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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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09 Aug 2019, 05:02
hi VeritasKarishmaThe product of 3 consecutive positive multiples of 4 must be divisible by each of the following EXCEPT: A)16 B)36 C)48 D)128 E)192 The OA is B but i don't understand why exactly. Arent all these numbers divisible by three consecutive multiples of 4



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13 Aug 2019, 20:41
Nums99 wrote: hi VeritasKarishmaThe product of 3 consecutive positive multiples of 4 must be divisible by each of the following EXCEPT: A)16 B)36 C)48 D)128 E)192 The OA is B but i don't understand why exactly. Arent all these numbers divisible by three consecutive multiples of 4 Say the 3 consecutive multiples are 4, 8 and 12. Their product = 4*8*12. This product is not divisible by 36. 36 = 4*9 The product does not have 9 in it (two 3s). It necessarily will have one 3 but it may or may not have two 3s. Hence it may not be divisible by 36. Three consecutive multiples of 4 will have three 4s, another 2 and a 3 necessarily. So the product will be divisible by 2^7 * 3 and all its factors. For more on this, check: https://www.veritasprep.com/blog/2011/0 ... cormath/https://www.veritasprep.com/blog/2011/0 ... hpartii/
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Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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25 Aug 2019, 03:38
Hiii VeritasKarishmahttps://gmatclub.com/forum/if64x5w ... 99799.htmlReferring to the question on this page i am really confused when they say " x=x " isn't the only possibility to this that x = 0??? because no number inside a mod sign can be negative as such



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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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25 Aug 2019, 18:34
VeritasKarishma wrote: tilak123  I am kicking off this post as per your request. This is a compilation of all links to the "Made Easy" series. Bunuel has already compiled links to all my posts on specific topics. I will provide links to those compilations: Statistics Made EasyInequalities Made EasySequences Made EasyCombinatorics Made EasyWeighted Avgs and Mixtures Made Easy Hi Karishma.. I request reference to Question 1 in your post "A group of 8 friends sit together in a circle. If A refuses to sit beside B unless C sits on the other side of A as well, how many possible seating arrangements are possible?" Even though i understood your explanation, can u please tell me whats wrong in my approach? My Approach: There are 2 scenarios possible, One where A sits next to both B & C and the other wherein A doesn't sit next to both B & C . In the former the no.of arrangements are: 1*2*1*5*4*3*2*1= 2*5! In the latter the no. of arrangements are: 1*5*4*5*4*3*2*1= 20*5! Thus total no. of arrangements are: 5!(20+2)= 22*5! However your answer is 32*5! Where am i missing??



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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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25 Aug 2019, 20:12
Nums99 wrote: Hiii VeritasKarishmahttps://gmatclub.com/forum/if64x5w ... 99799.htmlReferring to the question on this page i am really confused when they say " x=x " isn't the only possibility to this that x = 0??? because no number inside a mod sign can be negative as such Your assumption is that x is a positive number and hence, x is a negative number. But can x be a POSITIVE number? Think about it. What about the case when x = 5? Then x =  (5) = 5 What about the case when x = 23? Then x =  (23) = 23 x can be positive when x is negative. In these cases, x = x And hence, we define x as: x = x when x >= 0 x = x when x < 0
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25 Aug 2019, 22:33
A car travels from town A to town B at constant speed. If it increases speed by 20%, it will arrive 1 hour ahead of schedule.If it increased by 25% after travelling the first 120 km at usual speed,it will arrive 36 minutes ahead.Find the distance from A to B.
a.205 b.210 c.230 d.220 e.250
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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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26 Aug 2019, 02:06
Kamran129239 wrote: A car travels from town A to town B at constant speed. If it increases speed by 20%, it will arrive 1 hour ahead of schedule.If it increased by 25% after travelling the first 120 km at usual speed,it will arrive 36 minutes ahead.Find the distance from A to B.
a.205 b.210 c.230 d.220 e.250
Posted from my mobile device Given: 1. A car travels from town A to town B at constant speed. 2. If it increases speed by 20%, it will arrive 1 hour ahead of schedule. 2. If it increased by 25% after travelling the first 120 km at usual speed,it will arrive 36 minutes ahead. Asked: Find the distance from A to B. Let the distance from A to B be x km and normal speed of the car be v kmh x/1.2v = x/v  1 (1) x/v = 6 => 1/v = x/6 x = 6v (1a) \(\frac{x120}{1.25v} = \frac{x120}{v}  \frac{36}{60}\) (2) (x120)/v = 36*5/60 = 3 x 120 = 3v = x/2 x = 240 Distance from A to B = 240 km None of the choices match the correct answer.



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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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26 Aug 2019, 18:36
VeritasKarishma wrote: pankajpatwari wrote: VeritasKarishma wrote: tilak123  I am kicking off this post as per your request. This is a compilation of all links to the "Made Easy" series. Bunuel has already compiled links to all my posts on specific topics. I will provide links to those compilations: Statistics Made EasyInequalities Made EasySequences Made EasyCombinatorics Made EasyWeighted Avgs and Mixtures Made Easy Hi Karishma.. I request reference to Question 1 in your post "A group of 8 friends sit together in a circle. If A refuses to sit beside B unless C sits on the other side of A as well, how many possible seating arrangements are possible?" Even though i understood your explanation, can u please tell me whats wrong in my approach? My Approach: There are 2 scenarios possible, One where A sits next to both B & C and the other wherein A doesn't sit next to both B & C . In the former the no.of arrangements are: 1*2*1*5*4*3*2*1= 2*5! In the latter the no. of arrangements are: 1*5*4*5*4*3*2*1= 20*5! Thus total no. of arrangements are: 5!(20+2)= 22*5! However your answer is 32*5! Where am i missing?? How about the case when A sits next to C but not to B. A does NOT refuse to sit beside C unless B sits next to him on the other side. We can have AC sitting together in 2 ways, one of the other 5 (except B) sitting next to A on the other side and the other 5 people arranged in 5! ways. This can be done in 2*5*5! = 10*5! ways Thank You very much. I really appreciate



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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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31 Aug 2019, 04:42
Kamran129239 wrote: A car travels from town A to town B at constant speed. If it increases speed by 20%, it will arrive 1 hour ahead of schedule.If it increased by 25% after travelling the first 120 km at usual speed,it will arrive 36 minutes ahead.Find the distance from A to B.
a.205 b.210 c.230 d.220 e.250
Posted from my mobile device When distance travelled is the same, ratio of speeds is inverse of ratio of time taken. Case 1: When car increases its speed by 20% over the entire distance. Ratio of speeds = 5:6 (20% increase in speed), ratio of time taken = 6:5 The actual difference in time taken is 1 hr so in usual case, the car takes 6 hrs but with increased speed, it will take 5 hrs. Case 2: When car increases its speed by 25% over second part of the distance (ignore the first 120 kms for the time being) Over the distance that the car increases its speed by 25%, ratio of speeds = 4:5 (increase of 25%) Then ratio of time taken = 5:4 (a difference of 1 on ratio scale) The actual difference in time taken is 36 mins = 36/60 hrs = 3/5 hrs. So normally, the car takes 5*(3/5) hrs = 3 hrs over this distance. Now note that for the entire distance, the car usually takes 6 hrs. For the second part distance, the car usually takes 3 hrs. So this second part distance must be exactly 1/2 of total distance. Hence total distance must be 2*120 km = 240 km
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Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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18 Sep 2019, 09:49
Hi VeritasKarishmaWhen I have a PS problem such as \(3^{2x} + 3^{x}\)  12 = 0 how should I approach this problem? Second, related to the above, I have realized that I am oftentimes confused by \(2^{2+x}\) = \(2^2*2^x\) and \(2^{2x}\) = (\(2^{x}\))*(\(2^{x}\)). Usually, what I do is I make up numbers and variables to see how it works. However, this takes considerable time and is not exactly the best technique under pressure. Thus, I am wondering if you can recommend some drill or daily exercise that would help me to solidify these differences. Thank you very much in advance. I really appreciate your help.



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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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19 Sep 2019, 09:22
VeritasKarishma thank you very much for both the detailed explanation and the tips for exercises. I really appreciate it. Thank you!



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19 Sep 2019, 15:10
A bus stop is served by express buses, which run every hour at ten minutes past the hour, and local buses, which run every quarter of an hour, starting at five past the hour. What is the probability that a passenger arriving at the stop at random will have to wait more than six minutes for a bus?
A) 1/2 B) 31/60 C) 29/60 D) 1/15 E) 11/20 My confusion is that Say , If he arrives at 6:14 , then he will have to wait for 6 minutes as the next bus is at 6:20. Why is then 6:14 being included ? We need those timeinstants for which waitperiod is more than 6 minutes....right ?
Please help.



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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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19 Sep 2019, 19:01
sayan640 wrote: A bus stop is served by express buses, which run every hour at ten minutes past the hour, and local buses, which run every quarter of an hour, starting at five past the hour. What is the probability that a passenger arriving at the stop at random will have to wait more than six minutes for a bus?
A) 1/2 B) 31/60 C) 29/60 D) 1/15 E) 11/20 Express: 6:10, 7:10, 8:10, 9:10 ... Local: 6:05, 6:20, 6:35, 6:50, 7:05, 7:20, 7:35, 7:50 ... If passenger arrives in these time slots between 6:00 to 7:00, he will need to wait more than 6 mins. 6:10 to 6:14 (arrives just a microsecond before the clock turns 6:14 so the minute of 6:13 to 6:14 is also added) 6:20 to 6:29 6:35 to 6:44 6:50 to 6:59 This adds up to 31 mins out of the total 60 min in an hr so the probability = 31/60.
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19 Sep 2019, 19:13
VeritasKarishma wrote: sayan640 wrote: A bus stop is served by express buses, which run every hour at ten minutes past the hour, and local buses, which run every quarter of an hour, starting at five past the hour. What is the probability that a passenger arriving at the stop at random will have to wait more than six minutes for a bus?
A) 1/2 B) 31/60 C) 29/60 D) 1/15 E) 11/20 Express: 6:10, 7:10, 8:10, 9:10 ... Local: 6:05, 6:20, 6:35, 6:50, 7:05, 7:20, 7:35, 7:50 ... If passenger arrives in these time slots between 6:00 to 7:00, he will need to wait more than 6 mins. 6:10 to 6:14 (arrives just a microsecond before the clock turns 6:14 so the minute of 6:13 to 6:14 is also added) 6:20 to 6:29 6:35 to 6:44 6:50 to 6:59 This adds up to 31 mins out of the total 60 min in an hr so the probability = 31/60. Hi VeritasKarishma, That is what my question is. Why are you considering the timeinstant 6:14? From 6:14 to 6:20 ,it is 6 minutes..right ? And at 6:20 , local bus is available. I am okay with 6:13 as from 6:13 to 6:20 it's more than 6 minutes (7 minutes actually). Can you please explain a bit ? I apologise because of my inability to get your explanation. Posted from my mobile device



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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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20 Sep 2019, 21:41
sayan640 wrote: VeritasKarishma wrote: sayan640 wrote: A bus stop is served by express buses, which run every hour at ten minutes past the hour, and local buses, which run every quarter of an hour, starting at five past the hour. What is the probability that a passenger arriving at the stop at random will have to wait more than six minutes for a bus?
A) 1/2 B) 31/60 C) 29/60 D) 1/15 E) 11/20 Express: 6:10, 7:10, 8:10, 9:10 ... Local: 6:05, 6:20, 6:35, 6:50, 7:05, 7:20, 7:35, 7:50 ... If passenger arrives in these time slots between 6:00 to 7:00, he will need to wait more than 6 mins. 6:10 to 6:14 ( arrives just a microsecond before the clock turns 6:14 so the minute of 6:13 to 6:14 is also added)6:20 to 6:29 6:35 to 6:44 6:50 to 6:59 This adds up to 31 mins out of the total 60 min in an hr so the probability = 31/60. Hi VeritasKarishma, That is what my question is. Why are you considering the timeinstant 6:14? From 6:14 to 6:20 ,it is 6 minutes..right ? And at 6:20 , local bus is available. I am okay with 6:13 as from 6:13 to 6:20 it's more than 6 minutes (7 minutes actually). Can you please explain a bit ? I apologise because of my inability to get your explanation. Posted from my mobile deviceRead again the highlighted part above. 6:10 to 6:14 gives us 4 mins  from exact 6:10 to a microsecond before 6:11 (imagine the face of a clock with the second hand. The second hand starts from 12 at 6:10 and completes a full circle and just before it touches 12 again, a minute is over) from 6:11 to just before 6:12 from 6:12 to just before 6:13 from 6:13 to just before 6:14 This adds up to 4 mins. The moment of 6:14 is not included. 6:14 is a moment, not a minute. I have explained this on the question link too on which you tagged me. Check that out too.
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21 Sep 2019, 09:07
VeritasKarishma wrote: VeritasKarishma wrote: sayan640 wrote: A bus stop is served by express buses, which run every hour at ten minutes past the hour, and local buses, which run every quarter of an hour, starting at five past the hour. What is the probability that a passenger arriving at the stop at random will have to wait more than six minutes for a bus?
A) 1/2 B) 31/60 C) 29/60 D) 1/15 E) 11/20 Express: 6:10, 7:10, 8:10, 9:10 ... Local: 6:05, 6:20, 6:35, 6:50, 7:05, 7:20, 7:35, 7:50 ... If passenger arrives in these time slots between 6:00 to 7:00, he will need to wait more than 6 mins. 6:10 to 6:14 ( arrives just a microsecond before the clock turns 6:14 so the minute of 6:13 to 6:14 is also added)6:20 to 6:29 6:35 to 6:44 6:50 to 6:59 This adds up to 31 mins out of the total 60 min in an hr so the probability = 31/60. Hi VeritasKarishma, That is what my question is. Why are you considering the timeinstant 6:14? From 6:14 to 6:20 ,it is 6 minutes..right ? And at 6:20 , local bus is available. I am okay with 6:13 as from 6:13 to 6:20 it's more than 6 minutes (7 minutes actually). Can you please explain a bit ? I apologise because of my inability to get your explanation. Posted from my mobile deviceRead again the highlighted part above. 6:10 to 6:14 gives us 4 mins  from exact 6:10 to a microsecond before 6:11 (imagine the face of a clock with the second hand. The second hand starts from 12 at 6:10 and completes a full circle and just before it touches 12 again, a minute is over) from 6:11 to just before 6:12 from 6:12 to just before 6:13 from 6:13 to just before 6:14 This adds up to 4 mins. The moment of 6:14 is not included. 6:14 is a moment, not a minute. I have explained this on the question link too on which you tagged me. Check that out too.[/quote] VeritasKarishma maa'm, Why do you say "........from exact 6:10 ..." ? Why are you considering the "6:10 " moment....? There is an express bus at 6:10 ...right ?I apologize if I am overanalyzing. VeritasKarishma



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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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23 Sep 2019, 21:58
sayan640 wrote: VeritasKarishma maa'm, Why do you say "........from exact 6:10 ..." ? Why are you considering the "6:10 " moment....? There is an express bus at 6:10 ...right ?I apologize if I am overanalyzing. VeritasKarishma It's the moment right after 6:10. Say a nano second after 6:10. It is infinitesimally small.
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10 Oct 2019, 21:24
vanam52923 wrote: Done here: https://gmatclub.com/forum/ifxandya ... l#p2378680
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