Question

Two trains A and B start from two points P1 and P2 respectively at the same time and travel towards each other. The difference between their speed is 10 kmph and train A takes one hour more to cover the distance between P1 and P2 as compared to train B. Also by the time they meet, train B has covered 200/9 km more as compared to train A. What is the distance between P1 and P2?​

Answer 1

Answer:

The distance between P1 and P2 is 200 km

Step-by-step explanation:

Let the speed of the trains A and B be $V_A$ and $V_B$ respectively

$V_B-V_A=10$

Let the distance between P1 and P2 be d

Time taken by A - Time taken by B = 1

or, $\frac{d}{V_A}-\frac{d}{V_B}=1$

or, $d(\frac{V_B-V_A}{V_A.V_B})=1$

$\implies d(\frac{10}{V_A.V_B})=1$

$\implies V_A.V_B=10d$

By the concept of relative velocity, the time in which the trains meet

$t=\frac{d}{V_B+V_A}$

The distance covered by B in t time $=V_Bt$

The distance covered by A in t time $=V_At$

According to the question

$V_Bt-V_At=\frac{200}{9}$

$\implies t(V_B-V_A)=\frac{200}{9}$

$\implies t\times 10=\frac{200}{9}$

$\implies \frac{d}{V_B+V_A}=\frac{20}{9}$

$\implies V_B+V_A=\frac{9d}{20}$

Now

$(V_B-V_A)^=(V_B+V_A)^2-4V_A.V_B$

$\implies 10^2=(\frac{9d}{20})^2-4\times 10d$

$\implies 100=\frac{81d^2}{400}-40d$

$\implies 40000=81d^2-16000d$

$\implies 81d^2-16000d-40000=0$

$\implies d=\frac{16000\pm \sqrt{(16\times 10^3)^2+4\times 40000\times 81}}{2\times 81}$

$\implies d=\frac{16000\pm \sqrt{256\times 10^6+12960000}}{162}$

$\implies d=\frac{16000\pm \sqrt{10^4(25600+1296)}}{162}$

$\implies d=\frac{16000\pm \sqrt{10^4(26896)}}{162}$

$\implies d=\frac{16000\pm 16400}{162}$

$\implies d=\frac{400}{162}, \frac{32400}{162}$

$\implies d=\frac{400}{162}, \frac{32400}{162}$

$\implies d=2.5, 200$

But 200/9 is already greater than 2.5

Thus distance between P1 and P2 cannot be 2.5 km

Therefore, the distance between P1 and P2 = 200 km

Hope this helps.

Answer 2

Answer:

what is the ratio of the speed of a and b

2:1