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

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

kstv wrote:If x and y are +ve integers, are x & y squares of consecutive numbers?

A . √x + y = 7
B . √y + x = 11

Statement 1:

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.

harshavardhanc wrote:

kstv wrote:If x and y are +ve integers, are x & y squares of consecutive numbers?

A . √x + y = 7
B . √y + x = 11

Statement 1:

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.

Oh yes I totally forgot about the number 1 for statement one. Thanks , I shall edit my post

Thanks for explanation guys!!! you guys will surely nail 50 in Quant section

Thank you akahuja, lets keep our fingers crossed and hope we all get 50 in Quant

E

kstv wrote:If x and y are +ve integers, are x & y squares of consecutive numbers?

A . √x + y = 7
B . √y + x = 11

A new approach to the problem.
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 ?

pops wrote:

kstv wrote:If x and y are +ve integers, are x & y squares of consecutive numbers?

A . √x + y = 7
B . √y + x = 11

A new approach to the problem.
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 ?

In my view , its not wrong and its a different approach but definitely more tedious. You might cross 2 mins while solving