# find distance between 2 points

$A$ and $B$ run towards each other starting from $K$ and $L$ respectively with respective speeds of 2 kmph and 3 kmph. After meeting each other, to reach $L$, if $A$ takes 7 hours less than the square of magnitude of time (in hours) taken by $B$ to reach $K$, find the distance between $K$ and $L$?

The following is how I approached: Let distance covered by $A$ before meeting be $x$, and total distance be $D$, So, distance covered by $B = D-x$ Since time taken is same, $\frac{x}{2} = \frac{(D-x)}{3}$ Solving which we get $D = \frac{5x}{2}$ Also relation for time taken by both to compete the rest of the part is given as, $(x/3)^2= \frac{(D-x)}{2+7}$ Putting the value of $D = \frac{5x}{2}$ in this equation, I got the following equation, $4x^2 - 27x - 252 = 0$.

I am unable to solve further, Did I go wrong somewhere or should some other approach be taken?

Let the distance between $K$ & $L$ be $d \ km$ then

the distance $d_A$ covered by $A$ in time $t$ to meet $B$ $$d_A=\text{(speed)}\times \text{(time)}=2t$$ Similarly, the distance $d_B$ covered by $B$ in time $t$ to meet $A$ $$d_B=3t$$ For meeting each other at a point between $K$ & $L$, $A$ & $B$ together will cover the whole distance $d$, hence we have $$d_A+d_B=d$$ $$2t+3t=d\implies \color{red}{t=\frac{d}{5}}$$ This is the time taken by both $A$ & $B$ to meet each other at some point between $K$ & $L$

Now, the time taken by $A$ to reach $L$ by covering remaining distance $\left(d-\frac{2d}{5}=\frac{3d}{5}\right)$ $$t'_A=\frac{\frac{3d}{5}}{2}=\frac{3d}{10}$$ Similarly, the time taken by $B$ to reach $K$ by covering remaining distance $\left(d-\frac{3d}{5}=\frac{2d}{5}\right)$ $$t'_B=\frac{\frac{2d}{5}}{3}=\frac{2d}{15}$$ Now, applying the given condition that the time $t'_A$, taken by $A$ to reach $L$, is $7$ hours less than square of magnitude of time $t'_B$ taken by $B$ to reach $K$ as follows $$t'_A=(t'_B)^2-7$$ $$\frac{3d}{10}=\left(\frac{2d}{15}\right)^2-7$$ $$135d=8d^2-14\times 225$$ $$8d^2-135d-3150=0$$ $$\implies d=\frac{-(-135)\pm\sqrt{(-135)^2-4(8)(-3150)}}{2(8)}=\frac{135\pm 345}{16}$$ but, distance $d$ is positive hence the distance between $K$ & $L$ $$d=\frac{135+345}{16}$$$$=\frac{480}{16}=\color{blue}{ 30 \ km}$$

• You have used T'a' as 2d/10 instead of 3d/10 Commented Jul 24, 2015 at 20:44
• I mean, u derived the time taken by A to complete the remaining distance, t'a' = 3d/10 , but used 2d/10 in the equation, Commented Jul 24, 2015 at 21:03
• Yes. you are absolutely right. I spotted & corrected the error. you may check it. Commented Jul 24, 2015 at 22:08

Hint: If you're having trouble solving for $x$ in the final equation I would recommend noting that for a general quadratic equation $ax^2+bx+c$, we may use $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ (The Quadratic formula).

That should get you some mileage.