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For consecutive odd integers a and b, a>b>0. If a^2+b^2=202, what is

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For consecutive odd integers a and b, a>b>0. If a^2+b^2=202, what is [#permalink]

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New post 21 Feb 2017, 05:25
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

69% (01:22) correct 31% (01:12) wrong based on 63 sessions

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Re: For consecutive odd integers a and b, a>b>0. If a^2+b^2=202, what is [#permalink]

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New post 21 Feb 2017, 05:34
Bunuel wrote:
For consecutive odd integers a and b, a>b>0. If a^2+b^2=202, what is the value of a?

A. 7
B. 9
C. 11
D. 13
E. 15




a^2+b^2=202

Checking Options

A. 7 i.e. 7^2 + 5^2 = 49+25 which is not equal to 202 hence incorrect
B. 9 i.e. 9^2 + 7^2 = 81+49 which is not equal to 202 hence incorrect
C. 11 i.e. 11^2 + 9^2 = 121+81 which is equal to 202 hence CORRECT
D. 13
E. 15

Answer: Option C
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Re: For consecutive odd integers a and b, a>b>0. If a^2+b^2=202, what is [#permalink]

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New post 22 Feb 2017, 05:00
1
Option C

: For consecutive positive odd integers a & b (a > b > 0), \(a^2 + b^2 = 202\). Find a.

\(: b = (a - 2)\)
\(: a^2 + (a-2)^2 = 202\)
\(: a^2 - 2a - 99 = 0\)
\(: (a - 11)(a + 9) = 0\)

: a= 11 or -9
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Re: For consecutive odd integers a and b, a>b>0. If a^2+b^2=202, what is [#permalink]

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New post 23 Feb 2017, 10:50
Bunuel wrote:
For consecutive odd integers a and b, a>b>0. If a^2+b^2=202, what is the value of a?

A. 7
B. 9
C. 11
D. 13
E. 15


Since a is greater than b, and a and b are consecutive odd integers, we can say: a = b + 2.

Squaring both sides of our equation, we have:

a^2 = (b + 2)^2

a^2 = b^2 + 4b + 4

We are also given that a^2 + b^2 = 202, or a^2 = 202 - b^2; thus:

202 - b^2 = b^2 + 4b + 4

202 = 2b^2 + 4b + 4

2b^2 + 4b + 4 = 202

2b^2 + 4b = 198

Dividing the entire equation by 2 and moving the constant to the left side, we have:

b^2 + 2b - 99 = 0

(b + 11)(b - 9) = 0

b = -11 or b = 9

Since a and b must be positive, a = 11.

Answer: C
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Re: For consecutive odd integers a and b, a>b>0. If a^2+b^2=202, what is   [#permalink] 23 Feb 2017, 10:50
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