Time, Speed & Distance
Aptitude

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Q.

A train covers a distance in 100 min, if it runs at a speed of 48kmph on an average. The speed at which the train must run to reduce the time of journey to 40 min will be:

 A.

30 kmph

 B.

50 kmph

 C.

80 kmph

 D.

120 kmph

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Solution:
Option(D) is correct

Time = \(\dfrac{100}{60}=\dfrac{5}{3}\)hr

Speed =48 mph

Distance $=S \times T$

$= 48 \times \dfrac{5}{3}$

$=80 \text{ km}$

Now, as per the question, journey is to be reduced to 40 min.

So, new time,

$=40 \text{ min}= \dfrac{40}{60} \text{ hr}$

$=\dfrac{2}{3} \text{ hr}$

New speed,

$=\dfrac{\text{Distance}}{\text{New Time}}$

$=\dfrac{80}{2/3}$

$=80 \times \dfrac{3}{2}$

$= \textbf{120 kmph}$

Edit: For an alternative solution (without calculating the distance), check comment by Chirag Goyal.

Edit 2: For yet another alternative solution, check comment by Vejayanantham TR.

Edit 3: For yet another alternative solution, check comment by VENI.

Edit 4: For yet another alternative solution using Ratio method, check comment by Akshay.


(10) Comment(s)


Akshay
 ()

#AKSHAY we can do simply without using pen and paper previously it was 100 min and now it is 40 min so the ratio is 2/5. So speed is increased by 5/2 of the old speed therefore 5/2 *48 = 120



Bharat Bhushan
 ()

Since distance is same use S1T1 = S2T2. hence 100/40 *48 = speed required.



VENI
 ()

$t=100$ min $v=48$ kmph

$t=40$ min $v=?$

For the same distance,

$v=\dfrac{(48*100)}{40}$

$v=\textbf{120 kmph}$



Vejayanantham TR
 ()

 $\dfrac{S_1}{S_2} = \dfrac{T_2}{T_1}$

$T_1 = 100 \text{ min} = \dfrac{100}{60} = \dfrac{5}{3}$

$T_2 = 40\text{ min} = \dfrac{40}{60} = \dfrac{2}{3}$

So, as per formula,

$\dfrac{48}{S_2} = \dfrac{2}{5}$

$S_2 = 48* \dfrac{5}{2} = 120$



Chirag Goyal
 ()

Hello there,

We don't have to calculate distance Because it is same for both cases

and we can use this as following.

Let $S_1$ is speed of train which is $48km/hr$

$T_1$ is the time $(100\ min\ or\ \dfrac{5}{3}hr)$ to complete Certain Distance $D$

$S_2$ is required Speed of Train to cover the same distance $D$ in time $T_2$ which is $40\ min\ or\ \dfrac{2}{3}hr$.

$S_1\times{T_1}$=$S_2\times{T_2}$ as Distance $D$ is same

Solving this we get

$S_2=120\ km/hr$ is the required answer.



Saif Pathan
 ()

Initial speed =48kmph

Dist=?

Time = 100min

S=D/T

48=D/100

48*100=D

D=4800

Now

When Time=40kmph and D=4800

Using formula S= D/T

i.e. S= 4800/40

S= 120 kmph

Answer 120 kmph


Shubh
 ()

You have arrived at the result but I strongly advise against mixing the units.

You are multiplying 'kmph' with 'min' in the first step $(D=4800)$.

In the second step, its effect is nullified during the division operation $(S=D/T)$, when you have divided again by min'.

Very bad approach of solving questions this way though.


Rishabh Chaudha
 ()

the ans is 120 km/hr



Anujpachauri
 ()

How can you travel 80 km in 40 min with speed of 30 km?

and answer is 120kmph

sol is.

total distant is $\dfrac{100}{60} \times 48=80$ km

and

speed $=\dfrac{80}{40/60}$

$\dfrac{80}{40} \times 60=120$ kmph

hence and ans is $\textbf{120 kmph}$


Deepak
 ()

Thank you Anuj for pointing out the error. Solution is updated and correct option choice is pointed as a right answer now.