Hi
An MGMAT qn, the explanations for which I have seen are fairly elaborate algebra - wondering if anyone has a better approximate method or approach to solve this on real question.
I also did not grasp completely what is the underlying concept/key idea being tested.
Kindly help
If q, r, and s are consecutive even integers and q < r < s, which of the following CANNOT be the value of s2 - r2 - q2?
-20
0
8
12
16
OA follows....
OA 8
Thanks
kaps
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Consecutive integers- faster method anyone? and key idea
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GMATGuruNY
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I was able to solve very quickly by plugging in values.kaps786 wrote:Hi
An MGMAT qn, the explanations for which I have seen are fairly elaborate algebra - wondering if anyone has a better approximate method or approach to solve this on real question.
I also did not grasp completely what is the underlying concept/key idea being tested.
Kindly help
If q, r, and s are consecutive even integers and q < r < s, which of the following CANNOT be the value of s2 - r2 - q2?
-20
0
8
12
16
OA follows....
OA 8
Thanks
kaps
-4,-2,0:
0^2 - (-2)^2 - (-4)^2 = -20.
Eliminate A.
-2,0,2:
2^2 - 0^2 - (-2)^2 = 0.
Eliminate B.
0,2,4:
4^2 - 2^2 - 0^2 = 12.
Eliminate D.
Plugging in 2,4,6 -- or a set of greater consecutive even integers -- will yield a result greater than 12.
No way to yield a result of 8.
The correct answer is C.
For the skeptical:
2,4,6:
6^2 - 4^2 - 2^2 = 16.
Eliminate E.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
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saketk
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you can do this question like this as well -
let
q= 2a
r= 2a+2
s= 2a+4
where a is an integer number both positive and negative ( -1,0,1,2,3 etc )
now as per the equation given in the question
s2-r2-q2 -- we get,
4a2+16+16a - (4a2+4+8a) - 4a2
or, -4a2+8a+12 --
now, start plugging values,
a=1, gives 16
a=2, gives 12
a=3, gives 0
a=4, gives -20
Therefore, the correct answer is 8.
let
q= 2a
r= 2a+2
s= 2a+4
where a is an integer number both positive and negative ( -1,0,1,2,3 etc )
now as per the equation given in the question
s2-r2-q2 -- we get,
4a2+16+16a - (4a2+4+8a) - 4a2
or, -4a2+8a+12 --
now, start plugging values,
a=1, gives 16
a=2, gives 12
a=3, gives 0
a=4, gives -20
Therefore, the correct answer is 8.
