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MichelleSavina
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Hi cg0588,

In DS questions, every little piece of information matters, so you have to pay careful attention and take the necessary notes so that you don't miss anything.

Here, the prompt tells us that A and B are POSITIVE INTEGERS, so negatives, 0s and fractions are not possible.

Fact 1 tells us that A^2 - B^2 = 169

Since we're restricted to POSITIVE INTEGERS and A^2 - B^2 = 169, we know that A MUST be bigger than B, thus A > B.

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Ans A

A2-B2 can be written as (A-B)*(A+B)

Stat 1: A2-B2=169 =13*13 or 1*169

Comparing 13*13 with (A-B)*(A+B) we cannot write it in this form using integers
Comparing 1*169 with (A-B)*(A+B) we can write it as (85-84)*(85+84) and this is the only comb => A=85 and B=84 => A/B = 85/84

Stat 2: A-B =1
A can be 3 and B 2...or A=5,B=4...many combinations....so no unique A/B is possible

So Stat 1 alone is Sufficient
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MichelleSavina
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

Given information= a and b are positive integers
Question asked=a/b?

(1) a^2 – b^2 = 169

169 has 3 factors 1, 13, 169

(a+b) (a-b)= 13 X 13

Since 169 is +ve number and a and b both are positive (a-b) will be +ve, which means a>b

if a>b and both a and b are positive, then (a+b) and (a-b) both will have different values and those values can be 169 and 1 respectively

a+b= 169
a-b= 1

We can find the value of both a and b and a/b

Statement 1 is sufficient

2) a – b = 1
a and b can have multiple values (3-2 or 5-4 or 9-8 and so on)
Not sufficient

A is the answer
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Explanation:
Given that a and b are positive integers. ⇒ a + b and a − b will also be integers

Considering statement 1:
a2 − b2 = 169
⇒ (a − b)(a + b) = 169 = 13 × 13
Now there are 4 possibilities for a + b and a − b: 1, 169 or 13, 13 or −1, −169 or −13, −13.

Case 1:
a + b = 169 and a − b = 1
Solving 2 equations => a = 85 and b = 84

Case 2:
a + b = 13 and a − b = 13
Solving 2 equations ⇒ a = 13 and b = 0
Since a and b are positive integers, so these values can be rejected.

Case 3:
a + b = −1 and a − b = −169
Solving 2 equations ⇒ a = −85 and b = −84
Since a and b are positive integers, so these values can be rejected.

Case 4:
a + b = −13 and a − b = −13
Solving 2 equations ⇒ a = −13 and b = 0
Since a and b are positive integers, so these values can be rejected.

Since we are getting a definite answer from above statement (only Case 1), statement 1 itself is sufficient to provide the answer.

Considering statement 2:
a − b =1
As only known information is that a and b are positive integers , infinite values of a and b are possible for which a − b =1.

Since we are not getting a definite answer from above statement , statement 2 also itself is not sufficient to provide the answer.

Answer: A
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