time speed distance

Question

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.

Can you tell us a little bit about what you have tried so far? — BaronVT, Oct 25 '13 at 15:04
@BaronVT added..i dont think it will help, that is why i didnt write it earlier — RbG, Oct 25 '13 at 15:14
Thanks, and that's ok - mostly we will want to see that you have put some effort into the problem (instead of just 'dumping' it here), and it also helps us identify and address what is the main difficulty in your understanding. — BaronVT, Oct 25 '13 at 15:18
@BaronVT...will remind it from next time for sure...and please provide a solution — RbG, Oct 25 '13 at 15:21

score 2 · Accepted Answer · answered Oct 25 '13 at 15:21

2

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

answered Oct 25 '13 at 15:21

lab bhattacharjee

274,582

the answer must have numeric value ..not any variable...answer is 10..sorry but i want answer with numeric value(i.e. 10) – RbG Oct 25 '13 at 15:27
@RitabrataGautam, put $b=2a$ in the last expression to get the time $=600$ minutes – lab bhattacharjee Oct 25 '13 at 15:28
why $b=2a$ ..how r u deriving it???? – RbG Oct 25 '13 at 15:32
@RitabrataGautam, do you see its a quadratic equation in $b$? – lab bhattacharjee Oct 25 '13 at 15:33
i didnt understand your last two steps – RbG Oct 25 '13 at 15:34
yeah thats what i can understand... – RbG Oct 25 '13 at 15:36
@RitabrataGautam, then solve for $b$ in terms of $a$ – lab bhattacharjee Oct 25 '13 at 15:36
i have a querry..this is a aptitude question comes in an entrance test..and i am supposed to do it with in 1 min...is there any shorter way?? @lab – RbG Oct 25 '13 at 15:39
@RitabrataGautam, Wish I had a shorted solution ? Please notify me if you have one or a solution without algebra – lab bhattacharjee Oct 25 '13 at 15:40
@RitabrataGautam, here is one (http://pagalguy.com/forums/quantitative/official-quant-thread-cat-2013-t-88456/p-16608898/r-16610200) – lab bhattacharjee Oct 25 '13 at 15:45

BaronVT · Answer 2 · 2013-10-25T16:48:10.310

1

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.

edited Oct 25 '13 at 16:48

answered Oct 25 '13 at 15:38

BaronVT

13,613

am sorry but applying your method i am getting d/b = 5 , -5/3 ...but the ans is 10 :( – RbG Oct 25 '13 at 15:45
i think you committed a slight typing mistake ...it would be 50 instead of 25 – RbG Oct 25 '13 at 15:47
Yeah, you're right. Sorry for the confusion. – BaronVT Oct 25 '13 at 16:34
vt ...... please explain one thing please...time taking by slower is d/a...why u calculated d/b????? – RbG Oct 26 '13 at 08:59
Ah, you're right; I've made another mistake. Anyway, you can solve for $d/b$, then use $d/a = d/b + 5$. – BaronVT Oct 28 '13 at 14:43

score 0 · Answer 3 · answered Oct 25 '13 at 15:55

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.

time speed distance

3 Answers3