Please continue the discussions on the following thread

Official CAT 08 Quant Thread

All the best

Please continue the discussions on the following thread

Official CAT 08 Quant Thread

All the best

Guys thread mein no activity since long .. what happened

dashing SaysCan anyone solve this ?

clearly from the question we can see that as hemant is faster

he is the one who started from shallow end

let the length of pool be z

and speeds of hemant be x and that of ajay be y

then 18.5/y=(z-18.5)/x for the first meeting

and then for the second meeting

z/x+45+(z-10.5)/x=z/y+45+10.5/y

eliminating y from the second using first we get a very simple quadratic

(2z-10.5)/x=(z+10.5)(z-18.5)/18.5x

on solving further z=0 or z=55.5-10.5=45

gautamgomzi SaysQ. Hemant and Ajay start a two-length swimming race at the same moment but from opposite ends of the pool. They swim in lane and at uniform speeds, but hemant is faster than Ajay. They first pass at a point 18.5 metre from the deep end and having completed one length, each one is allowed to rest on the edge for exactly 45 seconds. After setting off on the return length, the swimmers pass for the second time just 10.5 m from the shallow end. How long is the pool?ANS 45 metrePlz help me with these probz...GK

Can anyone solve this ?

i have further explained the soln.. go have a look

hi implex thnaks for the solution

but could you please tell how you arrived at the equation

unable to understand that

basically we need to find the time saved from the other case...

let me explain once again ...

suppose ravi speed is x and his wife speed is 5x(cars speed)

and the distance between his home and station be y

suppose he arrives at the station z mins earlier than when his wife comes to pick him up... this is the time he would need to wait..

now he decides not to wait...

so he walks right..

now use little bit of relative velocity and stuff (5x+x)=6x and u will get the soln .. as done by me...

if not .. i will explain it further

gautamgomzi SaysQ. Rahim sets out to cross a forest. On the first day, he completes 1/10th of the journey. On the second day, he covers 2/3rd of the distance traveled the first day. He continues in this manner, alternating the days in which he travels 1/10th of the distance still to be covered, with days on which he travels 2/3rd of the total distance already covered. At the end of the 7th day he finds that 22 km more will see the end of his journey. How wide is the forest?ANS - 120kmGK

Total

First Day -- 1/10-------------- 1/10

Second --- 2/30 ----------1/6

Third -- 5/60 -----------------1/4

Fourth -- 1/6 -------------------5/12

Fifth -- 7/120 -------------------57/120

Sixth -- 38/120 ---------------95/120

Seventh -- 25/1200 -----------975/1200

Remaining 225/1200 of the distance which is equal to 22.5 km

So total = 120KM

Hey PuysThe thread u guys are using is "The Official CAT 2007 Quant Thread"One more thread by the name of "The Official CAT 2008 Quant Thread" has been created

would request you to please post all the questions and answers on that thread as it is tuff to keep track of both the threads..

Also would request the MODS to please close this thread

Cheers

Kartik

Please if u could do the same

Thanks

Kartik

gautamgomzi SaysQ. Two people A and B start from P and Q (distance = D) at the same time towards each other. They meet at a point R which is at a distance 0.4D from P. They continue to move to and fro between the two points. Find the distance from point P at which the fourth meeting takes place ?ANS- 0.8D************************************************Plz help me with these probz...GK

When they both are covering D distance together the difference in the distance travelled by both is 0.2D

When they together travel 2D the diff should be 0.4D

So the secnd time they meet at 0.8D from P

Third time they meet at 0.4D from P

and the Fourth time they meet at 0.8D from P

**ANSWER = 0.8D:splat:**