =20 therefore 1000/a×(b−22) =980., hence 1000b−22000 =980a, 100b−2200 =98a therefore 50b−1100 =49a therefore 50b−49a=1100—(Equation 2)
(Equation 1)×49 ⇒95×49/100*b−49a =14×49 ⇒46.55b−49a=686—(Equation 3) Now, (Equation 2) – (Equation 3) ⇒3.45b =414 ⇒b=414/3.45=120 seconds. Substituting this value of b in Equation 1⇒95×120/100 −a= 14⇒114−a=14 ⇒a=114−14=100
i.e., The time taken by A to run a km = 100 seconds.

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

It is decided to construct a 2 metre broad pathway around a rectangular plot on the inside. If the area of the plot is 96 sq.m. and the rate of construction is Rs.50 per square metre, what will be the total cost of construction?

Correct

Solution: Let length and breadth of the rectangular plot be l and b respectively.
Area of the rectangular plot =96 sq.m.
⇒lb=96
Width of the pathway =2, Length of the remaining area in the plot =(l−4)
Breadth of the remaining area in the plot =(b−4)
Remaining area in the plot =(l−4)(b−4)
Area of the pathway = Area of the rectangular plot – Remaining area in the plot =96−(l−4)(b−4) =96−(lb−4l−4b+16)= 96−(96−4l−4b+16) =4l+4b−16
We do not know the values of l and b and hence area of the pathway cannot be found out. So we cannot determine total cost of construction.

Incorrect

Solution: Let length and breadth of the rectangular plot be l and b respectively.
Area of the rectangular plot =96 sq.m.
⇒lb=96
Width of the pathway =2, Length of the remaining area in the plot =(l−4)
Breadth of the remaining area in the plot =(b−4)
Remaining area in the plot =(l−4)(b−4)
Area of the pathway = Area of the rectangular plot – Remaining area in the plot =96−(l−4)(b−4) =96−(lb−4l−4b+16)= 96−(96−4l−4b+16) =4l+4b−16
We do not know the values of l and b and hence area of the pathway cannot be found out. So we cannot determine total cost of construction.

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

A large tanker can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time and A and B fill it together for the other half?

Correct

Solution: Part filled by pipe A in 1 minute =1/60, Part filled by pipe B in 1
minute =1/40, Part filled by both pipes A and B in 1 minute
=1/60+1/40=(2+3)/120 =5/120=1/24, Suppose the tanker is filled in x minutes.
Then, to fill the tanker from empty state, B is used for x/2 minutes and A and B
together is used for the rest x/2 minutes.
x/2×1/40+x/2×1/24=1⇒x/2(1/40+1/24)=1 ⇒x/2×8/120 =1 ⇒x/2×1/15=1,
x=15×2=30 minutes

Incorrect

Solution: Part filled by pipe A in 1 minute =1/60, Part filled by pipe B in 1
minute =1/40, Part filled by both pipes A and B in 1 minute
=1/60+1/40=(2+3)/120 =5/120=1/24, Suppose the tanker is filled in x minutes.
Then, to fill the tanker from empty state, B is used for x/2 minutes and A and B
together is used for the rest x/2 minutes.
x/2×1/40+x/2×1/24=1⇒x/2(1/40+1/24)=1 ⇒x/2×8/120 =1 ⇒x/2×1/15=1,
x=15×2=30 minutes

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