made mistakes in st(2) and marked incorrect answer C, while it's E - explanation by expert Mitch in the following post.

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

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)}

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.

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

Look for combinations of values that satisfy both statements.

Let x=1.
Statement 2:
y = 1² - 1 + 1 = 1.
Thus, the combination x=1 and y=1 satisfies statement 2.

Plugging x=1 and y=1 into statement 1:
(1+1-1)! < 100
1! < 100
1 < 100.
Thus, the combination x=1 and y=1 satisfies statement 1.

In this case, x+y = 1+1 = 2.

Let x=2.
Statement 2:
y = 2² - 2 + 1 = 3.
Thus, the combination x=2 and y=3 satisfies statement 2.

Plugging x=2 and y=3 into statement 1:
(2+3-1)! < 100
4! < 100
24 < 100.
Thus, the combination x=2 and y=3 satisfies statement 1.

In this case, x+y = 2+3 = 5.

Since in the first case x+y = 2, and in the second case x+y = 5, the two statements combined are INSUFFICIENT.

The correct answer is E.

If x andy are positive integers, what is the value of (x + y)?
(1)(x + y- 1)!< 100
(2)y = x²-x + 1


Guyzzz honestly i did not solve the way you guyz did it...

Like statement 1 says that (x + y - 1)! < 100

All it means is that the factorial value should be less than 100, i.e the total...

Going by that I got 4 combinations of x and y...

[(3,2) ; (4,1) ; (2,3) ; (1,4)] as only 4! value is less than 100. So the total of x and y can

only be 5. So statement 1 gives me 4 options, hence not sufficient.


Statement 2: I converted the statement into :

(x+y) = (x^2 + 1)
there can be many values of x and y and so statement 2 by itself also not sufficient.


Combined: IF you put the values from statement 1 i.e the 4 options, then you see that (2,3) or

x=2 and y=3 is the only option that suffices the equation...

Hence the answer is C but the x + y = 5 ...


Please help me if there is some mistake...

Thanks in advance for doing so....

you are limiting the factorial to certain number, that is you *assume* that x+y is always 5, while it can be 4 or 3 or 2 as well in st(1). Hence you arrive to not quite correct conclusion about both statements combined being sufficient.

the correct answer and precise explanation can be found in the expert post by Mitch Hunt.

[email protected] wrote:If x andy are positive integers, what is the value of (x + y)?
(1)(x + y- 1)!< 100
(2)y = x²-x + 1


Guyzzz honestly i did not solve the way you guyz did it...

Like statement 1 says that (x + y - 1)! < 100

All it means is that the factorial value should be less than 100, i.e the total...

Going by that I got 4 combinations of x and y...

[(3,2) ; (4,1) ; (2,3) ; (1,4)] as only 4! value is less than 100. So the total of x and y can

only be 5. So statement 1 gives me 4 options, hence not sufficient.


Statement 2: I converted the statement into :

(x+y) = (x^2 + 1)
there can be many values of x and y and so statement 2 by itself also not sufficient.


Combined: IF you put the values from statement 1 i.e the 4 options, then you see that (2,3) or

x=2 and y=3 is the only option that suffices the equation...

Hence the answer is C but the x + y = 5 ...


Please help me if there is some mistake...

Thanks in advance for doing so....

I genuinely missed out the option of x=1 and y=1 and thus made a mistake.

Since there are 2 options that are possible and thus the answer is E.

THanx GMATGuru!!! for a wonderful explanation.....