Aptitude - Time and Distance - Discussion
Discussion Forum : Time and Distance - General Questions (Q.No. 13)
13.
It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the cars is:
Answer: Option
Explanation:
Let the speed of the train be x km/hr and that of the car be y km/hr.
| Then, | 120 | + | 480 | = 8  |
1 | + | 4 | = | 1 | ....(i) |
| x | y | x | y | 15 |
| And, | 200 | + | 400 | = | 25 | |
1 | + | 2 | = | 1 | ....(ii) |
| x | y | 3 | x | y | 24 |
Solving (i) and (ii), we get: x = 60 and y = 80.
Ratio of speeds = 60 : 80 = 3 : 4.
Discussion:
84 comments Page 7 of 9.
Prashant said:
7 years ago
Thanks @Rahul Singh.
Sushil said:
7 years ago
Very good explanation, Thanks @Kavitha.
(1)
Sai said:
6 years ago
How to find the x and y values can anyone help me?
(2)
Bansal Hitesh said:
6 years ago
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.
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.
(15)
Juwel said:
6 years ago
It takes 4 hours for 300 km whereof 60 km is covered by train and the rest by car. It takes 20 min more if 100 km is covered by train and rest by car then what is the ration. Please explain.
Sainath said:
6 years ago
How did we got 25/3?
Asmita said:
5 years ago
How to slove eq1 and 2? please anyone explain.
(1)
Vikky said:
5 years ago
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.
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.
(2)
Akanksha said:
5 years ago
I didn't understand the step of solving the equation 1 and 2. Please someone explain me.
(2)
Deepu said:
5 years ago
Please, show how to solve equation 1 and 2 step by step? Anyone explain, please.
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