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Retired Moderator Joined: 29 Apr 2015
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If x and y are positive and x^2+y^2=100, then for which of the followi  [#permalink]

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Question Stats: 60% (01:44) correct 40% (01:40) wrong based on 269 sessions

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If x and y are positive and $$x^2+y^2$$=100, then for which of the following is the value of x+y greatest?

A. x=10
B. x=9
C. x=8
D. x=7
E. x=6

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Math Expert V
Joined: 02 Sep 2009
Posts: 55188
Re: If x and y are positive and x^2+y^2=100, then for which of the followi  [#permalink]

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6
reto wrote:
If x and y are positive and $$x^2+y^2$$=100, then for which of the following is the value of x+y greatest?

A. x=10
B. x=9
C. x=8
D. x=7
E. x=6

$$x^2 + y^2 = 100$$ --> $$(x+y)^2-2xy=100$$ --> $$(x+y)^2=100+2xy$$ --> $$x+y=\sqrt{100+2xy}$$.

We need to maximize $$x+y$$, thus we need to maximize $$\sqrt{100+2xy}$$. To maximize $$\sqrt{100+2xy}$$ we need to maximize $$xy$$.

Now, for given sum of two numbers, their product is maximized when they are equal, hence from $$x^2 + y^2 = 100$$ we'll have that $$x^2y^2$$ (or which is the same xy) is maximized when $$x^2=y^2$$.

In this case $$x^2 + x^2 = 100$$ --> $$x=\sqrt{50}\approx{7}$$.

Answer: D.
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Math Expert V
Joined: 02 Sep 2009
Posts: 55188
Re: If x and y are positive and x^2+y^2=100, then for which of the followi  [#permalink]

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Bunuel wrote:
reto wrote:
If x and y are positive and $$x^2+y^2$$=100, then for which of the following is the value of x+y greatest?

A. x=10
B. x=9
C. x=8
D. x=7
E. x=6

$$x^2 + y^2 = 100$$ --> $$(x+y)^2-2xy=100$$ --> $$(x+y)^2=100+2xy$$ --> $$x+y=\sqrt{100+2xy}$$.

We need to maximize $$x+y$$, thus we need to maximize $$\sqrt{100+2xy}$$. To maximize $$\sqrt{100+2xy}$$ we need to maximize $$xy$$.

Now, for given sum of two numbers, their product is maximized when they are equal, hence from $$x^2 + y^2 = 100$$ we'll have that $$x^2y^2$$ (or which is the same xy) is maximized when $$x^2=y^2$$.

In this case $$x^2 + x^2 = 100$$ --> $$x=\sqrt{50}\approx{7}$$.

Answer: D.

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given-that-abcd-is-a-rectangle-is-the-area-of-triangle-abe-127051.html

Hope it helps.
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If x and y are positive and x^2+y^2=100, then for which of the followi  [#permalink]

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2
Bunuel wrote:
reto wrote:
If x and y are positive and $$x^2+y^2$$=100, then for which of the following is the value of x+y greatest?

A. x=10
B. x=9
C. x=8
D. x=7
E. x=6

$$x^2 + y^2 = 100$$ --> $$(x+y)^2-2xy=100$$ --> $$(x+y)^2=100+2xy$$ --> $$x+y=\sqrt{100+2xy}$$.

We need to maximize $$x+y$$, thus we need to maximize $$\sqrt{100+2xy}$$. To maximize $$\sqrt{100+2xy}$$ we need to maximize $$xy$$.

Now, for given sum of two numbers, their product is maximized when they are equal, hence from $$x^2 + y^2 = 100$$ we'll have that $$x^2y^2$$ (or which is the same xy) is maximized when $$x^2=y^2$$.

In this case $$x^2 + x^2 = 100$$ --> $$x=\sqrt{50}\approx{7}$$.

Answer: D.

I did this by plugging nos.

$$x^2 + y^2 = 100$$ => $$y^2 = 100 - x^2$$ => $$y = \sqrt{(10 + x)(10 - x)}$$ (didn't consider the negative value after taking sqrt as x & y are +ve)

Hence, $$x + y = x + \sqrt{(10 + x)(10 - x)}$$

Now plugging values,

A. $$x = 10 => x + y = \sqrt{(10 + 10)(10 - 10)} + 10 = 10$$
B. $$x = 9 => x + y = \sqrt{(10 + 9)(10 - 9)} + 9 = \sqrt{19} + 9 = 13.5(roughly)$$
C. $$x = 8 => x + y = \sqrt{(10 + 8)(10 - 8)} + 8 = \sqrt{36} + 8 = 14$$
D. $$x = 7 => x + y = \sqrt{(10 + 7)(10 - 7)} + 7 = \sqrt{51} + 7 = 14.1(roughly)$$(slightly over 14)
E. $$x = 6 => x + y = \sqrt{(10 + 6)(10 - 6)} + 6 = \sqrt{64} + 6 = 14$$

Hence answer is D
Manhattan Prep Instructor G
Joined: 04 Dec 2015
Posts: 729
GMAT 1: 790 Q51 V49 GRE 1: Q170 V170 Re: If x and y are positive and x^2+y^2=100, then for which of the followi  [#permalink]

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I plugged in the answer choices. Whoever wrote this problem deliberately made the answer choices easy to plug in - the solution is pretty elegant!

(A) x = 10 gives y = 0, for a sum of 10.

For (C) and (E), notice that this is basically the Pythagorean theorem written out! (6,8,10) is a Pythagorean triple. So, if x = 6, y = 8, and vice versa. Both (C) and (E) give us a sum of 6 + 8 = 14. You can conclude that neither of these can be the answer, since if one of them was right, the other one would also have to be right.

Also, since 14 is greater than 10, (A) can't be the right answer. Our only two contenders are (B) and (D). At this point, consider guessing.

IF you keep working:

(B) gives a sum of 9 + sqrt(100 - 81) = 9 + sqrt(19).
(D) gives a sum of 7 + sqrt(100 - 49) = 7 + sqrt(51).

Which of those is greater? sqrt(19) is definitely less than 5, so (B) is definitely less than 14. (B) can't be right, and you can stop here.

To prove (D) is correct, notice that sqrt(51) is more than 7, so (D) is definitely more than 14 and is the largest value.
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Re: If x and y are positive and x^2+y^2=100, then for which of the followi  [#permalink]

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Answer is D, since square root of 50 is not there in the options

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Veritas Prep GMAT Instructor D
Joined: 16 Oct 2010
Posts: 9224
Location: Pune, India
Re: If x and y are positive and x^2+y^2=100, then for which of the followi  [#permalink]

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1
reto wrote:
If x and y are positive and $$x^2+y^2$$=100, then for which of the following is the value of x+y greatest?

A. x=10
B. x=9
C. x=8
D. x=7
E. x=6

Maximum/minimum values often occur at extremes/transition points.

For the maximum value of x + y, there are two extremes:

- x and y are as far apart as possible (x is slightly less than 10 and y is slightly more than 0. Here x+y = about 10)
- x and y are as close as possible ($$x = \sqrt{50} = 7.something$$. Here x + y = 7.something + 7.something = 14.something)

Answer (D)
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Manager  B
Joined: 17 Aug 2015
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Re: If x and y are positive and x^2+y^2=100, then for which of the followi  [#permalink]

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Although I first solved it by simply walking over choices (first saw 10 and then went to 8..then went to 6 and understood the answer is 7).

Another interpretation could be that if x and y are sides of a triangle and there is a third side that is root of (x^2+ y^2) then that must be the hypotenuse. Note that the hypotenuse has to be 10. So let us maximize the perimeter of this triangle keeping one side fixed. It is quick to see that this will happen if the sides x and y are closest to each other. In choice A and B, one side is extreme. In C, D and E choices, choice D has most balance . If I have one side as 7 then other side is root(51) which is close to 7. This is our answer
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Posts: 2616
Schools: Boston U '20 (M)
GRE 1: Q169 V154 If x and y are positive and x^2+y^2=100, then for which of the followi  [#permalink]

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2
This is such a great Question.
I actually used Brute force here.
We can use estimation to check the values of x+y in each option.

Option 1-> x=10 => y=0 --> Not allowed as y>0
Option 2-> x=9=> y=4.something => x+y=> 13.something
Option 3-> x=8=> y=6 =>x+y=14
Option 4-> x=7 => y=7.something => x+y=14.something
Option 5=> x=6=>y=8=> x+y=14

Hence clearly Option 4 will give us the greatest value of x+y
Hence D.

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Re: If x and y are positive and x^2+y^2=100, then for which of the followi  [#permalink]

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The statement does not affirm that X and Y are positive integers, only affirms that they are positive numbers.

But if we consider positive integers we would have two correct alternatives C and E.

Therefore the criterion is not correct.

If the values for X, Y must be positive numbers, then the closer these numbers are, the greater their sum. X = 7 and Y = sqr (51), then X + Y> 14

Answer C

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Re: If x and y are positive and x^2+y^2=100, then for which of the followi  [#permalink]

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reto wrote:
If x and y are positive and $$x^2+y^2$$=100, then for which of the following is the value of x+y greatest?

A. x=10
B. x=9
C. x=8
D. x=7
E. x=6

Let’s analyze each answer choice:

A) When x = 10, y = 0 and x + y = 10.

B) When x = 9, y = √19 ≈ 4.4 and x + y ≈ 13.4.

C) When x = 8, y = 6 and x + y = 14.

D) When x = 7, y = √51 ≈ 7.2 and x + y ≈ 14.2.

E) When x = 6, y = 8 and x + y = 14.

We see that the sum of x and y will be the greatest when x = 7.

Answer: D
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Re: If x and y are positive and x^2+y^2=100, then for which of the followi  [#permalink]

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