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If a, b, and c are consecutive 2digit positive integers in that order
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09 Aug 2018, 00:30
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[ Math Revolution GMAT math practice question] If \(a, b,\) and \(c\) are consecutive 2digit positive integers in that order, and \(a+b+c\) is a multiple of \(10\), what is the value of \(c\)? 1) \(a\) is a prime number 2) \(c\) is a prime number
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If a, b, and c are consecutive 2digit positive integers in that order
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09 Aug 2018, 00:51
MathRevolution wrote: [ Math Revolution GMAT math practice question] If \(a, b,\) and \(c\) are consecutive 2digit positive integers in that order, and \(a+b+c\) is a multiple of \(10\), what is the value of \(c\)? 1) \(a\) is a prime number 2) \(c\) is a prime number Let a=x1, b=x, and c=x+1 where x>10 and an an integer. So, a+b+c=3x=10k (Given) So, \(x=(\frac{10}{3})*k\) We have (a,b,c) triplets when k=6,9,12,15,....27. So we have (a,b,c)=(19,20,21), (29,30,31),.............,(59,60,61),.....,(89,91,90) St1:\(a\) is a prime number. c could be 21,31 etc Insufficient. St2: \(c\) is a prime number c could be 31,41,61. Insufficient. Combined, we have two pair of triplets, (29,30,31) and (59,60,61), where c has two values. Insufficient. Ans. (E)
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If a, b, and c are consecutive 2digit positive integers in that order
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09 Aug 2018, 01:50
MathRevolution wrote: [ Math Revolution GMAT math practice question] If \(a, b,\) and \(c\) are consecutive 2digit positive integers in that order, and \(a+b+c\) is a multiple of \(10\), what is the value of \(c\)? 1) \(a\) is a prime number 2) \(c\) is a prime number \(a, b,\) and \(c\) are consecutive 2digit positive integers in order could be as follows a,a+1,a+2...........So a+a+1+a+2 = 3a+3= 3(a+1) = mulitple of 10....this will happen if a =19 or 29 or 39.............89..the question will be: Is a =19 or 29 or 39..........or 89?? Note it can't be 99 as b and c will be 3 digit numbers. 1) \(a\) is a prime numberMany values for a. It could be 19 or 29 or 59 or 79 Insufficent 2) \(c\) is a prime number Many values for c. If a =29.... then c= 31= prime If a =59......then c=61= Prime Combine 1 & 2 Take same examples above.............No clear value for a Answer: E



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Re: If a, b, and c are consecutive 2digit positive integers in that order
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09 Aug 2018, 07:52
MathRevolution wrote: [ Math Revolution GMAT math practice question] If \(a, b,\) and \(c\) are consecutive 2digit positive integers in that order, and \(a+b+c\) is a multiple of \(10\), what is the value of \(c\)? 1) \(a\) is a prime number 2) \(c\) is a prime number ST 1. 41 40 39 ST 2. 39 40 41 so C could be any value. hence E can someone confirm if my understanding is correct



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Re: If a, b, and c are consecutive 2digit positive integers in that order
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09 Aug 2018, 18:12
dave13 wrote: MathRevolution wrote: [ Math Revolution GMAT math practice question] If \(a, b,\) and \(c\) are consecutive 2digit positive integers in that order, and \(a+b+c\) is a multiple of \(10\), what is the value of \(c\)? 1) \(a\) is a prime number 2) \(c\) is a prime number ST 1. 41 40 39 ST 2. 39 40 41 so C could be any value. hence E can someone confirm if my understanding is correct dave13 , Yes C Can be any value when you combine both Statement 1 & 2 Ex: (29,30,31) and (59,60,61) 39 40 41 Doesn't hold ture, when you combine both Statements bcoz 39 is not a Prime
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Re: If a, b, and c are consecutive 2digit positive integers in that order
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13 Aug 2018, 05:21
=> Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. Since a = b – 1 and c = b + 1, we have a + b + c = ( b – 1 ) + b + ( b + 1 ) = 3b is a multiple of 10 and b is a multiple of 10. Thus, the possible triples ( a, b, c ) are ( 19, 20, 21 ), ( 29, 30, 31 ), ( 39, 40, 41 ), ( 49, 50, 51 ), ( 59, 60, 61 ), ( 69, 70, 71 ), ( 79, 80, 81 ) and ( 89, 90, 91 ). Since we have 3 variables and 2 equations ( a = b – 1, c = b + 1), D is most likely to be the answer. So, we should consider each of the conditions on their own first. Condition 1) The triples in which a is a prime number are ( 19, 20, 21 ), ( 29, 30, 31 ), ( 59, 60, 61 ), ( 79, 80, 81 ) and ( 89, 90, 91 ). Thus, c could be 21, 31, 61, 81 or 91. Since we don’t have a unique solution, condition 1) is not sufficient. Condition 2) The triples in which c is a prime number are ( 29, 30, 31 ), ( 39, 40, 41 ), ( 59, 60, 61 ), ( 69, 70, 71 ) and ( 89, 90, 91 ). Thus, c could be 31, 41, 61, 71 or 91. Since we don’t have a unique solution, condition 2) is not sufficient. Conditions 1) & 2): The triples in which both a and c are prime numbers are: ( 29, 30, 31 ), ( 59, 60, 61 ) and ( 89, 90, 91 ). Thus, c could be 31, 61 or 91. Since we don’t have a unique solution, both conditions are not sufficient, when considered together. Therefore, E is the answer. Answer: E If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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