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If 8x > 4 + 6x, what is the value of the integer x? (1) 6 [#permalink]
 18 Aug 2010, 10:06
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58% (00:31) wrong based on 3 sessions
If 8x > 4 + 6x, what is the value of the integer x? (1) 6 – 5x > -13 (2) 3 – 2x < -x + 4 < 7.2 – 2x Could anybody tell me when it is appropriate to add up equations? That's what I did and it didn't work...
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heyholetsgo wrote: If 8x > 4 + 6x, what is the value of the integer x?
(1) 6 – 5x > -13
(2) 3 – 2x < -x + 4 < 7.2 – 2x
Could anybody tell me when it is appropriate to add up equations? That's what I did and it didn't work... Given: x=integer and 8x>4+6x --> 2x>4 --> x>2. Question: x=?(1) 6-5x>-13 --> 19>5x --> \frac{19}{5}=3.8>x --> as x=integer and x>2, then x=3. Sufficient. (2) 3-2x<-x+4<7.2-2x --> take only the following part: -x+4<7.2-2x--> x<3.2 --> as x=integer and x>2, then x=3. Sufficient. Answer: D. So no need to add inequality in this case. But if you are interested: You can only add inequalities when their signs are in the same direction:If a>b and c>d (signs in same direction: > and >) --> a+c>b+d. Example: 3<4 and 2<5 --> 3+2<4+5. You can only apply subtraction when their signs are in the opposite directions:If a>b and c<d (signs in opposite direction: > and <) --> a-c>b-d (take the sign of the inequality you subtract from). Example: 3<4 and 5>1 --> 3-5<4-1. Hope it helps.
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Bunuel wrote: heyholetsgo wrote: If 8x > 4 + 6x, what is the value of the integer x?
(1) 6 – 5x > -13
(2) 3 – 2x < -x + 4 < 7.2 – 2x
Could anybody tell me when it is appropriate to add up equations? That's what I did and it didn't work... Given: x=integer and 8x>4+6x --> 2x>4 --> x>2. Question: x=?(1) 6-5x>-13 --> 19>5x --> \frac{19}{5}=3.8>x --> as x=integer and x>2, then x=3. Sufficient. (2) 3-2x<-x+4<7.2-2x --> take only the following part: -x+4<7.2-2x--> x<3.2 --> as x=integer and x>2, then x=3. Sufficient. Answer: D. So no need to add inequality in this case. But if you are interested: You can only add inequalities when their signs are in the same direction:If a>b and c>d (signs in same direction: > and >) --> a+c>b+d. Example: 3<4 and 2<5 --> 3+2<4+5. You can only apply subtraction when their signs are in the opposite directions:If a>b and c<d (signs in opposite direction: > and <) --> a-c>b-d (take the sign of the inequality you subtract from). Example: 3<4 and 5>1 --> 3-5<4-1. Hope it helps. Hi, can you explain stmt B please, why is that only the second part is considered?
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Hi Bunuel,
why have to taken only the second equation in Statement B not the first one? Plz explain briefly.
Thanks..
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ramana wrote: Bunuel wrote: Given: x=integer and 8x>4+6x --> 2x>4 --> x>2. Question: x=?
(1) 6-5x>-13 --> 19>5x --> \frac{19}{5}=3.8>x --> as x=integer and x>2, then x=3. Sufficient.
(2) 3-2x<-x+4<7.2-2x --> take only the following part: -x+4<7.2-2x--> x<3.2 --> as x=integer and x>2, then x=3. Sufficient.
Answer: D. Hi, can you explain stmt B please, why is that only the second part is considered? mission2009 wrote: Hi Bunuel,
why have to taken only the second equation in Statement B not the first one? Plz explain briefly.
Thanks.. Because to reach the answer we don't need the first part at all. The part which says -x+4<7.2-2x is enough to give necessary info: x<3.2 --> as x=integer and x>2, then x=3. Sufficient.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory
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. NEW!!!
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. NEW!!!
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Bunuel wrote: heyholetsgo wrote: If 8x > 4 + 6x, what is the value of the integer x?
(1) 6 – 5x > -13
(2) 3 – 2x < -x + 4 < 7.2 – 2x
Could anybody tell me when it is appropriate to add up equations? That's what I did and it didn't work... Given: x=integer and 8x>4+6x --> 2x>4 --> x>2. Question: x=?(1) 6-5x>-13 --> 18>5x --> \frac{18}{5}=3.6>x --> as x=integer and x>2, then x=3. Sufficient. (2) 3-2x<-x+4<7.2-2x --> take only the following part: -x+4<7.2-2x--> x<3.2 --> as x=integer and x>2, then x=3. Sufficient. Answer: D. So no need to add inequality in this case. But if you are interested: You can only add inequalities when their signs are in the same direction:If a>b and c>d (signs in same direction: > and >) --> a+c>b+d. Example: 3<4 and 2<5 --> 3+2<4+5. You can only apply subtraction when their signs are in the opposite directions:If a>b and c<d (signs in opposite direction: > and <) --> a-c>b-d (take the sign of the inequality you subtract from). Example: 3<4 and 5>1 --> 3-5<4-1. Hope it helps. Slightly off on statement 1, however the answer doesn't change. (1) 6-5x>-13 --> 19>5x --> \frac{19}{5}=3.8>x --> as x=integer and x>2, then x=3. Sufficient.
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Bunuel wrote: heyholetsgo wrote: If 8x > 4 + 6x, what is the value of the integer x?
(1) 6 – 5x > -13
(2) 3 – 2x < -x + 4 < 7.2 – 2x
Could anybody tell me when it is appropriate to add up equations? That's what I did and it didn't work... Given: x=integer and 8x>4+6x --> 2x>4 --> x>2. Question: x=?(1) 6-5x>-13 --> 19>5x --> \frac{19}{5}=3.8>x --> as x=integer and x>2, then x=3. Sufficient. (2) 3-2x<-x+4<7.2-2x --> take only the following part: -x+4<7.2-2x--> x<3.2 --> as x=integer and x>2, then x=3. Sufficient. Answer: D. So no need to add inequality in this case. But if you are interested: You can only add inequalities when their signs are in the same direction:If a>b and c>d (signs in same direction: > and >) --> a+c>b+d. Example: 3<4 and 2<5 --> 3+2<4+5. You can only apply subtraction when their signs are in the opposite directions:If a>b and c<d (signs in opposite direction: > and <) --> a-c>b-d (take the sign of the inequality you subtract from). Example: 3<4 and 5>1 --> 3-5<4-1. Hope it helps. One small doubt... As said X>2 and x>3.8 how can we say x=3 even it is integer.The 2nd one say X.3.8 so y can't we take x=4???? Thanks to help Thanks to help
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8x>4+6x
=>2x>4
=> x>2
x=?
1. Sufficient
6-5x>-13
-5x>-19 => x<19/5 => x<3.2 => x ={..........-1,0,1,2,3} as x is an integer
as x>2 and x<3.2=> x can only be 3
2. Sufficient
3-2x < - x+4 <7.2-2x
3-2x<-x+4 => -1 < x
-x+4 < 7.2-2x => x<3.2
=> -1<x<3.2 , but we know x>2 = 2<x<3.2 then x can only be 3 as it has to be an integer.
Answer is D.
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Manager
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heyholetsgo wrote: If 8x > 4 + 6x, what is the value of the integer x?
(1) 6 – 5x > -13
(2) 3 – 2x < -x + 4 < 7.2 – 2x
Could anybody tell me when it is appropriate to add up equations? That's what I did and it didn't work... Given equation can be simplified to 8x > 4 + 6x => x < 2 Given x is an integer 1) 6 – 5x > -13 => x < 3.2 and since x > 2 It means x = 3 Sufficient 2) -x + 4 < 7.2 – 2x => x < 3.2 3 – 2x < -x + 4 => x > -1 And we know x > 2 and x is an integer So x can be only 3 Sufficient. So D is the answer.
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