In the question above if we take x as 2 and y as 1/2, then the numerator would 2^5 and the denominator would have 2^3, simplifying out to be 2^2 or 4. However Greg took them both as 1 and the answer came out to be 16. What’s wrong with my approach?
Thanks!
I’m assuming this is a typo. Let us just use your case, the original expression would be
\cfrac{2^{ (\cfrac{5}{2})^2}}{2^{(\cfrac{3}{2})^2}} = \cfrac{2^{(\cfrac{25}{4})}}{2^{(\cfrac{9}{4})}}
= 2^{ (\cfrac{25}{4}-\cfrac{9}{4})} = 2^ {(\cfrac{16}{4})} =16
As you can see, the outcome still holds true by using your assumption numbers.
I personally think choosing number here is probably a quicker approach but also demonstrating disrespect for math as it is clearly an algebra-type question.
If x=2, y=1/2
Then (x+y)= 2+(1/2)= 5/2 so, (x+y)^2= 25/4
(x-y)= 3/2 so, (x-y)^2= 9/4
So, 2^(25/4)/2^(9/4)= 2^(16/4)= 2^4= 16
Hope, it helps.
When doing discrete quant instead of Quantitative comparison and In which the single choice is correct, we mostly need to check for only 1 case and in here x=1,y=1 is the easiest and quickest way to do this problem. On the other hand, if this was a QC or multiple choice correct then we need to test multiple cases along with the case above .
Thank you very much!
