shivani_bha

Hi Syphon Is the answer 7? or Cannot be determined? Shivani

Hi TestMagicians For those of you interested the answer to the 2nd problem When the bird flies from the first car to the second car it flies 12 km but the second car would have also moved forward by d kms so the bird flies 12 - d kms. The time taken for the bird to fly 12 - d km and the second car to move d kms will be the same Set up an equation = 12-d/120=d/60 Here d=4, the bird flies 8 km Now when the bird flies the remaining 4 km (12-8) the equation will be 4-d/120=d/60, where d= 4/3. Therefore the bird will have covered 4 - 4/3 kms = 2.66 Total kms covered = 10.66 As for the 3rd question the bird will fly infinite times since this string of equations will continue till the cars crash Shivani

Hi Testmagicians, Found the mistake I had made. 400 years have ONLY 97 leap years since 1700, 1800 and 1900 will not be leap years. Therefore I get 497 days which leaves no remainder. 23rd Jan + 2 = 25 Jan = Wed, Thursday. And you were all right, the correct answer is Thursday Shivani

Yes padmavathi has explained that correctly..2 cars crash in 1/10 hour, that is 60 km per hour for both cars

Hi Testmagicians, I would like to add to the above post 3 people can also be chosen as 3 husbands or 3 wives (3C1 + 3C1) = 6 additional ways Shivani

Hi Testmagicians, Am not sure, so correct me if I'm wrong. There are 10 people, 5 husbands and 5 wives. 3 people can be chosen as 2 Husbands and 1 wife or 2 wives and 1 husband. This can be done in 5C2*3C1 (excluding those two husbands wives) + 5C2*3C1 Hope this helps Shivani

Dear Testmagicians, A bird is sitting on top of a car. It sees another car approaching it ata distance of 12km. The speed of the two cars is 60 m/h each. The birds Start flying from the first car and moves toward the second car, reaches the second car and comes back to the first car and so on. if the speed at which the bird flies is 120 km/h then answer the following questions.Assume that the Two cars have a crash. 1) The total distance travelled by the bird before the crash is. a)6km b)12km c)18 km d)None of these 2) The total distance travelled by the bird before it reaches the second car for the second time is.. a)10.55km b)11.55km c)12.33km d)None of These 3)The total number of times that the bird reaches the bonnet of the second car is(theoratically): a)12 times b)18 times) c)Infinite times d)Cannot be determined I know the first answer = 12 km. Could you help me with the 2nd and the 3rd. Thanks, Shivani

Hi sambazone, I think the logic is to see the power of 7. Since red beads have 7 points and 3000 cannot be split further into any factors of 7, the answer will be 2. Shivani

Thanks thefish, You are right on both counts. This question is one of the toughest I have seen and I doubt the GMAT will want to ask a question like this. Shivani

Hi Gmatfordays, TestMagicians AP = Arithmetic progression where each term has a common difference eg a, a+d, a+2d where a= first term and d= common difference 10, 20, 30, the first term is 10, the second = 10+1.10 the third is 10+2.10 Rumen, I understand the solution but did you arrive at the solution by the trial and error method? eg 3x + 60 > 100 Thanks Shivani

2^10=1024 and 2^9=512. So, the number 2^99 = 1024*1024*1024*.....(nine tines)*512. If we divide 1024 by 100 we get 24 as remainder. Dividing 512 by 100 gives 12 as remainder. Adding all the remainders we get 9*24+12 = 228. Dividing 228 by 100 gives 28 as remainder which is same as that obtained by dividing 2^99 by 100. So, the tens digit is 2.

Hi Testmagicians, If a three digit number is divided into three two digit numbers and if all these three two digit numbers form an AP with a common difference of 20, how many three digit numbers satisfy this condition? 1. 54 2. 45 3. 46 4. 55 Thanks Shivani

Hi Everyone, Divisibility by 13 can be done using the positive root (don't ask me how that is arrived at, I don't know). The digital root of 13 is 4. Therefore take a no like 1001 Take the last digit of the no = 100 / 1 multiply by digital root (4) add to the balance = 104, continue 10+4*4=10+16=26 Which is divisible by 13 Take another example = 1223 =122+3*4=134=13+16=29 Not divisble by 13 You can use the same method for other difficult numbers No Digital Root 17 12 19 2 29 3 Shivani

Dear GMatfor days, The key here is that the commodities are sold at equal prices SP=100 The CP however will be different, CP=100/110% and 100/90% = 90.9+111.1=202 = While the selling price will be 100+100=200 Therefore there is a loss of 2 on 202 = .99% = rounded off and = 1% Is this clearer? Shivani

Thanks thefish, I've been agonising over this one but didn't think of using squares. Shivani

Hi Ashhad, Solve this using (interior angle) - exterior angle=120 =(2n-4)*90/n - 360/n=120 When you solve the above equation you get n=12 Shivani

Hi Ashhad, Just doing these kind of problems so I can help you. 2003-1603=400 years. The trick here is that every leap year has 2 odd days (an odd day is one excess of 7 eg 366/7 remainder=2) and a non-leap year has 1 odd day. Therefore in 400 years 100 leap years and 300 non-leap years. Therefore 500 odd days. 500/7 = whatever, remainder = 3. Therefore count forward from Tuesday = Wed, Thurs Fri. My answer is Friday. Is that correct? Shivani

Dear Gmatfordays, TestMagicians I think it 101 too. Generally for questions like this take a smaller number and experiment eg 10^1= 2 digits, 10^2=3 digits = power + 1 digit Shivani

Dear gmatfordays/ TestMagicians, I've got the correct method for this question..a long one but an answer all the same..you have the cycle both the units and the tens digit which will be another cycle. eg. 02 04 16 .... Shivani

Hi Gmatfor days, After pondering over this I have come to the conclusion that its the HCF we are talking about (the question may be incorrect) and not the LCM. That is the only way there can be an answer, 8mths = x, 4 mths = y, difference = 4, HCF = 4 This question was sent from a friend of mine by mail. You can access questions like this one at Ascent Education's group ascent4cat@yahoogroups.com - messages Shivani

Hi gmatfordays, truman, For 1 a) 3C3 is the no of ways three people will get at least one bill while the remaining 4 can be distributed in 4C3 ways. Am not sure, correct me if I'm wrong Shivani

Hi, Try to do this faster, scan the options and see that 280 is the only no divisble by 7 and that gives x.5 when divided by 8. Shivani

Hi, Try to do this faster, scan the options and see that 280 is the only no divisble by 7 and that gives x.5 when divided by 8. Shivani

Hi Testmagicians, what is the tenths digit in the number 2^99 The units digit can be solved using cyclicity but how does one find the tens digit? Thanks Shivani:(

Hi Testmagicians, Can't seem to solve this. When I met a friend today we observed that my birthday is x months away, while his birthday is y months away. The difference between our ages in less than a year. Also we observed that the difference of our ages in months is the LCM of x and y. Which of the following can be the difference of our ages. 1) 4 months 2) 6 months 3) 8 months 4) 10 months Any 2 nos that are co-primes of each other will give an LCM of greater than 12 (eg 7 and 3) while any no that is a multiple of the other no will give the greater no as LCM (therefore this cannot be the no) Any ideas on how to solve this? Thanks Shivani