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satisfied (‐15 is not within (‐3,4) interval.)d) . ‐‐> . We reject the solution because our condition is notsatisfied (‐1 is not more than 4)(Optional) The following illustration may help you understand how to open modulus at different conditions.Answer: 0Example #2Q.: . What is x?Solution: There are 2 conditions:a) ‐‐> or . ‐‐> . x e { , } and both solutionssatisfy the condition.b) ‐‐> . ‐‐> . x e { , } and both solutions satisfythe condition.(Optional) The following illustration may help you understand how to open modulus at different conditions.Answer: , , ,Tip & TricksThe 3‐steps method works in almost all cases. At the same time, often there are shortcuts and tricks that allowyou to solve absolute value problems in 10‐20 sec.I. Thinking of inequality with modulus as a segment at the number line.For example,Problem: 1<x<9. What inequality represents this condition?‐ 19 ‐GMAT Club Math Bookpart of GMAT ToolKit iPhone App
A. |x|<3B. |x+5|<4C. |x‐1|<9D. |‐5+x|<4E. |3+x|<5Solution: 10sec. Traditional 3‐steps method is too time‐consume technique. First of all we find length (9‐1)=8 andcenter (1+8/2=5) of the segment represented by 1<x<9. Now, let’s look at our options. Only B and D has 8/2=4 onthe right side and D had left site 0 at x=5. Therefore, answer is D.II. Converting inequalities with modulus into range expression.In many cases, especially in DS problems, it helps avoid silly mistakes.For example,|x|<5 is equal to x e (‐5,5).|x+3|>3 is equal to x e (‐inf,‐6)&(0,+inf)III. Thinking about absolute values as distance between points at the number line.For example,Problem: A<X<Y<B. Is |A‐X| <|X‐B|?1) |Y‐A|<|B‐Y|Solution:We can think about absolute values here as distance between points. Statement 1 means than distance between Yand A is less than Y and B. Because X is between A and Y, distance between |X‐A| < |Y‐A| and at the same timedistance between X and B will be larger than that between Y and B (|B‐Y|<|B‐X|). Therefore, statement 1 issufficient.PitfallsThe most typical pitfall is ignoring third step in opening modulus ‐ always check whether your solution satisfiesconditions.Practice from the GMAT Official Guide:The Official Guide, 12th Edition: PS #22; PS #50; PS #130; DS #1; DS #153;The Official Guide, Quantitative 2th Edition: PS #152; PS #156; DS #96; DS #120;The Official Guide, 11th Edition: DT #9; PS #20; PS #130; DS #3; DS #105; DS #128‐ 20 ‐GMAT Club Math Bookpart of GMAT ToolKit iPhone App
Page 2 and 3: For the latest version of the GMAT
Page 4 and 5: Number TheoryDefinitionNumber Theor
Page 6 and 7: • 1 (and ‐1) are divisors of ev
Page 8 and 9: other digits: 21‐(9+8+6+9)=‐11,
Page 10 and 11: Sum of n first positive integers:Su
Page 12 and 13: Converting Decimals to Fractions•
Page 14 and 15: Operations involving the same bases
Page 16 and 17: PERCENTDefinitionA percentage is a
Page 18 and 19: Absolute ValueDefinitionThe absolut
Page 22 and 23: FactorialsDefinitionThe factorial o
Page 24 and 25: AlgebraScopeManipulation of various
Page 26 and 27: There is a common misconception tha
Page 28 and 29: RemaindersDefinitionIf and are posi
Page 30 and 31: A. 2B. 4C. 8D. 20E. 45divided by yi
Page 32 and 33: Example #10 (hard)When positive int
Page 34 and 35: Word Problems OverviewThe Following
Page 36 and 37: Key words: less than (implies subtr
Page 38 and 39: Distance/Speed/Time Word ProblemsWh
Page 40 and 41: In the next section we will learn h
Page 42 and 43: Time(3) = Time(1) + Time(2) ‐‐
Page 44 and 45: 3*x = 5*(x ‐ 20) ‐‐> x = 50 m
Page 46 and 47: Work Word ProblemsWhat is a ‘Work
Page 48 and 49: Example 3:Working together, printer
Page 50 and 51: Advanced Overlapping Sets ProblemsS
Page 52 and 53: the writing club. If 6 students sig
Page 54 and 55: two of the courses and 1 employee h
Page 56 and 57: Consider this problem to be an over
Page 58 and 59: Median The median of a triangle is
Page 60 and 61: • The midsegment is always parall
Page 62 and 63: Equilateral triangle all sides have
Page 64 and 65: • A right triangle where the angl
Page 66 and 67: Then the triangles ABC, CHB and CHA
Page 68 and 69: Quadrilateral A polygon with four '
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Squares A 4‐sided regular polygon
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• Area ‐ The usual way to calcu
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• A circle is the shape with the
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ChordA line that links two points o
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angle.• An inscribed angle is exa
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chords.‐ This becomes the theorem
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Coordinate GeometryDefinitionCoordi
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As you can see, the distance formul
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Lines in Coordinate GeometryIn Eucl
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The equation of a straight line tha
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Positive slopeHere, y increases as
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A horizontal line has a slope of ze
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Solution: We first find the slope o
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To answer, we must find the slope o
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Example #2Q: Find the point of inte
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ParabolaA parabola is the graph ass
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Standard DeviationDefinitionStandar
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Standard deviation of is . Decrease
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ProbabilityDefinitionA number expre
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Therefore, the probability of getti
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2) Reversal combinatorial approach:
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Combinations & PermutationsDefiniti
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1. The total number of arrangements
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is the first termDefining Propertie
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Solution(1) a1=8, does not tell us
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CuboidA cube is the 3‐D generaliz
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ConeA cone is a 3‐D object obtain
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Volume =Surface Area w/o base =Surf
Page 126:
(2) : Sum of sides = . Not sufficie
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