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May 0510:00 AM PDT
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GMAT Inequalities is a high-frequency topic in GMAT Quant, but many students struggle because the concepts behave differently from standard algebra. Understanding the right rules, patterns, and edge cases can significantly improve both speed and accuracy.- May 06
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In Episode 3 of our GMAT Ninja Critical Reasoning series, we tackle Discrepancy, Paradox, and Explain an Oddity questions. You know the feeling: the passage gives you two facts that seem completely contradictory.... - May 08
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18 Jan 2018, 01:44
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[GMAT math practice question]
What is the product of all roots of the equation (x+1)^2=|x+1|?
A. -2
B. -1
C. 0
D. 1
E. 2
=>
Now,
(x+1)^2=|x+1|
⇔ |x+1|^2=|x+1|
⇔ |x+1|^2-|x+1|= 0
⇔ |x+1|(|x+1|-1) = 0
⇔ |x+1| = 0 or |x+1| = 1
⇔ x = -1 or x+1 = ±1
⇔ x = -1 or x = -1 ±1
⇔ x = -1, x= -2 or x = 0
The product of these solutions is (-1)*(-2)*0 = 0.
Therefore, the answer is C.
Answer: D
What is the product of all roots of the equation (x+1)^2=|x+1|?
A. -2
B. -1
C. 0
D. 1
E. 2
=>
Now,
(x+1)^2=|x+1|
⇔ |x+1|^2=|x+1|
⇔ |x+1|^2-|x+1|= 0
⇔ |x+1|(|x+1|-1) = 0
⇔ |x+1| = 0 or |x+1| = 1
⇔ x = -1 or x+1 = ±1
⇔ x = -1 or x = -1 ±1
⇔ x = -1, x= -2 or x = 0
The product of these solutions is (-1)*(-2)*0 = 0.
Therefore, the answer is C.
Answer: D
19 Jan 2018, 01:20
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[GMAT math practice question]
Two students were asked to solve the equation x^2+px+q=0. Alice’s answer was x=1 and 5. This answer is incorrect for the original equation, but correct for the equation x^2+px+r=0. Bob’s answer was x=-2 and -4. This answer is incorrect for the original equation, but correct for x^2+sx+q=0. What are the solutions of the original equation?
A. 1, 8
B. -1,-8
C. 2,4
D. -1,-5
E. 1,-5
=>
Using the factor theorem with Alice’s solution yields
(x-1)(x-5) = x^2 – 6x + 5 = x^2+px+r.
So, p = -6.
Using the factor theorem with Bob’s solution yields
(x+2)(x+4) = x^2 + 6x + 8 = x^2 +sx +q.
So, q = 8.
Thus, the original equation is x^2-6x+8 = (x-2)(x-4) = 0. Its roots are 2 and 4.
Therefore, the answer is C.
Answer: B
Two students were asked to solve the equation x^2+px+q=0. Alice’s answer was x=1 and 5. This answer is incorrect for the original equation, but correct for the equation x^2+px+r=0. Bob’s answer was x=-2 and -4. This answer is incorrect for the original equation, but correct for x^2+sx+q=0. What are the solutions of the original equation?
A. 1, 8
B. -1,-8
C. 2,4
D. -1,-5
E. 1,-5
=>
Using the factor theorem with Alice’s solution yields
(x-1)(x-5) = x^2 – 6x + 5 = x^2+px+r.
So, p = -6.
Using the factor theorem with Bob’s solution yields
(x+2)(x+4) = x^2 + 6x + 8 = x^2 +sx +q.
So, q = 8.
Thus, the original equation is x^2-6x+8 = (x-2)(x-4) = 0. Its roots are 2 and 4.
Therefore, the answer is C.
Answer: B
21 Jan 2018, 18:41
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[GMAT math practice question]
If 0<2x+3y<10 and -10<3x+2y<0, then which of the following must be true?
I. x<0
II. y<0
III. x<y
A. I only
B . II only
C. I & II
D.I & III
E. I, II, &III
=>
Label the inequalities as follows:
0<2x+3y<10 --- (1)
-10<3x+2y<0 --- (2)
We consider each statement individually.
Statement I:
Multiplying (1) by -2 yields -20 < -4x – 6y < 0, and multiplying (2) by 3 yields -30 < 9x + 6y < 0. Adding these inequalities gives -50 < 5x < 0 or -10 < x < 0.
This statement is true.
Statement II:
Multiplying (1) by 3 yields 0 < 6x + 9y < 30, and multiplying (2) by -3 yields 0 < -6x – 4y < 20. Adding these inequalities gives 0 < 5y < 50 or 0 < y < 10.
This statement is false.
Statement III:
Multiplying (1) by – 1 yields -10 < -2x – 3y < 0. Adding this to inequality (2) yields -20 < x – y < 0.
This implies that x < y, and statement III is true.
Therefore, the answer is D.
Answer: B
If 0<2x+3y<10 and -10<3x+2y<0, then which of the following must be true?
I. x<0
II. y<0
III. x<y
A. I only
B . II only
C. I & II
D.I & III
E. I, II, &III
=>
Label the inequalities as follows:
0<2x+3y<10 --- (1)
-10<3x+2y<0 --- (2)
We consider each statement individually.
Statement I:
Multiplying (1) by -2 yields -20 < -4x – 6y < 0, and multiplying (2) by 3 yields -30 < 9x + 6y < 0. Adding these inequalities gives -50 < 5x < 0 or -10 < x < 0.
This statement is true.
Statement II:
Multiplying (1) by 3 yields 0 < 6x + 9y < 30, and multiplying (2) by -3 yields 0 < -6x – 4y < 20. Adding these inequalities gives 0 < 5y < 50 or 0 < y < 10.
This statement is false.
Statement III:
Multiplying (1) by – 1 yields -10 < -2x – 3y < 0. Adding this to inequality (2) yields -20 < x – y < 0.
This implies that x < y, and statement III is true.
Therefore, the answer is D.
Answer: B
