A car is traveling at a uniform speed. The driver sees a milestone showing a 2-digit number. After traveling for an hour thedriver sees another milestone with the same digits in reverse order. After another hour the driver sees another milestone containingthe same two digits with a zero in between(0). What is the average speed of the driver.PLEASE EXPLAIN STEP WISE

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Answered by: swamidayalsrivastava

  • Sep 26th, 2007

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)
After solving, we get  ---->   y=6x  but since x and y are digits so only possible combination is x=1 and y=6,
  So average speed = 45 KM/HR

Showing Answers 1 - 10 of 10 Answers

swamidayalsrivastava

  • Sep 26th, 2007
 

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)
After solving, we get  ---->   y=6x  but since x and y are digits so only possible combination is x=1 and y=6,
  So average speed = 45 KM/HR

varun.pulyani

  • Jun 20th, 2008
 

Without using equations :-

1st distance = xy
2nd distance = yx
3rd = x0y or y0x
since first 2 valuess r two digits....avg speed can't be greater than 100.
Thus difference between 2nd and 3rd value is also less than 100.
Thus third distance is in 100's
thus x=1
now checking for possibility of y, y=6
Average speed = 45

hirenshirke

  • Dec 6th, 2009
 

Aree it has insufficient data
How can you assume that the distance travelled is same
assume 1st bechmark = 12
2nd benchmark = 21
3rd benchmark = 102
Now say how can tha distance are same and how can you assume that tga speed is uniform?
The equation is of the form speed=(109x-18y)/2
2 correponds to 2hrs, since it has been mentiond that he travls for 2hrs.

senthilrajait

  • Jan 29th, 2010
 

If the car travels in a paricular speed, suppose it want to travel from A to B.
Consider any point which is middle to the distance, then the distance specified in C should be less than that of A and distance specified in C should be less than that of B.
But in question they indicates that distance is increasing.
It is not possible. 'As time increases distance should decrease.'                      

soumit mukherjee

  • Jan 1st, 2011
 

Mathematically 45km/hr is evaluated... but logically it is impossible, coz if you see a milestone, it means you are going towards the destination, not opposite to that...
it is possible if first time : 106
2nd time :61
and 3rd time : 16

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