For consecutive odd integers a and b, ab0. If a^2+b^2=202, what is

Question

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

GMATinsight · Answer

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

RR88 · Answer

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

JeffTargetTestPrep · Answer

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

Archit3110 · Answer

pattern for square of any odd integer will give unit digits ; 1,9,5,9,1 
so to get 2 as unit digit the value of both a&b has to have unit digit as 1 
option C matches

HowdyPartner · Answer

Crucial to carefully read question stem here. Consecutive. Odd. While you may think that you need to find the factors of 202 -- this is not true and is a trap.

Key to focus on the fact that A and B are consecutive, odd integers that and the sum of each squared equals 202. Test cases here -- and you quickly find 9 and 11. (11 being a).

Also -- you should reference the answer choices to see what it could be (allows you to narrow down quite easily).

C is our answer.

bumpbot · Answer

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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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