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

Question

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

jokschmer · Accepted Answer

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.

jinxd123 · Answer

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

Thanks in advance for the Kudos I will receive :-D

akakad · Answer

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.

Envoyé de mon SM-A500F en utilisant Tapatalk

oryahalom · Answer

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

JeffTargetTestPrep · Answer

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.

Answer: B

adstudy · Answer

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.

Bunuel · Answer

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.

GyMrAT · Answer

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

Answer B.

Thanks,
GyM

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions