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If y>0>x, and (3+5y)/(x−1) < −7, then which of the following [#permalink]
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01 Jul 2013, 23:38
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If y>0>x, and (3+5y)/(x−1) < −7, then which of the following must be true? A. 5y−7x+4 < 0 B. 5y+7x−4 > 0 C. 7x−5y−4 < 0 D. 4+5y+7x > 0 E. 7x−5y+4 > 0 P.S: Bunuelalthough I got it right but I'm having confusion on the 'flip of sign' as we're to multiply this inequality with a negative quantity...Can you please explain in detail?
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Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following [#permalink]
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01 Jul 2013, 23:46



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Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following [#permalink]
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01 Jul 2013, 23:58
Bunuel wrote: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following must be true?
A. 5y−7x+4 < 0 B. 5y+7x−4 > 0 C. 7x−5y−4 < 0 D. 4+5y+7x > 0 E. 7x−5y+4 > 0
Since given that x is a negative number then x1 is also some negative number. Multiply both sides of the inequality by x1. When multiplying by negative flip the sig, so we'll have:
\(\frac{3+5y}{x1} < 7\) > \(3+5y>7(x1)\) > \(3+5y>77x\) > \(5y+7x4>0\).
Answer: B.
Hope it's clear. There I'm having confusion...as x1 is negative so multiplying with it both sides of the inequality should flip their signs.But for example, RHS remains with a negative sign! Is it because of the fact that x1 is itself negative so with 7 multiplied with x1 will yield positive (negative * negative =positive)...Same logic goes for LHS of inequality which is primnarily negative because of its negative denominator, so LHS multiplied with (x1) yields a positive. Let me know if this is correct...
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Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following [#permalink]
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02 Jul 2013, 00:03
debayan222 wrote: Bunuel wrote: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following must be true?
A. 5y−7x+4 < 0 B. 5y+7x−4 > 0 C. 7x−5y−4 < 0 D. 4+5y+7x > 0 E. 7x−5y+4 > 0
Since given that x is a negative number then x1 is also some negative number. Multiply both sides of the inequality by x1. When multiplying by negative flip the sig, so we'll have:
\(\frac{3+5y}{x1} < 7\) > \(3+5y>7(x1)\) > \(3+5y>77x\) > \(5y+7x4>0\).
Answer: B.
Hope it's clear. There I'm having confusion...as x1 is negative so multiplying with it both sides of the inequality should flip their signs.But for example, RHS remains with a negative sign! Is it because of the fact that x1 is itself negative so with 7 multiplied with x1 will yield positive (negative * negative =positive)...Same logic goes for LHS of inequality which is primnarily negative because of its negative denominator, so LHS multiplied with (x1) yields a positive. Let me know if this is correct... Flipping sign means flipping "<" sign to ">". Nothing else.
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Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following [#permalink]
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02 Jul 2013, 00:09
Bunuel wrote: debayan222 wrote: Bunuel wrote: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following must be true?
A. 5y−7x+4 < 0 B. 5y+7x−4 > 0 C. 7x−5y−4 < 0 D. 4+5y+7x > 0 E. 7x−5y+4 > 0
Since given that x is a negative number then x1 is also some negative number. Multiply both sides of the inequality by x1. When multiplying by negative flip the sig, so we'll have:
\(\frac{3+5y}{x1} < 7\) > \(3+5y>7(x1)\) > \(3+5y>77x\) > \(5y+7x4>0\).
Answer: B.
Hope it's clear. There I'm having confusion...as x1 is negative so multiplying with it both sides of the inequality should flip their signs.But for example, RHS remains with a negative sign! Is it because of the fact that x1 is itself negative so with 7 multiplied with x1 will yield positive (negative * negative =positive)...Same logic goes for LHS of inequality which is primnarily negative because of its negative denominator, so LHS multiplied with (x1) yields a positive.Let me know if this is correct... Flipping sign means flipping "<" sign to ">". Nothing else. Ok! But could you please let me know whether my reasoning above(as highlighted) is correct...?
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Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following [#permalink]
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02 Jul 2013, 00:15
debayan222 wrote: Bunuel wrote: debayan222 wrote: There I'm having confusion...as x1 is negative so multiplying with it both sides of the inequality should flip their signs.But for example, RHS remains with a negative sign!
Is it because of the fact that x1 is itself negative so with 7 multiplied with x1 will yield positive (negative * negative =positive)...Same logic goes for LHS of inequality which is primnarily negative because of its negative denominator, so LHS multiplied with (x1) yields a positive.Let me know if this is correct... Flipping sign means flipping "<" sign to ">". Nothing else. Ok! But could you please let me know whether my reasoning above(as highlighted) is correct...? Don't know why do you need this reasoning to solve this question, but anyway: 7*(x1) = negative*negative = positive; (3+5y)/(x−1) = positive/negative = negative. When multiplied by (x1)=negative, (x1) reduces and we get (3+5y)=positive. Hope it's clear.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following [#permalink]
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02 Jul 2013, 00:22



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Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following [#permalink]
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12 Aug 2013, 16:44
C. 7x−5y−4 < 0 why C is not always true, x is given negative, y is positive. then 7(ve)  5 (+ve) 4 < 0 ve  ve ve < 0 this also negative always.
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Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following [#permalink]
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13 Aug 2013, 08:29
PiyushK wrote: C. 7x−5y−4 < 0
why C is not always true, x is given negative, y is positive.
then 7(ve)  5 (+ve) 4 < 0
ve  ve ve < 0 this also negative always. If y>0>x, and (3+5y)/(x−1) < −7, then which of the following must be true? Here we have two equations. The 1st conforms (C) but the 2nd doesn't conforms (C). i mean we can't have (C) from the 2nd condition. we have to select a combine effect of both as the result......
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Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following [#permalink]
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13 Aug 2013, 10:40
Asifpirlo wrote: PiyushK wrote: C. 7x−5y−4 < 0
why C is not always true, x is given negative, y is positive.
then 7(ve)  5 (+ve) 4 < 0
ve  ve ve < 0 this also negative always. If y>0>x, and (3+5y)/(x−1) < −7, then which of the following must be true? Here we have two equations. The 1st conforms (C) but the 2nd doesn't conforms (C). i mean we can't have (C) from the 2nd condition. we have to select a combine effect of both as the result...... Ohhh IC, I got it. Thanks.
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Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following [#permalink]
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13 May 2016, 13:50
Bunuel wrote: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following must be true?
A. 5y−7x+4 < 0 B. 5y+7x−4 > 0 C. 7x−5y−4 < 0 D. 4+5y+7x > 0 E. 7x−5y+4 > 0
Since given that x is a negative number then x1 is also some negative number. Multiply both sides of the inequality by x1. When multiplying by negative flip the sig, so we'll have:
\(\frac{3+5y}{x1} < 7\) > \(3+5y>7(x1)\) > \(3+5y>77x\) > \(5y+7x4>0\).
Answer: B.
Hope it's clear. I tried plugging in sample values say y=2 x=1 into 5y+7x4 1074 < 0 What am i missing here ? Please explain



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Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following [#permalink]
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