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a, b, and c are three integers such that a and b are less than 100, an
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Updated on: 29 Jul 2018, 13:48
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26% (02:53) correct 74% (03:02) wrong based on 241 sessions
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Originally posted by Bunuel on 30 Aug 2015, 10:59.
Last edited by abhimahna on 29 Jul 2018, 13:48, edited 2 times in total.
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Re: a, b, and c are three integers such that a and b are less than 100, an
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01 Sep 2015, 19:42
dav90 wrote: Ans should be C, Unless we get values of a and k we cannot calculate sum, to get values we need both the statements Not really. Look below. In DS questions, you need to absolutely sure that you will not get an answer by using the statements alone. Sum of n terms of an arithmetic progression (a sequence in which each term after the 1st one is related to the previous term by a constant, k/11 in this case) =\(\frac{n*[2*a+(n1)*d]}{2}\) where, n =100, d = k/11 Substituting, we get, Sum of 100 terms = 50* [2a+9k] Per statement 2, after rearranging the terms, we get 2a+9k = 46. Thus this statement is sufficient to give the sum of 100 terms = 50*46 = a unique value. Per statement 1, 2a+11k = 46 > a = (4611k)/2 and as 'a' is a positive integer (>0), the only possible values of k to give integer value of 'a' are k = 4 or k =2 Thus we get , a=12 when k =2 or a=1 when k=4. Case 2: a=1 when k=4 , NOT possible as we are told that 1 of 'a' and 'k' is a prime number and as both 1 and 4 are NOT primes, this case is discarded. Case 1: a=12 when k =2 , possible as 2 is a prime number. Thus this is the only case possible and hence statement 1 is sufficient as well. Thus both statements are sufficient on their own. D is the correct answer.




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Re: a, b, and c are three integers such that a and b are less than 100, an
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Updated on: 31 Aug 2015, 12:13
Sum of the first 100 terms in the sequence = 50 * (2a + 9k) = 50 * (2a + 11k 2k) Statement 1. provides us the value of (2a + 11k) only. hence not enough. Statement 2. provides us the value of (2a + 9k). Hence enough. Hence, answer is B.



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Re: a, b, and c are three integers such that a and b are less than 100, an
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31 Aug 2015, 05:20
First term in a sequence is a, and each term thereafter is k/11 greater than the term before it. If a and k are both positive integers but only one of them is a prime number, what is the sum of the first 100 terms in the sequence? a + a + 99 (k/11) = 2a + 9k
100×(2a + 9k )/2 =50 (2a+9k)
(1) 2a + 11k = 46 We have 2 equations in which we can solve for each variable sufficient
(2) 3k = 14  2/3*a We have 2 equations in which we can solve for each variable sufficient
Answer: D



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Re: a, b, and c are three integers such that a and b are less than 100, an
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Updated on: 06 Sep 2015, 20:53
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution. The first term in a sequence is a, and each term thereafter is k/11 greater than the term before it. If a and k are both positive integers but only one of them is a prime number, what is the sum of the first 100 terms in the sequence? (1) 2a + 11k = 46 (2) 3k = 14  2/3*a ==> In the original condition, there are 2 variables (a,k) and we need 2 equations to match the number of variables and equations. Since there is 1 each in 1) and 2), C is likely the answer. Using 1) & 2) together, 1)=2) k=2, a=12 thus it is sufficient. Therefore the answer is D.
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Re: a, b, and c are three integers such that a and b are less than 100, an
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01 Sep 2015, 12:22
Using 1 is not sufficient Since we have s100= 50*462k Using 2 we have S100=50*42 So B?
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Re: a, b, and c are three integers such that a and b are less than 100, an
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01 Sep 2015, 19:26
Ans should be C, Unless we get values of a and k we cannot calculate sum, to get values we need both the statements
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Re: a, b, and c are three integers such that a and b are less than 100, an
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09 May 2017, 14:28
Wow, the question is really nasty. Rearanging 2 we get 2a+9k= 42. Both (12,2) and (3,4) satisfy the equation. However, if you rearrange condition given in the question using arithmetic progression formula, as Engr2012 did, you will see that 2 is sufficient.



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Re: a, b, and c are three integers such that a and b are less than 100, an
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10 May 2017, 01:34
I believe answer should be A If you evaluate the 2nd statement you get two sets of values for A and K i.e (a=3 and k=4 or a=12 and k=2) in this both solution leads to a single prime value but statement 1 there is only one possible solution with one prime number. Bunuel plz comment Thanks Varun



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Re: a, b, and c are three integers such that a and b are less than 100, an
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10 May 2017, 02:13
I myself marked A initially.
Condition 1 is sufficient Condition 2, we get 2 answers a=3,k=4 or a= 12,k=2 so I don't think it is sufficient,initially but I think there is alternative for 2nd condition, Sum of AP comes out to be 50*(2a+9k), value of 2a+9k is given as 42 as per second condition so it is also sufficient hence D



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Re: a, b, and c are three integers such that a and b are less than 100, an
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25 May 2018, 02:47
Bunuel, Could you please tell me how (i) is sufficient to solve this.
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Re: a, b, and c are three integers such that a and b are less than 100, an
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01 Jun 2018, 04:56
Bunuel wrote: The first term in a sequence is a, and each term thereafter is k/11 greater than the term before it. If a and k are both positive integers but only one of them is a prime number, what is the sum of the first 100 terms in the sequence?
(1) 2a + 11k = 46 (2) 3k = 14  2/3*a
Kudos for a correct solution. Shouldn't it be A? Since stmt. 2 has two valid solutions of (a,k), i.e. (a,k) = (3,4) or (2,12) From stmt.1 we can make sure (a,k) = (2,12) how is this wrong?
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Re: a, b, and c are three integers such that a and b are less than 100, an
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01 Jun 2018, 05:30
thefibonacci wrote: (2) 3k = 14  2/3*a Shouldn't it be A? Since stmt. 2 has two valid solutions of (a,k), i.e. (a,k) = (3,4) or (2,12) From stmt.1 we can make sure (a,k) = (2,12) how is this wrong? First of all not (2,12), but (12,2). You are asked to find the sum of 100 numbers of the set, which = 50*(2a+9k) In both cases (3,4) or (12,2), the Sum = 50*42. So B is sufficient too.




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