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Math Expert V
Joined: 02 Sep 2009
Posts: 58410
A lunar mission is made up of x astronauts and is formed from a total  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 27% (02:20) correct 73% (02:16) wrong based on 61 sessions

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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

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Math Expert V
Joined: 02 Aug 2009
Posts: 8007
Re: A lunar mission is made up of x astronauts and is formed from a total  [#permalink]

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Bunuel 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

Hi...

I take it that these p too are selected from existing 12 ..

When you make a team, you are actually dealing with COMBINATIONS..
So you can choose x out of 12 in 12Cx ways..
You can choose x+p out of 12 in 12C(x+p)..
But the # of missions are same so 12Cx=12C(x+p)..

But nCx=nC(n-x)..
So 12Cx=12C(12-x)=12C(x+p)...
So 12-x=x+p.....12=2x+p..
So p=12-2x=2(6-x)..
So our answer has to be an EVEN number..
ONLY C is odd..
Ans C..

Otherwise you can solve it ..
12=2x+P..
x=1, p=10
X=2, p=8..... Choice E
X=3, P=6..... Nothing but choice B. p=x+3....Also choice D
X=4, P=4..... Nothing but choice A. x=p
X=5, p=2

Only C that is 3 is not there

C
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Posts: 182
A lunar mission is made up of x astronauts and is formed from a total  [#permalink]

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Bunuel 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

I tried to solve it but Im not sure if I am right or not

Total = 12

so teams can be made as such
2 2 2 2 2 2 here x= 2
3 3 3 3 here x= 3
4 4 4 here x= 4
6 6 here x= 6

If p members are added then what CANT BE P?

1) x , so we need to put p= x ,

if we take x = 2 then p=2 cant be distributed equally among team of 2 as we need 6 members to add one in every team to keep same number of teams

if we take x = 3 then p=3 cant be distributed equally among team of 3 as we need 4 members to add one in every team to keep same number of teams

if we take x = 4 then p=4 cant be distributed equally among team of 4 as we need 3 members to add one in every team to keep same number of teams

if we take x = 6 then p=6 CAN be distributed equally among team of 6 as we need 2 members to add one in every team to keep same number of teams and we can distribute 6 members in 2 teams

2) x + 3 , so we need to put p = x+3 ,

if we take x = 2 then p=5 cant be distributed equally among team of 2 as we need 6 members to add one in every team to keep same number of teams

if we take x = 3 then p=6 cant be distributed equally among team of 3 as we need 4 members to add one in every team to keep same number of teams

if we take x = 4 then p=7 cant be distributed equally among team of 4 as we need 3 members to add one in every team to keep same number of teams

if we take x = 6 then p=9 cant be distributed equally among team of 6 as we need 2 members to add one in every team to keep same number of teams

3) p = 3

3 members can be divided equally when x = 4

4) p = 6

6 members can be divided equally when x= 2

5) p = 8

8 members can be divided equally when x= 3

SO ONLY B don't have all possible ways in which p can be equally divided .

Manager  B
Joined: 11 Feb 2017
Posts: 182
Re: A lunar mission is made up of x astronauts and is formed from a total  [#permalink]

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chetan2u wrote:
Bunuel 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

can u plz look at my solution above and tell me what I have done wrong?
Math Expert V
Joined: 02 Aug 2009
Posts: 8007
Re: A lunar mission is made up of x astronauts and is formed from a total  [#permalink]

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rocko911 wrote:
chetan2u wrote:
Bunuel 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

can u plz look at my solution above and tell me what I have done wrong?

Hi..
There is a problem in understanding of the problem.
Although not very clearly mentioned, the wordings do point towards..
There are 12 astronauts. A team of x will be send on a mission. If P more astronauts are added on a mission, now totalling x+p, and the number of missions possible remain the same.
It has to mean this because otherwise if 12 is added in group of x, and then in x+p, the number of group can NEVER be the same.
Example 12 divided with 4in each group makes 3group so if you add 1 more , the number of groups can never be 4.

But in the given Q, you have to find the combinations of x astronauts and x+P astronauts when TOTAL is 12...
So 12Cx=12C(x+P)..
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Re: A lunar mission is made up of x astronauts and is formed from a total  [#permalink]

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Bunuel 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 total number of possible lunar missions before p astronauts are added is 12Cx. The total number of possible lunar missions after p astronauts are added is 12C(x+p). Since total number of possible lunar missions remain unchanged, we have:

12Cx = 12C(x+p)

Recall that we have a formula: nCx = nC(n-x). Since x + (n-x) = n and apply this to our equation, we have:

x + (x+p) = 12

2x + p = 12

Now let’s check the given answer choices (notice that we are looking for a value that can’t be p):

A) p = x

2x + x = 12

3x = 12

x = 4

This is not the choice we are looking for.

B) p = x + 3

2x + x + 3 = 12

3x = 9

x = 9

This is not the choice we are looking for.

C) p = 3

2x + 3 = 12

2x = 9

x = 4.5

Since x has to be an integer, then x can’t be 4.5, which means p can’t be 3.

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