swerve wrote:A lunar mission is made up of x astronauts and is formed from a total of 12 astronauts. A day before the launch the commander of the program decides to add p astronauts to the mission. If the total number of possible lunar missions remain unchanged after the commander's decision, then which of the following cannot be the value of p?
A. x
B. x + 3
C. 3
D. 6
E. 8
The OA is C.
Source: Magoosh
Before the launch day:
The total number of possible lunar missions = 12Cx;
On the launch day:
The total number of possible lunar missions = (12 + p)Cx;
We are given that 12Cx = (12 + p)Cx
Note nCr = nC(n - r)
Thus, (12 + p)Cx = (12 + p)C(12 + p - x)
=> x = 12 + p - x
p = 2x - 12
Let's look at the options now.
A. x: Pluggin-in the value p = x in p = 2x - 12, we get x = 12, a valid value of x; thus, p can be x or 12
B. x + 3: Pluggin-in the value p = x + 3 in p = 2x - 12, we get x = 15, a valid value of x; thus, p can be x + 3 or 15
C. 3: Pluggin-in the value p = 3 in p = 2x - 12, we get x = 7.5, not a valid value of x since x is a positive integer; thus, p cannot be 3
Correct answer.
Though we got the answer, let's discuss option D and E.
D. 6: Pluggin-in the value p = 6 in p = 2x - 12, we get x = 9, a valid value of x; thus, p can be 6
E. 8: Pluggin-in the value p = 8 in p = 2x - 12, we get x = 10, a valid value of x; thus, p can be 8
The correct answer:
C
Hope this helps!
-Jay
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