time speed distance 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.
 A: 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
A: First, let's identify what you actually want to solve for, which is $\frac{d}{b}$. Solve for $a$ in your first equation: $a = 3/10 d - b$ and substitute into the second equation
$$
\frac{d}{\frac{3}{10} d - b} - \frac{d}{b} = 5\\
db- d\left(\frac{3}{10} d - b\right) = 5b\left(\frac{3}{10} d - b\right)\\
d\left(2b - \frac{3}{10}d \right) = \frac{3}{2}bd - 5b^2 \\
 \frac{3}{10}d^2- \frac{1}{2}d b -5b^2 = 0
$$
then, dividing by $b^2$
$$
\frac{3}{10}\left(\frac{d}{b}\right)^2 - \frac{1}{2}\frac{d}{b} - 5  =0 \\
3\left(\frac{d}{b}\right)^2 - 5\frac{d}{b} - 50  =0 
$$
which is a quadratic in the variable you want to solve for.
A: Let $s_i$ be the speeds of the horses, with $s_1>s_2$ for definiteness. Let $d$ be the total distance. Let $t_i$ be the time taken to cover $d$, that is, $t_i = \frac{d}{s_i}$.
Clearly we have $s_i >0, d>0$.
Since the horses meet after 200 mins., we have the total distance travelled to be $d$, that is, $200(s_1+s_2) = d$. This gives $200(\frac{s_1}{d}+\frac{s_2}{d}) = 200(\frac{1}{t_1}+\frac{1}{t_2}) =1$,
Since the slower horse takes 300 mins. longer to cover $d$, we have $t_2-t_1 = 300$.
Substituting $t_1 = t_2-300$ into $200(\frac{1}{t_1}+\frac{1}{t_2}) =1$ gives $200(\frac{1}{t_2-300}+\frac{1}{t_2}) =1$, multiplying across by $t_2(t_2-300)$ simplifies to $200(2t_2-300) = t_2(t_2-300)$, which reduces to $(t_2-100)(t_2-600) = 0$. Using $t_1 = t_2-300$ gives solutions $(-200,100), (300,600)$.
The only solution with $t_i >0$ is $(300,600)$, so the answer is $t_2 = 600$ mins.
