If x and y are positive integers, what is the value of (x + y)?

You are making a mistake here.

(X+y-1)!<100

But 4! Is less than 100, therefore,

(X+y-1)! can be equal to 24 as well which is 4!

I.e. x+y can be equal to 4+1 =5 as well


But I had made a small arithematic mistake earlier which I have corrected now and the answer of the question must be option E.

Yes realised my mistake.

Kuods. Thank you.

My Approach::
Given both x and y are positive.

(1) (x + y - 1)! < 100
-- We know only that (x+y-1) can be 1,2,3, or 4.
Since only for 1,2,3, and 4 we have the factorial value less than 100.
hence, we can have multiple values for x+y::
2,3,4 OR 5. (1)
So, Not Sufficient.

(2) y = x^2 - x + 1
-- from this we get, x+y=x^2 + 1.
so for x=1=> x+y= 2.
for x=2 => x+y=5
and so on.. x+y can have many values like 2,5,10,17,26 etc.(2)
So,Not Sufficient.

Combining both statements,
x+y can be 2,3,4 OR 5 from (1) and,
x+y can be 2,5,10,17,26 etc from (2).
So, we have two common values 2 and 5 for x+y.
Hence again Not Sufficient.
So, (E) answer.

It should be E

Given -> x and y are positive integers. and we need value of x+y.

statement 1:

(x+y-1)! <100.

this means x+y-1 can be 1,2,3,4
because 5! is 120.
thus x+y can be 2,3,4,5.

Insufficient


statement 2:-

y = x^2 -x +1

if x = 2 , y=3
if x=1 , y=1

not sufficient

combining 1 and 2
both the values of x,y i.e (1,1) and (2,3) satisfies both eqns

hence E

1: \(5! = 120\), so \(x+y-1<5\). Doesn't mean anything since if \(x=1,y=1,x+y=2;2-1<5\) or if \(x=2,y=1,x+y=3;3-1<5\). So not sufficient.
2: This doesn't mean anything alone since if \(x=1,y=1-1+1=1,x+y=2\) or if \(x=2, y=4-2+1=3, x+y=5\). So not sufficient.

Together: not sufficient, since as demonstrated above, if \(x=1,y=1-1+1=1,x+y=2;2-1<5\) and if \(x=2, y=4-2+1=3, x+y=5, x+y-1=4 < 5\). So the answer is E.

MANHATTAN GMAT OFFICIAL SOLUTION:

(1) INSUFFICIENT: We are told that ( x + y – 1)! is less than 100. To see the possible values of (x + y – 1), we list the factorials of the first few integers:
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120 (too large)
So, (x + y – 1) ≤ 4, and (x + y) ≤ 5.

(2) INSUFFICIENT: We can add x to both sides of y = x^2 – x + 1 to create (x + y) on one side of the equation:
y = x^2 – x + 1
x + y = x + x^2 – x + 1
x + y = x^2 + 1

The exact value of (x + y) is unknown, as it depends on the value of x^2, which could be any positive integer.

(1) AND (2) INSUFFICIENT: (x + y) ≤ 5 and x + y = x^2 + 1 combine to tell us that x^2 + 1 ≤ 5. There are two integer solutions: x = 1 or x = 2.

Checking our work with the original equation from Statement (2):
If x = 1, then y = 1^2 – 1 + 1 = 1 and x + y = 1 + 1 = 2
If x = 2, then y = 2^2 – 2 + 1 = 3 and x + y = 2 + 3 = 5

The correct answer is E.

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