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

Question

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

Bunuel · Accepted Answer

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.

EvaJager · Answer

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.

ashivapu · Answer

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.

zerosleep · Answer

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.

grant1377 · Answer

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

Bunuel · Answer

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.

nikhilsehgal · Answer

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

pqhai · Answer

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.

nikhilsehgal · Answer

Great Thanks

Missed on that one.

GMATGuruNY · Answer

Statement 1:  a^2-b^2=169 
Case 1: (a+b)(a-b) = 169*1, implying that a+b=169 and a-b=1
Adding together a+b=169 and a-b=1, we get:
2a = 170
a=85
Since a=85 and a-b=1, b=84.
Thus:
 \frac{a}{b} = \frac{85}{84}

Case 2: (a+b)(a-b) = 13*13, implying that a+b=13 and a-b=13
Adding together a+b=13 and a-b=13, we get:
2a = 26
a=13
Since a=13 and a-b=13, b=0.
Not viable, since a and b must both be positive.

Since only Case 1 is viable, SUFFICIENT.

Statement 2: a-b=1
Clearly INSUFFICIENT.

A

Bambi2021 · Answer

It is helpful here to know that only one difference of squares will be 169 (if both numbers must be positive).

For any difference of squares, the greater number in (a+b) will be half of "the difference of the squares plus one". For example:

169 = difference between 85^2 and 84^2; 170/2 = 85

7351 = difference between 3676^2 and 3675^2; 7352/2 = 3676

a = (diff + 1)/2

Posted from my mobile device

bumpbot · Answer

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

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