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Integers’ Squadron

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Integers’ Squadron

by sanju09 » Tue Apr 13, 2010 4:01 am
A rectangle has its perimeter and diagonal integers. Is the breadth of rectangle an integer?

(1) Area of the rectangle is an integer.

(2) Perimeter of the rectangle is 2 more than its area.
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by ymach3 » Tue Apr 13, 2010 7:53 am
Is it E?

S (i) says that the Length and breadth are reciprocal to each other... L=x and B=1/x or L=1/x and B=x. -Insufficient.

S (ii) says

Perimeter and Diognals are integers.

P=2(l+b)---> gives us a clue that if sum (l+b) were a fraction, then its denominator would definitely be 2.

Diognal = Sqrt(l^2 + b^2) - i dint find any useful info from this..

Therefore S(ii) is Insufficient.

Hence E.


Let me know if i missed to read or misread the info from the given clues...
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by eaakbari » Tue Apr 13, 2010 8:08 am
IMO D

Stem
We can conclude that for diagonal and perimeter to be integers, then both the length and breadth must be either integers or both decimals

Statement 1
Given area is integer requires length and breadth to be x and 1/x if either is non integer but we know from stem that both must be decimal or both integer. Hence both Integer
Hence Suff

Statement 2
Does imply that area is integer. Same logic as above
Hence Suff

Answer D
Last edited by eaakbari on Tue Apr 13, 2010 8:27 am, edited 1 time in total.
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by liferocks » Tue Apr 13, 2010 8:17 am
eaakbari wrote:IMO A

Stem
We can conclude that for diagonal and perimeter to be integers, then both the length and breadth must be either integers or both decimals

Statement 1
Given area is integer requires length and breadth to be x and 1/x if either is non integer but we know from stem that both must be decimal or both integer. Hence both Integer
Hence Suff

Statement 2
Does not say whether Area is integer or not, Hence Insuff
statement 2 also says that area is an integer
let length and breath be a and b..from 2 we get
2(a+b)-2=ab
LHS is integer hence RHS will be integer..but how did u conclude that either both will be integer or both will be decimal please explain
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by eaakbari » Tue Apr 13, 2010 8:26 am
liferocks wrote:
eaakbari wrote:IMO A

Stem
We can conclude that for diagonal and perimeter to be integers, then both the length and breadth must be either integers or both decimals

Statement 1
Given area is integer requires length and breadth to be x and 1/x if either is non integer but we know from stem that both must be decimal or both integer. Hence both Integer
Hence Suff

Statement 2
Does not say whether Area is integer or not, Hence Insuff
statement 2 also says that area is an integer
let length and breath be a and b..from 2 we get
2(a+b)-2=ab
LHS is integer hence RHS will be integer..but how did u conclude that either both will be integer or both will be decimal please explain

I am sorry , I read the second statement as perimeter is twice are. D should be the answer. Will edit it.
Well I kind of picked numbers and came to the logic. Take the diameter
d^2 = l^2 + b^2

For d to be an integer, d^2 will be an integer

The sum of 1 decimal and one integer can never give you a integer answer. Either both be decimal or both be integer.
And if l^2 and b^2 are integer or decimal l and b will be the same

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by kstv » Tue Apr 13, 2010 9:14 am
sanju09 wrote:A rectangle has its perimeter and diagonal integers. Is the breadth of rectangle an integer?
(1) Area of the rectangle is an integer.
(2) Perimeter of the rectangle is 2 more than its area.
if the sides are x and y.
Perimeter (P) = 2(x+y) = an integer k so (x+y) = k/2
Diagonal (D) = √x²+y²= a integer c , x²+y² = c² definitely an integer

(1) Area (A) = xy is an integer
(x+y)² = x²+y²+2xy from (P) and (A) x²+y² and xy are integers
so (k/2)² is an integer or k is a multiple of 2
so (x+y) is an integer
but cannot say whether x and y are integers
Insuff

(2) 2(x+y) - 2 = xy or
x+y-1 = xy/2
Insuff

Combining (1) and (2)

We know x+y - 1 is an integer
xy/2 is an integer
xy is a multiple of 2 x or y is an even no
x+y is an integer where one of them is a even no
so the other no whether it is x or y cannot be a fraction
taking too long .
Can you pick holes in the logic.
That will be faster.
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by liferocks » Tue Apr 13, 2010 5:14 pm
@eaakbari
to l^2+b^ an integer,l and b both do not have to be decimal or integer one can be decimal and other can be integer.Ex sqroot(3)and 1,sqroot(5),2 etc
we can argue that in these cases 2(l+b) will not be an integer but the list is so long i dont really think we can prove it beyond question by picking numbers. Let me know your thoughts.

@kstv
I like this approach.I was initially doing same but got confused at the end :( .Thanks for an elaborate explanation.
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by eaakbari » Tue Apr 13, 2010 10:24 pm
liferocks wrote:@eaakbari
to l^2+b^ an integer,l and b both do not have to be decimal or integer one can be decimal and other can be integer.Ex sqroot(3)and 1,sqroot(5),2 etc
we can argue that in these cases 2(l+b) will not be an integer but the list is so long i dont really think we can prove it beyond question by picking numbers. Let me know your thoughts.

@kstv
I like this approach.I was initially doing same but got confused at the end :( .Thanks for an elaborate explanation.
I definitely agree with you that irrational numbers do not hold to this argument, but then i thought since we are only tested for real numbers on the GMAT l and b have to be real. Or am i wrong here?
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by eaakbari » Tue Apr 13, 2010 10:24 pm
Do give the OA sanju
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by ymach3 » Tue Apr 13, 2010 11:10 pm
I Agree with eaakbari -


"The sum of 1 decimal and one integer can never give you a integer answer. Either both be decimal or both be integer.
And if l^2 and b^2 are integer or decimal l and b will be the same "


I missed to extract imp info from the Clue given...

-Thanks Ppl.
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by liferocks » Wed Apr 14, 2010 6:20 pm
eaakbari wrote:
liferocks wrote:@eaakbari
to l^2+b^ an integer,l and b both do not have to be decimal or integer one can be decimal and other can be integer.Ex sqroot(3)and 1,sqroot(5),2 etc
we can argue that in these cases 2(l+b) will not be an integer but the list is so long i dont really think we can prove it beyond question by picking numbers. Let me know your thoughts.

@kstv
I like this approach.I was initially doing same but got confused at the end :( .Thanks for an elaborate explanation.
I definitely agree with you that irrational numbers do not hold to this argument, but then i thought since we are only tested for real numbers on the GMAT l and b have to be real. Or am i wrong here?
irrational numbers are real numbers. imaginary numbers are the numbers which have a form ni where n is nonzero and real and i^2=-1
IMO irrational numbers should be considered for GMAT. Some one please correct if this is not the case.
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by sanju09 » Mon Apr 19, 2010 2:13 am
ymach3 wrote:Is it E?

S (i) says that the Length and breadth are reciprocal to each other... L=x and B=1/x or L=1/x and B=x. -Insufficient.

S (ii) says

Perimeter and Diognals are integers.

P=2(l+b)---> gives us a clue that if sum (l+b) were a fraction, then its denominator would definitely be 2.

Diognal = Sqrt(l^2 + b^2) - i dint find any useful info from this..

Therefore S(ii) is Insufficient.

Hence E.


Let me know if i missed to read or misread the info from the given clues...
how (1) says so?
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
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