If x andy are positive integers, what is the value of (x + y)?
(1)(x + y- 1)!< 100
(2)y = x²-x + 1
(1)(x + y- 1)!< 100
(2)y = x²-x + 1
st(1) implies (x+y-1)!=(x+y)!/(x+y) and this is less than 100. Restating (x+y)! < 100(x+y) {since both x and y are +ve values, we can multiply the sides of inequality by (x+y)}knight247 wrote:If x and y are positive integers, what is the value of (x + y)?
(1)(x + y- 1)!< 100
(2)y = x²-x + 1
This is Not Sufficient, as we are not restricted per any set of values, rather we are given range
x+y=5, 5!<100*5
x+y=4, 4!<100*4
and so on
st(2) can be regrouped as x+y=x^2+1. Since x can be any integer and x+y have many values this is Not Sufficient.
Combining statements (1&2): Sufficient, as we can have x=2 with st(2) -> (2-1)(2+1), then x+y=3 and apply this for st(1) x+y=3, 3!<100*3
BUT NOT x=3, st(2) -> (3-1)(3+1), then x+y=8 and applying for st(1) x+y=8, 8!<100*8 NOT TRUE
So x can be only 2 and x+y=3
c
because of previous mistake in st(2) didn't test x=1 as assumed restriction put on x>1 and got Sufficiency with both statements. This is not correct.
