Aptitude - Time and Distance - Discussion

train : car = time
120 : 480 = 8hr.
200 : 400 = 8hr 20min.

So, 80km =20min more by train.
480km dist. by train in 2hr.
Then the speed of the train is 60k/h,
Total time 8hr.
8 - 2 = 6hr.
Car speed = 480/6 =80.
T:C = 60:80.
= 3:4.

@All.

Here is my explanation.
120/x - 480/y = 8
by taking 120 commons we get
1/x - 4/y = 8/120,
1/x -4/y = 1/15 ---> (1)
200/x - 400/y = 8+20/60
min so div by 60.

By taking common 200 we get
1/x - 2/y = 1/24
Solving both equations.
1/x + 4/y = 1/15 --> (1)
1/x + 2/y = 1/24 --> (2)
By subtracting (1),(2)
4/y -2/y = 1/15 - 1/24,
2/y = 8-5/120,
2/y = 1/40,
Y = 80.

Sub y = 80 in eq (1)
X = 60
Therefore x : y.
60:80 div by 2,
And finally answer = 3:4 we get.

Let the speed af the train be T and car be C.

The time taken by train and car is:
120/T+480/C=8hrs is equation 1
200/T+400/C=8hrs 20 mins which is 200/T+400/C=25/3 is equation 2

Multiply equation (1)by 5 and equation (2) by 3 to cancel the fractions;
Therefore,
600/T+2400/C = 40
600/T+1200/C = 25

Now subtract equation (1) and (2);
Therefore,
(600/T + 2400/C = 40) - (600/T + 1200/C = 25).
1200/C = 15,
C = 1200/15,
C = 80.

Substitute C = 80 in equation (1)
600/T + 2400/80 = 40,
600/T = 40 - 2400/80.
600/T = 800/80,
T = 600/10.
T = 60.

So the ratio for train and car would be:
60:80
Therefore the ratio is 3:4.

How did 8 come in solution?

Please explain me.

Let the speed of the train be =x km/hr.
And the speed of the car be =y km/hr.

It takes eight hours(8hrs) for a 600
km journey, if 120;
km is done by train and the rest by car(600-120=480).

Convert the above word statement into mathematical form-
Then,
120x + 480y = 8.

Take the number common(120) from both sides of the equations and so it is removed.
∴1x+4y=1/15 ---> (a)

Also, given that; It takes 20.
minutes more, if 200.
km is done by train and the rest by car.

Convert minutes in an hour
∴200/x+400/y = 8 + 20/60.

Simplify the above equation;
⇒200/x+400/y = 8 + 1/3
⇒200/x + 400/y = 25/3.

Take common(200) from both sides of the equation and remove
∴1/x+2/y=1/24 ---> (b).

Take equations (a) and (b)
and find the value of the unknown.

By using the elimination method- subtract equation (b) from equation (a).
⇒4/y−2/y=1/15−1/24.
( 1x is removed in subtraction)
taking lcm.

⇒2/y = 8 - 5/120
2/y = 3/120,
y = 80km/hr.
Substitute value of “y” in equation (a)
∴1/x+4/80=1/15 ⇒1/x+1/20=1/15.

Make unknown “x” as the subject –
∴1x = 1/15−1/20,
⇒1x = 20−15/(15)(20),
⇒1/x = 5/300,
⇒1/x = 1/60,
⇒x = 60 km/hr.
Now the ratio of the speed of car and train is;
=x/y = 60.80 = 3/4.

@Suresh.

Why 120/x + 480/y = 8, How 8 come?
Please explain.

We got that,

Eq 1: 1/x+4/y=1/15 =>1/x=1/15-4/y.
Eq 2: 1/x+2/y=1/24 =>1/x=1/24-2/y.

Now equate the 1/x values.
i.e 1/15-4/y = 1/24-2/y =>2/y = 1/15-1/24.

So you get y = 80.
By submitting y in the eq 1: 1/x+4/80=1/15 and you get x =60.
X : Y = 60:80 =>3:4.

Question in given that,
Total Distance covered by train & car = 600 km.
By train covered distance: 120 Km.
By car covered distance: 600-120 = 480 Km.
Time taken covered by total distance both train and car = 8 hours.

Let,
Train speed is = x kmph.
Car speed is = y kmph.

So,
120/x+480/y=8 --> (1)
In second part given time more 20 minutes=8+20/60=8+1/3=25/3
200/x+400/y=25/3 --> (2)
In eq. 1 and 2 for equal x or y number equal for getting value x and y in one
(1) eq. multiply 5 and (2) equation 3, getting
600/x+2400/y=40 --> (3)
600/x+1200/y=75/3--> (4)
Subtracting eq. 3 in 4, getting;
1200/y = 40 - 75/3.
1200/y = 120 - 75/3.
1200/y = 45/3
15Y = 1200.
Y = 80kmph speed of Car.
Putting y value in eq. (2).
200/x+400/80 = 25/3.
200/x+5 = 25/3.
200+5x/x = 25/3.
200+5x = 25x/3.
200 = 25x/3-5x
200 = 25x-15x/3
200 = 10x/3
600 = 10x
x = 60kmph speed of the train.
the ratio of speed train: car.
= 60:80.
= 3/4.

Why 120/x + 480/y = 8, How 8 come? Please explain.

@Karan.

It's Given in the question;
Distance by train=200.
Remaining distance by car, So 600-200 = 400 is the distance covered by a car.