amanbaua

 Profile Answers by amanbaua 

• Sep 10th, 2010

 

This question can be solved in 2 parts
1) Find the original speed 2) Find the original time

1)
The difference of 40 min is in this 50 miles
Time taken when accident took at original place=50/3s/5

Time taken when acc took 50 miles later=50/s
We can say 50*5/3s -50/s=40/60

Which we get as s=50

2) Can we say speed decreases by 40%
Hence time increases by by 66.66%
Let us take orig time as x
2/3*x=2 hence x=3

Hence total distance=50*3=150km

hsirah

 Profile Answers by hsirah 

• Oct 1st, 2010

 

Total distance is 200 km.

Case 1: t1 + t2 = T +120
Case 2: t1 + t2_ + t3_ = T+80

t2 -(t2_+t3_) = 40

(t2-t3_)-t2 = 40

t2 = x2/(3/5*x1/60)
t3_ = x3_/(3/5*x1/60)

t2 = x2_/x1/60

Also x2 - x3_=50

you can solve for x1 with these eqns: x1 =50

also in case 1 : x2 = A-50
and T = A/x1/60

with this you can solve for A, Final answeer A =200 

Abhash SInha

 Profile Answers by Abhash SInha 

• Sep 4th, 2011

 

In this question, we have to solve for original time and original speed and then we will calculate the total distance.
Let x & y be the speed & time respectively.
Therefore, total distance=xy mile

FIRST CASE:
(1 hour + time taken to travel after problem) = (original time + 2 hours)
=> 1+ (xy-x)/(3/5x) = y+2
=> (5y-5)/3 = y+1
=> 5y-5 = 3y+3
=> 2y = 8
=> y=4 hour

SECOND CASE:
distance travelled in 1 hour = x mile
i.e. total time taken to travel the journey of xy mile = ( (x+50)/x + (xy-(x+50))/3/5x)
according to question if problem would have occurred after 50 mile then driver would have saved 40 min
=> (y+2) - ( (x+50)/x + (xy-(x+50))/3/5x) = 40/60
=> 6 - (18x-100)/3x = 2/3
=> 18x-18x+100 = 2x
=> 2x = 100
=> x=50 mile/hour

Thus, we have time=4 hour and speed=50mile/hour
Hence total distance= 4*50 mile = 200 mile