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# For consecutive even integers a and b, a > b. If a^2 + b^2 = 340 whicH

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Math Expert
Joined: 02 Sep 2009
Posts: 59587
For consecutive even integers a and b, a > b. If a^2 + b^2 = 340 whicH  [#permalink]

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06 Jul 2017, 23:26
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65% (hard)

Question Stats:

63% (02:02) correct 37% (02:20) wrong based on 282 sessions

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For consecutive even integers a and b, a > b. If $$a^2 + b^2 = 340$$, which of the following could be the value of a?

A. -14
B. -12
C. -8
D. 12
E. 16

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Re: For consecutive even integers a and b, a > b. If a^2 + b^2 = 340 whicH  [#permalink]

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10 Jul 2017, 11:58
1
4
Bunuel wrote:
For consecutive even integers a and b, a > b. If $$a^2 + b^2 = 340$$, which of the following could be the value of a?

A. -14
B. -12
C. -8
D. 12
E. 16

as a and b are consecutive even int we can write following equation as

$$a^2 + b^2 = 340$$

$$a^2 + (a-2)^2 = 340$$

$$a^2 - 2a- 168 = 0$$

(a-14)(a+12)=0

a = 14 or -12

So ans is B -12.
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Re: For consecutive even integers a and b, a > b. If a^2 + b^2 = 340 whicH  [#permalink]

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12 Jul 2017, 16:25
1
Bunuel wrote:
For consecutive even integers a and b, a > b. If $$a^2 + b^2 = 340$$, which of the following could be the value of a?

A. -14
B. -12
C. -8
D. 12
E. 16

We can test our answer choices using units digits:

A) -14

So, a = -14 and b = -16.

Since 4^2 = 16 and 6^2 = 36, the units digit of a^2 + b^2 is 2, and thus A cannot be the answer.

B) -12

So, a = -12 and b = -14.

Since 2^2 = 4 and 4^2 = 6, the units digit of a^2 + b^2 is 0, and thus B can be the answer.

C) -8

So, a = -8 and b = -10.

Since 8^2 = 64 and 0^2 = 0, the units digit of a^2 + b^2 is 4, and thus C cannot be the answer.

D) 12

So, a = 12 and b = 10.

Since 2^2 = 4 and 0^2 = 0, the units digit of a^2 + b^2 is 4, and thus D cannot be the answer.

E) 16

So, a = 16 and b = 14.

Since 6^2 = 36 and 4^2 = 16, the units digit of a^2 + b^2 is 2, and thus E cannot be the answer.

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Re: For consecutive even integers a and b, a > b. If a^2 + b^2 = 340 whicH  [#permalink]

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07 Jul 2017, 00:08
For consecutive even integers a and b, a > b. If a2+b2=340a2+b2=340, which of the following could be the value of a?

A. -14
B. -12
C. -8
D. 12
E. 16

Note that all the answers are values and none of them is like it cannot be determined. So we can put the values and test the values for answer
(P.S: I found it easier to solve this by testing the values rather than by solving algebraically)

A. a= -14, b= -16 since a > b
so, a^2+b^2 = 196 + 256 = 452
B. a= -12, b= -14 since a > b
so, a^2+b^2 = 144 + 196 = 340 .....Hence B is the ANSWER
C. a= -8, b= -10 since a > b
so, a^2+b^2 = 64 + 100 = 164
D. a= 12, b= 10 since a > b
so, a^2+b^2 = 144 + 100 = 244
E. a= 16, b= 14 since a > b
so, a^2+b^2 = 256 + 196 = 452

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Re: For consecutive even integers a and b, a > b. If a^2 + b^2 = 340 whicH  [#permalink]

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07 Jul 2017, 00:10
Answer B, a=-12 (b being -14)
A quick hint on this one is that 12^2 is quite common number which gives 144 and we would need 196 to reach the 340 using the immediate lower number (if we would have taken the positive that would have lead us to use 10 for b to end up with a total of 244) while for negatives -12>-14 and gives the right answer.

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Re: For consecutive even integers a and b, a > b. If a^2 + b^2 = 340 whicH  [#permalink]

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07 Jul 2017, 05:35
1
a and b are consecutive even integers while a>b. So we can write it as a>a-2 and solve the equation: a^2 + (a-2)^2=340 ----> a^2 + a^2 - 4a + 4=340 ----> 2a^2 -4a -336=0 ----> a^2 -2a -168 =0. Now we can use the quadratic equation and get that a equal to either 14 or -12. Given the answers -12 is the right option. Answer: B
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Re: For consecutive even integers a and b, a > b. If a^2 + b^2 = 340 whicH  [#permalink]

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17 Jul 2018, 00:55
Saw this question today in my veritas prep question bank. It was marked as sub 400. Here its labelled as 65% hard. I am perplexed.
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Posts: 59587
Re: For consecutive even integers a and b, a > b. If a^2 + b^2 = 340 whicH  [#permalink]

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17 Jul 2018, 00:59
Saw this question today in my veritas prep question bank. It was marked as sub 400. Here its labelled as 65% hard. I am perplexed.

The difficulty level of a question is calculated automatically based on the timer stats from the users which attempted the question. So our stats say that it's 700 level. Veritas says that it's easy.
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Re: For consecutive even integers a and b, a > b. If a^2 + b^2 = 340 whicH  [#permalink]

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17 Jul 2018, 22:27
Bunuel wrote:
For consecutive even integers a and b, a > b. If $$a^2 + b^2 = 340$$, which of the following could be the value of a?

A. -14
B. -12
C. -8
D. 12
E. 16

Given a & b are consecutive even integers & a > b

Hence b = a - 2

We get, $$a^2 + b^2 = 340$$

$$a^2 + (a-2)^2 = 340$$

simplifying, we get a*(a-2) = 168

Looking at answer choices only -12 will give last digit as 8

a = -12

Thanks,
GyM
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Re: For consecutive even integers a and b, a > b. If a^2 + b^2 = 340 whicH   [#permalink] 17 Jul 2018, 22:27
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# For consecutive even integers a and b, a > b. If a^2 + b^2 = 340 whicH

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