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Two natural numbers a and b are such that a = 3b + b^2
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06 Jan 2019, 09:21
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Two natural numbers a and b are such that a = 3b + b^2. Is 'a' a multiple of 5? (1) b is a multiple of 2 (2) b is a factor of 10
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Re: Two natural numbers a and b are such that a = 3b + b^2
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07 Jan 2019, 22:38
From statement 1: b is a multiple of 2. b can be 0, 2, 4, 6, etc. If b is 0 and 2, then a will be a multiple of 5. If b is 4, then a will not be a multiple of 5. Hence statement 1 is insufficient. From statement 2: b is a factor of 10. b is 1, 2, 5, 10. If b is 2, 5 or 10. Then a will be a multiple of 5. If b is 1, then a will not be a multiple of 5. Insufficient. Combining both gives the value of b as 2 or 10. For 2 or 10, a will be a multiple of 5. Hence sufficient. C is the answer.
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Re: Two natural numbers a and b are such that a = 3b + b^2
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08 Jan 2019, 00:46
Hi, Given, “a” and “b” are natural numbers. That is, “a” and “b” are positive integers. Such that, a = 3b + b^2. Question: Is “a” multiple of 5 ? i.e., whether a = 5, 10, 15…..? Statement I is insufficient:b is a multiple of 2. “b” is a positive integer and multiple of 2.. So, “b” can be 2,4,6,8,10… If b = 2, then a = 3b + b^2 = 6 +4 = 10 and is a multiple of 5. Answer to the question is YES. But if b = 4, then a = 3b + b^2 = 12 + 16 = 28 and it is not a multiple of 5. Answer to the question is NO. Statement II is insufficient:b is a factor of 10 So, “b” values are 1, 2, 5 or 10. If b = 2, then a = 3b + b^2 = 6 +4 = 10 and is a multiple of 5. Answer to the question is YES. But if b = 1, then a = 3b + b^2 = 3 + 1 = 4 and it is not a multiple of 5. Answer to the question is NO. Together it is sufficient.Values of “b” which satisfy both statement I and II is 2 and 10. So, If b = 2, then a = 3b + b^2 = 6 +4 = 10 and is a multiple of 5. Answer to the question is YES. And if b = 10, then a = 3b + b^2 = 30 + 100 = 130 and it is also a multiple of 5. Answer to the question is again YES. So together it is sufficient. Answer is C. Hope this helps. Regards, Junaid. Byjus GMAT Quant Expert
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Re: Two natural numbers a and b are such that a = 3b + b^2
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02 Feb 2019, 09:38
akurathi12 wrote: Two natural numbers a and b are such that a = 3b + b^2. Is 'a' a multiple of 5?
(1) b is a multiple of 2
(2) b is a factor of 10 Natural numbers are 1,2,3,4,5,6,7,8 from 1, b will be 2,4,6,8,10 a = 3b + b^2, just substitute the values to get Yes when b = 2 No when b = 4 from 2, b= 1,2,5,10 Yes when b=2,5,10 No when b=1 Combine both to get b= 2,10, both will give Yes C
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Re: Two natural numbers a and b are such that a = 3b + b^2
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27 Mar 2019, 22:42
Given , a=3b+b^2 or a=b(3+b) To find a/5 = Integer or b(3+b)/5 = Integer ?
Statement 1, b/2 = Integer Insufficient, since b can be 2 (sufficient) , but if b is 4 (insufficient)
Statement 2: B is a factor of 10 i.e 1 ,2, 5, 10 Alone insufficient, since b can be 1 also.
After combining : Sufficient to say Yes, since possible values of b can be 2 and 10.
Ans : C




Re: Two natural numbers a and b are such that a = 3b + b^2
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27 Mar 2019, 22:42






