kamrim322 wrote:
catty2004 wrote:
If a, b, c, and d, are positive numbers, is \(\frac{a}{b} < \frac{c}{d}\)?
(1) \(0 < \frac{(c-a)}{(d-b)}\)
(2) \((\frac{ad}{bc})^2 < \frac{(ad)}{(bc)}\)
Hello everyone,
I see many great explanations regarding this question but I still cannot understand how to know quickly that 1 is not sufficient. I understand that to satisfy the condition 0<c-a/d-b, either both must be negative or both must be positive. But from there I'm lost... I know that c-a can be positive or negative, d-b can also be positive or negative. But so what? How does this help to determine if this statement is sufficient? Can someone please explain? Thank you!
kamrim322 Genoa2000Let's solve once again
(1) \(0 < \frac{(c-a)}{(d-b)}\)
Either both numerator and denominator are positive or both are negative.
Case I: Let's take them positive. So, c > a and d > b
But if we divide
\(\frac{c}{d} > \frac{a}{b}\) (\(\frac{3}{4} > \frac{1}{2}\)) \(\implies\) bc > ad; YES
Case II: Both are negative. So, c < a and d < b
\(\frac{c}{d} < \frac{a}{b}\) (\(\frac{1}{3} < \frac{2}{4}\)) \(\implies\) bc < ad; NO
(Note that taking either both numerator and denominator are positive or both are negative, we are making an assumption which leads to other possibilities thus no unique solution)
Also, beware of dividing or multiplying by variable with negative sign. I have taken only positive values and still reached insufficiency of statement 1.INSUFFICIENT.
(2) \((\frac{ad}{bc})^2 < \frac{(ad)}{(bc)}\)
Here you can solve it for a unique solution:
Dividing the statement by \(\frac{(ad)}{(bc)}\)
\((\frac{ad}{bc}) < 1\)
ad < bc; YES
or
\((\frac{ad}{bc})^2 - \frac{(ad)}{(bc)} < 0\)
\(\frac{ad}{bc}(\frac{ad}{bc} - \frac{(ad)}{(bc)}) < 0\)
So, either \((\frac{ad}{bc}) < 0\) OR \((\frac{ad}{bc}) < 1\)
The former is not true as a, b, c and d all are positive.
Thus ad < bc.
SUFFICIENT.
Answer B.
HTH
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