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What is the largest possible value of y if (x+3)^2 + (y+6)^2 = 100?

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Joined: 02 Sep 2009
Posts: 49303
What is the largest possible value of y if (x+3)^2 + (y+6)^2 = 100?  [#permalink]

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New post 22 Aug 2018, 00:49
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

76% (01:18) correct 24% (02:25) wrong based on 33 sessions

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Joined: 18 Jul 2018
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Concentration: Finance, Marketing
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Re: What is the largest possible value of y if (x+3)^2 + (y+6)^2 = 100?  [#permalink]

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New post 22 Aug 2018, 01:00
Let's consider X = -3. Then the max value of Y can be 4.

B is the answer.

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What is the largest possible value of y if (x+3)^2 + (y+6)^2 = 100?  [#permalink]

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New post 22 Aug 2018, 01:02

Solution



Given:
    • Two variables x and y
    • \((x+3)^2 + (y+6)^2 = 100\)

To find:
    • Largest possible value of y

Approach and Working:
If y is largest, then \((y+6)^2\) must be largest, and simultaneously \((x+3)^2\) must be smallest.
    • Now, the minimum value of any square number is always 0.
      o Hence, if y is largest, we can say \((x+3)^2\) is 0.
      o Therefore, \((y+6)^2 = 100 = 10^2\)
      Or, y + 6 = 10 (ignoring the negative root, as we are maximising y)
      Or, y = 4

Hence, the correct answer is option B.

Answer: B

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Re: What is the largest possible value of y if (x+3)^2 + (y+6)^2 = 100?  [#permalink]

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New post 22 Aug 2018, 01:20
Bunuel wrote:
What is the largest possible value of y if (x+3)^2 + (y+6)^2 = 100?

A. -2
B. 4
C. 5
D. 8
E. 12



Given

(x+3)^2 + (y+6)^2 = 100

we are looking for the largest value of y. Therefore we need to minimize \((x + 3)^2\). x can be anything as there are no conditions related to x are given. any value other than - 3 is not accepted in this case.

\((- 3 + 3 )^2 + (y + 6)^2 = 100\)

\((0)^2 + ( 4 + 6 )^2 = 100\)
So, y = 4.

anyone can plug other values but these values won't work as we have even exponents .

The best answer is B.
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Re: What is the largest possible value of y if (x+3)^2 + (y+6)^2 = 100? &nbs [#permalink] 22 Aug 2018, 01:20
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