time taken to meet = 80 min.
let A in x hr, then B in x+2...
I think some info is missing in this question....

shankar.ashwin wrote:2 people A and B at opposite end of a road start walking towards each other at 10:00 a.m; they meet together at 11:20 a.m, A reaches the end of the road (point where B had begun) 2 hrs before B reaches the other end. Find the total time taken for travel of the entire journey by A.

this is an incomplete question.

is the answer 2 hours?

abhimanyu.tanwar wrote:is the answer 2 hours?

Question is incomplete. You cannot find the answer with the information provided.

time unit is an hour
distance = d
cross point time = 4/3
times: A=B-2
speeds: d/(B-2), d/B
[spoiler]combined speed: 2d(B-1)/B(B-2)[/spoiler]
equation => 4/3*[d/(B-2)+d/B]=d, 8d(B-1)/3B(B-2)=d, 8(B-1)/3B(B-2)=1, 8(B-1)=3B(B-2), 8B-8=3B^2-6B, 3B^2-14B+8=0, B(1,2)=[14 +- sqroot(196-96)]/6, B(1,2)=(14 +- 10)/6 <> (2/3;4)

since cross point time was 4/3 the time spent by B cannot be 2/3 (ignore). Admit 4. Time spent by A is (4-2)=2 hours

2 hours is my answer

shankar.ashwin wrote:2 people A and B at opposite end of a road start walking towards each other at 10:00 a.m; they meet together at 11:20 a.m, A reaches the end of the road (point where B had begun) 2 hrs before B reaches the other end. Find the total time taken for travel of the entire journey by A.

@ Pemdas

4/3*[d/(B-2)+d/B]=d

This equation is incorrect. On the RHS you have a distance unit on the LHS you have a Speed unit. Kindly re-evaluate your equation.

IMO the problem definitely lacks information.

let us assume, A travels at avg speed 'a' and B travels at avg speed 'b'. Since nothing is mentioned about the relative velocities of these two individuals we assume (a,b) to remain constant.

At 4/3 hrs, they meet.

Thus total distance of journey = 4/3*(a+b)
Now lets say A completes his journey in t1 time. thus B will have completed his in t1+2 hrs.
Since we are aware of Total Distance and assumed A's speed to be a we can formulate the total time taken by A to complete the journey, which is:

t1 = 4/3*(a+b)/a
= 4/3 +4/3*(b/a)

Since no ratio is provided to correlate both the speeds, we are lacking that info. Notice that if it was mentioned that B completes his journey in 2t1 as opposed to t1+2 then we could have determined t1 as t1 = 4/3 + 4/3*1/2 = 4/3+2/3 = 6/3 = 2hrs.

Guys apologies. There was a small mistake in the way I interpreted the problem.
This problem does have some data missing to determine a finite solution.

Now try this,

2 people A and B at opposite end of a road start walking towards each other at 10:00 a.m; they meet together at 11:20 a.m, Had A and B begun their journey at the same point at the same time, A would have reached the end of the road 2 hrs before B did. Find the total time taken for travel of the entire journey by A.

shankar.ashwin wrote:Guys apologies. There was a small mistake in the way I interpreted the problem.
This problem does have some data missing to determine a finite solution.

Now try this,

2 people A and B at opposite end of a road start walking towards each other at 10:00 a.m; they meet together at 11:20 a.m, Had A and B begun their journey at the same point at the same time, A would have reached the end of the road 2 hrs before B did. Find the total time taken for travel of the entire journey by A.

Hey,

Can you please tell me what is the source of this question ?

this question also does not have any absolute answer..

let me explain:- let the total distance between A and B is equal to x.

As per second part of the question-- If A takes 't' hours to complete X miles
then 'B' will take 't+2' hrs.

A's speed = x/t miles/hr
B's speed = x/(t+2) miles/hr

now let's go to first part of the question -

A and B meets at 11.20 am i.e after 1 hr 20 mins

or after 4/3 hrs

Distance covered by A = 4/3*(x/t)
Distance covered by B = 4/3*[x/(t+2)] -- this is the distance which A will have to cover

x/t hrs ---- 4/3 hrs
[x/(t+2)] ---- 4/3*(t/x)*[x/(t+2)] hrs
or ---------- 4/3*t/(t+2) hrs

Therefore total time taken by A to cover X miles--

4/3+ 4/3*t/(t+2) hrs

shankar.ashwin wrote:Guys apologies. There was a small mistake in the way I interpreted the problem.
This problem does have some data missing to determine a finite solution.

Now try this,

2 people A and B at opposite end of a road start walking towards each other at 10:00 a.m; they meet together at 11:20 a.m, Had A and B begun their journey at the same point at the same time, A would have reached the end of the road 2 hrs before B did. Find the total time taken for travel of the entire journey by A.

starting from opp points: Taking the time in mins
a = rate of A
b = rate of B
when they meet at 11:20am (80 mins from start time) they both must have covered the entire distance D
so D = 80a + 80b => D = 80(a+b) ---> eq1

starting from same points: Taking time in mins
let time taken by B to cover distance D be x mins
so time taken by A ot cover D is x-120 mins (given, A would have reached the end of the road 2 hrs before B did)
So D = bx ---> eq2
& D = a(x-120) ---> eq3
from 1 & 2, b = (80a)/(x-80)
1 & 3, a(x-120) = 80b
subs b in above eq => x = 40mins or 240mins
40mins does not satisfy, so 240mins or 4hrs is the ans.

please let me know if this is the correct ans

The OA is 2 hrs

My soln;

Let speed of A = s1
speed of B = s2
Total time taken by A = t
Distance between 2 points = d

s1 * t = d .... (1)
s2 * (t+2) = d ..... (2)
(s1 + s2) 4/3 = d ....(3) [Both would have covered the total distance when they meet after 1hr 20 mins]

From (1) and (3)

s1*t = (s1+s2) 4/3
s1(3t-4) = 4s2 ......(4)

From (2) and (3)

4s1=s2(2+3t) .......(5)

From (4) and (5)

4/(3t-4) = (2+3t)/4

3t^2 - 2t - 8 = 0

t = -4/3 (or) 2

So time taken by A = t = 2hrs.
Time taken by B = t+2 = 4hrs.

I guess its a fairly difficult problem, maybe 750+

@Proleefeek, on the right hand side I have a distance, that's correct.
on the left hand side I also have a distance. When A and B cover small distances towards each other, they will meat at some point and will definitely cover ONE whole distance upon the time of cross point, AGREE?

So, on the LHS and RHS is the same value - distance.

p.s. before judging a solution as correct or incorrect, plz weigh all pros and cons

Proleefeek wrote:@ Pemdas

4/3*[d/(B-2)+d/B]=d

This equation is incorrect. On the RHS you have a distance unit on the LHS you have a Speed unit. Kindly re-evaluate your equation.

IMO the problem definitely lacks information.

let us assume, A travels at avg speed 'a' and B travels at avg speed 'b'. Since nothing is mentioned about the relative velocities of these two individuals we assume (a,b) to remain constant.

At 4/3 hrs, they meet.

Thus total distance of journey = 4/3*(a+b)
Now lets say A completes his journey in t1 time. thus B will have completed his in t1+2 hrs.
Since we are aware of Total Distance and assumed A's speed to be a we can formulate the total time taken by A to complete the journey, which is:

t1 = 4/3*(a+b)/a
= 4/3 +4/3*(b/a)

Since no ratio is provided to correlate both the speeds, we are lacking that info. Notice that if it was mentioned that B completes his journey in 2t1 as opposed to t1+2 then we could have determined t1 as t1 = 4/3 + 4/3*1/2 = 4/3+2/3 = 6/3 = 2hrs.

@Pemdas

Sorry mayte, just that your equations were too cramped up. Cheers.

and the problem shows pretty solvable with the existing data.

Proleefeek wrote:@Pemdas

Sorry mayte, just that your equations were too cramped up. Cheers.

shankar.ashwin wrote:
I guess its a fairly difficult problem, maybe 750+

Thanks Shankar. Might you also appraise us with the source of this problem?