Bunuel wrote:
For all positive integers x and y in the expression x&y, the symbol & represents one of the following six operations between x and y in the order shown: addition, subtraction, multiplication, division, exponentiation, or rooting (in this case, x&y would represent x^(1/y) ). If the six operations each have an equal chance of being represented, which of the following points in the xy-coordinate plane would have exactly a 50% chance of yielding the same result in the expression x&y (with the other outcomes all different from this result as well as from each other)?
A. (1, 1)
B. (1, 2)
C. (2, 1)
D. (2, 2)
E. (3, 1)
Kudos for a correct solution.
MANHATTAN GMAT OFFICIAL SOLUTION:First, list out the six possibilities for the operation & acting on x and y:
Addition: x + y
Subtraction: x – y
Multiplication: xy
Division: x/y
Exponentiation: x^y
Rooting (as shown in the problem): x^(1/y)
Each of these six operations has an equal chance of being represented by the symbol &.
Now, we are asked “which of the following points in the xy-coordinate plane” does something with this symbol. Before we investigate exactly what that something is, we should recognize that this isn’t really a coordinate-plane problem. A point in the plane just represents a pair of numbers for x and y. So we are being asked to plug in the pairs of numbers given in the answer choices to see which one fits the condition we want.
This condition is that the expression x&y has a 50% chance of yielding the same number. Since there are six operations, this means that we want three (out of six) of the operations to yield the same number. We are also asked to ensure that the other three operations yield different numbers from this number and from each other.
So we now start testing:
A: (1, 1):
1 + 1 = 2
1 – 1 = 0
1*1 = 1
1/1 = 1
1^1 = 1
1^(1/1) = 1
Four out of six are the same, so we can cross out A.
B: (1, 2):
1 + 2 = 3
1 – 2 = –1
1*2 = 2
1/2 = 0.5
1^2 = 1
1^(1/2) = 1
Only two out of six are the same, so we can cross out B.
C: (2, 1):
2 + 1 = 3
2 – 1 = 1
2*1 = 2
2/1 = 2
2^1 = 2
2^(1/1) = 2
Four out of six are the same, so we can cross out C.
D: (2, 2):
2 + 2 = 4
2 – 2 = 0
2*2 = 4
2/2 = 1
2^2 = 4
2^(1/2) = square root of 2
Three out of six are the same, and the other three are all different. This is the condition we’re looking for. We should stop now, but if we really wanted to, we could check the last choice and find that it also failed to meet the condition.
The correct answer is D.