Square roots

[queenisabella]

please help me figure this out - I spent too much time on this one.

If x and y are positive, which of the following must be greater than 1/sqroot of (x+y)?

A) None
B) I only
C) II only
D) I and III
E) II and III

[neelgandham]

It is always easy to compare algebraic expressions with similar denominators. So, let us try and do that here.

1/√(x+y) = √(x+y)/(x+y).

So the question can be rephrased to
If x and y are positive, which of the following must be greater than (√(x+y))/(x+y)?

I)(√(x+y))/(2x)
II)(√x+√y)/(x+y)
III)(√x-√y)/(x+y)

A) None
B) I only
C) II only
D) I and III
E) II and III

Let us now compare the expression (√(x+y))/(x+y) to the expression (√(x+y))/(2x)
If x = 1 and y = 100, then (√(x+y))/(x+y) ~ 10/101, (√(x+y))/(2x) = 10/2, So (√(x+y))/(x+y) < (√(x+y))/(2x)
If x = 100 and y = 1, then (√(x+y))/(x+y) ~ 10/101, (√(x+y))/(2x) = 10/200, So (√(x+y))/(x+y) > (√(x+y))/(2x). Two different answers, Phew!, not the answer!

Let us now compare the expression (√(x+y))/(x+y) to the expression (√x+√y)/(x+y)
Since (√x+√y) is always > √(x+y)(x,y >0), (√x+√y)/(x+y) > (√(x+y))/(x+y). Perfect!

Let us now compare the expression (√(x+y))/(x+y) to the expression (√x-√y)/(x+y)
Since (√x-√y) is always < √(x+y)(x,y >0), (√x+√y)/(x+y) < (√(x+y))/(x+y). Oops, not the answer!

So answer is C - II only

[ArunangsuSahu]

There is one easier way to do it.

|a|+|b|>=|a+b| and |a|-|b|<=|a-b|

substitute a=sqrt(x) and b=sqrt(y)

(C) is the answer

[Anurag@Gurome]

please help me figure this out - I spent too much time on this one.

If x and y are positive, which of the following must be greater than 1/sqroot of (x+y)?

A) None
B) I only
C) II only
D) I and III
E) II and III

(I) 1/√(x + y) and √(x + y)/2x
Let us make the denominators of the two fractions equal.
Multiplying the numerator and denominator of 1/√(x + y) by 2x, we get, 2x/2x√(x + y)
Multiplying the numerator and denominator of √(x + y)/2x by √(x + y), we get, (x + y)/2x√(x + y)
Since the denominators of 2 fractions are the same, so we can compare the numerators now. 2x may or may not be > (x + y). So, this is ruled out.

(II) 1/√(x + y) and [√(x) + √(y)]/[x + y]
Multiplying the numerator and denominator of 1/√(x + y) by √(x + y), we get, √(x + y)/(x + y)
There is no need to multiply [√(x) + √(y)]/[x + y] since it's denominator is already (x + y)
Since the denominators of 2 fractions are the same, so we can compare the numerators now. √(x + y) is always < √(x) + √(y). So, 1/√(x + y) < [√(x) + √(y)]/[x + y]

(III) 1/√(x + y) and [√(x) - √(y)]/[x + y]
Multiplying the numerator and denominator of 1/√(x + y) by √(x + y), we get, √(x + y)/(x + y)
There is no need to multiply [√(x) - √(y)]/[x + y] since it's denominator is already (x + y)
Since the denominators of 2 fractions are the same, so we can compare the numerators now. √(x + y) may or may not be > √(x) - √(y). So, this is ruled out.

The correct answer is C.

[kul512]

In such type of questions, it is easy to use division-

We want to get the expression which is greater than 1/√(x+y),
so if we divide each expression by 1/√(x+y) i.e. multiply by √(x+y), and result is greater than 1, then that expression will be greater than 1/√(x+y).

1. ((√(x+y))/2x) *(√(x+y))= (x+y/2x)---------can't say if it is greater than 1.

2. (√x+√y)/(x+y) * (√(x+y))= (√x+√y)/√(x+y)=(x+y+2√(xy))/(x+y)=1+2 (√(xy)/x+y)-------which is surely greater than 1. So this is the answer.

3. (√x-√y)/(x+y) * (√(x+y))= (√x-√y)/√(x+y) = (x+y-2√(xy))/(x+y)=1-2 (√(xy)/x+y)---- which is surely less than 1.

So final answer is C.

[icemanKK]

I believe we can also do this problem by substituting any positive value for X and Y , cause of the following 2 reasons :

1. Question says that X and Y are positive (thus any positive value X=y,X>y,X<Y - can be assumed)

2. The answer choices to be validated against the question must satisfy the given question wherein it is mentioned that the answer choices must be "greater"

Note: It is easier to do such problems when numbers are substituted.