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chetan2u
The numerator of a fraction is multiple of two integers. One of the numbers is greater than the other by 2. If the greater number is smaller than the denominator by 4, what will be the minimum value of the fraction ?

(1) -1
(2) \(\frac{-1}{5}\)
(3) 0
(4) \(\frac{1}{5}\)
(5) 1

plug-in approach:
let x=lesser integer
x(x+2)/x+6=-1➡
x^2+3x+6=0
no, x can't be integer
x(x+2)/x+6=-1/5➡
5x^2+11x+6=0➡
(5x+6)(x+1)=0
x=-1
-1(-1+2)/(-1+6)=-1/5
B
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chetan2u
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The numerator of a fraction is multiple of two numbers. One of the numbers is greater than the other by 2. If the greater number is smaller than the
denominator by 4, what will be the minimum value of the fraction ?
(1) -1
(2) \(\frac{-1}{5}\)
(3) 0
(4) \(\frac{1}{5}\)
(5) 1


source - formatting, adding OA and reposting Q from other subforum

let the denominator be x, therefore the numerator will be (x-4)(x-6)
fraction = \(\frac{(x-4)(x-6)}{x}\)

what does (x-4)(x-6) tell you
1) It tells us that x<4 and x>6, both terms will be negative and the value will be positive.
2) At 4 and 6 it will be 0
3) AND at x as 5, the value will be NEGATIVE
so lowest value will be at x = 5
\(\frac{(x-4)(x-6)}{x}=\frac{(5-4)(5-6)}{5}=\frac{-1}{5}\)

B

If I let x = -1 in the equation from the quoted solution, then the constructed fraction works out to -35, which is smaller than any of the answer choices. Am hoping someone can help me understand my mistake.
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chetan2u But multiple of two integers is not necessarily same as product of two integers. For example, 48 may be a multiple of both 6 and 12 but not the product.
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chetan2u But multiple of two integers is not necessarily same as product of two integers. For example, 48 may be a multiple of both 6 and 12 but not the product.


hi..

there are two separate things :-
1) 48 is a multiple of 16. or 48 is a multiple of 3
2) but when we say that numerator is multiple of two integers, it would generally mean product itself. yes product could also have been used instead of multiple
3) otherwise multiple of two integers does not mean anything . ONLY prime numbers are multiple of just two integers
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So the fraction must be a proper fraction, where the numerator is less than the denominator.

Denominator= X, so

(X-4)(X-6)/X

and (X-4)(X-6)<X

So X^2-10X+24<X or

X^2-11X+24<0. Factoring:

(X-3)(X-8)<0.

So 3<X<8 ensures a negative numerator

So 5 and 7 can be tested only since 4 or 6 yield a 0 numerator

5:(5-4)(5-6)/5 = -1/5

7:(7-4)(7-6)/7 = 3/7

So -1/5 is minimum

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