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Re: If x and y are positive integers, what is the value of x^(1/2)+y^(1/2) [#permalink]

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21 Oct 2015, 22:19

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((x)^(1/2)+(y)^(1/2))^2 = x + y + 2 *(xy)^(1/2)

1. x+y= 15 Not sufficient

2. (xy)^(1/2) = 6 Not sufficient

Combining 1 and 2, we get x + y + 2 *(xy)^(1/2)= 15 + 2*6=27 => ((x)^(1/2)+(y)^(1/2))^2 = 27 => (x)^(1/2)+(y)^(1/2) =3*((3)^(1/2)

Sufficient Answer C
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If x and y are positive integers, what is the value of x^(1/2)+y^(1/2) [#permalink]

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22 Oct 2015, 23:43

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Bunuel wrote:

If x and y are positive integers, what is the value of \(\sqrt{x} + \sqrt{y}\)?

(1) x + y = 15 (2) \(\sqrt{xy}= 6\)

Kudos for a correct solution.

1) Gives you various single values for x and y. Therefore clearly insufficient. 2) If \(\sqrt{xy}= 6\), then xy = 36 which can be built with 3*12 or 6*6 ... insufficient.

1+2) Here we know, x+y = 15 and xy = 36, hence x, or y are splitted up as 12 and 3. It does actually not matter if x is 3 or 12 or y is 3 or 12. The sum of \(\sqrt{x} + \sqrt{y}\) will be the same.

Answer C.
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Re: If x and y are positive integers, what is the value of x^(1/2)+y^(1/2) [#permalink]

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23 Oct 2015, 04:42

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To find the value of \(\sqrt{x} + \sqrt{y}\) we need to know to have a value for x and a value for y.

Statement 1 : INSUFFICIENT x + y = 15 We have different possible values for x and y: x= 7 and y= 8 x= 9 and y= 6 x= 12 and y=3 All of these would yield different values for \(\sqrt{x} + \sqrt{y}\). Since we can't find a unique value, the statement is not sufficient.

Statement 2 : INSUFFICIENT If \(\sqrt{xy}=6\) then \((\sqrt{xy})^2=6^2\) and \(xy=\)36. Again, there are multiple values of x and y for which \(xy=36\): x=36 and y=1 x=6 and y=6 Since we can't find a unique value, the statement is not sufficient.

(1) + (2) = SUFFICIENT

We know that x+y = 15 and that xy=36, because x and y are positive integers we know that x=12 and y=3 OR x=3 and y=12 either way we will be able to calculate the value of \(\sqrt{x} + \sqrt{y}\) because it will not change the result.

If x and y are positive integers, what is the value of \(\sqrt{x} + \sqrt{y}\)?

(1) x + y = 15 (2) \(\sqrt{xy}= 6\)

Kudos for a correct solution.

Target question:What is the value of √x + √y?

Statement 1: x + y = 15 This statement doesn't FEEL sufficient, so I'll TEST some values. There are several values of x and y that satisfy statement 1. Here are two: Case a: x = 14 and y = 1, in which case √x + √y = √14 + √1 = √14 + 1 Case b: x = 9 and y = 6, in which case √x + √y = √9 + √6 = 3 + √6 Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: √(xy) = 6 In other words xy = 36 This statement doesn't FEEL sufficient either, so I'll TEST some values. There are several values of x and y that satisfy statement 2. Here are two: Case a: x = 1 and y = 36, in which case √x + √y = √1 + √36 = 1 + 6 = 7 Case b: x = 4 and y = 9, in which case √x + √y = √4 + √9 = 2 + 3 = 5 Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined Statement 1 tells us that x + y = 15 Statement 2 tells us that √(xy) = 6 Recognize that (√x + √y)² = x + 2√(xy) + y Rearrange to get: (√x + √y)² = 15 + 2(6) Evaluate: (√x + √y)² = 27 So, √x + √y = √27 Since we can answer the target question with certainty, the combined statements are SUFFICIENT

If x and y are positive integers, what is the value of \(\sqrt{x} + \sqrt{y}\)?

(1) x + y = 15 (2) \(\sqrt{xy}= 6\)

We need to determine the value of √x + √y.

Statement One Alone:

x + y = 15

If x = 1 and y = 14, then √x + √y = 1 + √14. However, if x = 4 and y = 11, then √x + √y = 2 + √11. We see that we don’t have enough information to determine a unique value of √x + √y.

Statement one alone is not sufficient to answer the question.

Statement Two Alone:

√(xy) = 6

If x = 6 and y = 6, then √x + √y = 2√6. However, if x = 4 and y = 9, then √x + √y = 5. We see that we don’t have enough information to determine a unique value of √x + √y.

Statement two alone is not sufficient to answer the question.

Statements One and Two Together:

Notice that (√x + √y)^2 = x + y + 2√(xy). From the two statements, we are given that x + y = 15 and √(xy) = 6, and thus (√x + √y)^2 = 15 + 2(6) = 27. Now, if we take the square root of both sides of the equation (√x + √y)^2 = 27, we have √x + √y = √27 = 3√3.

Answer: C
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If x and y are positive integers, what is the value of \(\sqrt{x} + \sqrt{y}\)?

(1) x + y = 15 (2) \(\sqrt{xy}= 6\)

Kudos for a correct solution.

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer and so we should consider 1) & 2) first.

\((\sqrt{x} + \sqrt{y})^2 = x + 2\sqrt{xy} + y = x + y + 2\sqrt{xy} = 15 + 2 \cdot 6 = 15 + 12 = 27\). Both conditions 1) & 2) are sufficient.

Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT 4(A).

Condition 1) Since \(y = 15 - x\), we have \(\sqrt{x} + \sqrt{y} = \sqrt{x} + \sqrt{15 - x}\). However, the condition 1) is not sufficient since we don't know \(x\).

Condition 1) Since \(xy = 36\) and \(y = \frac{36}{x}\), we have \(\sqrt{x} + \sqrt{y} = \sqrt{x} + \sqrt{\frac{36}{x}}\). However, the condition 1) is not sufficient since we don't know \(x\).

Therefore, C is the answer.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
_________________

Re: If x and y are positive integers, what is the value of x^(1/2)+y^(1/2) [#permalink]

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01 Jan 2018, 01:04

GMATPrepNow wrote:

Bunuel wrote:

If x and y are positive integers, what is the value of \(\sqrt{x} + \sqrt{y}\)?

(1) x + y = 15 (2) \(\sqrt{xy}= 6\)

Kudos for a correct solution.

Target question:What is the value of √x + √y?

Statement 1: x + y = 15 This statement doesn't FEEL sufficient, so I'll TEST some values. There are several values of x and y that satisfy statement 1. Here are two: Case a: x = 14 and y = 1, in which case √x + √y = √14 + √1 = √14 + 1 Case b: x = 9 and y = 6, in which case √x + √y = √9 + √6 = 3 + √6 Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: √(xy) = 6 In other words xy = 36 This statement doesn't FEEL sufficient either, so I'll TEST some values. There are several values of x and y that satisfy statement 2. Here are two: Case a: x = 1 and y = 36, in which case √x + √y = √1 + √36 = 1 + 6 = 7 Case b: x = 4 and y = 9, in which case √x + √y = √4 + √9 = 2 + 3 = 5 Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined Statement 1 tells us that x + y = 15 Statement 2 tells us that √(xy) = 6 Recognize that (√x + √y)² = x + 2√(xy) + y Rearrange to get: (√x + √y)² = 15 + 2(6) Evaluate: (√x + √y)² = 27 So, √x + √y = √27 Since we can answer the target question with certainty, the combined statements are SUFFICIENT

If x and y are positive integers, what is the value of \(\sqrt{x} + \sqrt{y}\)?

(1) x + y = 15 (2) \(\sqrt{xy}= 6\)

Kudos for a correct solution.

Target question:What is the value of √x + √y?

Statement 1: x + y = 15 This statement doesn't FEEL sufficient, so I'll TEST some values. There are several values of x and y that satisfy statement 1. Here are two: Case a: x = 14 and y = 1, in which case √x + √y = √14 + √1 = √14 + 1 Case b: x = 9 and y = 6, in which case √x + √y = √9 + √6 = 3 + √6 Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: √(xy) = 6 In other words xy = 36 This statement doesn't FEEL sufficient either, so I'll TEST some values. There are several values of x and y that satisfy statement 2. Here are two: Case a: x = 1 and y = 36, in which case √x + √y = √1 + √36 = 1 + 6 = 7 Case b: x = 4 and y = 9, in which case √x + √y = √4 + √9 = 2 + 3 = 5 Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined Statement 1 tells us that x + y = 15 Statement 2 tells us that √(xy) = 6 Recognize that (√x + √y)² = x + 2√(xy) + y Rearrange to get: (√x + √y)² = 15 + 2(6) Evaluate: (√x + √y)² = 27 So, √x + √y = √27 Since we can answer the target question with certainty, the combined statements are SUFFICIENT

But, isn't √x + √y = ± √27 which would not result in a single solution for the question?

The square root of a number (generally even root of a number) is non-negative: 0 or positive. \(\sqrt[even]{nonnegative \ number}\geq 0\). Thus, \(\sqrt{x} + \sqrt{y} = {nonnegative \ number} + {nonnegative \ number}= {nonnegative \ number}\), so it cannot equal to a negative number.
_________________

Re: If x and y are positive integers, what is the value of x^(1/2)+y^(1/2) [#permalink]

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01 Jan 2018, 09:32

Bunuel wrote:

sushforgmat wrote:

GMATPrepNow wrote:

If x and y are positive integers, what is the value of \(\sqrt{x} + \sqrt{y}\)?

(1) x + y = 15 (2) \(\sqrt{xy}= 6\)

Kudos for a correct solution.

Target question:What is the value of √x + √y?

Statement 1: x + y = 15 This statement doesn't FEEL sufficient, so I'll TEST some values. There are several values of x and y that satisfy statement 1. Here are two: Case a: x = 14 and y = 1, in which case √x + √y = √14 + √1 = √14 + 1 Case b: x = 9 and y = 6, in which case √x + √y = √9 + √6 = 3 + √6 Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: √(xy) = 6 In other words xy = 36 This statement doesn't FEEL sufficient either, so I'll TEST some values. There are several values of x and y that satisfy statement 2. Here are two: Case a: x = 1 and y = 36, in which case √x + √y = √1 + √36 = 1 + 6 = 7 Case b: x = 4 and y = 9, in which case √x + √y = √4 + √9 = 2 + 3 = 5 Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined Statement 1 tells us that x + y = 15 Statement 2 tells us that √(xy) = 6 Recognize that (√x + √y)² = x + 2√(xy) + y Rearrange to get: (√x + √y)² = 15 + 2(6) Evaluate: (√x + √y)² = 27 So, √x + √y = √27 Since we can answer the target question with certainty, the combined statements are SUFFICIENT

But, isn't √x + √y = ± √27 which would not result in a single solution for the question?

The square root of a number (generally even root of a number) is non-negative: 0 or positive. \(\sqrt[even]{nonnegative \ number}\geq 0\). Thus, \(\sqrt{x} + \sqrt{y} = {nonnegative \ number} + {nonnegative \ number}= {nonnegative \ number}\), so it cannot equal to a negative number.

With multiple books and reading a lot of content, I think I missed the basic point that you mentioned. Thanks, Bunuel.

I thought (1) alone was enough since if x+y=15 I could do the squared root of each term √x + √y = ± √15 Am I breaking some math rules?

Thank you in advance

If you take the square root from x + y = 15, you'll get \(\sqrt{x + y} = \sqrt{15}\), which is NOT the same as \(\sqrt{x}+\sqrt{y} = \sqrt{15}\). You see, generally, \(\sqrt{x + y} \neq \sqrt{x}+\sqrt{y}\). For example, \(\sqrt{2 + 2} \neq \sqrt{2}+\sqrt{2}\).
_________________

Yes it is perfectly fine and simpler. Square both sides and then take the square root of the resulting value and finally discard the negative value as x and y are positive

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gmatclubot

Re: If x and y are positive integers, what is the value of x^(1/2)+y^(1/2)
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23 Jan 2018, 18:45