kstv wrote:If x and y are +ve integers, are x & y squares of consecutive numbers?
A . √x + y = 7
B . √y + x = 11
IMO C
Deceptively simple
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eaakbari
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On rethinking I am a little confused, the question doesnt explicitly state whether x and y are squares. If they are not answer is E and if they are its A.
Let me assume they are
Statement one
Check squares of a number lesser than 7 that is 4 and 1 which makes options for x any y as 36 and 1 or 4 and 9 . hence Insuff
Statement two
There are 2 possibilities as there are 2 squares less than 11, x can be 4 or 9.Hence Insuff
Combining
4 and 9 is common possibility
Hence C
Let me assume they are
Statement one
Check squares of a number lesser than 7 that is 4 and 1 which makes options for x any y as 36 and 1 or 4 and 9 . hence Insuff
Statement two
There are 2 possibilities as there are 2 squares less than 11, x can be 4 or 9.Hence Insuff
Combining
4 and 9 is common possibility
Hence C
Last edited by eaakbari on Mon Apr 12, 2010 6:22 am, edited 1 time in total.
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harshavardhanc
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Statement 1:kstv wrote:If x and y are +ve integers, are x & y squares of consecutive numbers?
A . √x + y = 7
B . √y + x = 11
if X=36 and Y= 1 (X and Y are not squares of consecutive integers)
if X=9 and Y=4 (X and Y ARE squares of consecutive integers)
hence, this is insufficient.
Statement 2:
we have to consider following set of values for (Y,X) :
(100,1)(81,2)(64,3)(49,4)(36,5)(25,6)(16,7)(9,8)(4,9)(1,10)
for the emboldened set the answer is YES.
for all the others, the answer is NO.
Hence, insufficient.
combining these two : only (X=9 and Y=4) is common.
Hence, C is the answer.
Regards,
Harsha
Harsha
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eaakbari
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Oh yes I totally forgot about the number 1 for statement one. Thanks , I shall edit my postharshavardhanc wrote:Statement 1:kstv wrote:If x and y are +ve integers, are x & y squares of consecutive numbers?
A . √x + y = 7
B . √y + x = 11
if X=36 and Y= 1 (X and Y are not squares of consecutive integers)
if X=9 and Y=4 (X and Y ARE squares of consecutive integers)
hence, this is insufficient.
Statement 2:
we have to consider following set of values for (Y,X) :
(100,1)(81,2)(64,3)(49,4)(36,5)(25,6)(16,7)(9,8)(4,9)(1,10)
for the emboldened set the answer is YES.
for all the others, the answer is NO.
Hence, insufficient.
combining these two : only (X=9 and Y=4) is common.
Hence, C is the answer.
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akahuja143
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A new approach to the problem.kstv wrote:If x and y are +ve integers, are x & y squares of consecutive numbers?
A . √x + y = 7
B . √y + x = 11
To suffice the question condition:
(i) x=(k+1)^2 and y=k^2 or
(ii) x=k^2 and y=(k+1)^2
statement 1: plugging in x and y from (i) and (ii)
k+1+k^2 =7 or k+(k+1)^2=7 (plugging (i) and (ii))
k^2+k-6=0 or k^2+3k-6=0
k=3 or -2 or non integer values !
now since we are not able to nail down on one value this is insufficient
statement 2: plugging in x and y from (i) and (ii)
k+(k+1)^2=11 or k+1+k^2=11
k^2+3k-10=0 or k^2+k-10=0
k=5,-2 or non integer values !
now since we are not able to nail down on one value this is also insufficient
combining the above 2 statements we have one common value hence C!
does anyone see any problem with this method ?
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eaakbari
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In my view , its not wrong and its a different approach but definitely more tedious. You might cross 2 mins while solvingpops wrote:A new approach to the problem.kstv wrote:If x and y are +ve integers, are x & y squares of consecutive numbers?
A . √x + y = 7
B . √y + x = 11
To suffice the question condition:
(i) x=(k+1)^2 and y=k^2 or
(ii) x=k^2 and y=(k+1)^2
statement 1: plugging in x and y from (i) and (ii)
k+1+k^2 =7 or k+(k+1)^2=7 (plugging (i) and (ii))
k^2+k-6=0 or k^2+3k-6=0
k=3 or -2 or non integer values !
now since we are not able to nail down on one value this is insufficient
statement 2: plugging in x and y from (i) and (ii)
k+(k+1)^2=11 or k+1+k^2=11
k^2+3k-10=0 or k^2+k-10=0
k=5,-2 or non integer values !
now since we are not able to nail down on one value this is also insufficient
combining the above 2 statements we have one common value hence C!
does anyone see any problem with this method ?
Whether you think you can or can't, you're right.
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