Question

Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travel 60 km by train and the remaining by the bus. If she travel 100 KM by train and the remaining by bus she takes 10 minutes longer. Find the speed of the train and the bus separately.

Answer 1

let the speed of the train be x km/h
let the speed of bus be y km/h
60/x + 240/y = 4
100/x + 200/y = 4+10/60
100 x + 200 y = 25/6
let 1/x be a
let 1/y be. b
60a + 240b = 4. eq 1.
100 a + 200 b = 25/6
600 a + 1200 b = 25. eq 2
×10 in eq 1. 600a + 2400 b = 40
600 a + 1200 b = 25

1200b = 15
b = 15/1200
b = 1/80
put b= 1/89in eq 2 .
a= 1/60

so x=60
y=80

speed of train is 60km/h
speed of bus is 80km/h

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Answer 2

Given:-

• Roohi travels 300 km to home by train and partly by bus in takes 4 hr.
• and travels 60 km by train and the bus in travels 100 km.

To find:-

• Find the speed of the train and the bus separately.

Solutions:-

• Let the Speed of train be y and the speed of bus be x.
• Total times => 4 hours

=> 4 × 60

=> 240 mins

Now,

=> 60/y + 240/x = 240

=> (60x + 240y)/xy = 240

=> 60x + 240y = 240xy

=> 60(x + 4y) = 240xy

=> x + 4y = 240xy/60

=> x + 4y = 4xy ............(i).

=> 100/y + 200/x = 250

=> (100x + 200y)/xy = 250

=> 100x + 200y = 250xy

=> 50(2x + 4y) = 250xy

=> 2x + 4y = 250xy/50

=> 2x + 4y = 5xy ............(ii).

Subtracting Eq. (ii) and (i) we get,

${2x} \: + \: {4y} \: \: = \: \: {5xy} \\ {x} \: + \: {4y} \: \: = \: \: {4xy} \\ \underline{ - \: \: \: \: \: \: \: \: - \: \: \: \: \: \: \: \: = \: \: \: \: \: \: - \: \: \: \: \: \: \: \: \: } \\ \: {x} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: \: \: {xy}$

=> y = 1

Now, putting the value of y in Eq. (i).

=> x + 4y = 4xy

=> x + 4 × 1 = 4 × x × 1

=> x + 4 = 4x

=> 4 = 4x - x

=> 4 = 3x

=> x = 4/3

Thus, Speed of bus (x) = 4/3 × 60

=> 4 × 20

=> 80 km/hr.

Speed of train (y) = 1 × 60

=> 60 km/hr.

Hence, Speed of bus is 80 km/hr and speed of train is 60 km/hr.