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There is a simpler solution actually.

(a+1)/(b+1)=1.1a/b
Simplifying gives b=(1.1a)/(1-0.1a) .....(1)
We know that a and b are positive.

As a is positive, then 1-0.1a>0.
Thus, 0.1a<1

OR

a<10

Now, we know that a is prime. So, possible values of a are 2,3,5,7.

From Eqn (1),

For a = 2, we get b = 11/4, not an integer or a prime.

For a = 3, we get b = 33/7, not an integer or a prime.

For a = 5, we get b = 11, both an integer and a prime.

For a = 7, we get b = 77/3, neither an integer nor a prime.

Thus, there is only one solution.
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The set \(S\) contains fractions \(\frac{a}{b}\), where \(a\) and \(b\) are positive integers, which don’t share any prime factor.

How many of these fractions have the following property: when both numerator and denominator are increase by 1, the value of fraction is increased by 10%?

S = {a/b = k : where a and b are positive integers, which don’t share any prime factor e.g. co-primes}

\(\frac{a+1}{b+1} = \frac{1.1a}{b} = 1.1k\)

(a+1)b = 1.1a(b+1)
ab + b = 1.1ab + 1.1a
.1ab + 1.1a = b
ab + 11a = 10b
a(b+11) = 10b
\(a = \frac{10b}{b+11} = 10(b+11) - \frac{110}{b+11} = 10 - \frac{110}{b+11} \)

110 = 2*5*11 should be multiple of b+11; Since b>0; b+11>11

b+11 = {22,55,110}

Case 1: b+11 = 22; b=11; a = 10 - 110/22 = 10 -5 = 5; Fraction a/b = 5/11; a+1/b+1 = 6/12 = 1/2 = 1.1*5/11; Feasible
Case 2: b+11 = 55; b=44; a = 10 - 110/55 = 10 - 2 = 8; Fraction a/b = 8/44; Not Feasible since a & b are co-prime
Case 3: b+11 = 110; b=99; a = 10 - 110/99 ; Not feasible since a is a positive integer

IMO B
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