If x and y are positive integers less than 10, which of the following COULD be true?
i) √x + √y = √(x + y)
ii) y√x = x√y
iii) √x - √y = √(x - y)
A) i only
B) i and ii only
C) i and iii only
D) ii and iii only
E) i,ii and iii
Answer: D
Source: GMAT Prep Now
If x and y are positive integers less than 10, which of the following COULD be true?
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When I scan the statements, I see that ii and iii look easier, so I'll start with those.BTGModeratorVI wrote: ↑Sat Aug 08, 2020 7:01 amIf x and y are positive integers less than 10, which of the following COULD be true?
i) √x + √y = √(x + y)
ii) y√x = x√y
iii) √x - √y = √(x - y)
A) i only
B) i and ii only
C) i and iii only
D) ii and iii only
E) i,ii and iii
Answer: D
Source: GMAT Prep Now
ii) y√x = x√y
We can quickly see that, if x = y, then this statement is TRUE.
For example, if x = 1 and y = 1, we get 1√1 = 1√1, which is true.
So, statement ii COULD be true.
Check the answer choices......ELIMINATE A and C
iii) √x - √y = √(x - y)
Once again, this statement is TRUE when x = y.
For example, if x = 1 and y = 1, we get √1 - √1 = √(1 - 1), which is true.
So, statement iii COULD be true.
Check the answer choices......ELIMINATE B
i) √x + √y = √(x + y)
This is a tough one.
I can't think of any values for x and y that make this statement true, so now what?
Well, we might just conclude that statement i cannot be true.
However, we can do a bit of algebra to further convince ourselves.
Given: √x + √y = √(x + y)
Square both sides to get: (√x + √y)² = [√(x + y)]²
Expand to get: x + 2√(xy) + y = x + y
Subtract x and y from both sides to get: 2√(xy) = 0
This means that xy = 0, but xy cannot equal 0 since we're told that x and y are POSITIVE integers.
So, statement i cannot be true.
Answer: D
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Brent
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Solution:BTGModeratorVI wrote: ↑Sat Aug 08, 2020 7:01 amIf x and y are positive integers less than 10, which of the following COULD be true?
i) √x + √y = √(x + y)
ii) y√x = x√y
iii) √x - √y = √(x - y)
A) i only
B) i and ii only
C) i and iii only
D) ii and iii only
E) i,ii and iii
Answer: D
If x = y, we see that Roman numerals II and III are true. However, Roman numeral I can never be true when x and y are positive integers. That is because, when we square both sides of the equation, we have:
LHS = (√x + √y)^2 = x + 2√x√y + y and RHS = (√(x + y))^2 = x + y
We see that, when each side is squared, the LHS is 2√x√y more than the RHS. They still can be equal if either x or y is 0. However, we are given that x and y are positive; therefore, 2√x√y can never be 0. In other words, (√x + √y)^2 > (√(x + y))^2 or √x + √y > √(x + y). So √x + √y = √(x + y) is a false statement when x and y are positive.
Answer: C
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