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61. Two cars starts from A and B and travel towards each other at a speed of 50 kmph and 60 kmph respectively. At the time of their meeting the second car has travelled 120 kmph more than the first. The distance between A and B is:
62. The five tyres of a car (four road tyres and one spare) were used equally in a journey of 40,000 kms. The number of kms of use of each tyre was
Answer: Option C
Explanation:Total kilometers travelled by 4 tyre = 40000 * 4 = 1,60,000.
This has to be share by 5 tyres.
So each tyre capacity = 1,60,000 / 5 = 32, 000.
You have a doubt, after we travel 32,000 km, we are left with 4 worn tyes and one new tyre.
But If the tyres are rotated properly after each 8000 km,
All the tyres are equally used.
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63. Jake and paul start together to walk 10kms jake walks 1.5kmph faster than paul and arrives at his destination 1.5 hours earlier than paul jakes speed walking is
Answer: Option C
Explanation:Let speed of jake = x kmph.
So, speed of paul = ( x - 1.5 ) kmph.
Time difference =1.5hours
Total Distance = 10km
10/(x-1.5) -10/x = 1.5
By solving the above eq.
x = 4 kmph
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64. A horse starts to chase a dog that has left the stable two hours earlier. The horse runs at an average speed of 22km/hr. It crosses a 10 metre road, two small ponds 3 metres deep, and finally runs along two small streets of 200 metres long. After traveling 6 hrs, 2 hrs after sunset, it catches the dog. Compute the speed of the dog in Km/hr?
Answer: Option C
Explanation:We have the speed of horse = 22 kmph
And it ran for 6 hours.
Hence distance covered by it in 6 hours = 22 * 6
= 132 km
We know that dog has covered the same distance in 6 + 2 = 8 hours.
Hence speed of the dog = 132 / 8
= 16.5 kmph
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65. Ram and Shakil run a race of 2000 meters. First, Ram gives Shakil a start of 200 meters and beats him by one minute. If, Ram gives Shakil a start of 6 minutes Ram is beaten by 1000 meters. Find the time in minutes in which Ram and Shakil can run the races separately.
Answer: Option D
Explanation:Let x & y be the speeds of Ram & Shakil
Then acc. To the problem
2000/x=1800/y-1 .... (i) and 1000/x=2000/y-6 ..... (ii)
Solving the above equations we get x=250m/sec and y=200m/sec
So to cover a distance of 2000m separately they will take 2000/250 and 2000/200 secs respectively.
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66. George walks 36 kms partly at a speed of 4 kms per hour and partly at 3 km per hour If he had walked at a speed of 3km per hour when he had walked at 4 and 4 km per when he had walked at 3 he would have walked only 34 kms. The time (in hours) spent by George in walking was
Answer: Option D
Explanation:Let George walked "a" hours at 4 kmph, and "b" hours at 3 kmph.
Given, 4a + 3b = 36 - - - (1)
3a + 4b = 34 - - - (2)
Adding the above two equations and simplifying them, a + b = 10.
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67. George while driving along the highway saw road markers which are at equal distances from each other. He crosses the markers every 20 seconds. If he increases his speed by x meters per second, he crosses the markers at every 15 seconds. But if he increases his speed by y meters per second, he crosses the marker at every 10th second. If y-x = 40 meters per second, then what is the distance between two markers.
Answer: Option B
Explanation:Let speed be =z m/s then Distance= 20z m (z+x)15=20z; (z+y)10=20z
Also given that y - x = 40 solving we get 20z=1200
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68. One day, Eesha started 30 minutes late from home and reached her office 50 minutes late, while driving 25% slower than her usual speed. How much time in minutes does Eesha usually take to reach her office from home?
Answer: Option C
Explanation:Let us time taken by eesha daily = x
and usual speed = y
extra time taken = 50-30 = 20 min
So, (x+20)*(0.75*y) = x*y ----- (1)
0.75x+15 = x
=> 0.25x = 15
=> x = 60 min
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69. Tim and Elan are 90 km from each other, they start to move each other simultaneously, Tim at speed 10 and Elan 5kmph, if every hour they double their speed what is the distance that Tim will pass until he meet Elan?
Answer: Option B
Explanation:1st hour=tim 10kmph elan 5 the dist will be 90-15=75
2nd hr=tim->20kmph & elan->10 dist=75-30=45
3rd hr =tim->40 & elan->20..the totl dist for 3rd hr is 60.but we need only 45km to meet ..so we consider only 45mins instead of 3rd hr..
In 45mins =>tim ->30 & elan->15
totl dist travld by tim is 10+20+30=60
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70. An old man and a young man are working together in an office and staying together in a near by apartment. The old man takes 30 minutes and the young 20 minutes to walk from appartment to office. If one day the old man started at 10.00 AM and the young man at 10:05AM from the apartment to office, when will they meet?
Answer: Option A
Explanation:Ratio of old man speed to young man speed = 2:3
The distance covered by old man in 5 min = 10
The 10 unit is covered with relative speed=10/(3-2)=10 min
so, they will meet at 10:15 am.
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