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The product of 4 consecutive odd integers is negative. What is the lar

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New post 17 Jul 2018, 01:11
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[GMAT math practice question]

The product of 4 consecutive odd integers is negative. What is the largest odd integer?

1) The smallest integer is negative.
2) The third smallest integer is negative.

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The product of 4 consecutive odd integers is negative. What is the lar  [#permalink]

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New post 17 Jul 2018, 01:24
MathRevolution wrote:
[GMAT math practice question]

The product of 4 consecutive odd integers is negative. What is the largest odd integer?

1) The smallest integer is negative.
2) The third smallest integer is negative.


Let the odd integers be \(a,b,c,d\) in ascending order. We need the value of \(d\)

Given: \(a*b*c*d<0\). Hence either one or three numbers are negative

Statement 1: implies \(a<0\) but we know nothing about other numbers. if \(c<0\), then \(d=1\) but if \(b,c,d>0\), then \(d=5\). Insufficient

Statement 2: implies that \(c<0\), therefore \(a,b,c<0\) and \(d>0\) or \(d=1\). Sufficient

Option B
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Re: The product of 4 consecutive odd integers is negative. What is the lar  [#permalink]

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New post 17 Jul 2018, 01:42
MathRevolution wrote:
[GMAT math practice question]

The product of 4 consecutive odd integers is negative. What is the largest odd integer?

1) The smallest integer is negative.
2) The third smallest integer is negative.


4 consecutive ODD integers

Product is negative : ONLY when one or three of them are negative
A) only one...
-1,1,3,5
B) three of them
-5,-3,-1,1

Let us see the statements

1) the smallest integer is negative..
Both cases have smallest integer as negative..
Insufficient

2) the third smallest integer is negative.
Only possibility - case (B)
So largest number is 1
Sufficient

B
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Re: The product of 4 consecutive odd integers is negative. What is the lar  [#permalink]

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New post 17 Jul 2018, 02:15
MathRevolution wrote:
[GMAT math practice question]

The product of 4 consecutive odd integers is negative. What is the largest odd integer?

1) The smallest integer is negative.
2) The third smallest integer is negative.


Let the 4 consecutive odd integers be 2n-3, 2n-1, 2n+1, and 2n+3.

Smallest:- 2n-3
Largest:- 2n+3
Given, (2n-3)(2n-1)(2n+1)(2n+3)<0----(a)

St1:- (2n-3)<0
Or, n<1.5
So, we can have more than one value of 2n+3. (at n=-1, n=1)
Insufficient.

St2:- (2n+1)<0
Or, n< -0.5----------(b)
only n=-1 satisfies both (a) & (b).
So, largest: 2n+3=2(-1)+3=1
Sufficient.

Ans. (B)
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Re: The product of 4 consecutive odd integers is negative. What is the lar  [#permalink]

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New post 17 Jul 2018, 08:28
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MathRevolution wrote:
The product of 4 consecutive odd integers is negative. What is the largest odd integer?

1) The smallest integer is negative.
2) The third smallest integer is negative.


Target question: What is the largest odd integer?

Given: The product of 4 consecutive odd integers is negative.
It's worth taking the time to analyze what this is telling us.
If the product of 4 consecutive odd integers is negative, then there are only 2 possible scenarios:
Scenario #1: The numbers are -5, -3, -1, 1 (the product is -15, which is negative)
Scenario #2: The numbers are -1, 1, 3, 5 (the product is -15, which is negative)

Now let's check the statements:

Statement 1: The smallest integer is negative
Notice that BOTH scenarios meet this condition
In scenario #1, the answer to the target question is the largest integer is 1
In scenario #2, the answer to the target question is the largest integer is 5
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: The third smallest integer is negative
Only scenario #1 meets this condition
So, the answer to the target question is the largest integer is 1
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent
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Re: The product of 4 consecutive odd integers is negative. What is the lar  [#permalink]

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New post 19 Jul 2018, 02:26
=>

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.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

Since we have 1 variable (n) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.

Let the integers be \(2n – 3, 2n – 1, 2n + 1\) and \(2n +3.\)
Since \((2n-3)(2n-1)(2n+1)(2n+3) < 0,\) either one of the integers is negative, or three of the integers is negative. There are two possible lists of integers: \(-1, 1, 3, 5\) and \(-5, -3, -1, 1.\)

Condition 1)
The largest odd integers in the two possible lists are \(5\) and \(1\).
Since we don’t have a unique answer, condition 1) is not sufficient.

Condition 2)
If the third smallest integer is negative, then the integers are \(-5, -3, -1, 1.\)
The largest integer is \(1\).
Since we have a unique answer, condition 2) is sufficient.

Therefore, B is the answer.

Answer: B

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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Re: The product of 4 consecutive odd integers is negative. What is the lar &nbs [#permalink] 19 Jul 2018, 02:26
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