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 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby quaint lee - Tuesday, 1 May 2007, 01:23 AM Answers Q1. Number of colours needed for a n . n chessboard = sum of numbers 1+2+3 ... upto (n+1)/2 e.g.board 3.3 => 1+....+ (3+1)/2 => 1+ ... +2 => 3 colours board 7.7 => 1+....+ (7+1)/2 => 1+2+3+4 => 10 colours Q10. Outer surface area = 6(3)2 - 6(1)2 = 48; Inner surface area = 6*4(1)2 = 24Total surface area = 72m2
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Total Gadha - Tuesday, 1 May 2007, 01:36 AM
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Die Hard - Tuesday, 1 May 2007, 04:53 AM ans 1.     1+2+3+4+5......(n+1)/2 =(n+1)*(n+3)/8 ans 2.      9 ans 3.     64*3*5=960   28 factors ans 4.    107 ans5      ans 6    8 pm ans 7    49.5 litres ans 8    420 ans 9 ans 10  72 m*m
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Die Hard - Tuesday, 1 May 2007, 04:52 AM Q5. here 512=(100000000)base 2 513=100000001..........1023=111111111 so no. of 1s will be more than no. of 0s so prob. should be zero ie no chances of having more zeros than ones.........
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Die Hard - Tuesday, 1 May 2007, 12:11 PM ans 9.          ( 1, 3Â³, 3Â³) (1 ,1/3Â³, 1/3Â³).............till 48 terms multiplication will be 1 and last bracket consisting of 46th 47th and 48th terms will be (1 ,1/3Â³, 1/3Â³) now 49th 50th term will be 1,3Â³.........so multiplication will be 27 (of 50 terms) hence sum of first 2 terms will be 27+1=28 (ans)
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Total Gadha - Tuesday, 1 May 2007, 01:06 PM Hi Bharat, Post your solutions also. Let other users check if you're doing it right. I am not going to provide the correct answers yet. Let other students try their hands on it. Total Gadha
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Die Hard - Tuesday, 1 May 2007, 04:34 PM ans3    the no should have more no of 2 powers that will increase the no of factors the no which i found is 960=2pow6*3*5 ans 4  i name the bottom most blocks as a1,a2,a3.....a10 then moved up by adding the required blocks............. the equation of the top most block came as (a1)+3(a2)+3(a3)+3(a4)+2(a5)+6(a6)+(a7)+3(a8)+3(a9)+(a10) now for getting the least value i assinged 1 to the number having the greatest coefficient and so on eg a6=1..... the ans came out to be 107 ans6   the person took 12 hrs(9am to 9pm) to complete his journey..........and two hours in any country will be same.........so time taken in travelling the distance is 10 hours so he would have required 5 hrs to reach country x(10/2 aana jaana) the time in his country would be 9am+5 hrs=2pm and time in country x is 1pm so diff in local times is 1 hr ans 8pm ans 7  oh....the xtra large bottle of old monk(normal 720 ml) he will add water on 1st day=10 ml 2nd =20 ml...........99th day=990ml on 100th day he will finish the bottle so he would not add any water...... so effectively he drunk 1000 ml of rum and(10+20+30+40+............990)ml of water that adds to be 49.5 litres ans 8 i came out with 420 as ans because the cube route will be 7.sth as cube route of 512 is 8 now this no is divisible by 1,2,3,4,5,6,7(scope of a greater no is there..) ans10 calculated the area by length * breadth ans 2 i have found a mistake i will do it again
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Die Hard - Tuesday, 1 May 2007, 04:51 PM ans 2 is 49 19^92/92 361^46/92 (-7)^46/92 7^46/92 7*(343)^15 7*(-25)^15 -7*25*(625)^7 -7*25*(-19)^7 7*25*(19)^7 7*25*19*(361)^3 7*25*19*(-7)^3 7*25*19*25 625*19*7/92 -19*19*7 -361*7 -7*-7 49 ans i could not think of some better method to do this question........
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Die Hard - Tuesday, 1 May 2007, 04:56 PM hi tg atleast tell which questions i have done wrong
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Total Gadha - Tuesday, 1 May 2007, 07:35 PM Hi Bharat, Try question 3 and 8 again. Total Gadha
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby karunya bharadwaj - Tuesday, 1 May 2007, 08:00 PM Answer for Q 10 6*(3*3) + 6* (1*1) = 78
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Die Hard - Tuesday, 1 May 2007, 09:21 PM ans 3  is it (2^4)*(3^2)*5=720    30 factors
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Total Gadha - Tuesday, 1 May 2007, 09:31 PM ummm..try more? One hint- If I remember correct, the answers to question numbers 3 and 8 are the same..not sure though
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Die Hard - Tuesday, 1 May 2007, 09:38 PM i'll not bore u further, my last try 2^3*3*5*7=840 32 factors is it right...........
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Total Gadha - Tuesday, 1 May 2007, 09:50 PM yep. well done. Check the answer to 5th question also..I think your answer to 8th question is correct. Let me check. Btw, how were the questions?
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Die Hard - Wednesday, 2 May 2007, 11:31 AM hi tg , the questions were good and were quite new ,they force u to think.......and also have no standard methods or formulas to do them.........please post some more good questions regularly.............. thanks......
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Die Hard - Wednesday, 2 May 2007, 02:34 PM hi tg ....... i am not able to understand the 5th question ie what it is asking.......and what is the ans of question 8.........
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby chinni kiran - Wednesday, 2 May 2007, 06:48 PM hi frnds, how about 51/256 as the prob. of no.of zeroes > no.of ones? 1000000000(=512, enough digits to rep. upto 1023) 8c1+8c2+8c3(arranging no.of ones  above so that prob. of no.of zeroes > no.of ones satisfies)8 + 28 + 56/512
 Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Swastik Mishra - Wednesday, 2 May 2007, 05:57 PM Hi all,       just tried out the remainder problem and got the answer similar to bharat.My approach is as follows:Remainder of 19^92 / 92 Soln1: 19^92 / 92 = 19^92 / (4 * 23)                          we can write this as (19^92 / 23 ) * (1/4)  taking 1/4 out                          Simplifying this again we get (19^22 / 23 )*(19^22 / 23 )*(19^22 / 23 )*(19^22 / 23 )*(19^4 / 23)*(1/4)                          Now applying Fermat's Little Theorem...which states                          if p is a prime and a is an integer coprime to p, then   $a^{p-1} \equiv 1 \pmod{p}\,\!$                            Applying to the above expression (19^22 / 23 ) results in 1.                     We are left with (19^4 / 23)*(1/4)...                      This can be written as ( 361^2)/92                      i.e ( -7^2)/92                      i.e   49 I have also solved it by using Eulers Theorem which is easier than this method....Try it...
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby chinni kiran - Wednesday, 2 May 2007, 07:22 PM 23/128 ..small correction
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Total Gadha - Wednesday, 2 May 2007, 07:52 PM Good work Swastik and Kiran
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Die Hard - Thursday, 3 May 2007, 12:36 PM hi tg   Q5.  Let the set A = {512, 513, 514, â€¦ 1023}. What is the probability that these numbers, when written in base 2, have more 0â€™s than 1â€™s?  sir here it is talking about all the numbers written in base 2  so when we are talking about all the numbers the no of 1s are greater than no of 0s so prob shuld be 0 and if we r talking about choosing one no from the lot then the prob will be 23/128
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Total Gadha - Thursday, 3 May 2007, 01:24 PM Hi Bharat, The question is talking about the second scenario. I think I will phrase the question as "what is the probability that any number chosen from this set, when written in base 2, will have more 0's than 1's". Will this avoid the ambiguity? Total Gadha
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby chinni kiran - Thursday, 3 May 2007, 02:57 PM how come some posts were posted at 1 AM and 4 AM...mine is GMT + 5.5 hours(india time)...I think ppl burning midnight oil..... (or u in a diff country?)
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Die Hard - Friday, 4 May 2007, 12:27 AM yes sir u know better than i/me Eagerly waiting for ur next article............
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Die Hard - Friday, 4 May 2007, 01:28 AM hi tg Remainder of 19^92 / 92 Soln1: 19^92 / 92 = 19^92 / (4 * 23)                          we can write this as (19^92 / 23 ) * (1/4)  taking 1/4 out                          Simplifying this again we get (19^22 / 23 )*(19^22 / 23 )*(19^22 / 23 )*(19^22 / 23 )*(19^4 / 23)*(1/4)                          Now applying Fermat's Little Theorem...which states                          if p is a prime and a is an integer coprime to p, then   $a^{p-1} \equiv 1 \pmod{p}\,\!$                            Applying to the above expression (19^22 / 23 ) results in 1.                     We are left with (19^4 / 23)*(1/4)...                      This can be written as ( 361^2)/92                      i.e ( -7^2)/92                      i.e   49   i have a doubt in this method that how can we break the denominator .......... for eg    23/10 remainder is 3 now if i break the remainder into 5*2 then 1/5*(23/2)=1/5  remainder will be1 if this is the case in above method then why did the ans came same.......
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Die Hard - Friday, 4 May 2007, 10:29 AM hi tg my ans to 5th problem is 93/512  (1+8+28+56)/512=93/512
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Total Gadha - Friday, 4 May 2007, 02:29 PM Hi Bharat, No you cannot break the denominator like this. In the first case, you got the correct answer by breaking the denominator because 19^22 was giving remainder 1 by BOTH 23 and 4. Therefore, although you discounted 4, it didn't matter as the remainder was still 1. In the second, see that 23 will give different remainders when divided by 5 and 2. Therefore, the method won't work. Infact, the method is not valid at all. Total Gadha
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Total Gadha - Friday, 4 May 2007, 02:31 PM Hi Chinni, TG is alive 24 hours. There is no night and day on the internet. Total Gadha
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Die Hard - Friday, 4 May 2007, 05:48 PM hi tg........now i got it ........thanks a lot for taking pains in clearing our doubts...........please write articles on pand c and probability........because this topics will be very difficult for us to do at the last.......
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Deep Thinker Gadha - Saturday, 5 May 2007, 02:09 AM Hi Bharat, Can you explain Q.5 again. How you get probability 0 (Zero)? When From 512 to 960 we are always getting more number of Zeros than 1's. MyAns. is 0.876953125. Pls explain if i'm wrong. Regards, -Sandeep
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Die Hard - Saturday, 5 May 2007, 03:14 AM hi sandeep, Let the set A = {512, 513, 514, â€¦ 1023}. What is the probability that these numbers, when written in base 2, have more 0â€™s than 1â€™s?  i felt that here the question was asking about all numbers(no case of choosing 1 number) so when we add up all 1s and 0s from 512 to 1023  we will find no of 1s>no of zeros so chances of having  more 0â€™s than 1â€™s were zero tg rephrased the question as what is the probability that any number chosen from this set, when written in base 2, will have more 0's than 1's".  i think u have found out the numbers between 512 to 960 and divided by 512 consider 527( 1000001111)  543 (1000011111)(and more) i havent converted ur ans in fraction form i am just speculating......i think my ans is wrong 93/512 because i took 512 with only 8 zeros instead of 9 after that it was all sequence.......the best method which i think is to arrange 3 1s in 9 places+2 1s +1 1s+1(for 512)and dividing by 512.........
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Deep Thinker Gadha - Saturday, 5 May 2007, 12:57 PM Hi Guru,I have tried to solve these questions and i got following reslts :Q1. Since n-is odd and it's a nxn chess board so it needed n diffrent colors.Q3.Is this number is 840 (which has highest number of divisors)? 840 = (2x3x5x7)x4Q4. The smallest possible number assigned to top block is 134.144 = 31+52+61  { 31= 6+11+14, 52=11+19+25,61=14+22+25}                               {25=6+9+10, 22=5+8+9,19=4+7+8,......etc}Q5. Ans is 449/512. Since From number 512 to 960 all these numbers contains more number of zeros than 6 and only the numbers from 861 to 1023 contains equa or more number of 1's. so the ratio is 449:63 so probability is 449/512.Q6.Time in te country X is 8p.m. when local time is 9 p.m..Total journey time = 6hrs + 4 hrs. So ideally time to travel for single way = 5hrs hence Plane gains 1hr while journey towards X and Loses 1hr while returning from X. Therefore as it loses 1 hr while returning,There should be 8pm when local time is 9 pm.Q7. Ans-4950ml of waterHe is adding 10ml on day 1, 20ml on day 2,......till 990ml of water on day99 and on the 100th day he drunk whole bottle of mixture.Hence total water he drunk = 10+20+30+.........+990 = 99x100/2= 4950ml Q8. MyAns - N=2 (Since i have got only n=2 which satiesfies this condition)Q10. Total Surface Area = 72m2    External SA = 6x32 - 6x1 = 48m2    Internal  SA = 6X4x1 = 24m2    TotalSA = ExternalSA + InternalSA = 48+24= 72m2I have skipped Q2 & Q9 in this post.Guru, Please comment on the solutions and give  directions for further journey.Regards,-Sandeep
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Total Gadha - Sunday, 6 May 2007, 01:34 AM Hi Sandeep, Try solving questions 1, 4, and 8 once again.  Good solution to question 6. Total Gadha
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Deep Thinker Gadha - Sunday, 6 May 2007, 05:04 AM Hi Barat,Let me explain the solution. I have solved the problem likke this:We are looking for more no. of zeros than 1's i.e. as 1023 is a 10 digit binary no. to satiesfy this condition it should contain min 6 zeros. So, From 1000000000= (512)2 to 1111000000 =(960)2(which is the largest no. containing 6 zeros) we have all the no. having more no. of zeros than 1's and only no.s from 961(1111000001) to 1023(1111111111) contains more 1's than zeros.Hence the ratio is 449:63 so probability is 449/512.Regards,-Sandeep
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Deep Thinker Gadha - Sunday, 6 May 2007, 05:33 AM Hi Guru,I got it where i'm mistaking.Q1. The total number of colors for nxn chessbord = 1+2+3+.......+(n+2)/2as No. of colors for 3x3 = 1+(3+1)/2=3                                  5x5 = (5+1)/2 + 2+1=6......Q4. 107 Q5. ExplanationWe are looking for more no. of zeros than 1's i.e. as 1023 is a 10 digit binary no. to satiesfy this condition it should contain min 6 zeros. So, From 1000000000= (512)2 to 1111000000 =(960)2(which is the largest no. containing 6 zeros) we have all the no. having more no. of zeros than 1's and only no.s from 961(1111000001) to 1023(1111111111) contains more 1's than zeros. Hence the ratio is 449:63 so probability is 449/512. Regards, -Sandeep
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Die Hard - Sunday, 6 May 2007, 11:51 AM hi sandeep my previous post............i think u have found out the numbers between 512 to 960 and divided by 512 consider 527( 1000001111)  543 (1000011111)(and more) they do not contain 6 zeros...........and there are many no between 512 to 960 which contain equal no of 1 and 0 also we dont have to consider them........... bharat
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Myelin Ash - Monday, 7 May 2007, 04:54 PM Answer to Qs No. 6 is 8.00 P.M. We can be sure that Grihasti Lal is moving towards west, as the apparent time taken for onward journey is less than the return journey.  So, Onward Journey 13.00hrs -8.00AM (LOCAL TIMES for the country where he moved)= 5 hours (Flight time) Return Journey started at 1500 hrs local i.e., 1600 local for X Add 1600hrs to 5 hrs (flight time) we get 2100 hrs i.e, 9 PM.   Am i correct?
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Total Gadha - Monday, 7 May 2007, 10:54 PM Hi Vaseem, 8:00 pm is correct. See Sandeep's explanation also for Question no. 6. Total Gadha
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Deep Thinker Gadha - Tuesday, 8 May 2007, 11:13 AM Let me think once again......
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Long Last - Tuesday, 8 May 2007, 02:32 PM 10) answer is 72 unit square (9-1)*6 for cube external surface (1*1*1*1)*6 for internal open cubes
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby himanshu deshmukh - Wednesday, 9 May 2007, 07:39 PM Hi, Can you provide me the Correct answer and Solution to Q.5 ?
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby rakesh pallerla - Wednesday, 9 May 2007, 07:46 PM Sir i doubt the equation u got for Q 4... i got it as a1+a7+a10+ 3*(a2+a3+a4+a6+a8+a10)+6*a5 assigning samllest no to a5 and highest three to a1 a7 and a10 we get answer as 114 can u confirm regarding this...
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby himanshu deshmukh - Thursday, 10 May 2007, 12:06 PM Hi TG, I am getting answer for Q. 5 as  '107/256'.  (probability of no. of 0's greater than no. of 1's.) Is it correct??
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby himanshu deshmukh - Thursday, 10 May 2007, 02:46 PM Hi Bharat, I think for Question no 9, you have only checked the condition that  'multiplication of 50 terms give result 27 ', but you have missed the condition 'multiplication of All Numbers (here 100) also gives 27'. My answer to this Question is 12 (sum of first two numbers).   Series will be - 3,9,3,1/3,1/9,1/3,3,9,............,3(45th term),1/3(46th term),1/9(47th term),1/3(48th term),3(49th term),9(50th term),.................................................,3(93rd term),1/3(94th term),1/9(95th term),1/3(96th term),3(97th term),9(98th term),3,1/3   The sequence of numbers in red (series of 6 numbers) will repeat itself again and again.With this series all conditions ( including multiplication of first 50 give 27,multiplication of 100 numbers give 27 and each except first and last number is product of its neighbour) are satisfied.
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Total Gadha - Friday, 11 May 2007, 02:32 AM Hi All, I know you people are eager for the answers but please wait for some more time. I want some more attempts from students. Will post answers soon. Total Gadha
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Anirudh Kashyap - Saturday, 12 May 2007, 04:24 PM Hi TG, 1] Even I got the same answer as rakesh for Q. No. 4. The equation would be (b1+b7+b10)+3(b2+b3+b4+b6+b8+b9)+6(b5) = (8+9+10)+3(2+3+4+5+6+7)+6(1) = 114. 2] For Q. No. 3, as some people have already mentioned 840 has 32 factors and this is the maximum for a number upto 1000. But just to help people who haven't got it, the approach I used was:   i] Using maximum number of prime factors (Instead of a higher power of 1 or 2 prime factors)   ii] Trying to uniformly increase the power of each of them e.g.; 2*3*5*7*11 = 2310; 2*3*5*7 = 210; 2^2*3^2*5^2*7^2 = 44100; 2^2*3^2*5^2*7 = 6300; 2^2*3^2*5*7 = 1260; 2^2*3*5*7 = 420; But there is scope for one more multiplication by 2. So 2^3*3*5*7 = 840. Are these right? Thanks
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Anirudh Kashyap - Saturday, 12 May 2007, 04:38 PM Hi TG, I forgot to mention the approach I used for Q. No. 8. Multiply each no. from 1 with it's consecutive number as per divisibility requirements and check if it is less than the cube of the number after that. Find the highest number which satisfies this criteria. e.g. 1*2*3 = 6 < 64(4^3); 1*2*3*2(and not 4 as 2 is already present) = 12 < 125(5^3); 1*2*3*2*5*7(6 not required as 2 & 3 are already present) = 420 < 512(8^3). But 420*2(for divisibility by 8) = 840 > 729(9^3). So 420 is the highest such number. Thanks.
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby himanshu deshmukh - Friday, 1 June 2007, 07:46 PM Hi TG Sir, No new solutions............. No new doubts............   ab tho correct solutions bata do.........
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby abhinav trivedi - Tuesday, 12 June 2007, 12:10 PM can u tell me how did u solved this as i m not understanding ur method
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby kanika kwatra - Sunday, 17 June 2007, 04:58 PM hi swastik well u'v given a complete mathematical solution to dis problem. n its really gud.i was trying to understand it,but was stuck up wid one thing.. how did u express [(19^92)/23] as [19^22/23]*[19^22/23]*[19^22/23]*[19^22/23]*[19^4/23] ??? by doing this u'l get 23^5 in d denominator.. do tell me d solution for dis one as m quite weak in numbers.thanx kanika
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby varun soni - Thursday, 21 June 2007, 08:05 PM T.G...... I M DAMN TENSED..I DONNO A JACK ABT NE OF THESE QUESTIONS U HAV PUT UP...WAT TO DO..I REALLY WANNA SCORE BUT..PLS HELP ME OUT.MAIL ME HOW TO STUDY FOR THIS CAT...PLS..
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby shreyas nair - Sunday, 24 June 2007, 07:32 PM q1) minimum colours required are x*(x+1)/2, where x=(n+1)/2, where n is the length of the side. hence minimum number of colours required is: (n^2+3n+2)/8 u can check it out for any value of n (provided n is odd) ------------------ q2) 19^92/92 is same as 361^46/92= (368-7)^46/92.. the remainder of the same wil be remainder when 7^46/92.. 92=2^2*23 by euler's formula.. 7^(e(2^2))/2^2 % 1 = 7^2/4 % 1 similarly 7^22/23 % 1. therefore taking the lcm we get 7^22/23 % 1 7^46= 7^44 * 7^2. therefore it is same as dividing 7^2 ie 49 by 92 which gives a remainder 49. ------------------------ q3) highest number of divisors will be for a number with the maimum number of different factors for a number. therefore it will be the largest factorial less than 1000. which is 720. 720=6!= 2*3*4*5*6= 2^4*3^2*5^1 the number of factors for this number is 5*3*2=30 ---------------------------- q4) i cud not get a number for this problem. i just cud get a config for which 144 is the minimum answer. but i think there is a proper and correct solution.. tg sir please help. ---------------------------------------------- q5) 512=2*9= 1000000000 (in binary) while 1023 is 1111111111 apart from 1 there are 9 spaces. the number of zero's wil be more than the number of one's if there are 6,7,8 or 9 zero's ie 9C6 + 9C7 + 9C8 + 9C9= 84+36+9+1= 130. while total number of numbers between 512 and 1023 including them are 512. therefore probability is 130/512 = 65/256 -------------------------------------------------- q6) this one was pretty simple.. the total time wrt his home country was spent in the journey and the two hour stay. therefore 12= 2x+2.. hence time of travel in one direction is 5 hrs so he would have reached country x at 2 pm by his watch but the local time is 1 pm which is one hour back.. so the time when he reached home is 8pm by local time in country x ------------------------------------------------------ q7) the amount of fluid that raghav drank is 10+20+30+40+...+1000 thus the total amount of fluid that he drank is 10*(100*101)/2=50500 ml thus he drank 505 litres of which only one litre was rum. so the water he drank is 504 litres ---------------------------------------------------- q8) let n be the number.we should check for the largest number for which the product of the prime numbers less than its cube root. on checking by taking trails we get 1=1,2=2,3=3,4=2*2,5=5,6=3*2,7=7,8=2*2*2 (we dont need to check for 9 as the product of lesser primes will be greater than the cube of 9.) the ranges for numbers will be 1, 8,27,64,125,216,343,512. to be divisible by all numbers below 8 including 8 is 8*7*3*5=840. thus the largest such number is 840. this can be checked as all numbers less than (840)^1/3 ie 8,7,6,5,4,3,2 divide 840 perfectly. ------------------------------------------------------- q9) let first two numbers be x and y.. so the sequence will be x,y,y/x,1/x,1/y,x/y,x,y,.. the sequence repeates after 6 numbers. also the product of first six numbers is 1. so now, product of first 50 numbers will be xy=27 while product of first 100 numbers will be y^2/x=27. on solving we get y=9 and x=3. so their sum = x+y=12. ------------------------------------------------------ q10) area of small hollow cubes= 4*(1*1)=4 cu.m total small cubes =6. total arear for small hollow cubes= 6*4=24 cu.m surface area of central hollow cube is 2*(1*1)=2 cu.m area of external surface is ((3*3)-(1*1))=(9-1)=8 cu.m total external surface area is 8*6=48 cu.m total surface area = 24+2+48= 74 cu.m  ------------------------------ these are my solutions.. tg sir please verify and correct me wherever i am wrong   cheers, SHREYAS NAIR
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Sameer Rentala - Thursday, 19 July 2007, 05:37 PM HI TG answer for Q3 is 840 i suppose since 3*5*7*2=210 and out of remaining probable multipliers (2,3,4)since multiplying it with  4 gives max number of divisors Answer therefore is 840 AM i correct??? pls help
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Small Wonder - Friday, 20 July 2007, 03:19 PM Hi TG, I tried the following appraoch. Looked tempting at first sight but proceeded only to get entangled. Here is my approach- 1992 mod 92 = (23-4)92 mod (23*4) Expanding by binomial theorem we are left with the following expression- (492 +  2392) mod (23*4) This is where I couldn't proceed from to get an answer. I've tried a lot and had one doubt. I just want to verify that can we write the expression [(492 +  2392) mod (23*4)] as [(491) mod 23  + (2391) mod 4] Can we solve this problem without the help of Sir Fermat and Sir Euler? Please guide. Brgds, Small Wonder.
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Sameer Rentala - Friday, 20 July 2007, 05:40 PM answer 4-114
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Sameer Rentala - Friday, 20 July 2007, 05:48 PM Q.No 6 8PM
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Sanjay Varnwal - Monday, 20 August 2007, 03:57 PM Ans for Q No 10:Total initial surface area = 6*9 = 54After cut = 54 - 1*6 + 5*6 = 78Ans = 78There are five small surfaces in the groove.
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Amit Mittal - Tuesday, 4 September 2007, 05:11 PM Q.1 The number of NxN will have    Colors =   N    +      (N-1)/2    Put n = 1 you get only one color Put n = 3 get 4 color (see inner cubes of 3 size)
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Amit Mittal - Tuesday, 4 September 2007, 05:23 PM q.2 19^92 can be written as (19 ^2)46  = 361 ^41 = (92+269)^46 now if we expand it binomially we get 92 in every term except last term of 269 which when divided by 92 gives remainder =85
 Q:6by Amit Mittal - Tuesday, 4 September 2007, 05:29 PM Time in city X will be 7 PM
 Re: Q:1by sreedhar gali - Wednesday, 19 September 2007, 12:03 PM qn 1.  (n+1)/2 +(n-1)/2+(n-3)/2 + ..........(n-(n-2))/2  = n+1
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby pulkit mittal - Friday, 12 September 2008, 10:49 PM I got the same answer for question no 5. Answer should be 65/256 only.
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Maktoum Misfaq - Saturday, 27 September 2008, 12:02 PM ans 3: (2^3)*3*5*7=840the number of factors is 32
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Atul Mittal - Wednesday, 1 October 2008, 11:46 AM Hi Friends, Can you please clear me on the first question? How can there be asymmetric squares in a chess board of 5*5 of 25 unit sqares. I think all will be of same dimension and hence all symmetric. Please explain with a diagram if possible. Thanks
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Gaurav Talwar - Wednesday, 1 October 2008, 04:22 PM dude ... it is not js 269 but 269^92 ... the problem remains same ....   neways ... m getting the remainder as 19 ... hv cross checked it .... dunno where m going wrong ... but m going by the long way ....   19^92 = (361+7-7)^46 = 7^46 = (343)^15 * 7 = (25)^15 * 7  = (625)^7 * 25 * 7 = (625 )^7 * 83 = (19)^3 * 73 * 83 = 7 * 19 * 73 * 83 = 41 * 6059 = 41 *  79 = 3239 = 19   where m i going wrong ... waiting for the reply ....   Admin help me plz...
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Balaji Kunjeti - Tuesday, 21 July 2009, 02:26 PM 1) Sum of natural numbers till (n+1)/2 2) 3 3) 840; 32 divisors 4) 114 5) 130/512 6) 8:00 Pm 7) 49500ml 8) 420 9) 28 10) 72 Sq.m
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Balaji Kunjeti - Tuesday, 21 July 2009, 03:40 PM For 2nd Question, answer is 49
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Rakshit Mirg - Saturday, 26 June 2010, 05:02 AM hi friends plz explain me the 1st question about how colours are pattern acording to symmetry cudnt discern it.
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby Ankur Bansal - Friday, 2 July 2010, 05:47 PM Hi All,   When i was solving Q.7, i had solved through a different way but the final ans was same thn i thought my method was long and time consuming but then i realised that another ques can be developed from the way i solved this problem. Ques statement is similar to Q.7 just you have to calculate something different thing in this. Ques.11 For the same statement(Q. 7), how much water he had consumed till 7th day?
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby rakesh kumar - Thursday, 2 June 2011, 08:29 PM in q8n is three digit number or not???without mentioning number of digits we can't i think
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby TG Team - Friday, 3 June 2011, 05:20 PM Hi Rakesh What is the need of mentioning this be a three digit number. You have been asked about largest such number which is divisible by all of natural numbers less than its cube root.So say, if we take that number must be divisible by all natural numbers till 5 that means cube root of the number must be somewhere between 5 and 6 i.e. N must vary from 5³ = 125 to 6³ = 216. Also it must be multiple of LCM(1,2,3,4,5) = 60. So one such number we get in this range is 180.But is this the largest such number, may be no. So look further and get the highest such number.Hope it helps. Kamal Lohia
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby DeeScript .in - Sunday, 23 June 2013, 09:36 PM namaste sir!!!!!the answer for the second problem is 19soln: according to fermat's theorem.19^92-19 is divisible by 92.thas why the remainder of 19^92-19+19 when divided by 92 will be 19hence..19 is the answer.please tell me if i am wrong....
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby DeeScript .in - Sunday, 23 June 2013, 10:09 PM sir ...the  number having highest number of factors will be 840
 Re: Total Gadha's Quant Challenge for CAT 2007 Aspirantsby shubham badhe - Saturday, 9 January 2016, 10:59 AM Sorry tg,I think you are wrong. If we consider 959 I.e equal to 1110111111..I.e it has more 1's than zeros. I think the correct answer is 1/4.The logic is512=1000000000(9zeros) 1023=1111111111(10 ones). Exclude the first one we will consider it afterwards. So there are total 512 terms. Consider 513 and 1022. 513=000000001(8zeros) 1022=111111110(8 ones) This is exactly complementary. Note: we have excluded the first one. Same for 514 and 1021,515 and 1020,keep on doing this and you will find 256 numbers have more 1's and 256 has more 0's.now consider the first one. If we add it to both the 256 numbers the more ones will have no change while the ones having more zeros will be half. Half of it will conta have more zeros and half will conta have more ones. So the total numbers having more zeros will be 128 out of 512. So probability is 1/4.