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If p, q, and r are different positive integers such that p + q + r = 6

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If p, q, and r are different positive integers such that p + q + r = 6  [#permalink]

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New post 04 May 2015, 05:55
2
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A
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C
D
E

Difficulty:

  95% (hard)

Question Stats:

41% (02:26) correct 59% (02:47) wrong based on 253 sessions

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Re: If p, q, and r are different positive integers such that p + q + r = 6  [#permalink]

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New post 04 May 2015, 07:14
2
E

p,q and r can be any of the numbers 1,2 or 3

lets take p=1,q=2,r=3

1. x+x^2=2x^3

insuff as x can be 0 or 1

2 x+x^3=2x^2

insuf as x can be 0 or 1

taking togerther.

x can be 0 or 1

SO E
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Re: If p, q, and r are different positive integers such that p + q + r = 6  [#permalink]

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New post 11 May 2015, 06:10
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Bunuel wrote:
If p, q, and r are different positive integers such that p + q + r = 6, what is the value of x ?

(1) The average of x^p and x^q is x^r.

(2) The average of x^p and x^r is not x^q.


MANHATTAN GMAT OFFICIAL SOLUTION:

Before beginning work on the statements, note that p, q, and r are different positive integers that add up to 6. Only one set of three numbers satisfies that condition: 1, 2, and 3. Therefore, in some order, p, q, and r are 1, 2, and 3.

(1) INSUFFICIENT: Three cases are possible:

The average of x and x^2 is x^3;
The average of x and x^3 is x^2;
The average of x^2 and x^3 is x.
The first case gives the equation {x+x^2}/2=x^3 , or x+x^2=2x^3. This is a factorable quadratic equation:

2x^3-x^2-x=0
x(2x^2-x-1)=0
x(2x+1)(x-1)=0

This equation gives three solutions for x: x = 0, x = –1/2, and x = 1. Because there are multiple possible values for x, the statement is not sufficient. (There’s no need to investigate the other possible cases at this point, since you’ve just determined that this statement is insufficient.)

(2) INSUFFICIENT: Just about any value for x will make this statement true (that the average of x^pxp and x^rxr is not x^qxq). For instance, if p = 1, q = 2, and r = 3, then the only requirement is that the average of x and x^3x3 not be equal to x^2x2. Try random values of x: If x = 2, then the average of x and x^3x3 is (2 + 8)/2 = 5, which is not equal to x^2=4x2 = 4. If x = 3, then the average of x and x^3x3 is (3 + 27)/2 = 15, which is not equal to x^2=9x2 = 9. Because there are multiple possible values for x, the statement is not sufficient.

(1) AND (2) INSUFFICIENT: In order to determine the values of x that satisfy both statements, revisit the cases derived in statement 1, this time solving for all of them.

Case #1: If the average of x and x^2 is x^3, then, as shown above, x can be any of 0, –1/2, 1.

Case #2: If the average of x and x^3 is x^2, then {x+x^3}/2=x^2, or x+x^3=2x^2. This is also a factorable equation: x^3-2x^2+x=0 ––> x(x^2-2x+1)=0 ––> x(x-1)^2=0. Therefore, x can be either 0 or 1.
Case #3: If the average of x^2 and x^3 is x, then {x^2+x^3}/2=x, or x^2+x^3=2x. This too is a factorable equation: x^3+x^2-2x=0 ––> x(x^2+x-2)=0 ––> x(x+2)(x-1)=0. Therefore, x can be any of 0, –2, 1.

There are thus four possible values of x according to statement 1: x = –2, –1/2, 0, 1.

For the values x = 0 or 1, the average of two terms is equal to the third term in all 3 cases, so these two values can’t satisfy statement 2 (where the average of the first two terms cannot equal the third term). Discard the values 0 and 1.

On the other hand, if x = –2, p and q are 2 and 3 (in either order), and r = 1, then both statements are satisfied. The average of x^p and x^q is x^r: (4 + -8)/2 = -2, so statement 1 is true. In addition, the average of x^p and x^r is not x^q: (4 + -2)/2 does not equal -8, so statement 2 is true. Therefore, x could equal –2.

What about the final case? If x = –1/2, p and q are 1 and 2 (in either order), and r = 3, then both statements are also satisfied. The average of x^p and x^q is x^r: (–1/2 + 1/4)/2 = –1/8, so statement 1 is true. In addition, the average of x^p and x^r is not x^q: (–1/2 + –1/8)/2 does not equal 1/4, so statement 2 is true. Therefore, x can still be either –2 or –1/2. Both statements together are still not sufficient to answer the question.

The correct answer is (E).
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Re: If p, q, and r are different positive integers such that p + q + r = 6  [#permalink]

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New post 07 Jul 2016, 10:16
Bunuel wrote:
Bunuel wrote:
If p, q, and r are different positive integers such that p + q + r = 6, what is the value of x ?

(1) The average of x^p and x^q is x^r.

(2) The average of x^p and x^r is not x^q.


MANHATTAN GMAT OFFICIAL SOLUTION:

Before beginning work on the statements, note that p, q, and r are different positive integers that add up to 6. Only one set of three numbers satisfies that condition: 1, 2, and 3. Therefore, in some order, p, q, and r are 1, 2, and 3.

(1) INSUFFICIENT: Three cases are possible:

The average of x and x^2 is x^3;
The average of x and x^3 is x^2;
The average of x^2 and x^3 is x.
The first case gives the equation {x+x^2}/2=x^3 , or x+x^2=2x^3. This is a factorable quadratic equation:

2x^3-x^2-x=0
x(2x^2-x-1)=0
x(2x+1)(x-1)=0

This equation gives three solutions for x: x = 0, x = –1/2, and x = 1. Because there are multiple possible values for x, the statement is not sufficient. (There’s no need to investigate the other possible cases at this point, since you’ve just determined that this statement is insufficient.)

(2) INSUFFICIENT: Just about any value for x will make this statement true (that the average of x^pxp and x^rxr is not x^qxq). For instance, if p = 1, q = 2, and r = 3, then the only requirement is that the average of x and x^3x3 not be equal to x^2x2. Try random values of x: If x = 2, then the average of x and x^3x3 is (2 + 8)/2 = 5, which is not equal to x^2=4x2 = 4. If x = 3, then the average of x and x^3x3 is (3 + 27)/2 = 15, which is not equal to x^2=9x2 = 9. Because there are multiple possible values for x, the statement is not sufficient.

(1) AND (2) INSUFFICIENT: In order to determine the values of x that satisfy both statements, revisit the cases derived in statement 1, this time solving for all of them.

Case #1: If the average of x and x^2 is x^3, then, as shown above, x can be any of 0, –1/2, 1.

Case #2: If the average of x and x^3 is x^2, then {x+x^3}/2=x^2, or x+x^3=2x^2. This is also a factorable equation: x^3-2x^2+x=0 ––> x(x^2-2x+1)=0 ––> x(x-1)^2=0. Therefore, x can be either 0 or 1.
Case #3: If the average of x^2 and x^3 is x, then {x^2+x^3}/2=x, or x^2+x^3=2x. This too is a factorable equation: x^3+x^2-2x=0 ––> x(x^2+x-2)=0 ––> x(x+2)(x-1)=0. Therefore, x can be any of 0, –2, 1.

There are thus four possible values of x according to statement 1: x = –2, –1/2, 0, 1.

For the values x = 0 or 1, the average of two terms is equal to the third term in all 3 cases, so these two values can’t satisfy statement 2 (where the average of the first two terms cannot equal the third term). Discard the values 0 and 1.

On the other hand, if x = –2, p and q are 2 and 3 (in either order), and r = 1, then both statements are satisfied. The average of x^p and x^q is x^r: (4 + -8)/2 = -2, so statement 1 is true. In addition, the average of x^p and x^r is not x^q: (4 + -2)/2 does not equal -8, so statement 2 is true. Therefore, x could equal –2.

What about the final case? If x = –1/2, p and q are 1 and 2 (in either order), and r = 3, then both statements are also satisfied. The average of x^p and x^q is x^r: (–1/2 + 1/4)/2 = –1/8, so statement 1 is true. In addition, the average of x^p and x^r is not x^q: (–1/2 + –1/8)/2 does not equal 1/4, so statement 2 is true. Therefore, x can still be either –2 or –1/2. Both statements together are still not sufficient to answer the question.

The correct answer is (E).



I actually think the logic from Manhattan is backwards. If we would have arrived at the conclusion that one or both of the statements were sufficient on their own after choosing values for p,q, and r, then we would have to keep checking the other arrangement of values of p,q, and r to make sure that we weren't missing any cases. If we pick one arrangement, however, and determine that x could still have multiple values with both statements taken into account, then we can safely say that we need more information (as in hsbinfy's approach).
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Re: If p, q, and r are different positive integers such that p + q + r = 6  [#permalink]

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New post 08 Jul 2016, 22:01
There are 4 variables (p, q, r and x) and 1 equation (p+q+r=6) in the original condition. In order to match the number of variables and the number of equations, we need 3 equations. Since con 1) and con 2) each has 1 equation, there is a high chance that E is the correct answer. Using 1) & 2), from x+x^3=2x or x^3+x^2=2x, we get x=-1/2, -2. The answer is not unique and the condition is not sufficient. Hence, the correct answer is E.


- For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.
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Re: If p, q, and r are different positive integers such that p + q + r = 6  [#permalink]

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New post 23 Oct 2018, 08:11
how could be solved in 2 mins? It is tough
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Re: If p, q, and r are different positive integers such that p + q + r = 6   [#permalink] 23 Oct 2018, 08:11
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