Two horses start simultaneously towards each other and meet after $3h 20 min$. How much time will it take the slower horse to cover the whole distance if the first arrived at the place of departure of the second $5 hours$ later than the second arrived at the departure of the first.
MY TRY::
Let speed of 1st be a kmph and 2nd be b kmph
Let the distance between A and B be d km
d = 10a/3 + 10b/3
and
d/a – d/b = 5
now i cant solve it. 🙁
Spoiler: The answer is $10$ hours.
Best Answer
Let the distance from the meeting place of the departure of the first Horse $A$ is $a $ meter and the distance from the meeting place of the departure of the first Horse $B$ is $b$ meter
So, the total distance is $a+b$ meter
So, the speed of the first horse is $\displaystyle\frac a{200}$ meter/minute and that of the second is $\displaystyle\frac b{200}$ meter/minute
So, the first horse $A$ will need to cover $b$ meter more which it will take $\displaystyle\frac b{\frac a{200}}=\frac{200b}a$ minute
So, the total time taken by $A$ will be $\displaystyle200+\frac{200b}a$ minute
Similarly, the total time taken by $B$ will be $\displaystyle200+\frac{200a}b$ minute
If $A$ is slower than $B,$ $\displaystyle200+\frac{200b}a-\left(200+\frac{200a}b\right)=300\implies 2b^2-3ab-2a^2=0\implies b=2a$ (why?)
The total time taken by $A$ will be $\displaystyle\frac{a+b}{\frac a{200}}$ minute