Aptitude - Time and Distance - Discussion

For those who are facing problems solving equation 1 & 2.

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 u get x =60.
X:Y= 60:80 =>3:4.

Anyone please explain clearly.

Thanks @Ophi

How that 200 and 400 comes? Please explain me.

@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.

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

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.

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.

@Suresh.

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

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.