Question

Roohi travels 300 km to her home partly by train and partly by bus. She takes 4
hours if she travels 60 km by train and the remaining by bus. If she travels 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

Step-by-step explanation:

Let the speed of train be x

and speed of bus be y

Speed= Distance / Time

or

Time = Distance / Speed

Now, firstly we have to subtract 60km from300km=240km.

Now,

4=60/x + 240/y............eq.1

25/6 = 100/x + 200/y......eq. 2

Let 1/x = a, 1/y = b

60a+240b=4.......eq. 3

multiply eq. 3 by 6,we get,

600a+240b=40.......eq. 4

6(100a + 200b ) = 25

600a + 1200b = 25......eq. 5

By elimination method......

600a + 2400b = 40

600a + 1200 = 25

(-) (-) (-)

________________

1200b = 15

b = 15/1200

b = 1/80

putting the value of b in eq. 5,we get,

600a + 1200 × 1/80 = 25

600a = 25-15

600a = 10

a = 10/600

a= 1/60

1/x = a

1/x = 1/60

x = 60

1/y = b

1/y = 1/80

y = 80

Hence, the speed of train = 60km/h

and the speed of bus = 80km/h

Answer 2

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Let the speed of train and bus be u km/h and v km/h respectively.

According to the question,

$\frac{60}{u} + \frac{240}{v} = 4$....(i)

$\frac{100}{u} + \frac{200}{v} = 4 + \frac{10}{60} = \frac{25}{6}$....(ii)

Let $\frac{1}{u} = p\: and \frac{1}{v} = q$

The given equations reduce to:

60p + 240q = 4 ....(iii)

100p + 200q = $\frac{25}{6}$

600p + 1200q = 25....(iv)

Multiplying equation (iii) by 10, we obtain:

600p + 2400q = 40....(v)

Subtracting equation (iv) from equation (v), we obtain:

1200q = 15

q = $\frac{15}{1200} = \frac{1}{80}$

Substituting the value of q in equation (iii), we obtain:

60p + 3 = 4

60p = 1

p = $\frac{1}{60}$

:. p = $\frac{1}{u} = \frac{1}{60}$, q = $\frac{1}{v} = \frac{1}{80}$

u = 60 km/h , v = 80 km/h

Thus, the speed of train and the speed of bus are 60 km/h and 80 km/h respectively.

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