RE: hii....can anybody give me the solution of followi...
Hi Frenz,Trying to help u a little frm my side.(1) 50 minutes ago it was 4 times ..... ( For complete one, refer the 2nd question frm Shakuntala Devi's Puzzles to puzzle you)Let the time from now(Z) to 6 pm be x minsThen from the question, time from 3pm(X1) to the time that was fifty mins ago(Y) = 4x minsConsider the below fig X1--------------------------------Y------------------------------Z-------------------------------X2 3pm <----4x-----> 50mins <-------x--------> 6pm 4x + 50 + x = 180==> 5x = 130 ; x = 26 mins(2) Odd man out.3 10 20 27 54 61 _ 3 10 20 27 54 61 _____ +7 *2 +7 *2 +7 (s/b *2)So the answer is 61 * 2 = 122 ( The previous reply was right, except that a small calc mistake was done with the answer )Hope the above solution is made clearer. I'm ready to solve any other questions to the extrent I can.Frenz, I'm preparing for my dream company. Please pray for my success.Good luck to all job seekers.My sincere suggestion, " Have confidence on you and have faith in God". If you believe that you have done your part well, then leave rest to God. Prayers do have answers but only if we are correct in our part.So good luck to all my dear friends.Regards,Sats
RE: hii....can anybody give me the solution of followi...
Hi,,,The ans for 1). Fifty minutes ago if it was four times as many minutes past three o,clock,how many minutes is it to six o,clock?sol:The trick to this is understanding the questn properly... the question often misleads most ppl...Thanks to shakuntala devi(puzzles 2 puzzle u page 1)U shud understand that 50 mins ago it was 4 times the mins (reuired to be 6 o'clock) past 3 o'clock.let x be the mins to 6 oclock.Now we know mins betn 3 to 6 are 180.These 180 would comprise of x+50+4x (as it was 4 times mins required to b 6)thrfore.. x+50+4x=180i.e 5x = 180-50=130x=130/5 = 26 mins
RE: hii....can any body give me the solution of followi...
Consider all the Odd positioned number as one series and all the Even positioned numbers as another seriesEven Series: 10 27 61 (they can be written as 10 10+17*1(27) 27+17*2(61).....then ODD: 3 20 54 __ (it is similar to 3 3+17*1(20) 20+17*2 (54) 54+17*3(104)Ans: 104... better to go by the alternatives given for the problem...
RE: hii....can anybody give me the solution of followi...
Ans to Q#4 would be this.............. Consider the 1st trains leave from both sides,viz, Bglr and Mdrs. Let us focus on the 1st train leaving from Bglr. [ train leaving from Bglr= Bglr train ; train leaving from Mdrs= Mdrs train] The 1st Bglr train, meets the 1st Mdrs train after 2.5hrs from the beginning.Now, the 2nd Mdrs train is 1 hr behind the 1st mdrs train,as it is given. So, 1st Bglr train meets the 2nd Mdrs train 0.5 hr after meeting the 1st Mdrs train.( Pls dont forget that even our 1st bglr train is moving) Hence 1st Bglr train meets the 2nd Mdrs train after 3 hrs from the beginning.The 3rd train will be 2hrs behind the 1st Mdrs train =>the 1st bglr train meets the 3rd Mdrs train (2hrs/2) after meeting the 1st mdrs train, i.e., after 3.5 hrs from the beginning.............Going on like this, we can form the following table: index of the Mdrs Train met Time elapsed from the beginning 1 2.5 hrs 2 3 hrs 3 3.5 hrs 4 4 hrs 5 4.5 hrs 6 5 hrsThe above table is w.r.t. the 1st Bglr train.Notice that this train meets the 6th Mdrs train( which will be starting towards Bglr) at its end point.Hence the 1st Bglr train meets 5 trains during its journey.( excluding the train whose start coincides with the end point of this train) If we go further...................The 2nd Bglr train meets 6 trains , the 3rd - 7 trains, the 4th- 8trains,the 5th and the rest all meet 9 trains. [ In all these cases the coinciding start and end point of a pair of trains, is not considered as meeting during the journey]So any train, other than the 1st 5 trains from either side, meets totally 9 trains during their journey.This is in my opinion. What do you all say?
1 
0