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If x > y > 0, then what is the value of \(\frac{x^2+2xy+y^2}{x^2-y^2}\)?
(1) \(x = 4*y\)
(2) \(x=16*y^2\)
(1) \(x = 4*y\)
(2) \(x=16*y^2\)
18 Feb 2020, 23:01
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If x > y > 0, then what is the value of \(\frac{x^2+2xy+y^2}{x^2-y^2}\)?
(1) \(x = 4*y\)
(2) \(x=16*y^2\)
(1) \(x = 4*y\)
(2) \(x=16*y^2\)
Solution
Step 1: Analyse Question Stem
- • \(x > y > 0\),
- o \((x-y)>0\)
- o \(\frac{(x+y)^2}{(x+y)(x-y)} = \frac{(x+y)}{(x-y)}\) …(i)
Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE
Statement 1: \(x = 4*y\)
On substitute this value in equation (i), we get
- • \(\frac{(4y+y)}{(4y-y)}=\frac{5y}{3y}=\frac{5}{3}\)
Statement 2: \(x = 16*y^2\)
On substituting this value in equation (i), we get
- • \(\frac{(16y^2+y)}{(16y^2-y)}=\frac{y(16y+1)}{y(16y-1)} =\frac{(16+y)}{(16-y)}\)
- o In this case we need the value of \(y\), which we don't know.
Note:- Few students will mark C as the answer, because they will think that we need the value of \(x\) and \(y\) to solve the given expression, and by both the statements we can find the value of \(x\) and \(y\), but do we really need that, No we don’t need the value of \(x\) and \(y\), We just need to find the ratio of their sum and difference. Therefore, proper analysis is important in DS questions.
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