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25 Apr 2019, 02:18
Kudos
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
27% (02:51) correct
73%
(02:32)
wrong
based on 118
sessions
History
Date
Time
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Not Attempted Yet
Consider the set of all fractions x/y, where x and y are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by 1, the value of the fraction is increased by 10%? (Two integers a and b are said to be relatively prime, mutually prime, or coprime if the only positive integer (factor) that divides both of them is 1.)
(A) 0
(B) 1
(C) 2
(D) 3
(E) Infinitely many
(A) 0
(B) 1
(C) 2
(D) 3
(E) Infinitely many
25 Apr 2019, 11:11
Kudos
Bookmarks
Fraction is X/Y
Add 1 to both Num and Denom, \(\frac{(X+1)}{(Y+1)} = \frac{110}{100}*\frac{X}{Y}\) (As the new fraction is 1.1 times the previous fraction)
Solving gives, 10Y = X(Y+11)
As X and Y are co-prime, we can write Y as X+1. As the two consecutive integers are always co-prime
Put Y = X+1 in above equation, we get
\(10X+10 = X^2+12X\)
\(X^2+2X-10 = 0\)
As we don't get ay real roots, such a fraction doesn't exist.
A is the answer.
Add 1 to both Num and Denom, \(\frac{(X+1)}{(Y+1)} = \frac{110}{100}*\frac{X}{Y}\) (As the new fraction is 1.1 times the previous fraction)
Solving gives, 10Y = X(Y+11)
As X and Y are co-prime, we can write Y as X+1. As the two consecutive integers are always co-prime
Put Y = X+1 in above equation, we get
\(10X+10 = X^2+12X\)
\(X^2+2X-10 = 0\)
As we don't get ay real roots, such a fraction doesn't exist.
A is the answer.
(a+1)/(b+1)=(11/10)*(a/b)
10(ab+b) = 11(ab+a)
11a=10b-ab
b= 11a/(10-a)
b will be positive integer whenever a/(10-a) is positive integer
a/(10-a) is positive integer at a=5,8 and 9
At a=5 b=11 Both are coprimes
At a=8 b=44 Both are not coprimes
At a=9 b=99 Both are again not coprimes
Only one solution is possible
Bunuel You must change the OA
10(ab+b) = 11(ab+a)
11a=10b-ab
b= 11a/(10-a)
b will be positive integer whenever a/(10-a) is positive integer
a/(10-a) is positive integer at a=5,8 and 9
At a=5 b=11 Both are coprimes
At a=8 b=44 Both are not coprimes
At a=9 b=99 Both are again not coprimes
Only one solution is possible
Bunuel You must change the OA
