Latest IBM Aptitude Question SOLUTION: If a car starts from A towards B with some velocity. Due to some problem in the engine after traveling 30km, the car goes with 4/5 th of its actual velocity The car reaches B 45 mi

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If a car starts from A towards B with some velocity. Due to some problem in the engine after traveling 30km, the car goes with 4/5 th of its actual velocity The car reaches B 45 min later to the actual time. If the car engine fails after traveling 45km, the car reaches the destination B 36min late to the actual time What is the initial velocity of car and what is the distance between A and B in km

Solution

Let the distance between A and B is x and the velocity initially is v
then according to the given conditions
[30/v]+[(x-30)/(4v/5)] = [x/v]+3/4 here 3/4 is for 45 minute
[45/v]+[(x-45)/(4v/5)] = [x/v]+3/5 here 3/5 is for 36 minute
on solving we will get (x-30)/v = 3 and (x-45)/v = 12/5
or x-30 = 3v and x-45 = 12v/5 or x = 105 and v = 25

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Asked In IBM Aditi Sharma (13 years ago)

Solved HARMEET KAUR Read Solution (4)

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Read Solution (4) :

HARMEET KAUR  (13 years ago)

25km/hr and 105 kms.

Is this solution Helpfull? Yes (15) | No (15)  

Leelanand Kumar Yalla  (10 years ago)

Distance = 105 kilometers.. velocity- speed= 25 km/hour.....

Is this solution Helpfull? Yes (5) | No (4)  

Poulami Dey  (7 years ago)

Ans. let x is distance from A to B
and y is initial speed.
30/y+(x-30)5/4y -x/y = 3/4
=> 4x-12y=120 ----(1).
45/y + (x-45)5/4y -x/y=3/5
=> 5x-12y=225 -------(2).
From equ (1) and equ (2) we will get.
x=25 and y=105
so initial speed is 25 km/hr
and Distance From A to B is 105 km

Is this solution Helpfull? Yes (2) | No (3)  

  (6 years ago)

THis sum can be solved easily by propotionality
the vilocity is 4/5 th of its actual so s1:s2 =5:4 then speed inversely propotion to time so time t1:t2=4:5
the gap is 5-4 is 1 and train is 45 min late. so gap 1=45 so 45*4:45*5=180:225
so original time taken after 30 km is 180 mint.
now after 45 km by same process t1:t2=4:5 we already know and this time 36 mint late so
4*36:5*36=144:180. so this time 144 mint is actual time.
after 30 km 180 mint total after 45 144 minute. so common distance is 15km 45-30 and common time difference is 36 minut
now speed=15/36 nd we have to convert in hour so devide by 60 and it will
15*60/36=25km/hr
now we know speed total time 180 minut =3 hr so 25*3=75 +30 km before distance cover =105.
Its seems like hard but if u ill understand propotionalityy it can be solved in a minute.

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