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 3 of 9.
Abi said:
9 years ago
200/x + 400/y = 25/3.
200 (1/x + 2/y) = 25/3.
1/x + 2/y = 25/3 * 200 (multiply 3 * 200).
1/x + 2/y = 25/600.
1/x + 2/y = 1/24.
By solving this, we get 3 : 4.
200 (1/x + 2/y) = 25/3.
1/x + 2/y = 25/3 * 200 (multiply 3 * 200).
1/x + 2/y = 25/600.
1/x + 2/y = 1/24.
By solving this, we get 3 : 4.
(1)
Namita said:
9 years ago
How, from where that 1/15 came?
(1)
Ophi said:
1 decade ago
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.
(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.
(1)
Asmita said:
5 years ago
How to slove eq1 and 2? please anyone explain.
(1)
Hrushikesh said:
2 decades ago
Its not 8*(20/60). Its 8+(20/60) that gives us 25/3.
(1)
Hari said:
1 decade ago
Is there any short method for this?
(1)
Manthan Patel said:
4 years ago
Anyone please explain clearly.
(1)
Prasad said:
1 decade ago
Train speed = 80/20*60 = 240km.
Car speed = 600-240 = 360km.
Ratio 240:360 or 3:4.
Car speed = 600-240 = 360km.
Ratio 240:360 or 3:4.
(1)
Mathi said:
1 decade ago
Here to solve short cut method....
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
15y+60x=24y+48x
60x-48x=24y-15y
12x=9y
x/y=9/12 --> x/y=3/4
hence 3:4
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
15y+60x=24y+48x
60x-48x=24y-15y
12x=9y
x/y=9/12 --> x/y=3/4
hence 3:4
(1)
Rajeshwari said:
1 decade ago
Hi kishore,
The first eqn 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
The first eqn 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
(1)
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