Aptitude - Time and Distance - Discussion

For people who still has confusion with the 150*x/100:

Why don't you remember the LCM formula?.. if speed of car is x then speed of train is more than speed of car by 50% of it. so after we calculate 50% of speed of car, we should not forget to add the result of it to the actual speed of car, i.e-
[50% of x]+x
= [(50/100)*x]+x
= [50x/100]+x... now, this is actually [50x/100]+[x/1] the two denominators '100' and '1' here are 100, and 1 and the LCM of the two is 100. This makes 100 the common denominator. now following the rule of LCM, divide the common denominator by the first actual denominator, i.e 100/100, the result is '1', and now multiply the first numerator by this result '1' that keeps 50x*1 that is the same, so basically the first numerator does not change. now, do the same with the second denominator, multiply this common denominator 100 by the second actual denominator, i.e 100/1, the result is '100', and now multiply the second numerator by this result '100' that makes it 100*x that is 100x.. so after the LCM now it looks like:
(50x+100x)/100.

Now take x as common and comprehend this numerator as-
x(50+100)/100
=x(150)/100
=150x/100
DID U GET THIS NOW? THE TRICK LIES IN THE LCM FORMULA !!

This can again be derived to 3x/2, applying the ratio formula.

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Here is another simpler method.. when we calculate speed of train as 50% more than that of the car, in a lay-man's language it simply means speed of train is more than speed of car by half of car's speed, so instead of 50% of car's speed, we can also do half of car's speed, and if car's speed is 'x', then train's speed is

(x/2)+x, applying the same LCM formula, take 2 as the common denominator and the first numerator remains 'x' and the second numerator becomes '2*x'
= (x+2x)/2, now if u remember, 'x' is equivalent to '1*x', as per basic algebra logic- so,
= 3x/2.

GOT THE SAME ANSWER ??

Let suppose car speed = x km/h

As per problem train speed 50% more than car so = x+50/100x = 3/2x.

The Train lost 12.5 min to reach the destination.By this we understood that train took more time than car to reach the destination. So the car time should be subtracted from train's time and the result we get is -120. As speed will not be negative, the answer is 120. Is this right one?

Thanks Santhiraju :).

I am not understanding the main calculation part i.e. steps after finding speed pf the train.
Somebody please explain me the steps of main calculation

Let the speed of the car be 100% and the train is 50% more
100%+50%=150%
that is 150/100

Let Speed of the car=x.

Train Speed=Speed of car+50% (Speed of car).

=> x + 50/100 (x).

=> x + x/2.

=> 3x/2.

Say, the speed of the car be 1 km/hr (0. 5 +0. 5).

And if there is 50% increase in speed means (1 +0. 5) = 1. 5 km /hr.

That's how 3/2 came.

It's a very easy question
suppose train running 12.5 min. & lost 12.5 mint in station
s=d/t====>
s=(75/25*60)km/s
s=1/20km/s change in m/sec
s=1000/20 m/sec =50 m/sec
change in km/hr
50*18/5=180km/h train speed.....in option of this question
only 120 car speed 50% is satisfy train speed 120+120/2=180 that's it......

HOW 150 comes in the initial step ?