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What will be the value of a/b ? Given that a and b are positive integers (1) a^2 – b^2 = 169 (2) a – b = 1

I am unable to prove statement 1 is sufficient here. The only thing i could think of first is the 5-12-13 triplet but that would hold good only for a^2 + b^2 = 169. please help.

What will be the value of a/b ? Given that a and b are positive integers

(1) a^2 – b^2 = 169 --> as given that \(a\) and \(b\) are positive integers then \(a>b\). Next, \(a^2-b^2=(a-b)(a+b)=169=1*169=13*13\) --> again as \(a\) and \(b\) are positive integers and \(a>b\) then: \(a-b=1\) and \(a+b=169\). Solving gives: \(a=85\) and \(b=84\) --> \(\frac{a}{b}=\frac{85}{84}\). Sufficient.

What will be the value of a/b ? Given that a and b are positive integers (1) a^2 – b^2 = 169 (2) a – b = 1

I am unable to prove statement 1 is sufficient here. The only thing i could think of first is the 5-12-13 triplet but that would hold good only for a^2 + b^2 = 169. please help.

\(169 = 13^2\). \(a^2-b^2=(a+b)(a-b)\). Since \(a\) and \(b\) are positive integers, \(a+b>a-b\) and \(a+b>0\), so the only possible factorization is \(a+b=169\) and \(a-b=1\), from which \(a\) and \(b\) can be deduced, as well as the ratio \(a/b\). Therefore, (1) is sufficient. (2) obviously not sufficient.

Answer A. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: What will be the value of a/b ? Given that a and b are [#permalink]

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21 Aug 2013, 04:16

Can we also say that: a^2 - b^2 = 13^2, This is in the form of a pythagorean triplet. Since a and b are given as integers, there can be only one integral value for a and b which will satisfy such a condition. Thus this is a sufficient condition.

Re: What will be the value of a/b ? Given that a and b are [#permalink]

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22 Aug 2013, 04:58

ashivapu wrote:

Can we also say that: a^2 - b^2 = 13^2, This is in the form of a pythagorean triplet. Since a and b are given as integers, there can be only one integral value for a and b which will satisfy such a condition. Thus this is a sufficient condition.

Hi, Since it is already given both a and b are positive integers and a^2 - b^2 = 169 (POSITIVE), w can say a > b. If we as assume 169 as 13^2 we are contracdicting our assumption becasue a and b cant be equal. I hope it helps.

What will be the value of a/b ? Given that a and b are positive integers (1) a^2 – b^2 = 169 (2) a – b = 1

I am unable to prove statement 1 is sufficient here. The only thing i could think of first is the 5-12-13 triplet but that would hold good only for a^2 + b^2 = 169. please help.

\(169 = 13^2\). \(a^2-b^2=(a+b)(a-b)\). Since \(a\) and \(b\) are positive integers, \(a+b>a-b\) and \(a+b>0\), so the only possible factorization is \(a+b=169\) and \(a-b=1\), from which \(a\) and \(b\) can be deduced, as well as the ratio \(a/b\). Therefore, (1) is sufficient. (2) obviously not sufficient.

Answer A.

Can you please expand on why we know that the only possible factorization of a+b = 169? Is this because 169 is a square of a prime number and only has factors of 1 - 169, and 13-13? Hence, since a+b>a-b, the factors cannot be 13 * 13 and must be 169 * 1?

Therefore, if 169 was another number with varying positive factors, (1) would not be sufficient?

What will be the value of a/b ? Given that a and b are positive integers (1) a^2 – b^2 = 169 (2) a – b = 1

I am unable to prove statement 1 is sufficient here. The only thing i could think of first is the 5-12-13 triplet but that would hold good only for a^2 + b^2 = 169. please help.

\(169 = 13^2\). \(a^2-b^2=(a+b)(a-b)\). Since \(a\) and \(b\) are positive integers, \(a+b>a-b\) and \(a+b>0\), so the only possible factorization is \(a+b=169\) and \(a-b=1\), from which \(a\) and \(b\) can be deduced, as well as the ratio \(a/b\). Therefore, (1) is sufficient. (2) obviously not sufficient.

Answer A.

Can you please expand on why we know that the only possible factorization of a+b = 169? Is this because 169 is a square of a prime number and only has factors of 1 - 169, and 13-13? Hence, since a+b>a-b, the factors cannot be 13 * 13 and must be 169 * 1?

Therefore, if 169 was another number with varying positive factors, (1) would not be sufficient?

Thanks

Yes, 169 can be broken into the product of two multiples in two ways: 13*13 and 1*169. Since we know that one multiple is greater than the other, then only 1*169 is OK. _________________

Re: What will be the value of a/b ? Given that a and b are [#permalink]

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28 Aug 2013, 12:42

1

This post received KUDOS

nikhilsehgal wrote:

Hi Bunnel,

As per the question stem - we just know that a and b are >0 and we have to find the value of a/b.

how can we say that a>b - though that we came to know from the 2nd statement only i.e. a = b+1

Now a^2 - b^2 = 169 - (a+b) (a-b) = 13*13 or 169 *1 Now over here how can we say that a>b??

Can you explain!!

Thanks Nikhil

Hi nikhil

a & b are positive integers ==> (a + b) MUST be positive (a - b)(a +b) = 169 ==> (a - b) MUST be positive too ==> a > b

We know that (a-b)(a+b) = 1x169 (It cannot be 13x13 cause there are no two positive integers that have equal sum and deduction) ==> (a - b) = 1 ==> (a + b) = 169 ==> a = 85, b = 84

Hope it helps.

PS:Sorry Bunuel for jumping in. I just want to share what I'm thinking. _________________

Please +1 KUDO if my post helps. Thank you.

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Re: What will be the value of a/b ? Given that a and b are [#permalink]

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30 Sep 2014, 07:56

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