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Math Expert V
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(n-x)+(n-y)+(n-z)+(n-k) What is the value of the expression above?  [#permalink]

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Difficulty:   25% (medium)

Question Stats: 69% (01:22) correct 31% (01:39) wrong based on 148 sessions

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(n-x)+(n-y)+(n-z)+(n-k)

What is the value of the expression above?

(1) The average (arithmetic mean) of x, y, z, and k is n.

(2) x, y, z, and k are consecutive integers.

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Bunuel wrote:
(n-x)+(n-y)+(n-z)+(n-k)

What is the value of the expression above?

(1) The average (arithmetic mean) of x, y, z, and k is n.

(2) x, y, z, and k are consecutive integers.

$$(n-x)+(n-y)+(n-z)+(n-k) = 4n -(x+y+z+k)$$ ------------(1)

Statement 1: implies that $$x+y+z+k=4n$$

so equation (1) becomes $$= 4n-4n=0$$. Sufficient

Statement 2: this does not gives the value of x, y, z, k and n. Hence insufficient

Option A
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Re: (n-x)+(n-y)+(n-z)+(n-k) What is the value of the expression above?  [#permalink]

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Ques: (n-x)+(n-y)+(n-z)+(n-k)=?
Info: nothing

Before proceeding to the statements' evaluation, simplifying the expression, we get:
4n-(x+y+z+k)=?

St (1): The average (arithmetic mean) of x, y, z, and k is n.
Statement implies that $$\frac{x+y+z+k}{4}$$=n, or,
x+y+z+k=4n
Putting this in the question stem, we get 4n-4n=0.
Therefore, St (1) is sufficient to determine the value of the given expression, hence, answer choices AD remain [BCE eliminated].

St (2): x, y, z, and k are consecutive integers.
This statement implies that y=x+1, z=x+2, and k=z+3, and thus, x+y+z+k becomes 4x+6.
Now, putting in the question stem, we get, 4n-4x+6=?
As there's no information provided about n & x, they both can assume any value and we will get different results for the given expression.
Therefore, St (2) is insufficient, hence, answer choice A.
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Re: (n-x)+(n-y)+(n-z)+(n-k) What is the value of the expression above?  [#permalink]

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Bunuel wrote:
(n-x)+(n-y)+(n-z)+(n-k)

What is the value of the expression above?

(1) The average (arithmetic mean) of x, y, z, and k is n.

(2) x, y, z, and k are consecutive integers.

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 5 variables and 0 equation, E is most likely to be the answer.
As E is the most likely answer, we should consider both conditions 1) and 2) together before considering each of them individually. If they are not sufficient when taken together, E is the answer.

Conditions 1) & 2)

$$\frac{(x+y+z+k)}{4} = n$$
$$x + y + z + k = 4n$$
Thus $$( n - x ) + ( n - y ) + ( n - z ) + ( n - k ) = 4n - ( x+ y + z + k ) = 4n - 4n = 0$$.
Both conditions together are sufficient.

Since this question is a statistics question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
Actually, the condition 1) only is used in the conditions 1) & 2).

$$\frac{(x+y+z+k)}{4} = n$$
$$x + y + z + k = 4n$$
Thus $$( n - x ) + ( n - y ) + ( n - z ) + ( n - k ) = 4n - ( x+ y + z + k ) = 4n - 4n = 0$$.

The condition 1) only is sufficient.

Condition 2)
We can't derive anything $$4n - ( x+ y + z + k )$$ from the condition, since we don't know $$n$$.
The condition 2) is not sufficient.

Therefore, the answer is A.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
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Re: (n-x)+(n-y)+(n-z)+(n-k) What is the value of the expression above?  [#permalink]

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Bunuel wrote:
(n-x)+(n-y)+(n-z)+(n-k)

What is the value of the expression above?

Simplify the expression

4n - (x+y+z+k) = ?

Quote:
(1) The average (arithmetic mean) of x, y, z, and k is n.

(2) x, y, z, and k are consecutive integers.

S1) $$\frac{(x+y+z+k)}{4}$$ = n
=> x + y + z + k = 4n
Insert the value in the original expression
=> 4n - 4n = 0
Sufficient.

S2) x, y,z,k are consecutive integers
=> x, x+1, x+2, x+3
Inset the value in the original expression
=> 4n - (4x + 6)
Insufficient.

A is the answer.
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Re: (n-x)+(n-y)+(n-z)+(n-k) What is the value of the expression above?  [#permalink]

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