Aptitude - Time and Distance - Discussion

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.

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.

Hi manasa,


let the speed of the train be x kmph
let the speed of the car be y kmph
it is given that 120km is done by train,so D=120
by the formula D=S*T
T=120/X
and also given the rest is done by the car,
so D=600-120=480 km
T=480/y
then,120/x+480/y=8(read the question given problem clearly,in pbm it is given that to complete 600km journey,it will take 8hrs).
==>1/x+4/y=1/15...............eq1
similarly in case 2:
time for
train=200/x
car=400/y
and also given in pbm 20 min more so,
200/x+400/y=25/3(8hrs+20min=25/3)
==>1/x+2/y=1/24...............eq2
for solving these two equations as mathi said
then u will get 3:4



Hope u understand.
Bye

The first equation is:

(120/x)+(480/y) = 8.
120[(1/x)+(4/y)] = 8.
(1/x)+(4/y) = 8/120.
(1/x)+(4/y) = 1/15.

Similarly in the second eqn:

(200/x)+(400/y) = 25/3.
200[(1/x)+(2/y)] = 25/3.
(1/x)+(2/y) = 25/(3*200).
(1/x)+(2/y) = 1/24.

Now you want to equate both the equations so try and get the common values of xy for both equations.

1/x+4/y = 1/15 --> take xy as multiples.
(y+4x)/xy = 1/15 --> take 15 to left hand side.
15y + 60x = xy.

Similarly,
1/x+2/y = 1/24.
24y+48x = xy.

Now easy to equate, substitute value of xy.
15y+60x = 24y+48x.
60x-48x = 24y-15y.
12x = 9y.
x/y = 9/12 --> x/y = 3/4.

Hence 3:4.

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.

Train covers 80Km more and car covers 80Km less, as a result 20min more is required. Suppose train covers 240Km(i.e. 3*80) more and car covers 240km(i.e. 3*80) less than it tooks 60min(i.e. 3*20) more.

Similarly, if train covers 480km more and car covers 480km less than time will increase by 120min i.e. 2hr. It means it the whole journey is covered by train only than it takes 10 hrs(8+2) to complete it. so speed of train = 600/10 = 60km/hr.

Now, time taken by train = 120/60 = 2 hrs.
So, time taken by car = 8 - 2 = 6 hrs.
And, speed of car = 480/6 = 80km/hr.
Speed of train : speed of car = 60:80 = 3:4.

Distance =600 km
Time take =8 hrs.
Distance =Speed *time Then
Time = Distance/Speed
SO
Total time = Total Speed.
Suppose train speed = X, And Car speed = Y.

Then 120/X +480/Y = 8 hrs.
Then 120 take common,
120[1/X+4/Y] =8 hrs.
1/X+4/Y = 8/120
1/X+4/Y = 1/15 ----> (1) equation.

Total time is 8 hrs.20 min. mean(8*1/3)
8*1/3 =25/3
200/X+400/Y =25/3 Then
200 Common.

200[1/X+2/Y] = 25/3.
1/X+2/Y =25/3*200.
1/X+2/Y =1/24 ---> (2) equation.

Solve both (1),(2) eqn.

We get 2/Y =1/15-1/24,
When solve this we get 2/Y =1/40,
Y=80 put this value on the first eqn.
You get X = 60
So the ratio of X:Y 3:4.

Hello.


Here we can solve the equation 1 & 2 by multiplying equation 1 & 2 by number 2.

i.e. multiply eq 1 with 2 it becomes,
2/x + 8/y = 2/15,
Now multiply eq 2 with 2 we get,
2/x + 4/y = 2/24.
Now, solving eq 1 & 2.
i.e. subtract eq 2 from eq 1 we, get,
Eq 1 - Eq 2.
2/x + 8/y = 2/15.
-( 2/x + 4/y ) = -2/24.
By cancellation and sloving we get:


4/y = 2/15 - 2/24.
By taking L.C.M we, get;

4/y = (8-5)/60,
4/y = 1/20.
i.e.. y= 80 then substitute the value of y in eq1 or eq 2 u will get the value of X as 60
Then their ratio is 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.

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.