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# Arithmetic Aptitude :: Time and Distance

Home > Arithmetic Aptitude > Time and Distance > General Questions

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51. Maria drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 9 hours. When Maria drove home, there was no traffic and the trip only took 4 hours. If her average rate was 40 miles per hour faster on the trip home, how far away does Maria live from the mountains?

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Explanation:

Time taken for trip from home to mountains=9h
Time taken for trip from mountains to home=4h
Let distance from home to mountains=x miles
let average speed from home to mountains=y miles/h
Given avg speed on the trip home is 40m/h faster than speed from home to mountains=y+40 m/h
=> (x/9)=y and (x/4)=y+40
by solving this, we get y=32 m/h and the distance x=32*9=288 miles.

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Tags:  Wipro

52. Vikas can cover a distance in 1hr 24min by covering 2/3 of the distance at 4 kmph and the rest at 5kmph.the total distance is?

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Explanation:

Lets assume total distance = x km.
vikas covered 2/3 of x = 4kmph*84min
so,
x = (4kmph*84min*3)/(2*60) = 8.4 km.

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Tags:  Wipro

53. Driving her car to work 4 days per week one costs \$7.00 a day for parking and \$36.00 a week for fuel. How much would she save week by taking a bus at a fuel at a fare of \$1.30 each way?

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Explanation:

as per the question (4 days per week one costs \$7 )
So, total on fuel for week= 7*4=\$28
and she expend on partking for week = \$36
Total she expend per week = \$(28+36) = \$64
Per day by bus both side = 2*1.30= \$2.60
Total cost 4 days of week = 4*2.60 = \$10.40
So she save = (\$64-\$10.40) = \$53.60

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Tags:  IBM

54. A car departs at 11:10am. It travels at 110km/hr until 2:30 then it reduces its speed by 5%for the remainder of trip. It reaches destination at 7:30. How much kilometer did he travel?

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Explanation:

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Tags:  IBM

55. A cycled from P to Q at 10 kmph and returned at the rate of 9 kmph. B cycled both ways at 12 kmph. In the whole journey B took 10 minutes less than A. Find the distance between P and Q.

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Explanation:

Lets assume distance between P and Q = d km.
Time taken by A in both side = d/10 + d/9 = 19d/90 hrs.
Time taken by B in both side = 2d/12 = d/6 hrs.
B tool 10 min. or 1/6 hrs. less than A.
So, 19d/90 - d/6 = 1/6
=> (19d-15d)/90 = 1/6
=> 4d/90 = 1/6
=> d = 15/4 km = 3.75 km.

Workspace

Tags:  Accenture

56. A man traveled from the village to the post-office at the rate of 25 kmph and walked back at the rate of 4 kmph. If the whole journey took 5 hours 48 minutes, find the distance of the post-office from the village.

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Explanation:

Average speed = (2*a*b)/(a + b) here a = 25 b = 4
Average speed = 2*25 * 4/(25 + 4) = 200/29 kmph.
Distance covered in 5 hours 48 minutes = Speed * time
Distance = (200/29)*(29/5) = 40 kms
Distance covered in 5 hours 48 minutes = 40 kms
Distance of the post office from the village = (40/2) =20 km.

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Tags:  Accenture

57. A car starts from rest and accelerates uniform upto speed 90 km/hrs in 5 sec. Total distance covered by car in this time will be

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Explanation:

Speed of car at last instant of given interval
= 90*5/ 18 m/s = 25 m/s
so average speed = 0+25 / 2
Therefore distance = 5 * 25/2 = 125/2 = 62.5 m

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Tags:  Accenture

58. How many rotations per minute must a wheel of diameter 1.5m make to achieve a speed of 45 kmph?

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Explanation:

For 45kmph, distance to cover in one hour = 45 * 1000 = 45000 m.
Distance to be covered per minute (d) = 45000 / 60 = 750 m/min
Distance covered in 1 rotation of wheel =Pi*diameter=Pi*1.5 m.
So rotation /minute Required =750/(Pi*1.5)=750/[(22/7)*1.5]=159.09 ~ 159

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Tags:  Capgemini

59. Juan is a gold medalist in athletics. In the month of May, if Juan takes 11 seconds to run "y" yards. How many seconds will it take him to run "x" yards at the same rate?

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Explanation:

Juan takes 11 sec. to run "y" yards.
So, to cover 1 yard he need 11/y sec.
So to cover x yard he need 11x/y sec.

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Tags:  TCS

60. Two vehicles A and B leaves from city Y to X. A overtakes B at 10:30 am and reaches city X at 12:00 pm. It waits for 2 hrs. and return to city Y. On its way it meets B at 3:00 pm and reaches city Y at 5:00 pm. B reaches city X, waits for 1hr and returns to city Y. After how many hours will B reach city Y from the time A overtook him for the first time?

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Explanation:

Vehicle A overtaken B at 10.30 am and reached X at 12 pm.  It started at 2 pm and met B at 3 pm at Q. It means, Vehicle A took one hour to cover distance 'n', So it should be at Q at 11 pm.  It is clear that Vehicle A takes 0.5 hour to cover distance 'm'.

Now vehicle B travelled from 10.30 am to 3 pm to meet A. So it took 4.5 hours to cover m.  So Speeds ratio = 4.5 : 0.5 = 9 : 1.

Now Vehicle A took a total of 1.5 + 3 = 4.5 hours to travel fro P to Y. So It must take 4.5 * 9 + 1 = 41.5 hours

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Tags:  TCS