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Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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08 Oct 2018, 03:38
Hi karishma can you help me with following question/ general approach for following type of questions
The probability of the occurrence of event A is 0.50. The probability of the occurrence of event B is 0.40. What is the range of probability A will not happen AND B will not happen?
A. 0.1≤p≤0.5 B. 0.4≤p≤0.5 C. 0.5≤p≤0.6 D. 0.1≤p≤0.6 E. 0.2≤p≤0.5



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08 Oct 2018, 03:49
Hello Veritaskarishma,
If you can suggest me A proper Plan/strategy to ace Quant(PS and DS) (i.e >90% accuracy)from scratch,it would be great.Like how to start,which resources to use, practice questions etc... Thanks



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08 Oct 2018, 23:46
benignfiend wrote: Thank you Karishma for taking your time out and helping us.
I recently gave my GMAT and got a very low GMAT score. I was more disappointed with my Q 47. On the test day, I took more than 2 mins on tougher questions especially DS. I had problems solving DS since the start of my prep. GMAT club quant tests helped me a lot, but I always fear on DS that I might be missing something else. "Should I plug more ? Did I plug in enough ?" And that attitude ultimately killed my test.
Even though I practiced many questions on this club on Functions and Inequalities, as I am uncomfortable with those topics, on test day, GMAT threw surprises and I succumbed.
Shed some light on to how to stop being overcautious on DS and few tips on handling Functions and Inequalities questions.
Posted from my mobile device Here is the thing about DS questions  people love to solve them by plugging in numbers and I think that strategy often backfires. Until and unless one is super fast in one's calculations and has an intuitive understanding of number properties and how the behaviour of numbers changes at each transition point, one should not use number plugging. But then, if one does have all those skills, number plugging is not required. Hence, all in all, in my opinion, it is a loselose proposition. I am not saying that number plugging is useless:  I agree that for easier questions, number plugging could work (though it might take more time than the more logical approach).  Also, for harder questions, it can help you understand the problem and figure out the logic/pattern.  It also helps you weed out special cases such as n = 0. But the dependence we often have on this approach since it seems easy to follow is unwarranted. Focus on the logic being tested in a question and the issue of "being overcautious" or "missing something" will not arise. Send me some links of functions and inequalities questions that gave you a hard time and I will tell you what I am talking about. Have you seen my posts on functions and inequalities? If no, here are some links: https://www.veritasprep.com/blog/2015/0 ... songmat/https://www.veritasprep.com/blog/2015/0 ... questions/https://www.veritasprep.com/blog/catego ... equalities
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09 Oct 2018, 00:07
nitesh50 wrote: HI Karishma It really is helpful to read all of your posts. I have made it a habit to go through your posts everyday, and it really helps me a lot. So The problem I am facing right now is time management. I have my Gmat on the 4th of November. My lowest score on the Quant gmat club tests was 44 and the highest was 49. I mostly stay around 474849. According to my targets, I have to score Q50. The only topics on which I have not solved 700 level questions are Probability, P and C, Statistics and coordinate geometry. I am slightly under confident in Geometry. So out of these topics, what should I ideally cover? I would like to believe that my basics on these topics are not bad. Also it would be great if you could help me from where I could cover any of these topics efficiently. Regards Hey Nitesh, Glad to hear that! As for the topics mentioned by you: Step 1: Work on Geometry. Be brilliant at basic concepts such as 306090 and great at others such as polygons inscribed in circle etc. Coordinate geometry is gaining ground in GMAT so it is a good idea to be comfortable with it. I think it is extremely easy once you understand the basic things such as drawing lines, how small changes in slope and intercept change the line etc, relation between these concepts, distance of points from centre and its relation to pythagorean theorem etc. I have discussed quite a few Geometry concepts on my blog (if you wish to check them out) though you will need to look for them in order starting from the oldest page. https://www.veritasprep.com/blog/catego ... dom/page/9Step 2: I love Stats! I know that is not a good enough reason to focus on it but the point is that it is quite conceptual and questions on it are fun! I have discussed most of the Stats relevant topics on my blog: https://www.veritasprep.com/blog/2014/1 ... tatistics/https://www.veritasprep.com/blog/2015/0 ... thegmat/Step 3: Combinatorics  Too much time spent here may not lead to much. Say if you get a 600700 level question on it, you should not miss it. But a 700+ level question on it, give it your best shot and move on. There are tiny things, one word, that could change the question entirely from what you might have come across before. The entire approach would change in that case. Hence, I feel that the topic is just a wee bit too fickle to invest a lot of time in.
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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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09 Oct 2018, 01:43
apurva908 wrote: Hi karishma can you help me with following question/ general approach for following type of questions
The probability of the occurrence of event A is 0.50. The probability of the occurrence of event B is 0.40. What is the range of probability A will not happen AND B will not happen?
A. 0.1≤p≤0.5 B. 0.4≤p≤0.5 C. 0.5≤p≤0.6 D. 0.1≤p≤0.6 E. 0.2≤p≤0.5 Solution: P(A) = .5 P(B) = .4 You need to find P(Neither will happen). Think about the corresponding sets concept. Neither is given by Total  n(A or B) P(Neither will happen) = 1  P(A or B) = 1  [ P(A) + P(B)  P(A)*P(B) ] We don't know the relation between the events A and B so we don't know what P(A and B) is. Try to think of this in terms of the venn diagram. To minimise neither, we need to maximise P(A or B). So A and B should have minimum overlap. Assuming A and B are mutually exclusive events, P(A and B) = 0 P(Neither will happen) = 1  [ P(A) + P(B)  P(A and B)] = 1  [.5 + .4  0] = 0.1 To maximise neither, we need to minimise P(A or B). So A and B should have maximum overlap. Assuming B is a subset of A , P(A and B) = 0.4 P(Neither will happen) = 1  [ P(A) + P(B)  P(A and B)] = 1  [.5 + .4  .4] = 0.5 Answer (A)
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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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09 Oct 2018, 01:47
apurva908 wrote: Hello Veritaskarishma,
If you can suggest me A proper Plan/strategy to ace Quant(PS and DS) (i.e >90% accuracy)from scratch,it would be great.Like how to start,which resources to use, practice questions etc... Thanks Apurva, how to ace Quant is a very generic question  the answer would need a thesis of 25,000 words! I will try to condense the main points for you in a few days!
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17 Nov 2018, 01:54
Request to all forum users: I am one of the forum experts on the PS forum. If you would like me to take a look at any question on this forum, please mention it on this thread. Since I receive 100s of mails everyday in my mailbox, unfortunately, it becomes very difficult to follow up on every query even if I am tagged on it. I do take a look at this thread on a daily basis and hence will certainly get to your query if you mention here. Thank you!
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30 Nov 2018, 22:24
Hi VeritasKarishmaI had a query wrt your articles on the made easy series. Q. IN how many ways can A,B,C,D,E,F be arranged in a circular table provided that A cannot sit next to D or E? This question is a slight variation in comparison to your question on the Combinations Article 3: Circular Arrangements. I recon that it is slightly complicated. Here was my approach: If only 1. D/E is selected: 2 ways * 6 ways( 3c2 ways of choosing people to be beside A and 2! ways of arranging them) * 2 ways( the other people arrangement) 2. D and E both are selected: 3c2(out of remaining 3 people, 2 are selected)* 2 ways (those 2 people are arranged) * 2 (D,E are arranged in remaining places) 3. A is not selected: 4! SO total possibilities: 24+ 12+24= 60 ways. Am i correct in my reasoning ? and is there any simpler and faster way to do this question? Looking forward to your reply! Regards Nitesh



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01 Dec 2018, 01:31
vishuvashishth wrote: Sure, I will help you out on any questions you might have. Note that not every question given by you will involve making equations. But I will go one line at a time and show you how to evaluate it to arrive at the answer. Here is the first one: https://gmatclub.com/forum/ofthestude ... l#p2182147
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01 Dec 2018, 01:49
vishuvashishth wrote: No. 3 https://gmatclub.com/forum/inorderto ... l#p2182154This doesn't involve forming equations either. In fact, there are very few questions in which you will need to actually form an equation and then solve it. Notice the way you should evaluate this sentence to get to the answer.
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Question 2: A group of 10 people consists of 2 married couples and 6 bachelors. A committee of 4 is to be selected from the 10 people. How many different committees can be formed if the committee can consist of at most one married couple? Solution: We have to select 4 people out of: 6 bachelors and 2 married couples. The number of ways of selecting any 4 people out of 10 is 10*9*8*7/4! = 210 (Note here that we are just selecting 4 people. We are not arranging them so we divide by 4!) The people will get selected in various ways: 1. Four bachelors 2. One from a couple and three bachelors 3. Two from two different couples and two bachelors 4. One couple and two bachelors 5. One couple, one person from a couple, one bachelor 6. Two couples If we add the number of committees possible in each of these cases, we will get 210. Out of all these cases, only the last one (two couples) has more than one married couple. Instead of calculating the number of different committees that can be formed in each of the first five cases, we can calculate the number of committees in the last case and subtract it from 210. How many different committees can be formed such that there are 2 couples? Only one since we have only 2 couples. We will have to select both the couples and we will get 4 people. Number of different committees of 4 people such that there is at most one married couple = 210 – 1 = 209. Just for practice, let’s see how we can calculate the different number of committees that can be formed in each of the first five cases. The sum of all these cases should give us 209. 1. Select 4 bachelors from 6 bachelors in 6*5*4*3/4! = 15 different committees 2. Select 1 person out of the two couples (4 people) in 4 ways and 3 bachelors from 6 bachelors in 6*5*4/3! = 20 ways. So you select the 4 people in 4*20 = 80 different committees 3. Select 2 people from 2 different couples in 4*2/2! = 4 ways and 2 bachelors from 6 bachelors in 6*5/2! = 15 ways. So you select the 4 people in 4*15 = 60 different committees 4. Select 1 couple in 2 ways and 2 bachelors from 6 bachelors in 6*5/2! = 15 ways. So you select the 4 people in 2*15 = 30 different committees 5. Select 1 couple in 2 ways, 1 person from the remaining couple in 2 ways and 1 bachelor from 6 bachelors in 6 ways. So you can select the 4 people in 2*2*6 = 24 different committees The sum of all these five cases = 15 + 80 + 60 + 30 + 24 = 209 different committees Hi VeritasKarishmaFirst of all,Thanks for making these articles. Now I have a slight doubt related to above question . 4. Select 1 couple in 2 ways and 2 bachelors from 6 bachelors in 6*5/2! = 15 ways. So you select the 4 people in 2*15 = 30 different committees 5. Select 1 couple in 2 ways, 1 person from the remaining couple in 2 ways and 1 bachelor from 6 bachelors in 6 ways. So you can select the 4 people in 2*2*6 = 24 different committeesWhen I was doing the question, I considered 5 and 6 to be symmetrical cases. I calculated it as follows: 2c1(1 married couple selected) + 2c1(One member of the other married couple selected) + 7c2(selected 2 members out of a group of 7) = 84 ways Now i do realise that I am counting something twice. But I can't exactly point out where. Can you please tell me where I am going wrong, and If I were to proceed with my method, what should I have subtracted to get a correct answer. Hope to hear from you soon! Regards Nitesh



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01 Dec 2018, 01:53
vishuvashishth wrote: I have already solved no 2 by the variation method. I have also provided the link to my relevant post there. You can also solve it using unitary method as shown by some people there (if 10 workers make 50 chairs, 1 worker will make 5 chairs so 18 workers will make 18*5 chairs etc). But I prefer the simple and straight forward method shown by me. Once you understand it, it takes a few seconds to implement.
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01 Dec 2018, 03:37
nitesh50 wrote: Hi VeritasKarishmaI had a query wrt your articles on the made easy series. Q. IN how many ways can A,B,C,D,E,F be arranged in a circular table provided that A cannot sit next to D or E? This question is a slight variation in comparison to your question on the Combinations Article 3: Circular Arrangements. I recon that it is slightly complicated. Here was my approach: If only 1. D/E is selected: 2 ways * 6 ways( 3c2 ways of choosing people to be beside A and 2! ways of arranging them) * 2 ways( the other people arrangement) 2. D and E both are selected: 3c2(out of remaining 3 people, 2 are selected)* 2 ways (those 2 people are arranged) * 2 (D,E are arranged in remaining places) 3. A is not selected: 4! SO total possibilities: 24+ 12+24= 60 ways. Am i correct in my reasoning ? and is there any simpler and faster way to do this question? Looking forward to your reply! Regards Nitesh Hey Nitesh, I am not sure what your question is. There are two variations possible: Q. IN how many ways can A,B,C,D,E,F be arranged in a circular table provided that A cannot sit next to D and E at the same time. (Implying that A sitting next to D if E is far away is ok) or Q. IN how many ways can A,B,C,D,E,F be arranged in a circular table provided that A can sit next to neither D nor E? (Implying that both D and E should be far away from A) The first question is discussed as question 2 in my post: Question 2: There are 6 people, A, B, C, D, E and F. They have to sit around a circular table such that A cannot sit next to D and F at the same time. How many such arrangements are possible?
Solution: Total number of ways of arranging 6 people in a circle = 5! = 120
Now, A cannot sit next to D and F simultaneously.
Let’s first find the number of arrangements in which A sits between D and F. In how many of these 120 ways will A be between D and F? Let’s consider that D, A and F form a single unit. We make DAF sit on any three consecutive seats in 1 way and make other 3 people sit in 3! ways (since the rest of the 3 seats are distinct). But D and F can swap places so the number of arrangements will actually be 2*3! = 12
In all, we can make A sit next to D and F simultaneously in 12 ways.
The number of arrangements in which A is not next to D and F simultaneously is 120 – 12 = 108.The second question is discussed as Question 3 in my post: Question 3: There are 6 people, A, B, C, D, E and F. They have to sit around a circular table such that A can sit neither next to D nor next to F. How many such arrangements are possible?
Solution: In the previous question, A could sit next to D and F; the only problem was that A could not sit next to both of them at the same time. Here, A can sit next to neither D nor F. Generally, it is difficult to wrap your head around what someone cannot do. It is easier to consider what someone can do and go from there. A cannot sit next to D and F so he will sit next to two of B, C and E.
Let’s choose two out of B, C and E. In other words, let’s drop one of B, C and E. We can drop one of B, C and E in 3 ways (we can drop B or C or E). This means, we can choose two out of B, C and E in 3 ways (We will come back to choosing 2 people out of 3 when we work on combinations). Now, we can arrange the two selected people around A in 2 ways (say we choose B and C. We could have BAC or CAB). We make these three sit on any three consecutive seats in 1 way.
Number of ways of choosing two of B, C and E and arranging the chosen two with A = 3*2 = 6
The rest of the three people can sit in three distinct seats in 3! = 6 ways
Total number of ways in which A will sit next to only B, C or E (which means A will sit neither next to D nor next to F) = 6*6 = 36 ways https://www.veritasprep.com/blog/2011/1 ... tsparti/Let me know if this answers your question.
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vishuvashishth wrote: No 4 A cook went to a market to buy some eggs and paid $12. But since the eggs were quite small, he talked the seller into adding two more eggs, free of charge. As the two eggs were added, the price per dozen went down by a dollar. How many eggs did the cook bring home from the market? A. 8 B. 12 C. 15 D. 16 E. 18 Say price per dozen was P and it became (P  1) due to the extra 2 eggs. Say he bought D dozens and the extra 2 eggs made it (D + 2/12) dozens. Now note that total paid by him in both cases is $12 PD = 12 and (P  1)(D + 1/6) = 12 PD + P/6  D  1/6 = 12 Putting PD = 12 from above we get: P  6D = 1 Putting P = 12/D from above, we get: 12/D  6D = 1 Now solve for D to get: 12  6D^2 = D 6D^2 + D  12 = 0 6D^2 + 9D  8D  12 = 0 3D(2D + 3)  4(2D + 3) = 0 (2D + 3)*(3D  4) = 0 So D =  3/2 or 4/3 Since D must be positive, it should be 4/3. He brought home (4/3)*12 + 2 = 18 eggs
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vishuvashishth wrote: No 5: Mac can finish a job in M days and Jack can finish the same job in J days. After working together for T days, Mac left and Jack alone worked to complete the remaining work in R days. If Mac and Jack completed an equal amount of work, how many days would have it taken Jack to complete the entire job working alone? (1) M = 20 days (2) R = 10 days Rate of work of Mac = 1/M (since it takes him M days to finish 1 work and Rate = Work/Time) Rate of work of Jack = 1/J (since it takes him J days to finish 1 work and Rate = Work/Time) Mac worked for T days and did half the work so as per Work = Rate*Time, 1/2 = T/M Jack worked for (T + R) days and did half the work so as per Work = Rate*Time, 1/2 = (T+R)/J We need the value of J. (1) M = 20 days 1/2 = T/20 T = 10 days 1/2 = (10 + R)/J Two unknowns and 1 equation. Not possible to solve. (2) R = 10 days 1/2 = (T+10)/J Two unknowns and 1 equation. Not possible to solve. Using both, 1/2 = (10 + 10)/J J = 40 days Sufficient
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Article on Combinations made easy Question 1: There are x children and y chairs in a room where x and y are prime numbers. In how many ways can the x children be seated in the y chairs (assuming that each chair can seat exactly one child)?
Statement 1: x + y = 12 Statement 2: There are more chairs than children.
Solution:
There are x children and y chairs.
x and y are prime numbers.
Statement 1: x + y = 12
Since x and y are prime numbers, a quick run on 2, 3, 5 shows that there are two possible cases:
Case 1: x=5 and y=7
There are 5 children and 7 chairs.
Case 2: x=7 and y=5
There are 7 children and 5 chairs
At first glance, they might look like two different cases and you might feel that statement 1 is not sufficient alone. But note that the question doesn’t ask you for number of children or number of chairs. It asks you about the number of arrangements.
Case 1: x=5 and y=7
If there are 5 children and 7 chairs, we select 5 chairs out of the 7 in 7C5 ways. We can now arrange 5 children in 5 seats in 5! ways.
Total number of arrangements would be 7C5 * 5!
Case 2: x = 7 and y = 5
If there are 7 children and 5 chairs, we select 5 children out of the 7 in 7C5 ways. We can now arrange 5 children in 5 seats in 5! ways.
Total number of arrangements would be 7C5 * 5!
Note that in both cases the number of arrangements is 7C5*5!. Combinatorics does not distinguish between people and things. 7 children on 5 seats is the same as 5 children on 7 seats because in each case you have to select 5 out of 7 (either seats or children) and then arrange 5 children in 5! ways.
So actually this statement alone is sufficient! Most people would not have seen that coming!
Statement 2: There are more chairs than people.
We don’t know how many children or chairs there are. This statement alone is not sufficient.
Answer: A This question is discussed HERE.
We were tempted to answer the question as (C) but it was way too easy. Statement 1 gave 2 cases and statement 2 narrowed it down to 1. Be aware that if it looks too easy, you are probably missing something!
Now, what if we alter the question slightly and make it:
Question 2: There are x children and y chairs arranged in a circle in a room where x and y are prime numbers. In how many ways can the x children be seated in the y chairs (assuming that each chair can seat exactly one child)?
Statement 1: x + y = 12 Statement 2: There are more chairs than children HI VeritasKarishmaThis question is part of your articles on P/c. My doubt: If we are asked to arrange 7 children in 5 chairs, then this case is not possible.(provided that every chair can have 1 child only) There is no possibility in which all 7 children can be arranged in 5 chairs. Similarly, when I was doing Q1 stated above, I inferred from the question statement that the number of children has to be less than/equal to the number of chairs.
Hence the case in which we have 7 children and 5 chairs becomes impossible. Similarly in Q2, we cannot arrange 7 children in 5 chairs arranged in a circular manner. Hence the answer should be A. Is my stated reasoning correct? Or am I missing something? Regards Nitesh



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02 Dec 2018, 23:22
vishuvashishth wrote: Last two questions: https://gmatclub.com/forum/achessplay ... l#p2183020https://gmatclub.com/forum/ifxaandb ... l#p2183015
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02 Dec 2018, 23:32
nitesh50 wrote: When I was doing the question, I considered 5 and 6 to be symmetrical cases. I calculated it as follows: 2c1(1 married couple selected) + 2c1(One member of the other married couple selected) + 7c2(selected 2 members out of a group of 7)
= 84 ways
Now i do realise that I am counting something twice. But I can't exactly point out where. Can you please tell me where I am going wrong, and If I were to proceed with my method, what should I have subtracted to get a correct answer.
Hope to hear from you soon!
Regards Nitesh The two cases are different. A couple and 2 bachelors vs a couple, one from another couple and a bachelor. In your calculation, you are selecting 5 people  1 married couple selected (2 people) + One member of the other married couple selected + selected 2 members out of a group of 7 = 5 people selected Also, from where did you get the group of 7?
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02 Dec 2018, 23:50
nitesh50 wrote: Article on Combinations made easy Question 1: There are x children and y chairs in a room where x and y are prime numbers. In how many ways can the x children be seated in the y chairs (assuming that each chair can seat exactly one child)?
Statement 1: x + y = 12 Statement 2: There are more chairs than children.
Solution:
There are x children and y chairs.
x and y are prime numbers.
Statement 1: x + y = 12
Since x and y are prime numbers, a quick run on 2, 3, 5 shows that there are two possible cases:
Case 1: x=5 and y=7
There are 5 children and 7 chairs.
Case 2: x=7 and y=5
There are 7 children and 5 chairs
At first glance, they might look like two different cases and you might feel that statement 1 is not sufficient alone. But note that the question doesn’t ask you for number of children or number of chairs. It asks you about the number of arrangements.
Case 1: x=5 and y=7
If there are 5 children and 7 chairs, we select 5 chairs out of the 7 in 7C5 ways. We can now arrange 5 children in 5 seats in 5! ways.
Total number of arrangements would be 7C5 * 5!
Case 2: x = 7 and y = 5
If there are 7 children and 5 chairs, we select 5 children out of the 7 in 7C5 ways. We can now arrange 5 children in 5 seats in 5! ways.
Total number of arrangements would be 7C5 * 5!
Note that in both cases the number of arrangements is 7C5*5!. Combinatorics does not distinguish between people and things. 7 children on 5 seats is the same as 5 children on 7 seats because in each case you have to select 5 out of 7 (either seats or children) and then arrange 5 children in 5! ways.
So actually this statement alone is sufficient! Most people would not have seen that coming!
Statement 2: There are more chairs than people.
We don’t know how many children or chairs there are. This statement alone is not sufficient.
Answer: A This question is discussed HERE.
We were tempted to answer the question as (C) but it was way too easy. Statement 1 gave 2 cases and statement 2 narrowed it down to 1. Be aware that if it looks too easy, you are probably missing something!
Now, what if we alter the question slightly and make it:
Question 2: There are x children and y chairs arranged in a circle in a room where x and y are prime numbers. In how many ways can the x children be seated in the y chairs (assuming that each chair can seat exactly one child)?
Statement 1: x + y = 12 Statement 2: There are more chairs than children HI VeritasKarishmaThis question is part of your articles on P/c. My doubt: If we are asked to arrange 7 children in 5 chairs, then this case is not possible.(provided that every chair can have 1 child only) There is no possibility in which all 7 children can be arranged in 5 chairs. Similarly, when I was doing Q1 stated above, I inferred from the question statement that the number of children has to be less than/equal to the number of chairs.
Hence the case in which we have 7 children and 5 chairs becomes impossible. Similarly in Q2, we cannot arrange 7 children in 5 chairs arranged in a circular manner. Hence the answer should be A. Is my stated reasoning correct? Or am I missing something? Regards Nitesh Nitesh, the case of 7 children and 5 chairs is no different. You select 5 of the 7 children and make them sit on the 5 chairs. 2 children will be left standing. Just like in the case of 7 chairs and 5 children, you select 5 of the 7 chairs and arrange the 5 children in the 5 chairs. 2 chairs will be vacant. Both cases are about selecting 5 out of distinct 7 and matching them with other distinct 5.
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Karishma Veritas Prep GMAT Instructor
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