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Post subject: GMAT Algebra (Data Sufficiency) Posted: Fri Feb 24, 2012 7:12 pm 

Joined: Sun May 30, 2010 3:15 am Posts: 424

Is (x – 4)(x – 3)(x + 2)(x + 1) > 0 ? (1) 3 > x (2) x > 1
A. Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not. B. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not. C. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient. D. Either statement BY ITSELF is sufficient to answer the question. E. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that further information would be needed to answer the question.
(C) The expression (x – 4)(x – 3)(x + 2)(x + 1) is composed of four factors. It will equal 0 if at least one of the factors is 0. It will be positive if all the four factors are positive, if all the four factors are negative, or if two of them are negative and the other two are positive. Otherwise the epxression will be negative.
Statement (1), 3 > x, implies that the factors (x – 4) and (x – 3) are negative. The signs of the other two factors, (x + 2) and (x + 1), are not defined. E.g. they both can be positive if x = 1. Or one of them can equal 0 if x = 1 or x = 2. Or (x + 2) can be positive and (x + 1) can be negative if x = 1.5, etc. Therefore the original expression can be positive, negative or 0 and we can NOT give a definite answer to the original question. Statement (1) by itself is NOT sufficient.
Statement (2), x > 1, implies that the factors (x + 2) and (x + 1) are positive. The signs of the other two factors, (x – 4) and (x – 3) , are not defined. E.g. they both can be positive if x = 5. Or one of them can equal 0 if x = 3 or x = 4. Or (x – 3) can be positive and (x – 4) can be negative if x = 3.5, etc. Therefore the original expression can be positive, negative or 0 and we can NOT give a definite answer to the original question. Statement (2) by itself is NOT sufficient.
If we use the both statements together, statement (1) implies that factors (x – 4) and (x – 3) are negative. Statement (2) implies that factors (x + 2) and (x + 1) are positive. Therefore the original expression must be positive (2 negative factors × 2 positive factors). The both statements taken together are sufficient to answer the question. The correct answer is C.
Alternative method: You may solve the original inequality first and then compare the solution with the inequality (1), inequality (2) and a system of inequalities (1) and (2) using the number line. 
Between 1 & 3 the expression acquires negative as well as positive values.


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Gennadiy

Post subject: Re: t.1, qt. 13: inequalities. data sufficiency Posted: Fri Feb 24, 2012 7:23 pm 

Joined: Sun May 30, 2010 2:23 am Posts: 498

Quote: Between 1 & 3 the expression acquires negative as well as positive values. It's x that is between 1 and 3, not the range of values of the whole expression. If x > 1 then ( x + 2) and ( x + 1) are positive. If x < 3 then ( x – 4) and ( x – 3) are negative. Therefore the product of all factors, ( x – 4)( x – 3)( x + 2)( x + 1), is positive for all values of x between 1 and 3.


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questioner

Post subject: Re: t.1, qt. 13: inequalities. data sufficiency Posted: Fri Mar 23, 2012 3:52 pm 

Joined: Sun May 30, 2010 3:15 am Posts: 424

1st won't work  check for x = 3.5 and 7  they will give opposite signs. 2nd won't work  check for x = 1 and 5  they will give opposite sings.
Combining the both won't work  check for x = 1 and 1.5  they again give opposite signs.
Hence the answer is E.


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Gennadiy

Post subject: Re: t.1, qt. 13: inequalities. data sufficiency Posted: Fri Mar 23, 2012 4:24 pm 

Joined: Sun May 30, 2010 2:23 am Posts: 498

Quote: 1st won't work  check for x = 3.5 and 7  they will give opposite signs. The both values do not fit in statement (1) 3 > x. Quote: 2nd won't work  check for x = 1 and 5  they will give opposite sings. x = 1 and 5 both give the same sign. (1 – 4)(1 – 3)(1 + 2)(1 + 1) = (3)(2)(3)(2) = 36 > 0 (5 – 4)(5 – 3)(5 + 2)(5 + 1) = 1 × 2 × 7 × 5 = 70 > 0 Quote: Combining the both won't work  check for x = 1 and 1.5  they again give opposite signs. The value 1.5 does not fit in statement (2) x > 1.


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baphna

Post subject: Re: t.1, qt. 13: inequalities. data sufficiency Posted: Sat Mar 24, 2012 2:15 am 

Joined: Sat Mar 24, 2012 2:10 am Posts: 1

OKk..how about this.. 1st won't work  check for x=1 and 1.5  they will give opp.signs and satisfies x<3 2nd won't work  check for x=3.5 and 5  they will give opp.sings and satisfies x > 1
combining both won't work check for x=1 and 1.5  they again give opp.signs and satisfies 1<x<3
hence ans is E


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Gennadiy

Post subject: Re: t.1, qt. 13: inequalities. data sufficiency Posted: Sat Mar 24, 2012 2:32 pm 

Joined: Sun May 30, 2010 2:23 am Posts: 498

Quote: 1st won't work  check for x = 1 and 1.5  they will give opp.signs and satisfies x < 3 That is correct. Quote: 2nd won't work  check for x = 3.5 and 5  they will give opp.sings and satisfies x > 1 That is correct as well. Quote: combining both won't work check for x = 1 and 1.5  they again give opp.signs and satisfies 1 < x < 3 That is NOT correct, because 1.5 does NOT satisfy 1 < x < 3. 1.5 < 1 < x < 3 Here is the proof of why 1 < x < 3 works: If x > 1 then ( x + 2) and ( x + 1) are positive. If x < 3 then ( x – 4) and ( x – 3) are negative. Therefore the product of all factors, (x – 4)(x – 3)(x + 2)(x + 1), is negative × negative × positive × positive = positive for all values of x between 1 and 3.


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questioner

Post subject: Re: t.1, qt. 13: inequalities. data sufficiency Posted: Mon May 07, 2012 4:36 am 

Joined: Sun May 30, 2010 3:15 am Posts: 424

For x = 3/2 the answer is ive while for x = 0 it will be +ive. Thus even if we use both A and B, the value for the expression is not certain. For range 1 to 2 the expression is ive For range 2 to 3 the expression is +ive
Thus answer should be E


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Gennadiy

Post subject: Re: t.1, qt. 13: inequalities. data sufficiency Posted: Mon May 07, 2012 6:46 am 

Joined: Sun May 30, 2010 2:23 am Posts: 498

(1) 3 > x(2) x > 1 The both statements combined result in 1 < x < 3. Quote: For x = 3/2 the answer is ive … x = 3/2 is less than 1 and thus is NOT within the range. Quote: For range 1 to 2 the expression is ive For range 2 to 3 the expression is +ive The both ranges are NOT within the range 1 < x < 3 .


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questioner

Post subject: Re: t.1, qt. 13: inequalities. data sufficiency Posted: Sat Jul 14, 2012 4:07 pm 

Joined: Sun May 30, 2010 3:15 am Posts: 424

Can you explain the alternative method in a little more detail.


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Gennadiy

Post subject: Re: t.1, qt. 13: inequalities. data sufficiency Posted: Sat Jul 14, 2012 4:31 pm 

Joined: Sun May 30, 2010 2:23 am Posts: 498

questioner wrote: Can you explain the alternative method in a little more detail. We solve the inequality ( x – 4)( x – 3)( x + 2)( x + 1) > 0 first. Mark the points 4, 3, 2 and 1 on the number line: The number line is divided into 5 segments. In each segment each factor does not change its sign (positive or negative). So the whole expression in the left side is of the same sign as any value within that segment. Buy picking numbers 3, 1.5, 0, 3.5, and 5 you can see that the solution of the inequality is: x < 2 , 1 < x < 3 , x > 4 Now, when you solved the original inequality, compare the solution to each statement. (1) 3 > xAs you can see, within this range there are values from the solution ( x < 2 , 1 < x < 3 ) as also values that are not in the solution ( 2 ≤ x ≤ 1 ). So we do not have a definite answer. (2) x > 1 As you can see, within this range there are values from the solution ( 1 < x < 3 , x > 4 ) as also values that are not in the solution ( 3 ≤ x ≤ 4 ). So we do not have a definite answer. (1) 3 > x AND (2) x > 1 yield 1 < x < 3 . All the values within this range are from the solution. So we have the definite answer "If (1) and (2) are both true (given) then the original inequality holds true."


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