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MBA.Aspirant
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Source: Beat The GMAT — Problem Solving |
MBA.Aspirant Posted Wed Aug 17, 2011 4:04 am
I get x>y, then how to go about it?
I think the answer should be (E).
I reckon this is how:
Let us assume following values of x and y:
1)x=-1, y=-2 2)x=2, y=1
x^5--- x^4--- x^3--- x^2--- x^1--- y^1--- y^2--- y^3--- y^4 (taken from the question itself)
-1 1 -1 1 -1 -2 4 -8 16 (for 1)x=-1, y=-2)
32 16 8 4 2 1 1 1 1 (for 2)x=2, y=1)
For the above table, one should be able to make out that only condition in option (E) holds true.
Hope this helps.
Thanks.
I get x>y, then how to go about it?
I think the answer should be (E).
I reckon this is how:
Let us assume following values of x and y:
1)x=-1, y=-2 2)x=2, y=1
x^5--- x^4--- x^3--- x^2--- x^1--- y^1--- y^2--- y^3--- y^4 (taken from the question itself)
-1 1 -1 1 -1 -2 4 -8 16 (for 1)x=-1, y=-2)
32 16 8 4 2 1 1 1 1 (for 2)x=2, y=1)
For the above table, one should be able to make out that only condition in option (E) holds true.
Hope this helps.
Thanks.
Hi,
Q1)x=0, x=3 both should satisfy.
Q3)y=x+5
So, x+y = 2x+5
If you are getting 23, you are assuming that x is an integer and hence max(x) = 9 which leads to max(2x+5) = 23.
Now, 5<x<10. So, 10<2x<20
10+5<2x+5<20+5 => 15<x+y<25
So, 24 is the greatest integer value of (x+y).
Q1)x=0, x=3 both should satisfy.
Q3)y=x+5
So, x+y = 2x+5
If you are getting 23, you are assuming that x is an integer and hence max(x) = 9 which leads to max(2x+5) = 23.
Now, 5<x<10. So, 10<2x<20
10+5<2x+5<20+5 => 15<x+y<25
So, 24 is the greatest integer value of (x+y).
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
