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

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17 Oct 2012, 03:01
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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
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17 Oct 2012, 03:06
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GMATBaumgartner wrote:
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.

(2) a – b = 1. Clearly insufficient.

Answer: A.
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17 Oct 2012, 03:07
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GMATBaumgartner wrote:
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.
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Re: What will be the value of a/b ? Given that a and b are [#permalink]

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19 Aug 2013, 08:12
Bumping for review and further discussion.
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Re: What will be the value of a/b ? Given that a and b are [#permalink]

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21 Aug 2013, 03:16
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.
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22 Aug 2013, 03:58
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.
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27 Aug 2013, 17:44
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EvaJager wrote:
GMATBaumgartner wrote:
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
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28 Aug 2013, 08:58
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grant1377 wrote:
EvaJager wrote:
GMATBaumgartner wrote:
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.
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Re: What will be the value of a/b ? Given that a and b are [#permalink]

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28 Aug 2013, 11:26
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
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28 Aug 2013, 11:42
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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.
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28 Aug 2013, 11:45
Great Thanks

Missed on that one.
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