Question

A train runs at 60 km per hour for 20 minutes, 90 km per hour for 10 minutes and 40 km per hour for 15 minutes.

(i) Find out the average speed of the train.
(ii) How much distance it has travelled from the 15th minute to 40th minute.
​

Answer 1

Solution :-

Case ❶ :-

→ Speed = 60km/h.

→ Time = 20min = (20/60) = 1/3 hour.

→ Distance Covered = S * T = 60 * (1/3) = 20km.

Case ❷ :-

→ Speed = 90km/h.

→ Time = 10min = (10/60) = 1/6 hour.

→ Distance Covered = S * T = 90 * (1/6) = 15km.

Case ❸:-

→ Speed = 40km/h.

→ Time = 15min = (15/60) = 1/4 hour.

→ Distance Covered = S * T = 40 * (1/4) = 10km.

Adding All we get :-

→ Total Distance covered = 20 + 15 + 10 = 45km.

→ Total Time = 20min. + 10min. + 15min. = 45min. = (45/60) = (3/4) Hour.

So,

☛ Average Speed = ( Total Distance covered ) / ( Total Time )

☛ Average Speed = (45) / (3/4) = 45 * (4/3) = 60km/h. (Ans).

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Part Second is Not Clearly written now.

It should be :- A train runs at 60 km per hour for first 20 minutes, 90 km per hour for next 10 minutes and 40 km per hour for next 15 minutes.

now,

From case 1 we get , for first 20 min. Train cover 20km. So,

→ from 15th min to 20thmin , Train covered = 6km (1km in Each min. for 6 min. )

Now, From 20th min to 30th min. for next 10min , from case 2 ,

→ Train covered = 15 km.

Now, from 30th to 40th min. train speed was 40km/h.

So,

→ Distance Train covered in Next 10min = S * T = 40*(10/60) = 6.67km.

Hence,

☛ Total Distance covered by Train from 15th min to 40min = 6 + 15 + 6.67

= 27.67km. (Ans).

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Answer 2

The train runs at,

• $\sf{60\ km\,h^{-1}}$ for $\sf{\dfrac{20}{60}=\dfrac{1}{3}}$ hours

• $\sf{90\ km\,h^{-1}}$ for $\sf{\dfrac{10}{60}=\dfrac{1}{6}}$ hours

• $\sf{40\ km\,h^{-1}}$ for $\sf{\dfrac{15}{60}=\dfrac{1}{4}}$ hours

Then the total distance travelled by the train is,

$\displaystyle\longrightarrow\sf{s=60\times\dfrac{1}{3}+90\times\dfrac{1}{6}+40\times\dfrac{1}{4}}$

$\displaystyle\longrightarrow\sf{s=20+15+10}$

$\displaystyle\longrightarrow\sf{s=45\ km}$

And the total time taken by the train is,

$\displaystyle\longrightarrow\sf{t=(20+10+15)\ min}$

$\displaystyle\longrightarrow\sf{t=45\ min}$

$\displaystyle\longrightarrow\sf{t=\dfrac{45}{60}\ h}$

$\displaystyle\longrightarrow\sf{t=\dfrac{3}{4}\ h}$

Hence the average speed of the train is,

$\displaystyle\longrightarrow\sf{v=\dfrac{45}{\left(\dfrac{3}{4}\right)}\ km\,h^{-1}}$

$\displaystyle\longrightarrow\sf{v=45\times\dfrac{4}{3}\ km\,h^{-1}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{v=60\ km\,h^{-1}}}}$

• For the first 20 minutes, i.e., from 1st minute to 20th minute, the train travels at $\sf{60\ km\,h^{-1}.}$ Hence the speed of the train from 15th minute to 20th minute is $\sf{60\ km\,h^{-1}.}$

• For the next 10 minutes, i.e., from 21th minute to 30th minute, the train travels at $\sf{90\ km\,h^{-1}.}$

• For the last 15 minutes, i.e., from 31th minute to 45th minute, the train travels at $\sf{40\ km\,h^{-1}.}$ Hence the speed of the train from 31th minute to 40th minute is $\sf{40\ km\,h^{-1}.}$

Distance travelled by the train from 15th minute to 20th minute is,

$\displaystyle\longrightarrow\sf{s_1=60\times\dfrac{20-15+1}{60}}$

$\displaystyle\longrightarrow\sf{s_1=20-15+1}$

$\displaystyle\longrightarrow\sf{s_1=6\ km}$

Distance travelled by the train from 21th minute to 30th minute is,

$\displaystyle\longrightarrow\sf{s_2=90\times\dfrac{30-21+1}{60}}$

$\displaystyle\longrightarrow\sf{s_2=90\times\dfrac{1}{6}}$

$\displaystyle\longrightarrow\sf{s_2=15\ km}$

Distance travelled by the train from 31th minute to 40th minute is,

$\displaystyle\longrightarrow\sf{s_2=40\times\dfrac{40-31+1}{60}}$

$\displaystyle\longrightarrow\sf{s_2=40\times\dfrac{1}{6}}$

$\displaystyle\longrightarrow\sf{s_2=\dfrac{20}{3}\ km}$

Hence the total distance travelled by the train from 15th minute to 40th minute is,

$\displaystyle\longrightarrow\sf{s=6+15+\dfrac{20}{3}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{s=\dfrac{83}{3}\ km}}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{s=27.67\ km}}}$