GMAT Prep Course 2

1. Geometry Introduction: about 1/3 of the math problems on the test involved geometry. There are no proves. Fortunately, the figures on the tests are drawn. The scale since you can check your work in some cases, even solve a problem by eyeball in the drawing, we'll discuss this technique in detail later. Falling is a discussion of the basic properties of geometry. You probably know many of these properties memorize any that you do not know. 2. Geometry Lines and Angles 1: lines and angles. When two straight lines meet at a point, they form an angle. The point is called the vertex of the angle, and the lines are called the sides of the ankle. The angle shown can be identified in three ways as either angle. X angle B or angle A, B, C or angle. C B A. When two straight lines intersect at a point, they formed four angles, the angles opposite each other called vertical angles and their congruent equal in the figure shown a equals B and see angle C equals angle D. 3. Geometry Lines and Angles 2: Ingalls heir. Measured in degrees. By definition, a circle contains 360 degrees, so an angle could be measured by its fractional part of a circle. For example, an angle that is 1 360th of the arc of a circle is one degree, and an angle that is 1/4 of the arc of a circle is 1/4 of 360 which is 90 degrees. So here we have the representation of one degree of a circle, which is 1 360th of the AARP. Of the circle, in a right angle is 90 degrees of the arc of a circle, and 240 degrees is 2/3 of 360 degrees. 4. Geometry Lines and Angles 3: There are four major types of angle measures, and acute angle has measured less than 90 degrees. A right angle has measured off 90 degrees, and a two single has measure greater than 90 degrees. A straight angle has measure 180 degrees, so this straight angle or straight line from here to here has 180 degrees. 5. Geometry Example 1: Since Angles A and B form a straight angle, there's some is 180 degrees, so a plus b equals 1 80 that we're told. The ratio or the quotient of A and B is seven has so a divided by B is equal a seven divided by two. Solving this equation for a we get and plugging this result into the other equation. Now multiplying through by two we get seven B plus to be is equal to 3 60 or nine. B equals 3 60 Divided by nine we get be equals 40. It's the answer is C. 6. Geometry Lines and Angles 4: two angles are supplementary if they're angle. Some is 180 degrees in the drawing. This single here 45 degrees plus 1 35 as upto one turn 80 degrees and forms a straight angle. Two angles or complementary if they're angle sums 90 degrees in this growing that 30 degrees plus 60 degrees adds up to 90 informs the right angle. 7. Geometry Lines and Angles 5: perpendicular lines meet at right angles, as shown in the figure two lines in the same plane or parallel if they never intersect. Parallel lines have the same slope. When parallel lines are cut by a trans versatile three important angle relationships exist . Alternate interior angles are equal. The angle. These two angles are equal, as are these two corresponding angles are equal and interior angles on the same side of the trans versatile. In this case, A and B are supplementary that is the add up to 180 degrees. 8. Geometry Lines and Angles 6: the shortest distance from a point toe. A line is along the new line that passes through the point as perpendicular to the original line. Here we have a point, not on a line, and the shortest distance is this perpendicular distance. Because this line here is I had partners of a right triangle. They're forced longer than either of the other two sides, and that will occur. No matter where you draw this line. It will always be longer than that line. 9. Geometry Triangles 1: Triangle's a triangle containing a right angle is called a right triangle. The right angle is denoted by a small square. They try and go with equal sides is cool. I saw sleaze, the angles opposite equal size air called base angles, and they are congruent equal. A trying with all three sides equal is called equal lateral, and each angle measure 60 degrees. A triangle with no equal sides and therefore no legal angles is called scaling. 10. Geometry Triangles 2: the altitude to the base of a nice sauce. Elise, or equal lateral triangle, bisects the base and by sex vertical angle. So these bisected bases are equal and the vertical angle is also bisected. The angle some of a triangle is 180 degrees, so a plus B plus C equals 1 80 11. Geometry Example 2: since X and 1 50 Former straight angle There's some is 1 80 solving for X. We get X equals 30. Now the angle. Some of this triangle here will be Z plus X plus 90 because of the right angle here equals 1 80 Plugging in X equals 30 and solving this for Z we get Z equals 60 now why and ZR vertical angles. So why is also 60? And now finding the angle Some of this triangle we get w plus why plus 90 again because it's a right angle equals 1 80 and we have wise 60. So we will get W plus 60 plus 90 equals 1 80 and solving this for W we get w equals 30. It's the answer is a 12. Geometry Triangles 3: the area of a triangle is 1/2 the base times the height. Sometimes the base has to be extended in order to draw the altitude. As in the third drawing shown below here, we had the extend the base out this distance here so that we can drop a perpendicular to it . And the former for the area of a triangle is very common on the test. 13. Geometry Triangles 4: in a triangle. The longer side is opposite the larger angle and vice versa. For this triangle, since 50 is greater than 30 degrees, we know that side B is greater than side A, and since 100 is greater than 50 or 30 we know that side, See is the longest side of the triangle. 14. Geometry Triangles 5: Pythagorean Thorough. Now this applies the only right triangles, so you have to see a little square or something else to indicate that you have a right triangle. The square of the high pot new C is equal to the sum of the squares of the two legs A and B Pythagorean triples the numbers 34 and five can always represent the sides of a right triangle, and they appear very often. Five squared is equal to three squared, plus four sward. Another but less common Pythagorean triple is 5 12 and 13 13. Squared, which is 1 69 is equal to 25 plus 1 44 two triangles, or similar same shape and usually different size if their corresponding angles are equal. If two triangles are similar, their corresponding size of proportional. 15. Geometry Triangles 6: if two angles of a triangle are congruent to two angles of another triangle. The triangles are similar in the figure of the large and small. Triangles are similar because both contain a right angle and they share angle. A. So the large triangle here here in here is similar to the small triangle here here in here because they're both right triangles and they both contain this. The angle a two triangles incongruent, identical if they have the same size and shape in a triangle and exterior angle is equal to the sum of its remote interior angles. So E is equal to a plus B and therefore greater than either of them, since all angles are positive and he trying all the some of the lengths of any two sides is greater than the length of the remaining side. So X plus why is greater than Z and why Plus Z is greater than acts and X plus Z will be greater than why 16. Geometry Example 3: since two X plus 60 is, an exterior angle is equal to the sum of the remote interior angles, namely X plus 90. Solving this equation, we subtract X from both sides and 60 from both sides, which gives us X equals 30. Hence, the answer is a. 17. Geometry Triangles 7: in a 30 60 90 degree triangle. The sides have the falling portions. Now you can multiply each of these numbers by acts, and the proportions will be preserved. And that will give us this drawing. Here in a 45 45 90 right triangle, The sides can be s Yes, controlling. And then the hypothesis will be route two times s. 18. Geometry Quadrilaterals 1: quarter laterals. The angle some of a quadrilateral, is 360 degrees. You can view a quadrilateral as being composed of 2 180 degree triangles. So for this quadrilateral here, you can draw a diagonal across here, and then it divides it into two triangles, each of which has 180 degrees, so the some would be 360 degrees. A parallelogram is a quadrilateral in which opposite sides or both parallel and congruent. Its area is the base times the height. So for this figure here, the area would be the base. Be times the height age. The diagrams of parallelogram bisect each other as shown in the drawing. A parallelogram with four right angles is a rectangle. If W's A with an L is the length of a rectangle than its area is L Times W and the perimeter. The distance around it is to W plus two l. So if this were a say a swimming pool and you were walking around it, you would walk l here then W than another l to give you two owls and then another w to give you two W's 19. Geometry Example 4: at the following line to the drawing. Since the legs of the right triangle formed are of lengths three and four, we must have a 345 right triangles. So this length here is five. And since this is a rectangle, the opposite side is also five. Now, the perimeter of the object is simply the sum of all the sides. So we get starting here we get four plus five plus four again plus another four. And then finally a three. Adding these numbers up, we get 20. Hence, the answer is D. 20. Geometry Quadrilaterals 2: if the opposite sides of a rectangle are equal, it is a square in its area is s squared and his perimeter is for s Where s is the length of aside the diagonals of a square bisect each other and are perpendicular to each other as shown in the figure. 21. Geometry Quadrilaterals 3: 1/4 lateral with only one pair of parallel sides is a trap is oId. The parallel sides are called bases and the non parallel sides or cold legs. So in this drawing, the horizontal lines are parallel as indicated by the little arrows, and the vertical lines are not parallel. The area of a trapezoid is the average of the bases times the height. So for this trap is oy. We add up the two bases, beasts of one and beasts up to and divide by two to form the average and then multiply by the height age. 22. Geometry Volume 1: the volume of erecting their solid A box is the product of the length, width and height the surface areas of some of the area the six faces. So our volume here is the length times the with times the height. If the length within height of erecting their solid a box are the same. It is a cube. Its volume is the cube of one of its sides, and surface area is a some of the areas of the six faces. The volume of a cylinder is pi r squared age and the lateral surface area. Excluding the top in the bottom is two pi R H, where R is the radius and H is the height. 23. Geometry Example 5: let e be the length of an edge of a cube. Now recall that the volume of a cube is e cubed. In the surface area is six e square, and we're told that the surface area is equal to the volume of the Cube. Hence, E cubed is equal to six p square. It's attracting six e squared from both sides of this equation. We get factoring out e squared, no set each factor equal to zero. So he squared equals zero and e minus six equals zero Taking the square root here we get eagle zero and adding six both sides. Here we get eagle six. Now we reject equals zero because if the length of the side was zero, we won't even have a cube. Hence e equal six. And the answer is a 24. Geometry Circles 1: a line segment from a circle to its center is a radius. A line segment of both ends. On the circle is accord. A core passing through the center of the circle is the diameter. A diamond can view be viewed as to radio I and hints. A diamonds length is twice that of a radius. A line passing through two points on a circle is the C camp. A piece of the circumference is an ark. The area bounded by the circumference at an angle with a vertex. At the center of the circle is a sector. 25. Geometry Circles 2: a tangent line to a circle intersects a circle at only one point. The radius of the circle is perpendicular to the tangent line at the point of tangent C, as showing this figure to tensions to a circle from a common exterior point of the circle are congruent. So here, lined A B is congruent to line A C because they're both external tangents. From a common point, an angle inscribed in a semi circle is a right angle. 26. Geometry Circles 3: a central angle has by definition the same measure as it's intercepted arc. For this drawing, the arc length here is 60 degrees. Therefore, the central angle is also 60 degrees by definition, and inscribed angle has 1/2 the measure of its intercepted arc. Here, the intercepted arc is 60 degrees. Therefore, the inscribed angle is 1/2 that, or 30 degrees. 27. Geometry Circles 4: the area of a circle is pi r squared and it's a conference or perimeter. Is two pi r where R is the radius on the test. Pi is equal to three. It's efficient approximation for pi you don't need pi equals 3.14 28. Geometry Example 6: these, the covers of a circle is two pi R. Since we get to pie times two did The radius of the circle is, too, which gives us four pi. Now a central angle has, by definition the same degree measure as it's intercepted arc. Hence the Ark A C B is 60 degrees, and there are 360 degrees in a circle. So the fraction of that circle that the Ark comprises is 60 divided by 3 60 or 1/6. Hence the length of arc is 1/6 of four pi, which is 2/3 of a pie. Hence, the answer is B. 29. Geometry Shaded Regions 1: shaded regions to find the area of the shaded region of a figure subtracted area of the unshaded region from the area of the entire figure. 30. Geometry Example 7: to find the area of the shaded region, subtract the area of the circle from the area of the rectangle. Now the area of the rectangle is three times five or 15 and the area of the circle is pie R squared in our is one which gives us Popeye. Hence the answer is B. 31. Geometry Example 8: since we're not given the radio of the circles the promise independent of the length of the radio I as long as they are as long as one is three times the other. Let the outer radius be three and let the inner radius B one in the area. The larger circle is pie R squared or nine pie. In the area of the smaller circle is pi times one squared or just pie. Hence, the area of the shaded region is nine pie minus pi or a pie. Now, for me, the fraction of the shaded region, which is ate pie to the area of the smaller circle, which is pie canceling the pies, we get 8/1. Hence the answer is C. 32. Geometry 'Birds Eye' View 1: bird's eye view. Most geometry problems on the test require straightforward calculations. How are some problems? Measure your insight into the basic rules of geometry. For this type of problem, you should step back and take a bird's eye view of the problem. The following example will illustrate. 33. Geometry Example 9: the diagonals of a square are equal and Sophie draw in O. R. It will equal sp o r equals SP now O R is the radius of a circle and we're told that the radius is too. So are is too. It's SP also has a length of two. Thus the answer is D. 34. Geometry Eye Balling 1: eyeballing. Surprisingly, on the test, you can often solve geometry problems by merely eyeballing the given drawing even on problems whose answers you can't get directly by looking, you can often eliminate a couple of the answer choices. All figures are drawn to scale. Insipid angle looks like it's about 90 degrees. It is. If one figure looks like it's about twice as large as another figure it is. The examples in this section were sold before. Now we will solve them by eyeballing the drawings. 35. Geometry Example 10: by eyeballing the drawing, we can see that why is less than 90 degrees? It looks like it's somewhere between 65 85. But the only answer choice offered in that range is de since the answer is D. 36. Geometry Example 11: the area of the larger triangle is 1/2 the base times the height in the basis to as is the height, which gives us two. Now, the shaded region looks to be roughly half the area of the larger triangle, so it's area should be about 1/2 of to, which is one and the closest answer. Toe one is 7/8. Hence the answer is C. 37. Geometry Problem 1: the little square tells us that we have a right triangle. Instead, Pythagorean theorem applies. That is the high pot. News six squared is equal to the sum of the squares of the legs. So why squared? Plus three. Swear in performing the operations we get 36 equals. Why square plus nine subtracting nine. We get 27 equals y squared. Taking the square root of both sides we get why equals the square root of 27. Hence the answer is B. 38. Geometry Problem 2: since the diameter of circle P is to its radius is one in this area Pie are square, is pi times one squared or just pie. And since the diameter of circle que is one, its radius is 1/2. Therefore, this area is pi r square So pi times 1/2 square which is pi over four. Now the area of the shaded region will be the difference between the larger circle and a smaller circle. So we get pie for the larger circle and pi over four for the smaller circle and getting common denominator of four, we get three force poet. Hence the answer is a 39. Geometry Problem 3: we're told that each arc is the arc of a circle with its center out of Vertex. Yes, we have four arcs of circles. Taken together, we have one full circle, and from the drawing we can tell that the length of the radius of the circle is three. It's the area of the circle, which is the area of the four arcs is pie R squared or pi times three sward, which is nine pie Now. The square is six inches, or six units on a side three plus three. So the area of the square is six squared or 36. It's the area of the square minus. The area of the circle will give us the area shaded region, which will be 36 minus nine pie. Hence the answer is C. 40. Geometry Problem 4: since the area of each circle is two pi, we get the area. The circle pyre square is equal to two pi cancer. The pies we have our squared equals two taking the square root of both sides. We get r equals radical too. So the radius of the circle is radical too Which will be this distance here, This distance here, here, in here. So the length of the side of the square will be four radical too. Now the area The square is its side square which is four radical to square and four scored a 16 and radical two squared is too. So we get 32 hence the answer is E. 41. Geometry Problem 5: Oh, yes, and o t are congruent because they're both radi i of the circle Hence angle T is also 51 degrees. Recall that the angles some of a triangle is 180 degrees. We get s Ingle s plus angle T plus angle y equals 1 80 but s is 51 degrees T s 51 degrees. Solving this equation for why we get why equals 78. And the answer is D. 42. Geometry Problem 6: let's extend these horizontal lines to make the drawing a little clearer. Notice that this is 90 degrees here because of the square and this is 90 degrees and this trans Ercel adds upto 180 degrees. The interior angles on the same side of the trans versatile as up to 80 degrees. Therefore, these two horizontal lines are in fact parallel and extending this trans versatile here we noticed that A and the angle 29 are alternate interior angles, so a is also 29 degrees and of course, B is also 90 degrees because it's a supplement to this right angle. Here, therefore be is 90 and we have a is 29 and B is 90 for a total of 1 19 Since the answer is B. 43. Geometry Problem 7: since lie Nelson. One is parallel to lie Nelson too s and X or corresponding angles and therefore congruent. Now about any point. There are 360 degrees. So we get five X plus s equals 3 60 replacing s with X Because they are congruent we get five x plus X equals 3 60 six x is 3 60 and dividing by six we get X equals 60 and the answer is C. 44. Geometry Problem 8: since O P and o que are ready I of the circle, they are congruent. Yes, we have a nice Aussies triangle There, for this angle is also 59 degrees. Now the angle Some of the triangle is 180 degrees. So angle oh plus 15 Angle P, which is 59 plus angle Q, which is also 59 at up to 1 80 Solving this equation for angle Oh, we get Anglo equals 62 degrees, So the single here is 62 degrees. Now the largest side of a triangle is opposite the largest angle. Hence, P Q is the largest side of the triangle. Therefore, it will be greater than either of the other two sides. In particular, it will be greater than o que against. The answer is a 45. Geometry Problem 9: since X is the radius of the larger circle. The area of the larger circle is pie X squared. Instant sex is the diameter of the smaller circle. 1/2 of it is the radius of the smaller circle, so the area the smaller circle is pi times X over two square or pine X squared over four. Now forming. The difference between these two will give the shaded region, which gives us pi X squared minus hi X squared over four. Getting a common denominator of four. We get four hair minus one here for a total of three pi X square over the common denominator of four. Hence, the answer is a. 46. Geometry Problem 10: Since the site of the square is six units long areas 36 equals six squared. Since the side of the square is six, the diameter of the circle is also six. Hence the radius is three. So the area of the circle is night pie, which is pied times three square. It's a practice area The circle from the area of the square, we get 36 minus nine pie. This gives the combined areas all four here, here, here, in here. But only two of those four regions are shaded. So we divide this by two. Hence the answer is C. 47. Geometry Problem 11: the length of beside the triangle R P is eight three plus five. Now, applying the Pythagorean theorem to the large triangle, we get a squared plus P s square equals tens where or 64 plus PS square equals 100. Taking the square root of both sides of this equation, we get eight. So PS is eight now, looking at this smaller right triangle here we play the Pythagorean theorem again and that gives us five squared plus eight squared equals Q s squared. Solving this equation for Q s. We get the square root of 61. Hence the answer is B. 48. Geometry Problem 12: Since angle P o que is 70 degrees, we get that angle y plus angle X plus angle 20 must add up to 70 degrees. So why plus X Plus 20 equals 70. Solving this equation for why we get why equals 50 minus X. Since we're given that X is greater than 15 we can see intuitively that this expression must be less than 35. But let's prove this. What we'll do is take this inequality here and try to create this expression out of it. So we have X is greater than 15. Multiply through by negative one because we have a negative here that will flip the direction of the inequality. And since we have a 50 here will add 50 each term and 50 minus 15 is 35. So we have proven that that the expression 50 minus acts is, in fact less than 35. Hence, the answer is B 49. Geometry Problem 13: since Lines L and K or Parallel, we have corresponding angles here which are equal. Hence why equals two y minus 75. Subtracting two wife from both sides. We get negative. Y equals 75 native 75 and multi through by negative one we get Why equals 75 inst answers d . 50. Geometry Problem 14: since the height and the base of the larger triangle are the same. The slope of the high part news is 45 degrees, so the base of the smaller triangle will be the same as the height. Find the area of the shaded region. We find the area of the larger triangle and subtract from it the area of the smaller triangle. The area, the larger triangles, 1/2 the base, which is to times the height, which is also to Nice area. The smaller triangle, which is the base when half the base, which is three halves times the height, which is also three have. And that gives us two minus 98 which is 7/8 hence. The answer is C. 51. Geometry Problem 15: since we're not giving the radio I of the circles, I don't told only that the larger circle is twice that of the smaller circle. We can choose any convenient numbers to work with, as long as one is twice the other that the radius of the smaller circle be one, and then the radius of larger circle would be to. Therefore, the area of the larger circle is pi times its radius squared, which gives us four pi in the area of the smaller circle, is going to be pie times its radius, which is one squared, and that gives us pie. Therefore, the area of the shaded region is for pied minus pi or three pie. Now we formed the racial of the shaded region, which is three piles toothy area of the smaller circle, which is pie canceling the pies. We get three over one or in roi ratio notation three colon one. Hence, the answer is C 52. Geometry Problem 16: since Triangle PST is that I saw sleaze, right? Triangle its legs air Congrats. P t is congruent to t us and let's call those lengths X just so we don't have to write so many letters. Now, applying the Pythagorean theorem to this triangle, we get X squared plus X squared equals I partners squared ps No, you're told that PS has a value of two. So we get two X squared is equal to two squared or four divided by two. We get X squared equals two or X equals radical too. Having this information to the figure, we get radical too radical too. And since this is a square, we have 111 and one. Now the area of the shaded region equals the area of triangle P S. T, which is the larger triangle minus the area of triangle are for me up pr you, which is a smaller triangle here, which gives us 1/2 radical two times radical to for the area the larger triangle and for the smaller triangle, the bases one in the highest one because this is a right angle here. So we get radical two times Radical to is too divided by two is one minus 1/2 which is 1/2 The answer is E 53. Geometry Problem 17: the area of Triangle P que s is 1/2 the base, which is five times the height, which is six gives us 15 now the area of the large triangle P Q R, which we're told us 40 minus the area of the triangle Peak U S that we just calculated will give the area of the other triangle and 40 minus 15 is 25. Hence, the answer is D. 54. Geometry Problem 18: since the figures a square and this site has length for we know that peak you has a length of four since I m is the midpoint, this length here between M and Q is too and against this is a square. We have a right angle here. So the area of this triangle is 1/2 the base, which is to times the height, which is four, which is for And a similar analysis shows that the area this triangle is also for so. The total area of the Unshaded region is 84 plus four. Now the area of the square is 16 which is four squared in subtracting the unshaded region. From the area of the square, we get 16 minus e, which is a hence. The answer is a 55. Geometry Problem 19: since the area of the circle is nine pie we get pi r squared equals nine high. Canceling the pies You get R squared equals nine or are take the square root. We get R equals three. Now the circumference of the circle is two pi R, which we've calculated to be three or six pie. Since the central angle is 30 degrees the length of the Ark P. Our Q is 30% of the number of degrees in the circle, which gives us 1/12 time. See and C. We've completed to be six pie, which gives us 1/2 pie, adding all this information to the drawing. We get pi over two and then the radius again is three. So this is three and this is three. So the perimeter of that sector is three plus three plus pi over two or six plus hi over to hence, the answer is B 56. Geometry Problem 20: the circumference of a circle is two pi Oh, are now the two and the pyre constant. So in order to solve this problem, we have to find the radius. They were told that a denotes that area of the circle, so a equals pi r squared. Solving this for our will substituted into this other equation. So the Bible says about pie and we get r squared is equal to a over pie and take the square root of both sides. We get our equals radical, a over pie, plugging that into this expression, we get to pie radical a over pie, which is answer choice E. 57. Geometry Problem 21: the straw picture depicting the situation ship. Why will be here heading in this direction and ship X will be here heading in this direction. We have a right angle here. Let the distance be between ship. Why and the point of collision B D then said Ship X is one mile closer to that point, its distance will be D minus one and originally the ships are five miles apart. So that's the hypothesis. Applying the Pythagorean theorem to this right triangle, we get five squared. The hypothesis squared equals this side squared plus this side square. So we have 25 equals D squared plus D squared minus to D plus one, combining like terms we get to d squared minus two D and bring the 25 over subtracted from from one. We get minus 24 divide you know to and let's set zero on the right side, where it's more natural. Factoring we're looking for factors of 12 whose summer difference is one and that would be four and three. The four takes the negative because the middle term is negative and three takes the positive setting. Each factory with zero we get D equals four and d equals three ***. Three. We reject Mega three cause we're dealing with distances. Hence the answer is D. 58. Geometry Problem 22: since the square has length, for we know this distance here is four. Now the radio of the circle are congruent. Hence the triangle is I saw sleaze, so it's labeled the base Ingles X. Now the angle sub of a triangle is 1 80 So x plus X plus 60 equals 1 80 and solving this equation for X, we get X equal 60. Yes, we have unequal lateral triangle. Therefore, the radius of the circle is four. Hence, the circumference of the circle is two pi r, which we've just calculated to before, which gives us a pie. Now the portion of the perimeter formed by the circle there was this region here has a length of 3 60 the total distance around the circle minus the art which is 60 degrees divided by 3 60 time. See which gives us 56 times a pie, calculated the circumference to be a pie, and that reduces to 20 thirds pi now adding the three sides of the square 44 and four To this expression, we get the total perimeter of the object. Hence the answer is D 59. Geometry Problem 23: before meal for the perimeter of a rectangle is twice the length plus twice the with and were given. The length is six him and the width is for him, which gives us 20 a. M. Now the former for the perimeter of a square is four x where X is the length of the square and we're told the prisoner of the square is equal to the perimeter of the rectangle. It's four x equals 20 a. M. Solving this equation for X we get X equals five them. Hence the answer is C. 60. Geometry Problem 24: the formula for the circumference of a circle with diameter D is two pi r. Let's put the two in the are together. Since the diameter is twice the radius. This gives his pie times D. Hence, the ratio of the circumference of the circle to the diameter, which is what we're being asked to calculate, is equal to Heidi over DE, which gives us pie. It's the answer is a 61. Geometry Problem 25: were given That anklet is 10 degrees greater than angle B, so ankle a equals angle B plus 10. And we're also given that angle B is 10 degrees greater than angle see, so Angle B is equal angle C plus Tim in the triangle. The some of the three angles is 180 degrees, so we get angle a plus angle B plus single see equals 1 80 to solve the system of three equations for solve the second equation for C, which gives us see angle see equals angle B minus 10. Now substitute thes equation for angle A and the equation for ankle see into the bottom equation. We get Angle B Plus 10 and we'll in case it parentheses, just to show that that it is a group. But there's no mathematical purpose for that. And likewise for angles, See will replace in case it parentheses as well. That's solving this equation. For Angle B. We get angle B equals 60. Hence, the answer is D 62. Geometry Problem 26: the area of a square of side s is a square. When joining to such squares, the resulting area will be twice the area of either square. So we get to s squared. Yes. The answer is B. 63. Geometry Problem 27: the path taken by the person can be represented by the following diagram. Let DB the distance between his initial position and his final location. Since a person traveling due north has to turn 90 degrees to travel Dewey's ankle, ABC is a right angle. Yes, we can apply the Pythagorean thorough, which gives D Square equals 12 square plus 16 square. Solving this equation for D we get d equals 20 against Answer is D. 64. Geometry Problem 28: Triangle Peak you are is a right triangle with base PR equal to four and height p. Q. The area of the triangle is 1/2 the base times the height, and we're told that's equal to six. Substituting the values into this equation, we get 1/2 the base, which is for times the height, which is P Q equal. Six. Solving this equation for P. Q. We get three now applying the Pythagorean theorem to the triangle. We get P Cube Square, plus PR squared equals I partners Q R Square. Substituting the values P. Q. Is three p. R. Is four. Solving this equation for q R. We get Q r equals five. Hence, the answer is he? 65. Geometry Problem 29: to find the Y intercept, which is point A. We replace X was zero in the formula. So we get why equals negative 5/3 time zero plus 10 which gives a zero plus 10 or two. It's the height of the triangle is 10. And to find the base from O to be, we replaced the y, coordinate with zero and then solve the equation. So we have zero equals negative 5/3 X plus tune. Solving this equation for X gives X equal six. So the area of the triangle is 1/2 the base, which is six times the height, which we've conflated to be 10 which gives us 30 hence, the answer is B. 66. Geometry Problem 30: first, let's add the information to the drawing. Ingle Axes 54 An angle wise 72 since single x an angle b o d are vertical angles. Angle B o. D is also 54 degrees since single. Why an angle? A e o r. Vertical angles. We know that angle A e O is also 72. Since a straight angle has a turn 80 degrees. We know that angle Z plus angles 72 plus angled 54 add up to 1 80 solely for Z we get one, we get 54 against. The answer is a 67. Geometry Problem 31: were given that one of the sides in the figure has a length of three and see their X equals three or X plus six equals three. If X plus six equals three, then subtracting six from both sides. We get X equals negative three, and this is impossible, since a length cannot be negative since we know that excess three. Now the area of a triangle is the price of two successive sides, so we have X Times X Plus six and X. We have calculated to be three, so we get three times nine, which is 27 since the answer is D. 68. Geometry Problem 32: since A B and C are interior angles of a triangle. Their angle some is 1 80 I noticed that A and why are vertical angles So they're congruent likewise for being Z and also for CNX. So replacing a with wine, be with Z and see with X. We get that X plus y plus Z equals 1 80 hence the answer is C. 69. Geometry Problem 33: in a triangle. The some of the interior angles is 180 degrees. Applying this to Triangle a D. C. We get angle. A plus 45 plus 90 equals 180 solving for angle A. We get 45 triangle. ABC is I saw sleaze because to the sides have length 10. Hence the base angles be and C are congruent. We've already arrived. That angle see, is 45 degrees, so Angle B is also 45 degrees now for a triangle. ABC. The interior angles also add up to 1 80 so we get angle a plus angle B, which is 45 plus single C, which is also 45 at upto 1 80 Solving this equation for angle A. We get 90 degrees in strangle. ABC is a right triangle with right angle at a. Therefore, the area can be calculated by taking 1/2 the base. Either side could be taken to be the base. Either of the two congruent sides could be taken as a base, and either of the two congruent sides can be taken. Is the height and come complete this expression we get 50. Hence, the answer is B 70. Geometry Problem 34: first note that Angle B is 40 plus 40 or 80 and the English some of a triangle is 180 degrees. So angle A, which is 50 plus angle B, which we calculated to be 80 plus angle. X equals 1 80 Solving this for X we get backs equals 50 since the answer is D. 71. Geometry Problem 35: Let's at an altitude to Triangle ABC by extending side BC as shown in the following figure . Now the formula for the area of a triangle is 1/2 the base times the height. Hence the area of the triangle ABC is 1/2 bc times a F and length BC is two plus one or three plugging that into the formula. We get 1/2 times three times a F, and this equals 30 because we're told that area of Triangle ABC is 30. Solving this equation for a F gives a F equals 20. Now the area of Triangle A. D. C is equal to 1/2 D C. Times A F, which gives us 1/2 times one from the drawing D. C is one times a F, which we've calculated to be 20 which gives us 10. Hence, the answer is B 72. Geometry Problem 36: notice that angle. Why plus 30 is an extra angle hence by the exterior angle. Thuram. It's equal to the sum of this remote interior angles, which is why I minus 15 and why plus 15. Adding up like terms. The fifteens cancel we get too wide equals. Why plus 30 subtracting wife on both sides. We get 30 equals. Why, it's the answer C. 73. Geometry Problem 37: The figure shows that the circle is located between horizontal lines. Why equals four? And why equals negative floor in the circle of symmetric about the X axis? From this, we make two observations. The center of the circle is on the X axis, and the diameter of the circle is the distance between the two lines, which is a so it's the center of the circle is on the X axis. The 0.20 and the point X zero are diametrically opposite points on the circle that is, they are in points of a diameter of this circle. Hence the distance between them, which is X minus two, must equal eight, which is the length of the diameter. Adding to to both sides of this equation, we get X equals tip. Hence the answer is D. 74. Geometry Problem 38: We're told that the ratio of extra why is to so X divided by why is to most by both sides of this equation by why we get X equals two now in a straight line. There are 180 degrees. So we get why plus X plus why equals 1 80 replacing the X here with too wide we get wide plus two y plus why equals 1 80 or for wine equals 1 80 Dividing my four we get Why equals 45 and stance er is de. 75. Coordinate Geometry Introduction 1: on a number line. The numbers increase in size to the right and decrease in size to the left. So, for example, negative five is smaller than negative four because negative five is too left of negative four on the number line. 76. Coordinate Geometry Introduction 2: if we draw a line through the 0.0 perpendicular to the number line. In other words, this line here we'll form a grid. The fix horizontal line in the bubble diagram is called the X axis and the thick vertical line again. This line here is the is called the Y axis, the point at which the axes meat 00 is called the origin on the X axis. Positive numbers are to the right of the origin and increase in size to the right, so larger this way. Further negative numbers are to the left of the origin and decrease in size to the left on the Y axis. Positive numbers are above the origin and ascend in size. Further, negative numbers are below the origin and decent in size, so these numbers get smaller as you go down further and further. 77. Coordinate Geometry Introduction 3: as shown in the diagram. The point represented by the ordered pair X Y, is reached by moving X units along the X axis from the origin and then moving. Why units vertically in the ordered pair, X Y X is called the obsessive and why is called the ordinate? Collectively, they are called the coordinates. The X and Y axes divide the plane into four quadrants, numbered one to three and four. 78. Coordinate Geometry Introduction 4: note. If X is not equal, why, then x y and why acts represent different points on the coordinate system. The points to common three Mega three comma one negative for common negative for and for common negative who are plotted in the falling coordinate system To get to the point to three from the origin you count over one two for the ex change and then up three for the white change. 79. Coordinate Geometry Example 1: since the white cornet of Point B is four line segment A B is also for, since the figure is a square, the distance from a toe Oh is also for now be is in the second quadrant. It's the X coordinate will be negative four. And the answer is D. I will be careful not to choose a because exes theme coordinate of point B. It is not the distance from the Y axis to be. 80. Coordinate Geometry Distance Formula: distance formula. The distance formula is derived by using the Pythagorean theorem north. In the figure below that, the distance between points X Y and A B is the high part news of a right triangle. The difference? Why minus B is the measure of the height of the triangle in the difference, X minus A is the measure of the base of the triangle. Applying the Pythagorean theorem we get. The high Patna squared is equal to the sum of the lengths of the legs where take the square root. Herbal size of this equation gives us the distance formula. 81. Coordinate Geometry Example 2: Since the circle is centered at the origin, it passes through the 0.0 negative three. The radius of the circle is three now, if any other point is on the circle the distance from that point to the center of the circle, the radius must also be three. Look at choice. Be using the distance formula to calculate the distance between the point B, and the origin will give notice that the distance in this expression is three. Hence, the point is on the circle and the answer is B. 82. Coordinate Geometry Midpoint Formula: midpoint formula. The midpoint M between points, X y and A B is given by the following formula. In other words, to find the midpoint simply average the corresponding coordinates. Of the two points. Here we have the average of the X coordinates, and here we have the average of the white cordons. 83. Coordinate Geometry Example 3: since Point are is on the X axis, it's why corn it is zero further, since the figure is a square and the ex accorded of Q is too. The ex accorded of our is to and the X coordinate of tea is too now. T is the midpoint of this side, and the site has a length of two. Therefore, by sex the side into two equal plains of one each, it's the white corner of tea is one, and the answer is D. 84. Coordinate Geometry Slope Formula: the slope of the line measures the inclination of the line. By definition is the ratio of the vertical change to the horizontal change, as shown in this figure. Here we have a vertical change of why minus B because it's a difference between the white coordinates, Why and B and the exchange or distance is X minus. A. Because it's a difference between the X coordinates, which are ex in a forming. The rise over the run gives us why minus B over X minus a. 85. Coordinate Geometry Example 4: using the slope formula, we get n equals. That difference in the UAE values four minus two over the difference in the X values five minus one, which gives us 2/4, which cancels the 1/2. Hence the answer is C. 86. Coordinate Geometry Slope Intercept Formula: slope intercept form well, point both sides of the equation by X minus a yield. This equation here Now, if the line passes through the Y axis at zero B, then the equation becomes by replacing a with Cyril and keeping be as is and then dropping this girl. We get this and then finally adding, Be to both sides, we get, we get why equals MX plus B. This is called a slope intercept form of the equation of a line where M is a slope and B is the Y intercept. This form is convenient because it displays the two most important bits of information about a line. It's slope, and it's why intercept? 87. Coordinate Geometry Example 5: since the equation isn't slope intercept form, that is why equals MX plus B form. We know that the slope him is 9/10 now the slope is the rise over the run and a rise is B O . And the run is a Oh, so we have b o over a Oh, which again we know is nine tails now multiplying both sides of this equation by a oh, we get b o equals nine tense of a Oh, and this says that a o is larger than B O. Hence the answer is a 88. Coordinate Geometry Intercepts 1: intercepts The X intercept right here is the point where the line crosses the X axis. It is found by setting y equals zero and solving the resulting equation. The Y intercept here is the point where the line crosses the Y axis. It is found by setting X equals zero and solving the resulting equation. 89. Coordinate Geometry Intercepts 2: Let's graph the equation. X minus two y equals four by finding the Excel My intercepts to find the X intercept replace why was zero? And that gives us X minus two times zero or X equals four right here. So the X intercept is 40 to find the Y intercepts at X equals zero in the equation, which gives us why equals negative two. So the Y intercept is at zero negative, too, and then conduct these points with a straight line. 90. Coordinate Geometry Areas and Perimeters: areas and perimeters. Often you will be given a geometric figure drawn on a coordinate system and will be asked to find its area or perimeter. In these problems, use the property to the coordinate system to deduce the dimensions of the figure and then calculate the area or perimeter. For complicated figures, you may need to divide the figure into simpler forms. Such a squares and triangles, a couple examples will illustrate. 91. Coordinate Geometry Example 6: If the quadrilateral is divided horizontally through the line, Y equals two to congruent. Triangles are formed as the figure shows the top triangle has height to and base four. Hence its area is 1/2 base. Times height equals 1/2 times four times two, which gives us four. The area the bottom triangle is the same. So the area of the hold quadrilateral is four plus four, which equals C inst answers d. 92. Coordinate Geometry Example 7: point A has co ordinates 04 point B has coordinates three zero and point C has coordinates five one using the distance formula to calculate the distances between points A, B, A, C and B C yields. Now, adding these lengths gives the perimeter of the triangle so we get a B plus A C plus B c equals five plus radical 34 plus radical five. And now notice that this is answer choice a. 93. Coordinate Geometry Problem 1: since the circle is centered at the origin and passes through the point negative three comma zero. The radius of the circle is three. Hence the area of the circle, which is pi R Square, equals pi times three squared or nine pie instead answers E. 94. Coordinate Geometry Problem 2: whatever the coordinates of Point p r. The line Opie is the hypothesis of a right triangle, with the sides being the absolute value of the X and Y. Coordinates in Soapy is greater than the white corner of Point P and answers a this label the point P X y, and drop a perpendicular to the X axis. That is the wide distance. And this will be the X distance here, since Opie again is the high pot news of a right triangle, it's greater than either X or a while. This problem brings up an issue of how much you can assume when viewing a diagram. We're told that he is appointing the coordinate system and that it appears in the second quadrant. Could PB on one of the axes or in another quadrant? No. Although P could be anywhere in Quadrant two, it could be over here, not necessarily where it's displayed. He could not be on the Y axis because the position of points angles, regions etcetera can be assumed to be in the order shown. If people are on the y axis that it would not be to the left of the Y axis, as it is in the diagram, that is, the order would be different 95. Coordinate Geometry Problem 3: since the point b zero b comma zero is the X intercept of the line. It must satisfy the given equation. That is, when X is be the wind must be zero. So we get the equation. Zero equals PB plus A. It's attracting a from both sides and then dividing by B the bees cancel and get RP is equal to negative a over B. Hence, the answer is C. 96. Coordinate Geometry Problem 4: since the line passes through Negative four negative five and the origin, which is 00 it's slope is the rise over the run or the change. And why over the change in X Delta stands for change and we get negative five minus zero over negative four minus zero, which is negative, which is positive. Five. Force. This shows that the change in the Y value five is always greater than the change in the X value, which is four. Hence the white corn. It is greater than the X coordinate, and the answer is B. 97. Coordinate Geometry Problem 5: in Quadrant two. All Ex coordinates are negative in all white coordinates or positive. Hence the X coordinate. A point P is negative. Therefore, A is negative. And in the fourth quadrant, all X coordinates or positive and all white coordinates or negative. Hence the Y coordinate a point. Q is negative, which is B hits the point. A B is a negative and the negative, and that occurs only in Quadrant three. Hence, the answer is C. 98. Coordinate Geometry Problem 6: let's write the equation of line using slope intercept form. That is why equals Imex plus B. Now, since the light passes through the origin, it goes through the 0.0 So the wider set the B is Earl and we get why equals MX. Now we just have to calculate the slope going back to the line. We knows that goes through the point to one and 00 So the slope is the change in the UAE over the change in the axe. The change of the Y is one minus zero, and the change in the X is two minus zero, which gives 1/2. So the equation becomes, Why equals 1/2 acts and we're told that X is four. Substituting that end, we get 1/2 of four, which is to 99. Coordinate Geometry Problem 7: the shaded region lies entirely within the third quadrant. Now, both coordinates of any point in the third quadrant are negative, and the only answer choice with both quadrants negative is Choice D or for the answer is D . 100. Coordinate Geometry Problem 8: for a point to be within a circle its distance from the center of the circle and must be less than the radius of the circle. The distance between 68 and the origin 00 is the radius of the circle. Using the distance formula, we get 10. Hence, the radius of the circle is 10 not looking. Answer. Choice Be its distance from the origin is the difference in the excess squared, plus the difference in the Y squared, which gives us the square root of 49 plus 49 or the square root of 98. Now, since 98 is smaller than 100 and the square root of 100 is 10 the square root of 98 is less than 10 hence the point native seven Common seven is within the circle, and the answer is B. 101. Coordinate Geometry Problem 9: Let's divide the polygon into triangles and squares by drawing two vertical lines here in here. Now the area of this first triangle is 1/2 the base times the height, which is 1/2 the base, which is 12 times the height, which is also to which gives us two. The area of the square in the middle here has a length of two on a side. Hence, its area is two squared or four in the last triangle has an area of 1/2 the base, which is only one unit times the height, which again is to. So we get one. Adding up the three numbers 12 and four. We get seven, since the answer is a 102. Coordinate Geometry Problem 10: from the distance formula. The distance between 0.41 and cue is radical, too. We can also get this result by used interpreting this as a right triangle here, with the distance being the high pot news and the legs having length one. So C squared is equal to one squared plus one squared or C squared is equal to two, so C is equal to radical, too. Now, using the distance formula for the distance between point P and 0.0.41 we get the square root of four minus one squared, plus one minus four squared, which is the square root of three squared plus three squared or the square root of two times three squared, which is three square root of two. Hence, the answer is D. 103. Coordinate Geometry Problem 11: dropping a vertical line from Point B perpendicular to the X axis will form a square of side, too, in a triangle with side one. Because this quarter, according here, is to in the court in here is three and their differences one. So the area of the triangle is 1/2 the base, which is one times the height, which is to which gives is one in the area of the square is the side squared, So we get two squared or four. It's the total area is four plus one or five, and the answer is B. 104. Coordinate Geometry Problem 12: since the product of the two numbers is negative, the numbers must have opposite signs that is, one must be positive and the other negative and that occurs in Quadrant two. We have a negative and a positive and occurs in Quadrant four, a positive and negative. It does not occur in Quadrant one because both are positive and it doesn't occur in Quadrant three because both are negative. Yes, we have two quadrants in which is true. Therefore, the answer is D. 105. Coordinate Geometry Problem 13: calculated the distance between V and the origin. We get the square root, that difference of the exes, plus the difference in the wise. So we get to root. Two. Since the square is rotated about the origin in a clockwise direction, this is between the origin and the point V is fixed. It's the new white quarter. V is negative to root. Two. The following diagram shows the final position after the rotation. Hence the answer is B. 106. Coordinate Geometry Problem 14: using the distance formula to calculate the distance of each point from the origin yields. Now just noticed that Radical 18 is the largest number listed. Hence the answer is B. 107. Coordinate Geometry Problem 15: point A has coordinates zero to point B has coordinates to zero, and Point C has coordinates. 12345 and 1234 So, according to 54 using the distance formula to calculate the distances between points A, B, A, C and B C yields heading these lanes gives the perimeter of the Triangle ABC. So we get a B plus A C plus B C equals too radical, too, plus radical 29 plus five. Instead, answer is B. 108. Elimination Strategies 1: on hard problems if you're asked to find the least or greatest number than eliminate the least or greatest answer choice. This rule also applies to easy and medium problems. When people guess on these problems, they most often choose either the least or the greatest number. But if the least or the greatest number were the answer, most people would answer the problem correctly, and therefore it would not be a hard problem. For example, suppose we were given a problem that asked for the maximum number of points common to the intersection of a square in a triangle. If no two sides coincide, then using above rule, we couldn't immediately eliminate nine the greatest number listed. 109. Elimination Strategies 2: on hard problems. Eliminate the answer. Choice. Not enough information. When people cannot solve a problem, they most often choose the answer choice, not enough information, but if this were the answer, then it would not be a hard problem. 110. Elimination Strategies 3: on hard problems. Eliminate answer choices that merely repeat numbers from the problem. For example, suppose were given the fault following problem we could immediately eliminate. Answer choice Be because eight is simply repeated from the problem, and by the second strategy, we would also eliminate e not enough information. Be careful if choice be contained. More than just the number eight supposed contain a plus radical, too, that it would not be eliminated by the above rule. 111. Elimination Strategies 4: on heart problems eliminate answer choices that can be derived from elementary operations. Suppose were given the following problem. We could immediately eliminate 24 because it's merely the product of eight and three that appear in the drawing. Further using strategy to weaken, eliminate choice e not enough information and know that 12 was offered as an answer choice because some people will interpret the drawing as a rectangle tilted halfway on its side and therefore expected to have 1/2 its original area. 112. Elimination Strategies 5: after you have eliminated as many answer choices as you can choose from the more complicated or more unusual answer choices remaining. For example, suppose you were given the following answer choices. In this case, Choices A or B are by far the most complicated, so if you could not solve the problem, you would choose either A or B. 113. Elimination Strategies 6: We have been discussing hard problems but have not mentioned how to identify a heart problem. Most of the time, we haven't intuitive feel for whether a problem is harder, easy but on tricky problems, problems that appear easy but are actually hard. Our intuition can fail us on the test. Your first question will be of medium difficulty. If you answer correctly, the next question will be a little harder if you again answer it correctly, the next questionably harder still and so on. If your math skills are strong and you're not making any mistakes, you should reach the medium part or hard problems by about the fifth problem. Although this is not very precise, it could be quite helpful. Once you pass the fifth question, you should be alert to any subtleties and senior only simple problems. 114. Elimination Strategies Problem 1: clearly there are more than 43 by three squares in the checkerboard. Hence we eliminate a next eliminate be, since it merely repeats a number from the problem further limited ee. Since it's the greatest, this leads choices C and D. If you count carefully, you will find 16 3 by three squares in the checkerboard, since the answer is D. 115. Elimination Strategies Problem 2: Since we're given the greatest value of em, we eliminate e also limb 85 because it is repeated from the problem. Now, since we're looking for the largest number, start with the greatest number remaining and work towards the smallest number. The first number that works will be the answer to that and let em equal three. And our expression becomes recall that P is the 1st 5 positive. Integer soapy is one to three four five, divided by 10 to the third power for choice. See, this reduces to 1 20 over 1000 which reduces further to 3/25 which is not an integer. So we eliminate C next, turning the choice be we get and squared on the bottom in the same product on the top one times two times three times four times five, which gives us 1 20 over 100 which reduces the 6/5. And still that is not an integer. So let maybe therefore by the process of elimination. The answer is a 116. Elimination Strategies Problem 3: $20 is too large since the discount was only 20%. Since we eliminate choice A. Both D and E are impossible since they are less than the selling price. Since we live in eight DNE number 12 choice See is the eye catcher, since 20% of 10 is too and 10 plus two equals 12. But that would be too easy for this heart problem, its elimination that the answer is B. 117. Elimination Strategies Problem 4: we can eliminate 50 the mirror average of 40 and 30 since that would be to elementary. Now the average must be closer to 40 than to 60 because the car travels for longer time at 40 MPH. But 48 is the only number given that is closer to 40 than to 60. Hence, the answer is A. We could also solve this problem mathematically. We call that the average speed is equal to the total distance divided by the total time. Now a car traveling at 40 MPH will cover 120 miles in 33 hours in a car traveling at 60 MPH , we'll cover the same one heard 20 miles in two hours, so the total traveling time is five hours, so we get total distance. Each weighs 120 and the total time is two plus three. And this reduces the 48. So again, the answer is a 118. Elimination Strategies Problem 5: we eliminate a because it repeats a number in the problem, namely 10% we could eliminate. Be because it's derive herbal from an elementary operation 30 minus 10 and Choice de similarly can be DR from Elementary Operation 30 plus 10 and Choice E can be written as or derived as 10 times 10. And this leaves only choice. See, so the answer is C. But let's also check this problem directly. The claws W is 10%. Lesson X can be written as w equals X minus 10% which is 100.10 x and this simplifies to one minus 10.1, which is 0.9 x next. The clause why is 30% less than Z can be written as why equals Z minus 0.3 z and one z minus 10.3 z is 0.70. Now multiply in these two equations here in here we get w wine equals 0.9 x times 0.7 z, which is 0.63 xsi, which in turn could be written as xz minus 0.37 Exit E. And this shows that w Y is 37% less than Exit E 119. Elimination Strategies Problem 6: Since we are looking for the greatest number of spaces from which not all eight moves are possible, we can eliminate the greatest number, which is 56 now. Clearly not all eight moves are possible from the outer squares, and there are 28 outer squares, not 32. Also, not all eight moves are possible from the next two outer squares, and there are 20 of them, not 24. All eight moves are possible from the remaining squares, since the answer is 28 plus 20 equals 48 which is Choice D. Notice that 56 which can be written as 32 plus 24 is given as an answer choice to catch those who don't add carefully. 120. Elimination Strategies Problem 7: Clearly there are more than three color combinations possible. This eliminates choices A and B. We can also eliminate C and E because they are both multiples of three. And that would be too ordinary, too easy to be the answer. It's by process of elimination. The answer is D. 121. Elimination Strategies Problem 8: first, let's calculate the expression to to the fourth. Power is 16 and 16. Squared is 2 56 minus one gives us 2 55 since the question asked for the greatest prime factor we can eliminate. 19. The greatest never given. Now we start with the next largest number and work our way up the list. The first number that divides into 2 55 evenly will be the answer. So starting with 17 we divided into 2 55 goes in once and subtracted 17 from 25. We get eight, bringing down the five we get 85 17 goes in 85 five times hence 17 is the largest prime factor of the number. 122. Elimination Strategies Problem 9: Since this is a hard problem, we can eliminate choice e not enough information. And because choice see is too easily derive eight equals four plus four. We can eliminate it for the week. Eliminate choice A. Because enter choices B and D Former more complicated set. At this stage, we cannot apply anymore elimination rules. So if we could not solve the problem, would guess either B or D the answer. This problem is actually be. 123. Elimination Strategies Problem 10: since the number five is merely repeated from the problem we can eliminate. Be further. Since this is a hard problem, we can eliminate e not enough information now, since five is prime, that's only factors are one and five itself, so see must have a value of five. Multiplying this expression out by the foil method, we get X squared plus five X plus one X plus five, combining like terms we get X squared plus six X plus five and notice six is in case position. Therefore, the answer is C k must equal six. 124. Inequalities Introduction: inequalities are manipulated. Algebraic lee the same way as equations, with one exception. Multiplying or dividing both sides of an inequality by a negative number reverses the direction of the inequality. That is, if X is greater than why and C is less than zero than see Times X is less than C times. Why, for example, see was the negative to then we would get negative, too. X will now be less than negative to why and to solve this inequality. Here we treat it like an equation. We would subtract six x from both sides and for X minus six X is two acts and hear the six X's cancel. Next, we would subtract three from both sides. The threes cancel and we get negative 11 and then finally dividing both sides by negative, too. We would reverse the direction of the inequality 125. Inequalities Positive and Negative Numbers 1: a number greater than zero is positive on a number line. The positive numbers are to the right of zero. A number less than zero is negative, and on the number line negative numbers are to the left of zero. Zero is the only number that is neither positive nor negative. It devised the two sets of numbers on the number line numbers increased to the right and decrease to the left. The expression X greater than why means X is greater than why or, in other words, excess to the right of why. Usually we have no trouble. Determine which of two numbers is larger when both are positive or one is positive and the other negative. For example, five s greater than two and 3.1 is greater than negative, too, because all positive numbers are greater than all negative numbers. However, sometimes we hesitate when both numbers are negative. For example, negative two is greater than negative four. When in doubt, think of the number line. If one numbers to the right of the other number, then it is larger, as the number line below illustrates negative. Two is to the right of negative four 0.5. Instinctive to is larger than native, 4.5 126. Inequalities Positive and Negative Numbers 2: the product or quotient of positive numbers is positive. The product or quotient of a positive number and a negative number is negative. For example, four divided by negative, too, is negative. Two. The product, or quotient of an even number of negative numbers is positive. For example, negative three times negative, too times a negative one times negative four is a product of four negatives, which is an even number. Negatives. Hence the result is positive. The product quotient of an odd number of negative numbers is negative. For example, negative one times a negative one over negative, too, is negative 1/2 because there's an odd number three of negatives, some of negative any number of negative numbers is negative, and a number raised to an even exponents is greater than or equal to zero. For example, negative too raised to an even exponents like such as four gives us 16 and 16 is greater than zero 127. Inequalities Example 1: since a number raised to an even exponents is greater than or equal to zero. We know that why squared is positive. It cannot be zero because then this product here would be zero, which would contradict this inequality. Yes, we can divide both sides of the inequality by why square And that gives us X Times e is less than zero, which is statement one which eliminates a C and E. Now Stephen to is not necessarily true because the following will equals negative 12 and the Z in this case is the three and three is not less than zero in statement, too is false or not necessarily true, which eliminates de. Therefore, the answer is B. 128. Inequalities Absolute Value: the absolute value of a number is its distance on the number line from zero. Since distance is a positive number, absolute value of a number is positive. For example, the distance between three and zero is three, and the distance between *** three and zero is also three. Students rarely struggle with the absolute value of numbers. If the number is negative, simply make it positive, and if it is already positive, leave it as is. For example, since negative 2.4 is negative, the absolute value of Nega 2.4 is positive 2.4, and since 5.1 is positive, the absolute value of 5.1 is simply 5.1 Further, students rarely struggle with absolute value of positive variables if the variables positive simply dropped the absolute value. For example, if we're given that X is greater than zero, then the absolute value of X is simply X. However, negative variables can cause students much consternation if X is negative in the absolute value of X equals negative acts. This often conceives his students because the absolute value should be positive. But the native acts appears to be negative is actually positive. It is the negative of a negative number which is positive. To see this more clearly, let X equals negative K where K is a positive number, then X is a negative number. So the absolute value of X is negative acts and then replace X with negative K. But we have taken Kate to be positive. It's the absolute value. Is positive. Another way to view this is the absolute value of X equals and negative X, which can be written as negative one times acts which is negative one times a negative number because X itself is negative and that gives us a positive number. 129. Inequalities Example 2: statement. One could be true because it does satisfy the equation. Negative absolute value of zero is equal to negative positive zero dropping that 2000. We put a plus there, and a *** times a positive is the negatives. We get negative zero and negative zero is just zero. So it does satisfy the equation. Statement to could be true because the right side of the equation is always negative. For example, negative absolute value of X is equal to the negative of a positive number, which is a negative number now. If one side of an equation is always negative, then the other side must always be negative. Otherwise, the opposite sides of the equation would not be equal, since statement three is the opposite of statement to it Must be false. It's the answer is D. 130. Inequalities Higher Order Inequalities 1: these inequalities have variables whose exponents are greater than one. For example, X squared plus four is less than two, and X cubed minus nine is greater than zero. The number line is often helpful in solving these types of inequalities that saw this inequality here with by using a number line first replaced the inequality symbol with an equal symbol and then solve this equation, being bringing this six X in the five over to the left side of the equation, and then the factors of five or five and one in there. Some six. What factors into this here? Using our zero product property, we set each factor equal to zero, and then we get X equals and negative five and negative one. Now the only numbers at which the expression can change. Sign or negative five and negative one. So negative five and negative one. Divide the number line into three intervals, interval one interval to an interval. Three. Now we merely choose a number on each interval to see whether it works or not. For Interval one, let's choose negative six, and then the original inequality appear becomes 36 greater than 31. That's a true statement there is therefore, everything smaller than negative five will make it true. Now this choose a number in the interval to let's choose negative three. Then the inequality becomes nine greater than 13 and this is false. It's no number and interval tuba work. Now this choose a number in Interval three. Let's choose zero cause it's easy to calculate with. Then the original inequality become. Zero is greater than negative five and this is true. It's all numbers bigger than negative. One will work and we get the following graph of our inequality. Note. If the original equality had included greater than or equal symbol, then the solution would have included the negative five, the negative one and we would have filled in the circles to indicate that. 131. Inequalities Higher Order Inequalities 2: summary a steps or solving higher order inequalities. First replaced inequality symbol with the equal symbol. Then move all terms to one side of the equation, usually the left side, then factor the equation and set the factors equal to zero. To find the zeros. Choose test points on either side of those zeros. If a test point satisfies the inequality that all numbers in that interval satisfy the inequality. Similarly, if a test point does not satisfy the inequality, then no numbers in that interval satisfied the inequality. 132. Inequalities Transitive Property: the transit property says that if X is less than why and why is in turn less than Z than exes, Lessons E. And this could be written more compactly as X less than why and why in turn less than Z, then X will be Lessons e. For example, one is less than two, which is less than pie. Therefore, one is less than pie. 133. Inequalities Example 3: Our goal here is to manipulate the given inequality so that it will look like one of the given answer choices. Now one over Q is given to be greater than one, and we know that one is greater than zero. So by the transitive property, we know that one over Q is greater than zero. It's que itself is greater than zero. So the smoke by both sides of the given inequality by Q, which will not change direction of the inequality because we just established that Q is positive cancer in the queue we get one is greater than Q but notice it's not offered as any of the answer choices but knows all the answer choices. Have a square in them. So this multiply both sides of this inequality by Q to create a square and we get que is greater than Q Square, which again does not change the direction of the inequality. Now, combining these two inequalities we get one is greater than Q, which again is greater than Q Square. So by the transitive property we have, one is greater than Q squared. And now just notice that that is answer choice. See 134. Inequalities Like Inequalities Can Be Added: like inequalities can be added. And usually when we do, we align the inequalities vertically. So x less than why and w less than Z. And then we just add the the columns X plus w will still be less than why plus city. 135. Inequalities Example4: notice that each answer choice involves the expression X minus y. So let's multiply this inequality by negative one to introduce the negative y, and then we have to flip the directions that inequalities. Now we can write this any quality more naturally, with the smaller number on the left, in the larger number on the right, and adding this to the other inequality we get negative. Three is less than X minus wine, which is less than two. And this is what statement a states. Hence, the answer is a. 136. Inequalities Problem 1: let's work on the right side of the Inequality. X is less than why. Now we're told that both X and Y are greater than one. So both are positive. So we can divide both sides of this inequality by X. And we could also divide both sides of this inequality by why combining these inequalities weaken right. That why over acts which is greater than one as a double inequality now by the transitive property, we know that X over Y is less than why over X, hence the answer is B. 137. Inequalities Problem 2: notice that the all the answer choices have X minus y. So let's multiply this inequality by negative one to create the negative. Why? So we get negative. Three is now greater than negative. Why which in turn is greater than negative seven. Now let's rewrite this inequality in a more natural form, with the smaller number negative seven on the left and the larger number negative three on the right. We won't change anything mathematically, just rearranging the order so we have negative. Seven is less than negative. Why? Which is less than native three. And let's combine that with the first inequality. Adding these inequalities we get negative. 10 less than X minus y less than negative four. Now divided. Return by two to create the X minus y over to term in the middle and the left side becomes a negative five. And the right side becomes a negative, too, and then just noticed that this is answer choice. A. So the answer is a 138. Inequalities Problem 3: since Why is less than nine and why equals they give to X minus eight. We get now at eight. Each side that we get minus two X is less than 17 and divide that by negative two. We get X is now greater than negative 17 has, which is negative 8.5. Now we're asked for the least value of X, for which wise less than nine. And since X is an integer, the first image, or bigger than negative 8.5, will be the answer, and that is negative. Eight. So the answer is B. 139. Inequalities Problem 4: Let's solve the inequality. Start by multiplying both sides by negative to to clear the fraction. Now this will flip the directions that inequality, because we multiply by a negative number, now distribute the negative, too. Now subtract four X from both sides of inequality, and we get eight acts and add six. Both sides make it eight, then divide both sides by eight and we get X is greater than or equal to one. And there's will only one inequality that matches that, and that's choice D. 140. Inequalities Problem 5: Since we're not told the length of the segments in the problem, the problem must be independent of the lanes. So let's choose a convenient number for the length of segment A. D sledded equal length of four. So the distance from here to here is four and M one is the midpoint. Therefore, each of these distances is too. Now I m two is the midpoint between M one and D. So that cuts these segments here and here in tow. Length one Now from the drawing, the length between em, someone and D is to and the length from A to M sub two from here to here is three. Therefore, the ratio is to over three, which is choice be 141. Inequalities Problem 6: since both X and Y are less than negative one both x and wire. Negative. Now the sum of two negative numbers X plus y is negative and the quotient of two negative numbers. Why over acts is positive. Negative, divided by a negative is a positive just as a negative times a negative is a positive. Hence why over X is greater than the negative X plus y and the answer is B. 142. Inequalities Problem 7: to solve inequality will set up a eggs Ilary equation to find the critical numbers, the numbers at which the inequality could change. Sign we get X squared equals two X subtracting two extra. Both sides kisses factory down in X and now set. Each factor individually equals zero, so we get X equals zero and two. So let's draw a number line with those critical numbers on it and take and take test points in each interval. Let's look at negative one, one and three plugging negative one in tow. Original inequality we get one is less than negative, too. And that's a false statement. So no number smaller than zero will work for one we get. And this is a true statement. One is less than two, so everything between zero and two will make the inequality true. And let's checked interval bigger than two plugging in three weeks. Nine less than six and this is a false statement. Hence, the answer is B 143. Inequalities Problem 8: since access to the left of zero. We know it is negative. And since why is to the right? Is there a. We know it is positive now the product of a negative number or the quotient of a negative number and a positive number is negative. Therefore, statement one is false and statement to is true regarding statement three. We know that Why? Because it's positive is greater than X because it's negative. So x minus y will be less than zero not greater than zero. So statement three s falls and the answer is B. 144. Inequalities Problem 9: Since X is raised to an even exponents, namely four. It is greater than or equal to zero further since X to the fourth times. Why is less than zero? We know it cannot equal zero. Yes, we know that neither X nor Y is equal zero. Otherwise, this expression would equal zero. Hence, we can divide both sides of inequality by X to the fourth and cancelling the terms we get. Why is less than zero cause anything divided into zero is zero. Likewise weaken Divide the second inequality by y to the fourth cancelling the white of force, we get X greater than zero. Since X is greater than zero, it is positive. And since why is less than zero, it is negative. And since all positive numbers are bigger than all negative numbers, we know that X is bigger than why. Hence, the answer is a 145. Inequalities Problem 10: Let's convert the right side of this inequality into a fraction. 0.1 is the same as one over 100 not a clear fractures both by both sides of this inequality by three to the end times 100 and that gives us 100 is less than three to the end. Now we just plug in the answer choices, starting with the smallest until we find one that works, and that will be our answer. For two. We get nine and nine is not greater than 100. For three, we get three cute, which is 27 again. It's not bigger than 100. For four, we get three to the fourth, which is 81. Still, not bigger than 100 and three to the fifth is to 43 and tooth 43 is bigger than 100. There for the answer is D 146. Inequalities Problem 11: translating the claws into an inequality We get. Some of the numbers divided by the number of numbers is the average, and we're told that's greater than or equal to eight and less than or equal to 12. Now we merrily saw this equation. Inequality for in multiplying through by three gives us, then subtracting 24 from each term. We saw that equality and the inequality states that the least possible value of N is zero. Hence, the answer is C. 147. Inequalities Problem 12: Let's manipulate this inequality. So it looks like this expression here so we can compare it to five. Since it's the X is being multiplied by negative. Three will multiply through my negative three, and that gives us three greater than *** three acts, which is greater than or equal to native six than add to each term. And that gives us five is greater than Tu minus three acts, which is greater than or equal to negative four. Now this flip this inequality over. We won't change anything mathematically, but we'll just write it in a more natural way, with a smaller number on the left and the larger number on the right. So we have two minus three X is less than five. Hence, the answer is B. 148. Inequalities Problem 13: first the spring. The negative up in the expression. Distribute the negative we get why minus X over Z. To make this expression a smallest possible, we need to make both wide minus X and Z as small as possible to make y minus sex. A smallest possible Let Weyco one at X equal 19. Then why minus X is equal to one minus 19 which is negative. 18. With these choices for why and exa smallest remaining value for Z is too. This gives. In this case, we made the numerator of smallest possible. Now let's make the denominator a smallest possible to that end. Choosy to be one. Why to be two and X equal to 19. Then we get negative 6 17 hence the answer is B. 149. Inequalities Problem 14: since X is greater than zero, the absolute value of X is just acts. So the equation becomes X equals one over X, multiplying both sides. By axe, we get X squared is equal to one. Taking the square root of both sides of this equation. We get X equals plus or minus one. And again, since we're given, the X is greater than zero. We take the positive route, which is one there for the answer is C. 150. Inequalities Problem 15: combining these two inequalities we get C is greater than a which in turn is greater than D . Since B is equal to two. The remaining numbers A, C and D must be the numbers 13 and four, not necessarily in that order. To satisfy this inequality, Seamus before a must be three and D must be one. Hence, the answer is C. 151. Inequalities Problem 16: note that the product of R and T is one. Now. The product of two numbers is positive only if both numbers air positive or both numbers or negative. Since our times t equals one and our is greater than tea, there are two possibilities. Case one. Both are negative, so our would be greater than negative one in less than zero and T would be less than negative one in case, too. Both positive would give us tea between zero and one and are greater than one. But this contradicts the fact that our is given to be less than one hence case to is impossible, and the answer is B. 152. Inequalities Problem 17: let's add given inequalities first, aligning him vertically we get and adding we get X Plus p is greater than why, plus Q. And both are greater than zero now, since why is greater than zero and Q is greater than zero. Why plus Q will be greater than zero, so he could divide both sides of this inequality by why, plus Q. It will not change the direction of the inequality, and this cancels and gives his one. And this is the expression in column A. Hence, the answer is a. 153. Inequalities Problem 18: Let's align the inequalities vertically so we get two. X plus y is greater than him and to wine plus X is less than in now. We can't add these inequalities because their point in opposite directions. So multiply the bottom one by negative one. We get negative. X minus two wide will now be greater. We have to switch the directions inequality than negative in. Now we can go ahead and add the two inequalities, and that gives us X minus Y is greater than him minus in. And we're looking for the answer choice for which X minus y is greater, and we've just shown that it is answer choice be.