If x andy are positive integers, what is the value of (x + y)?
(1)(x + y- 1)!< 100
(2)y = x²-x + 1
Sum of x and y
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made mistakes in st(2) and marked incorrect answer C, while it's E - explanation by expert Mitch in the following post.
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.
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.
Last edited by pemdas on Sun Feb 12, 2012 6:38 pm, edited 2 times in total.
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Look for combinations of values that satisfy both statements.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
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.
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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....
(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....
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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.
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....
Success doesn't come overnight!
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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.....
Since there are 2 options that are possible and thus the answer is E.
THanx GMATGuru!!! for a wonderful explanation.....
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LEARNING, APPLICATION AND TIMING IS THE FACT OF GMAT AND LIFE AS WELL... KEEP PLAYING!!!
Whenever you feel that my post really helped you to learn something new, please press on the 'THANK' button.
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Whenever you feel that my post really helped you to learn something new, please press on the 'THANK' button.