x does not have to be an integer

x = sqrt(2) y = 47/5

x=3 y=8


hence E

nolf wrote:Hi

i faced a problem in DS and here it is..

if x(2) + 5y = 49.. is y an integer (read as x square)...

a) 1<x<4

b) x(2) is an integer....


Now 1st option says that x lies between 1 and 4..so 2 and 3 are the options right??

THanks

Harsha

We have to be very careful about the assumptions that we make.

Nowhere does it say that x must be an integer. So, x could be 1.1, 1.2, 1.3, .... - there are lots of numbers between 1 and 4 other than 2 and 3.

So, for both (1) and (2) to be true, we could have:

x = root2
x = root3
x = root4 = 2
x = root5
x = root6
x = root7

all the way up to

x = root15