Question

Mansi travels 300 kms to her native partly by train and partly by bus. She takes
4 hours, if she travels 60 kms by train and the remaining by bus. If she travels 100 kms
by train and the remaining by bus, she takes 10 minutes longer. Find the average
speed of the train and the bus separately.​

Answer 1

Answer:

80km/hr

Step-by-step explanation:

Let the speed of the train be x km/hr and the speed of the bus is y km/hr.

So according to question and using

Time = \frac{Distance}{speed}Time=

speed

Distance

Total distance =300 km

Mansi travels 60 km by train and 300−60=240 by bus in 4 minute,

\frac{60}{x} + \frac{240}{y} = 4

x

60

+

y

240

=4

and 100 km by train, 300−100=200 by bus, and takes 10 minutes more,

\frac{100}{x} + \frac{200}{y} = 4 + \frac{1}{6} = \frac{24 + 1 }{6} = \frac{25}{6}

x

100

+

y

200

=4+

6

1

=

6

24+1

=

6

25

Now, let

\frac{1}{x} = a \:

x

1

=a

and

\frac{1}{y} = b

y

1

=b

then 60a+240b=4.............(1)

100a+200b=25/6----(2)

multiply (1) by 5 and (2) by 6 we get

300a+1200b=20..........(3)

600a+1200b=25...........(4)

Subtracting (3) and (4) we get

−300a=−5

a = \frac{1}{60}a=

60

1

Putting the value of a in (1) we get

60 \times \frac{1}{60} + 240b = 460×

60

1

+240b=4

\begin{gathered}240b = 3 \\ b = \frac{1}{80}\end{gathered}

240b=3

b=

80

1

Now ,

\begin{gathered}\frac{1}{x} = a \\ a = 60 km/h\end{gathered}

x

1

=a

a=60km/h

\begin{gathered}\frac{1}{y} = b \\ b = 80 km/h\end{gathered}

y

1

=b

b=80km/h

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

Answer 2

$\huge\bf\purple{\mid{\fbox{\underline{\underline{Answer}}}}\mid}$

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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