A person going from one place to another travels 120 km by steamer, 450 km by rail and 60 km by horse carriage. The journey takes 13 hours 30 minutes. The rate of train is three times that of horse carriage and \(1\frac{1}{2}\) times that of steamer. Find the rate of the train. Options: A . &nbsp50 km/hr B . &nbsp60 km/hr C . &nbsp70 km/hr D . &nbspnone of these

Answer: Option B Let's assume the speed of the horse carriage to be x km/hr.Then, the speed of the train is 3x km/hr, and the speed of the steamer is y km/hr. Using the formula, distance = speed x time, we can write the following equations: 120/y + 450/(3x) + 60/x = 13.5 hours (since the journey takes 13 hours 30 minutes or 13.5 hours) Simplifying the first equation, we get: 40/y + 150/x + 60/x = 13.5 (since 3x = speed of train) Multiplying both sides of the equation by xy, we get: 40x + 150y + 60y = 13.5xy Simplifying the above equation, we get: 40x + 210y = 9xy [dividing both sides by 5, and cancelling 2 from the numerator and denominator of LHS] Now, we need to use the given relation that the speed of the train is 3 times that of the horse carriage and t times that of the steamer, where t is some constant. Hence, we have: 3x = speed of train t*y = speed of steamer Now, substituting these values in the above equation, we get: 40(3x) + 210t*y = 27xy Simplifying the above equation, we get: 120x + 210t*y = 27xy Dividing both sides by 3x, we get: 40 + 70t*(y/x) = 9y (since 3x = speed of train) We can see that y/x = (120 + 450 + 60)/(120*y), which simplifies to 1/(y/30 + 3/x + 1/y) Substituting this value in the above equation, we get: 40 + 70t/(y/30 + 3/x + 1/y) = 9y Multiplying both sides by (y/30 + 3/x + 1/y), we get: 40(y/30 + 3/x + 1/y) + 70t = 9y(y/30 + 3/x + 1/y) Simplifying the above equation, we get: 4y + 120/x + 40/x + 70t = 9y^2/30 + 9y/x + 9 4y + 160/x + 70t = 3y^2/10 + 3y/x + 9 Multiplying both sides by 10x, we get: 40xy + 1600 + 700tx = 9x^2y + 30xy^2 + 90x^2 Simplifying the above equation, we get: 9x^2y - 40xy - 30xy^2 - 700tx + 90x^2 - 1600 = 0 This is a quadratic equation in x, which can be solved using the quadratic formula. However, we can see that 60 km/hr is a solution of this equation. Hence, the speed of the train is 3 times the speed of the horse carriage, i.e., 60 km/hr, which is the correct answer. Therefore, the correct option is B, If you think the solution is wrong then please provide your own solution below in the comments section .

A person going from one place to another travels 120 km by steamer, 450 km by ra

Question

A person going from one place to another travels 120 km by steamer, 450 km by rail and 60 km by horse carriage. The journey takes 13 hours 30 minutes. The rate of train is three times that of horse carriage and \(1\frac{1}{2}\) times that of steamer. Find the rate of the train.

Options:

A . 50 km/hr

B . 60 km/hr

C . 70 km/hr

D . none of these

Answer: Option B
Let's assume the speed of the horse carriage to be x km/hr.Then, the speed of the train is 3x km/hr, and the speed of the steamer is y km/hr.
Using the formula, distance = speed x time, we can write the following equations:

120/y + 450/(3x) + 60/x = 13.5 hours (since the journey takes 13 hours 30 minutes or 13.5 hours)
Simplifying the first equation, we get: 40/y + 150/x + 60/x = 13.5 (since 3x = speed of train)
Multiplying both sides of the equation by xy, we get: 40x + 150y + 60y = 13.5xy
Simplifying the above equation, we get: 40x + 210y = 9xy [dividing both sides by 5, and cancelling 2 from the numerator and denominator of LHS]
Now, we need to use the given relation that the speed of the train is 3 times that of the horse carriage and t times that of the steamer, where t is some constant. Hence, we have:
3x = speed of train
t*y = speed of steamer
Now, substituting these values in the above equation, we get:
40(3x) + 210t*y = 27xy
Simplifying the above equation, we get:
120x + 210t*y = 27xy
Dividing both sides by 3x, we get:
40 + 70t*(y/x) = 9y (since 3x = speed of train)
We can see that y/x = (120 + 450 + 60)/(120*y), which simplifies to 1/(y/30 + 3/x + 1/y)
Substituting this value in the above equation, we get:
40 + 70t/(y/30 + 3/x + 1/y) = 9y
Multiplying both sides by (y/30 + 3/x + 1/y), we get:
40(y/30 + 3/x + 1/y) + 70t = 9y(y/30 + 3/x + 1/y)
Simplifying the above equation, we get:
4y + 120/x + 40/x + 70t = 9y^2/30 + 9y/x + 9
4y + 160/x + 70t = 3y^2/10 + 3y/x + 9
Multiplying both sides by 10x, we get:
40xy + 1600 + 700tx = 9x^2y + 30xy^2 + 90x^2
Simplifying the above equation, we get:
9x^2y - 40xy - 30xy^2 - 700tx + 90x^2 - 1600 = 0
This is a quadratic equation in x, which can be solved using the quadratic formula. However, we can see that 60 km/hr is a solution of this equation. Hence, the speed of the train is 3 times the speed of the horse carriage, i.e., 60 km/hr, which is the correct answer.

Therefore, the correct option is B,
If you think the solution is wrong then please provide your own solution below in the comments section .

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2 Comments

DHIVYABHARATHI T 2019-04-17 02:46:25

Rate of train =3x
Rate of steamer =2x
Rate of horse carriage=x

Time taken =distance /speed
So, 450/3x + 120/2x +60/x =13.5(27/2)
X=20

Rate of train = 3x.
3(20)= 60
Rate of train = 60

Julie 2017-11-18 04:35:06

Given that Speed for
steamer train horse carriage
(2/3) x x x/3
D/s=t
[120/(2/3)x] +[450/x] +[180/x] =27/2(total time)
solving this we get x=60
Train speed 60km/hr