Transcripts
1. Quantitative Comparisons Text 1: quantitative comparisons are the most common math type questions. This is good news because they are the easiest problems to improve on generally quantitative comparison questions require, much less calculating, then do multiple choice questions. But they are trickier. Substitution is very effective with quantitative comparison problems, but you must plug in all five major types of numbers. Positives, negatives, fractions zero and one. Test the numbers 012 negative two and 1/2 in that order.
2. Quantitative Comparisons Text 2: the following principles can greatly simplify quantitative comparison problems. You can add or subtract the same term or number to both sides of a quantitative comparison problem, and you can multiply or divide both sides of a quantitative comparison problem by the same positive number. But be careful. You have to make sure that that number could never be negative or zero. You can think of a quantitative comparison problem as an inequality or equation. Your job is to determine whether the correct symbol with which to compare the columns is less than equal or greater than or that it cannot be determined. Therefore, all the rules that apply to solving inequalities apply to quantitative comparisons. That is, you can always add or subtract the same term to both columns of the problem. If the term is always positive, then you can multiply or divide both columns by it. The term cannot be negative because then it would invert the inequality. And, of course, it cannot be zero if you are dividing
3. Quantitative Comparisons Example 1: don't solve this problem by adding the fractions in each column. That would be too time consuming, since the L C. D is 120. Instead, merely subtract 1/5 and 1/8 from each column. So the 1/5 will cancel as with 1/8 and that leads is with 1/3 and column A and 1/4 and column B. And 1/3 of course, is larger than 1/4. Therefore column A is larger.
4. Quantitative Comparisons Text 3: if there are only numbers, that is no variables in a quantitative comparison problem, then not enough information cannot be the answer. This is from the Tri Khatemi Law in mathematics, which says that between two numbers, one must be larger than the other or smaller than the other or equal to each other.
5. Quantitative Comparisons Example 2: first cancel y to the fourth from both columns, and that gives us why cute and call him a and negative two y squared and column B now knows that why is positive. Therefore, we can safely divide by y squared, both called by Y squared without worrying about division by zero, and not without worrying about switching the direction of inequality. Call him a becomes why and column B becomes a negative to. And again, we're told that why is greater than zero, therefore column A is larger.
6. Quantitative Comparisons Example 3: since X is greater than one. We can subtract one from both sides of this inequality, and we get X minus. One is greater than zero. It's both expressions X and X minus one or positive. It's week. Most play both columns by their product to clear fractions, and that gives us X minus one and call him A and just X and column B. And now we can subtract X from both columns, which leads with leases with negative one and column A and zero and column B and zero is bigger than negative one. It's column B is larger.
7. Quantitative Comparisons Example 4: since in is a positive integer weaken. Divide both columns by N Squared because it cannot equal zero that reduces. Call him a tow one over X and calling be toe one. There were also told that X is greater than zero so we can vote by both Collins by acts to eliminate the fraction and that gives is and column a one and in column B acts. But we're also told that X is smaller than one hence column A is larger.
8. Quantitative Comparisons Text 4: Although we have already mentioned that you cannot multiply or divide both columns of a quantitative comparison problem by a negative number or zero, we're doing it again to just solidify that restriction. The following example illustrates the false results that can occur if you don't guarantee that the number you're multiplying or dividing by is positive. In this problem, we're tempted to divide both columns by X squared. Doing so would give X versus one, and we're told that X is smaller than one. It's calling B is larger, but we're also given that X could possibly equals zero here. And if it does, then the to call going back to the original expression. The two columns would be equal because zero cubed is zero and zero squared is also zero. Yes, we have a double case, and the answer is D
9. Quantitative Comparisons Text 5: caution. Don't cancel willy nilly. Although these techniques were studying are powerful, you still have to be a little careful in using them. Some people are attempted to cancel the X squares in this problem here, but you can't cancel because they have opposite signs. This one is positive, even though it's not written and this one is negative.
10. Quantitative Comparisons Text 6: You can square both sides of a quantitative comparison problem to eliminate square roots. For example, if call him a were radical five and call him be radical two plus radical three. Then squaring both sides gives us five and squaring the radical to we get to plus twice their product and squaring the radical three We get three now two plus three is five and now just noting that radical three radical two and two are all positive. Hence it's clear now that column B is larger.
11. Quantitative Comparisons Example 5: Let's square both columns to eliminate some of the radicals and call and be the radical cancels entirely, and we get just eight in column able foil it, and that will give us three plus two. Route three, Route five and then plus five and adding up like terms we get eight plus two. Route three, Route five. What we have here. Eight. Plus a positive number, which is greater than eight itself. Instead, answer is a
12. Quantitative Comparisons Example 6: Let's multiply both columns by 15 to clear the fractions and call. May we get five times route to and call and be the five or cancel in the 15 3 times and two times three or six Now this square, both columns to eliminate the radical. That gives us 25 times two versus 36 or 50 versus 36 and 50 is greater than 36. Therefore column A is larger.
13. Quantitative Comparisons Text 7: We already studied this method in the section substitution. Here we will practice more and learn a couple special cases in a problem with two variables , say X and Y. You must check the case in which X equals why this often gives a double case. Many students forget or don't realize that different variables can represent the same number, and often dio and the writers of the tests often capitalize on this oversight.
14. Quantitative Comparisons Example 7: we're told that X and Y are positive. So this choose X equal? Why? To equal one. Remember, different variables can represent the same number. And also remember the average of two numbers. Is there some divided by two. So call him a becomes one and call him be is also one. In this case, the columns are equal. Now let's choose X two equal. Why? To equal to then column A is equal to to and call them be is eight. In this case, calling B is larger. Hence we have a double case and the answer is D.
15. Quantitative Comparisons Example 8: If X does not equal, why then column A. Will be larger than column B, and you should plug in a few numbers just to see that this is the case. But if X is equal, why and equal to one, then the columns will equal each other for call. May we get to to the one plus one or two squared, which is four. And for calling B, we get to the one plus to the one, which is also four, and therefore we have a double case. An answer is D.
16. Quantitative Comparisons Text 8: when you are given X is less than zero, you must plug in negative whole numbers, negative fractions and negative one. Choose the numbers negative one negative to and negative 1/2 in that order.
17. Quantitative Comparisons Example 9: Let's solve this problem by substitution. Since K is less than zero, we have to look at the numbers negative one negative, too, and negative 1/2 in that order. If K equals negative one or negative to thank, Allah may well be the product of two positive squares and therefore we bigger than call him be. Now, if K equals negative 1/2 then call May will be zero and negative. 1/2 plus 1/2 is zero and zero squared. Zero anything multiply by zero is zero. In this case, the columns are equal. Whereas before column A was larger. Therefore, we have a double case and the answer is D.
18. Quantitative Comparisons Text 9: Sometimes you have to plug in the 1st 3 numbers, but never more than three numbers from a class of numbers. For example, you might be told that X is an integer and it's greater than five. Then you would have to check six, seven and eight, but you will not need to go any further.
19. Quantitative Comparisons Example 10: since X is both an injure and greater than one, we only have to plug in the 1st 3 Energy is bigger than one, which would be 23 and four. We don't have to look at five because the writers of the jury will not check change the result on the fourth number from a class of numbers plugging to into column A. We get now. The smallest positive factor of two is to itself, and column B is is, and the smallest part positive factor of B is also to. In this case, the columns are equal. For three, we get three and column A and for column B, we get 27 and the smallest common. The smallest positive factor, 27 is three. So the calls or get equal again. And finally, for four, the smallest positive factor of four is too now, for cute is 64 and the smallest positive factor of 64 is too. So in all three cases the columns are equal. Hence the answer is C.
20. Quantitative Comparisons Problem 1 : for this problem. All we have to do is calculate the value of column A and see whether it's larger. They call him be smaller than column B or equal to column B. The answer cannot be d not enough information because we don't have any variables following the rules of operations, we form what's inside the parentheses first and in there we have in addition and multiplication. So we do. The multiplication first, which gives us 10 20 and two plus 20 is 22 and three times 22 is 66. And, of course, 66 is bigger than 60. So the answer is calling a
21. Quantitative Comparisons Problem 2: 1/9 of 10 is 1/9 times 10 which is 10 nights and 10 nights is greater than one. Similarly, 1/10 of nine is nine tense and 19 says less than one because the numerator nine is smaller than in ometer. Therefore, column A is larger and the answer is a.
22. Quantitative Comparisons Problem 3: noticing column made that we have three negatives, and the product of an odd number of negatives is negative. So call him A is a negative number and column B is zero times negative one and zero times. Anything is zero. Now all zero is bigger than all negative numbers. Therefore, column B is larger than column a.
23. Quantitative Comparisons Problem 4: since X ranges between negative one and one exclusive of zero. We're dealing with Onley with negative and positive proper fractions. So all we need to check our negative 1/2 and positive 1/2 or any other negative and positive fraction in that range. Plug in a negative 1/2 into column A. The square will destroy the negative and we get positive 1/4. And similarly, when we plug in positive, 1/2 will also get positive. 1/4 and number one and the number one is greater than 1/4. Therefore, column B is larger than calling a
24. Quantitative Comparisons Problem 5: Let's solve this problem by substitution. Remember that different variables can be represented can represent the same number. So let's choose X equals. Why equals one forming the average and column A. Remember that the average is the some of the in numbers divided by the number of numbers. So we'll have one plus one divided by two, which is to over two or one and column B will give us one squared plus one squared, divided by two, which is to over to which is also one. In this case, the two columns are equal. Now let's choose X to be to and why it also equal to then the average and column one is two plus two over to or two, and the average and column B is going to be two squared plus two squared, and you can tell it's gonna be larger now. That will give us four plus four divided by two or eight House, which is four now Calling B is larger. Yes, we have a double case, and the answer is D
25. Quantitative Comparisons Problem 6: remember that you can view it any a quantitative comparison problem as an inequality so you can add or subtract inning term from both columns, and you can multiply or divide both columns by any non zero number and not affect inequality. Yes, we have two X in both columns so we can eliminate it eliminated by subtracting from both columns and call. May we get just why and column B We get zero, but we're told that why is greater than zero. Therefore, call him a is larger.
26. Quantitative Comparisons Problem 7: Let's solve this problem by substitution. Since we're given the A is less than zero. Let's choose a couple negative numbers. Start with a equal, the negative one, and then we'll try a equals negative, too. Plugging in negative one. We get negative. One squared, plus negative one cubed negative. One squared is one and negative. One cube is negative one and one minus 10 In this case, the columns are equal. Now let's try native to negative two squared plus negative two Cubed is positive for minus eight, which is negative. For now. Call May is smaller than calling B. This is a double case. Therefore, the answer is D.
27. Quantitative Comparisons Problem 8: since in is a class of numbers in particular positive imagers way need only check the 1st 3 positive imagers in that class. In other words, one, two and three choosing in to be one. We get negative one to the one which is negative. One and negative one is smaller than zero. And if we choose in to be negative Apartment two B two, we get negative ones square, which is positive. One in positive. One is greater than zero. Yes, we have a double case and the answer is D.
28. Quantitative Comparisons Problem 9: Let's solve this problem by substitution. Suppose X equals Y equals two, then column A becomes eight and column B becomes two times two times two, which is also ate. In this case, the two columns are equal. If you choose any other value for extra why the columns will not be equal, and you should check a few to convince yourself. Therefore, the answer is D not enough information to decide.
29. Quantitative Comparisons Problem 10: don't solve this problem by multiplying the terms in the columns that would take too long, it said. Simply divide boast columns by 10. Call him a will. Give us 35 times 54 and call and be. We'll also give us 35 times 54 hence the columns are equal and the answer is D.
30. Quantitative Comparisons Problem 11: notice that both A and B or less than zero. Therefore both A and B are negative now. The just as the product of two negative numbers returns a positive number. The quotient of two negative numbers returns a positive number, so call him a is positive. And of course, we're given that A is less than B, which is less than zero. So a is negative and all positive numbers are bigger than all negative numbers. Therefore, calling a is larger.
31. Quantitative Comparisons Problem 12: first cancel. Why Cube from both columns and that gives us why Squared versus negative y. Now we're given that Why is a positive number so we can divide both columns by why, without affecting the direction of the inequality. So call him one becomes why and call him be becomes a negative one. Now we're given that again. That why is greater than zero at all. Positive numbers are bigger than all negative numbers there for calling a is larger.
32. Quantitative Comparisons Problem 13: At first glance, it appears that column B would be larger because it involves three different positive numbers and call him A involves only two different positive members. So sleep. See if we can find an exception. This may call MBIA's smallest possible, so let's choose the numbers 13 and five. The product is 15 now, if we choose to and four for calling May. Those are even imagers. Positive and less. Lesson 10. Then we get eight. But if we choose, say, six and eight, we get 48 now for two and four column A is smaller and for six and eight column A is greater than 15. This is a double case. Therefore, the answer is D.
33. Quantitative Comparisons Problem 14: Let's clear fractions by multiplying both Collins by the LCD, which is 90 and call him a. The nine will cancel into the 90 10 times. So we get 10 times a or 80 in a column B. The 10 cancels into 99 times when we get nine times nine, which is 81 and anyone's greater than 80. Therefore, the answer is B.
34. Quantitative Comparisons Problem 15: first cancel X cubed from both columns by subtracting X cubed from both columns. That gives his negative five for column A and negative 15 for Colin Be and negative five is greater than negative 15. Therefore, the answer is a.
35. Quantitative Comparisons Problem 16: Since X is a positive integer, we need to look at only the 1st 3 positive imagers 12 and three. If X is one, then column A is one, and there are no prime miniatures and call him be is also one. So there again, there are no common no prime imagers there for the The answer for each of these is zero, so they're equal for two. We get the number two, so there is one prime factor in column A. And for calling B, we get X cubed equals two. Cubed, which is a now to cube, has only one prime in urge er, namely the number two. So again the columns are equal, and for three we have three, which is prime. So there's one prime number and call him A and column B becomes 27 which is three cubed. But again, there's only one prime number. That's the number three. So again, the columns are equal. So for all three cases, the answer is the columns are equal, So the answer is C.
36. Quantitative Comparisons Problem 17: Let's multiply both columns by radical five to clear the fraction and call him A. We get 10 and in column B, we get two times five, which is 10 as well against the columns are equal, and the answer is C.
37. Quantitative Comparisons Problem 18: start by canceling the threes and the sevens from both columns. So calm a reduces it just X and column B to just 11. Now we're given that X is greater than or equal 11. Well, effects were 11 then the columns would be equal and effects were greater than 11. Then call him a would be greater than calling B, and this is a double case. Therefore, the answer is D.
38. Quantitative Comparisons Problem 19: since X is greater than why we know that the difference between X and Y is greater than zero. Therefore, we can vote by both columns by X minus wide to clear the fraction. So I called a We have X squared minus two x y plus y squared and in column B. Let's go ahead and foil this expression. We'll play the 1st 2 We get X squared, the outer to give us minus X y the inner too. Give us minus X y again and the last to give us positive. Why square and adding up these middle terms you get minus two x y. And now we just noticed that this is the same expression as in column A. Therefore, the columns are equal and the answer is C.
39. Quantitative Comparisons Problem 20: since A is greater than zero weaken divide both Collins by a without affecting the direction of the inequality. So call him a becomes one over 10.3 and column B becomes too. Now we'll clear fractions. Multiply both calls by 0.3, cancels and column A and we get one and column B becomes 10.6, and the number one is bigger than 10.6, So the answer is a.
40. Quantitative Comparisons Problem 21: Let's solve this problem by substitution. If X equals Y equals one, then both X and wire, positive integers and calm A becomes one and column B becomes to over two, which is one as well. So for this case, the calls are equal. But if X does not equal, why, then the columns will not equal each other. For example, if X is equal to one and why is equal to 1/2 then call him. One will be 1/2 and calm be will be one plus one over 1/2 squared plus one which is to over 1/4 plus one, or to over five force, which is two times 4/5 which finally is 8/5 which doesn't equal 1/2. Therefore, we have ah, double case, and the answer is D.
41. Quantitative Comparisons Problem 22: first canceled to the two and the three from both columns, Then column. Reduce it. A reduces to P and Colin be reduces to eight. They were given that P is less than or equal to eight. So if P is equal eight, the columns are equal. And if P is less than eight, then colon B is larger. This is a double case. Therefore, the answer is D.
42. Quantitative Comparisons Problem 23: notice that too is in excess position and *** three is and wise position. Therefore, replace X with two. And why, with negative three performing the operation inside the absolute value, we get two plus three because a negative time to negative is a positive, which gives us the absolute value of fire. And this negative here is unaffected by that absolute value. So it stays. And the absolute value of five, of course, is just five, which gives us negative five. Hence, the columns are equal because both equal negative five.
43. Quantitative Comparisons Problem 24: this. Introduce a right triangle to this drawing by drawn a vertical line perpendicular to the X axis and label it Q. Now in a right triangle, the high pot news, the site opposite the right angle is the longest side. Therefore, Opie is longer than O que and O que is the X coordinate of Point P. Therefore, calling B is greater than column A.
44. Quantitative Comparisons Problem 25: when you are given a geometric growing check, whether other drawings or possible in this case, we're not given either dimensions of the triangle nor the measure of its angles. Hence, other drawings are possible in the given drawing. Clearly, X is greater than why plus Z. But if we drawn equilateral triangle in which all three angles are equal, so this angle is equal that one which is equal that one. Now the Y plus Z will be twice as big as acts, and this is a double case. Therefore, the answer is D.
45. Quantitative Comparisons Problem 26: Let's solve this problem by substitution, choosing into equal one call him a becomes negative. One squared, which is positive one and one is greater than zero. So for this choice of end, call him a is larger. Now let's try another number. How bout in equals two. So we get negative one cube, which is negative one, but negative one is less than zero. So we have a double case, and the answer is D.
46. Quantitative Comparisons Problem 27: Let's square both columns to eliminate the radical squaring column. A Kisses two. Because the square cancels with the radical and squaring column B gives us three squared over two squared, which is nine force and four goes into 92 times with the remainder of one. So we get to and 1/4 which of course, is greater than two. Therefore, the answer is B.
47. Quantitative Comparisons Problem 28: the product of any number of sixes ends with the digit six. For example, six times six is 36 and six times six times six is to 16 and both numbers in with the digit six. So call in a is six and column B the product of any number of five ends and five, for example, five times five is 25. If you multiply that by five, you get 1 25 So the digits always in in five. So calling B is five. Therefore column A Is greater than column B.
48. Quantitative Comparisons Problem 29: If the numbers in each column were always positive, then clearly the average and column B would be greater than the average in column A. But since negative numbers are not excluded, we can make column a greater by choosing the following numbers. Let's choose for column a one Let's choose for. Call him a The following Numbers one, three and five and for column B list, Choose negative 20 zero and 20. Then the average recall may is one plus three plus five divided by the number of numbers, which is three. And that gives us 9/3, which is three. And you might notice intuitively, the average here is gonna be zero. But let's go ahead and calculate it. We get 20 plus zero plus 20 divided by three, and negative 20 plus 20 is zero. So we get is there on the top and three on the bottom, which is zero. So call him a now is is larger. This is a double case. Therefore, the answer is D
49. Quantitative Comparisons Problem 30: since lying BP passes through the center of the circle. It is a diameter now. The largest court of a circle is a diameter. Therefore, BP is greater than the court PC. That is, BP is greater than the cord PC, multiplying both sides of this inequality by the positive number aapi, which is the length of the other cord we get. This therefore column A is bigger than column B, and the answer is a.
50. Quantitative Comparisons Problem 31: since X is an integer greater than one. We only have to look at the 1st 3 energies greater than one, which would be 23 and four for two. We get two plus two Cube, which is to plus eight, which is 10 versus two to the fourth Power, which is 16. In this case, calling B is larger. For three, we get 30 and three to the fourth. Power can be written as three squared times Three squared, which is nine times nine or 81 and again, calling B is larger. And for four, we good 68 and four to the fourth. Power is to 56 and once again the column B is larger. So in all three cases, calling B is larger. There for the answer is B.
51. Quantitative Comparisons Problem 32: suppose the parallelogram were a rectangle. Now this side of the rectangles three and this side is five. In the area of a reckoning, of course, is length times with so we have five times three or 15 which is colon. Be now we can view the given parallelogram A B C D as this rectangle tilted to the right 20 degrees because this is a right angle here, this Inglis 70. So the single here is 20. Now the question is, as we tilt the rectangle to the right, does the area increase or decrease? Let's take an extreme case. Let's tilt the rectangle 80 degrees, so this angle here down would be 10 degrees. And clearly this parallelogram here has less area than the original rectangle. Therefore, as we tilt the rectangle, the area decreases. So the area of the parallelogram A, B, C, D and colon A is less than 15 because the rectangle has area 15 hits. Column B is larger, and the answer is B
52. Quantitative Comparisons Problem 33: since the range of P and the range of Q overlap, there is not enough enough information to answer the question. For example, if P were equal 2.49 and Cure equal 2.45 then P would be greater than Q or call him a would be called greater than cold be. But if P is 0.49 and Cue is 0.9, then the 0.9 is bigger than 4.9. Therefore, column B now is larger, so we have a double case. An answer is D. We can also illustrate this with a number line. It's put zero here 0.4 0.5 and one so P is between zero and 00.5. Soapy varies from here to here, and cue buries from 0.4 21 and this region in here, the to variables overlap so they could have different sizes. So again, the answer is D
53. Quantitative Comparisons Problem 34: no. So you're given variables x You? Why? And V And in the columns we had the same variables. But they're all negatives. Solicit built by these inequalities by negative one. And when we do so well, flip the direction than equality. So negative x will be greater than negative. You and this problem this inequality will give us negative Wine is greater than V adding these inequalities vertically. We get negative. X minus y is greater than negative view minus V. But this is column A and this is calling B therefore column A is larger than call and be.
54. Quantitative Comparisons Problem 35: There are only two numbers that are both the square root and the square of the same imager . And those numbers are zero and one zero is a square oven imager, namely itself because zero equals zero squared and zero equals the square root of zero. So it's also the square root of an energy. Likewise, one is the square of an energy itself, and one is the square root of an integer, the square root of one itself. In both cases, zero and one, they're less than three. Therefore, calling B is larger.
55. Quantitative Comparisons Problem 36: were given that next divided by why is greater than zero. Therefore, the quotient is positive. This can occur only if both x and y are positive or both X and wire negative. So let's do this problem in cases. Assume both X and y are positive. Then, from the equation, X squared equals y squared. We take the square root and get backs equals plus or minus. Why now? If X were to equal negative, why then X would be negative, contradicting our assumption that X is positive. Therefore, X must equal Why now assume that ex n y are both less than zero. Then from the same equation, we get X equals plus or minus y Now, assuming X equals and negative wine, then this would be positive. And even though it looks like it might be negative because of the negative, there we are, assuming that why itself is negative. So we actually have a negative times another negative, which, of course, is a positive. Since this is a contradiction, X must equal why again. So in both cases, X equals y. Therefore, column B is equal to zero and the two columns are equal. Therefore, the answer is C
56. Quantitative Comparisons Problem 37: The key to this problem is to notice that ex Commie negative Suppose X is equal to positive 1/2 and then X equals negative 1/2 for positive. 1/2 we get 1/1 half, which is one times would have flipped over, which gives us two and column B becomes 1/2 1/1 half plus one, which is 1/3 House, which is 2/3 in this case column A is larger now. If X equals negative 1/2 then call him A will be negative, too, and column B will be won over negative 1/2 plus one, which is one over 1/2 which is to now call them be is larger. Therefore, we have a double case, and the answer is D.
57. Quantitative Comparisons Problem 38: were given That X is greater than one hints both X and X plus one or greater than one. Therefore, we can multiply both calls by X Times X plus one to clear the fractions. Call him a becomes X plus one and column B becomes one hence column. A is bigger than calm Be because X is bigger than one. Therefore, the answer is a.
58. Quantitative Comparisons Problem 39: since we have a straight angle here, this angle the supplement is 1 80 minus X, but we're told that that is equal. Why there for this angle is why? And that angle is why it's in spite of the theater appearance of the drawing the triangles actually a sauce. Elise, this is this size. Conclude that and that angle is congruent to that angle. Well, in this drawing, P are is greater than p que. But that's not necessarily the case. We could draw it much taller then the PR would be smaller than the peak you and this is a double case. Therefore, the answer is D.
59. Quantitative Comparisons Problem 40: it's since. Why equals e? We have Ah, I softly is triangle. Therefore the opposite sides are congruent. Hence PR is congruent to P Q. Therefore, the columns are equal and the answer is C.
60. Quantitative Comparisons Problem 41: Let's simplify column A From the formula a baby squid squared equals a squared B squared column A becomes now knows we have a common factor of X squared in the numerator so we can factor it out. That leases with y squared minus. Now we have this right. Something here to hold the position that X was X squared was in. And even though we don't write, write it. It is assumed to be a one in fronts. We're left with the one and now we can cancel the Y squared minus one. And now we just noticed that this is in fact call and be Therefore the columns are equal in the answer is C.
61. Quantitative Comparisons Problem 42: Let's solve this problem by substitution, choosing X to be nine and why to be three. These numbers will satisfy the given conditions because nine equals three times three and nine is greater than three. Then call him a becomes nine plus three or 12 which is greater than eight. Now let's choose X to be one. Pour me. Choose XP three and wide to be one. Then again, the equations. Air said. If this equation is satisfied, three does equal three times one and three is bigger than one. Now column A becomes three plus one, which is four. And for his lesson eight. Therefore, we have a double case, and the answer is D.
62. Quantitative Comparisons Problem 43: Let's solve this problem by substitution. Suppose in equals one, then two times one is to is a positive integer, and the square root of one is one is an integer, so it satisfies the conditions of the problem. And in this case, column B is larger. Now let's choose in to be four another convenient, perfect square to work with, then the square root of it. Four is too, which is an integer and two times for is a which is a positive integer. So again, this value of end satisfies the given conditions. But now call him a is four and calling B is still three. So now calm a is larger and we have a double case. Therefore, the answer is D.
63. Quantitative Comparisons Problem 44: this expression tells us on Lee that X is positive. And that's because if if X were negative on the bottom and on the top, it would always be positive, because the absolute value of a number is positive on a positive divided by a negative isn't negative, and no negative number can equal positive one. So that means that X could have any value up any positive values. So if you chose X to be one, the calls would equal each other. If you chose X to be to column B would be larger and two. It also satisfy this condition because the absolute value of 2/2 is equal to 2/2, which is equal to one. This is a double case, therefore, the answer is D.
64. Quantitative Comparisons Problem 45: now foreign nine or the only imagers that satisfied the given conditions. For example, 1/4 is less than one over the square root of four, which is 1/2 and 1/2 is less than 2/3 and 1/4 is smaller than one over the square root of nine, which is 1/3 which in turn is smaller than 2/3. Therefore, the ends and both and both four and nine are greater than three. Therefore, calling B is larger.
65. Quantitative Comparisons Problem 46: Let's solve this equation by factory. Now four times seven is 28 and negative for minus seven is negative. 11. So this equation factors into X minus four and X minus seven. Setting each factor equal to zero. We get X equals four and X equal seven. Now if excess seven, then the columns are equal because calling a of seven and if X equals four, then the columns are not equal. This is a double case. Therefore, the answer is D.
66. Quantitative Comparisons Problem 47: since X Square could never be negative when we moved by both sides of the inequality by two x squared to clear the fractions, the inequality will not reverse direction. This gives us two is less than X squared. Now let's rewrite this inequality with X squared on the left and two on the right, we haven't changed Anything. Mathematically just changed the direction in which the the air inequality is pointing, but it still states the same thing. Take the square root herbal sides we get or X less than negative. Radical, too, or tax greater than radical, too. Now we're given that X is negative so you can reject X greater than radical, too, and negative radical to is less than negative one. Therefore, column B is larger.
67. Quantitative Comparisons Problem 48: translating this sentence into an equation gives. Now, let's see if we can manipulate this equation to make it look like column B to clear fractions. We multiply each term by P, and that will give us P cubed. Here the piece cancel and we get plus one, and here we have negative to pee. Then add to pita both sides of this equation because we have a plus to be here, and we might as well subtract one at the same time. And the ones canceled here in the two peas cancelled here and we get P cubed plus two p is equal of negative one, which is the Value column A. Since the columns are equal in the answer is C.
68. Quantitative Comparisons Problem 49: notice that we have vertical angles here. So this angle is also 43 degrees. Now we're told that P. O. R. Is a right angle. Therefore, the single here is 90 minus 43 or 47. Now this is also a right triangle here, therefore and there are 100 and 80 degrees in a triangle. So this angle is 1 80 minus the 47 and then minus the right angle here, which is 90. And that gives us 43. So this angle here is larger than this angle here. Therefore, Q R is greater than O. R.
69. Quantitative Comparisons Problem 50: let's solve these equations to determine the possible values of X and Y Taking the cue bridgeable sides of this equation we get Why equals two and taking the fourth root of this equation we get plus and minus two. Now, if we choose X to be to than the two columns will equal each other. And if you choose X to be negative too, then call and be will be larger. This is a double case. Therefore, the answer is D.
70. Quantitative Comparisons Problem 51: first, let's factor the expression underneath the radical. We can treat it as a quadratic expression z to the fourth Z squared. So instead of a Z here we get a Z squared and another Z squared. And the factors air 16 who some is eight are negative for negative four. This gives us Z squared minus four. The quantity squared, and you might notice that that is the absolute value expressions so he can write that a Z squared minus four in absolute value. Now this expression is always non negative. That is, it must be either positive or zero, whereas column B can be both positive or negative. If you choosy to be zero, you get negative four. And if you choose easy to be bigger than four like five, it will give us a positive number. So we have a double case. Sometimes calling B is larger and sometimes calling be a smaller. So the answer is D
71. Quantitative Comparisons Problem 52: let be stand for the number of acres of farmland farmer Bob has. Then Farmer John has B plus X acres, and we're told that the two together have 200 acres, so we get be plus B plus acts equals 200. And there's no mathematical purpose for these parentheses. They're just there to show that that is a group adding up the light terms we get to be plus X equals 200. And those this equation is looking similar to call him be. So let's make it. Let's turn it into column B subtract X from both sides. We get to be equals 200 minus X, which is precisely calling B. And this says that Farmer Bob has twice the number of acres as this expression here, which is precisely what column A says. So the answer is C.
72. Quantitative Comparisons Problem 53: If the two chords forming the sectors of this circle passed through the center of the circle, then all four sectors would be congruent. But we can't assume that they passed through the center of the circle. We could have something like this in this case, the sectors. Each sector has a different size. Therefore, the column B does not necessarily equal column A and the answer is D.
73. Quantitative Comparisons Problem 54: remember that the percent increase is the increase divided by the original amount. Let's let extend for the number of products sold in 2014 then, since 2.1 million it's sold each year. The increases 2.1 divided by X. That would be for 2015. Now for 2016 there are a total of X plus 2.1 items sold in 2015. So this is the number of the original Mt for 2016. Other words. On the first day of 2016 For minutes percentage increase, we get 2.1 over X plus 2.1 now, since the numerator zehr the same and this denominator is larger, this fraction is smaller. Therefore column A is larger and the answer is a
74. Quantitative Comparisons Problem 55: Let's solve this problem by substitution. Suppose a is negative. 1/2 B is negative one and C is negative. Two. That will satisfy this condition here, and their product is negative one. And for calling B, we get now working from inside the parentheses. Here we have the same result as in column ace. That will just be negative. One cube, which is negative one. Now let's choose a to be negative one. Be to be negative, too, and see to be negative. Three. And we get calling A is negative six and column B will be negative. Six. Cute, which doesn't equal negative six. Therefore, we have a double case, and the answer is D.
75. Quantitative Comparisons Problem 56: Let's factor this expression for Z notice that it's quadratic not in X, but in X squared, because X to the fourth is X squared. Factory it We get X squared plus two and another X squared plus two so Z can be written as X squared plus two. The whole quantity squared because there's two identical terms here. This tells us that Z must be zero or a positive number. Is it possible for Zito actually equals zero? Well, that would occur only when X squared plus two equals zero. But since X is either zero or positive, it cannot be negative. This expression will always be greater than or equal to two so it can't possibly equals zero. So therefore Z must be a positive number. She is greater than zero and call him a is zero. Therefore, column B is larger.
76. Quantitative Comparisons Problem 57: we call that the area of a circle is pie R squared, and we're told that the radius is too so the top of the fraction we get pie times two squared and we're told the diameter goes to in the bottom of the fraction and the diameter is twice the radius. So we get two times two, which gives us four pi over four or pie. No pious, slightly larger than three. It's 3.14 Therefore, colony is larger.
77. Quantitative Comparisons Problem 58: Let's solve the equation to determine the possible values of acts on the left side, we get three x squared. This square destroys the negative because the negatives inside the parentheses and the same thing occurs on this side. But we also get three square because it's inside the parentheses. X Square and now this a practice three x squared from both sides. That gives us zero equals six x where divide by six you get X squared and finally take the square root herbal sides, which gives us X equals zero, the square root of zero Israel and column A zero. Therefore, the columns are equal, and the answer is C.
78. Quantitative Comparisons Problem 59: notice that these air vertical angles here so the singles also be and likewise this angle here is C. Now recall that the exterior angle to a triangle angle a is equal to the sum of its remote interior angles. Therefore, a equals B plus C. Now this manipulate this equation to make it look like calling a subtract C from both sides . Because column A involves a minus C and that gives us a minus. C equals B. So Colin A is B. Therefore, the columns are equal and the answer is C.
79. Quantitative Comparisons Problem 60: At first glance, it might look like column A Is larger than Colin be because 21 is bigger than 11. But there actually isn't enough information to decide. Supposing COLUMN A. We have the numbers 18 19 20 and 21 then their average would be 19.5. And if you choose for column B, the numbers eight, nine, 10 and 11 then there averages 9.5. For this case, we get what we expect. Call May is larger, however, keeping the same four numbers for column B and choosing new numbers for Kuala May, such as 123 and 21. Now the average drops dramatically down to 6.75 which is now smaller than 9.5. So we have a double case, and the answer therefore his D.
80. Quantitative Comparisons Problem 61: notice that each number on the left of the equation is a multiple of three writing that out explicitly. Be good. Three squared for nine three cube for 27 and three to the fourth power for 81. Adding up the experts, we get 259 and since the basis of the same, the exponents have to equal each other. So nine must equal X minus y, which is precisely calling A and 10 of course, is larger than nine, so calling B is larger.
81. Quantitative Comparisons Problem 62: notice the left side of the top equation is this perfect square. Try no meal. I will point it out. We get X squared plus two x y plus Why square? No notice We haven't X square plus a y squared They were given that there some is 12. So substituting that and we get X two x y plus 12 equals 24 and no notice. We have the expression x y that we haven't call him a solving for X y. We get divided by two gives us x y equals six and six is bigger than five. Therefore, calling a is larger.
82. Quantitative Comparisons Problem 63: the units. Digit of six raised to any power is always six. For example, six squared is 36 and six. Cube is to 16 and the units digit of five. Raised to any positive power is is five. For example, five squared is 25 and five cubed is 1 25 since six is greater than five. Column A is larger than calling B.
83. Quantitative Comparisons Problem 64: in column a notice that, too, is in excess position, and one is in wise position. So substituting those values into the right side of the form that we get to minus one or the absolute value of one and that's a value of one is positive one and a negative times. A positive is a negative, so we get negative one for column A and for calling B, we get X five is in excess position and sixes and wise position, so we get the absolute value of negative one. The negative outside stays where it is. If you drop the absolute value, you have to change that negative into a positive and a negative times. A positive isn't is a positive, so calling B is also equal the negative one. Hence the columns are equal, and the answer is C.
84. Quantitative Comparisons Problem 65: notice that column B is the denominator of column A. So let's calculate the fractions in column a then substance and column B and substitute him into column A. Getting a common denominator and call him be. We get three plus two over six, which is 56 Plugging that into column A. We get one over Main Division Bar 56 Now the divide means to in first and multiply. So we get one times 6/5 flipping the 56 over, which is six fists or, if you wish, one in 1/5 which is bigger than 56 hence COLUMN A is larger.
85. Quantitative Comparisons Problem 66: recall that the interior angles of a triangle add up to 180 degrees. So x plus four x plus five x equals 1 80 or 10 X equals 1 80 Dividing by 10 we get X equals 18. Turning to this angle, we substitute in 18 for X and we get five times 18 which is 90. Therefore, this is a writing right angle here, so we have a right triangle and the Pythagorean theorem applies, namely, the thigh pot. News squared is equal to the sum of the legs squared, but this is precisely calling B. Therefore, the columns are equal in the answer is C.
86. Quantitative Comparisons Problem 67: Let's reduce column a notice that 3 to 18th is embedded in 3 to 19. Digging that out, we get rewriting this as three to the 18th power times, three to the one and hair were using the property that if you multiply bases, you add exponents and factoring out three to the 18th from each term we get three minus something, and that something is a one you can always think of a number any number of being multiplied by one. And this gives us three to the 18th power times two notice. Each column has a factor of 3 to 18th power and call him A is multiplied by two, and calling B is multiplied by four. Therefore, calling B is larger.
87. Quantitative Comparisons Problem 68: were given that X is a non negative integer, so X could be 0123 etcetera. If x zero, then we get in column a 0.60 which is one by definition. If X is one, then we get 10.6 and FX is, too. We get 0.6 squared, which is 0.36 in all these cases. Column A is larger than calling because 1/3 is 0.33 indefinitely and then finally 4.6 cubed , we get point to 16 but 2.216 is smaller than 0.333 So we have a double case. Therefore, the answer is D.
88. Quantitative Comparisons Problem 69: let X be the side of the squares. And since it is a square, this is a right angle here. So the Pythagorean theorem applies, so we get X squared. Plus X squared is equal to six square or two. X squared is equal to 36. Divided by two gives X squared is equal to 18 but X squared is precisely the area of the square. Since the columns are equal, 18 equals 18 and the answer is C.
89. Quantitative Comparisons Problem 70: since X is an integer and greater than one, we're dealing with the numbers 234 etcetera. If X is three, then call him a becomes two to the three plus two or two to the Fifth Power, which is 32 and column B becomes three to the third power, which is 27. In this case, calling May's larger. Now, if X is four, we get to to the four plus two, which is to the six power, which is 64 and call them be becomes three to the fourth Power, which is 81 now 81 is greater than 64 so we have a double case and the answer is D.
90. Quantitative Comparisons Problem 71: Let's simplify the columns for column A. The 27 can be written as nine times three and nine is a perfect square, so it comes out as a three distributing the three on each term we get nine radical three plus 12 and then the street and Colin be distributing the nine we get canceling. The three we have three times four is 12 plus nine times Route three, and these expressions are the same. Therefore, the answer is C.
91. Quantitative Comparisons Problem 72: noticing COLUMN A. We are square in a proper fraction, namely a fraction between zero and one and call him be. We're square rooting the same fraction. Now recall that square squaring a fraction between Zeron one actually makes it smaller, and taking the square root of it makes it larger. Therefore, calling B is larger than call him a.
92. Quantitative Comparisons Problem 73: first simplifying the columns will use the property of exponents that X to the A over X to the B equals X to the A minus. Bi perform that rule and calling, May we get minus the bottom most. You have to encase it in parentheses. Could it's minus the entire quantity and the peace. Cancel, and we get text to the negative one or simply one over X. And for calling B, we get canceling the peas gives us X to the one or just X now using substitution. If we plug in X equals one, the two Collins will equal each other and for any other numbers, such as to the columns wanting to each other, this is a double case there, for the answer is D.
93. Quantitative Comparisons Problem 74: Since we're not given the dimensions of the rectangle nor the semi circle, the solution must be independent of their dimensions. Let's choose the radius of the semi circle to be one which is an easy number to calculate with. Then the with of the rectangle would be one because this is a radius as well. And both of these line segments are ready. I so the length of the rectangle is too. So the area of the rectangle ace of our is to my two times one or two. Now the area of a circle is pi r squared. And here we have a semi circle. So it's gonna be 1/2 pi r squared and again the radius is one. So we get pi over two. So the area of the shaded region is the area, the whole rectangle minus the area of the circle, and the whole rectangle is to in the area. The semi circle is pi over two. Now to form the percent and column made the percent of the rectangle that is shaded. We take the part and divided by the whole. The part is to minus pi over two from here, and the whole. In other words, the whole rectangle is, too. And if we do, these calculations will find that we get four minus pi over four. Now pie is bigger than three, so this is less than four minus three over four, which, of course, is 1/4 hence, column B is larger.
94. Quantitative Comparisons Problem 75: Let's solve the equation to determine the value of acts multiplying each term by the LCD, which is four will give. And here the four cancels entirely and we have to encase the expression in parentheses cause it's minus the entire expression. Now distributing we get six x minus two minus X minus two equals adding up like terms. We get five X minus four, and on this side we get negative X plus two, adding next both sides. We get six sex, adding four to both sides. We get six dividing by six, we get X equals one, so call him a becomes one. And of course, column B is one because X is one. So the columns are equal and answer is C.
95. Quantitative Comparisons Problem 76: when comparing two sides of a triangle, the side opposite the larger angle is the law longer side. Let this angle here be acts. Then, since the angles some of the triangles 1 80 we get subtracting 93 from both sides. We get X equals 87 so this angle is 87. Since angle B is slightly larger than angle see Side A C is larger than side side A B hence column B is larger.
96. Quantitative Comparisons Problem 77: were given that the some of the three numbers and each diagonal is the same. So adding up the numbers down this diagonal we get equals the sum of the numbers down this diagonal. Now subtracting why from both sides of this equation and subtracting three from both sides , we get X. This cancels in this castle's equals Z the wise cancel here and seven minus three is four now. This equation says that we have to add four to Z to make it as large as acts hence. X is larger than Z and column A is larger than be.
97. Quantitative Comparisons Problem 78: recall that the average is the some of the terms divided by the number of terms which can be written as the sum divided by the number of of terms for call may we get nor three terms here. So we divide by three and this reduces to to for calling B, we get four and four is greater than two, therefore, column B is larger.
98. Quantitative Comparisons Problem 79: that sold this problem by substitution that X equal y equals zero. Remember, different variables can represent the same number with these values column A becomes zero and column B becomes zero as well. For this choice of X and y, the columns are equal. Now list. Keep X at zero and change. Why Toe one. Then call him a becomes negative one and calling B becomes now. When we square the negative one, we get positive one and positive one is bigger than negative one. So now column B is is larger. This is a double case there, for the answer is D.
99. Quantitative Comparisons Problem 80: that's so this problem By substitution if X equals one, then column A becomes nine tense and column B becomes 10 nights now. 10 nights is bigger than nine tense, so call him B in this case is larger. And if X equals two, then column A becomes and column B becomes now 18 tents is an improper fraction. Therefore, it's greater than 28th so we have a double case and answer is D.
100. Quantitative Comparisons Problem 81: Since X is equal to P times the number Q that is bigger than one X is greater than P. We could also solve this problem by multiplying both sides of this inequality here by P so you get P Q is greater than P, and then equality course does not flip because Pia's positive, it's greater than one. And now just notice that P. Q is X so X is using the transitive property greater than P. So again, call him a is larger than calling B, and the answer is a.
101. Quantitative Comparisons Problem 82: we're given The radius of the circle is 2.5. So this radius here is 2.5, as is this and noticed that the diameter C B is simply there. Some, which is five now. We have a right triangle here so he can use the Pythagorean theorem, which says that the some of the legs squared a C squared, plus the other leg. A B, which is four square, equals I partners squared, which is five. So a C squared plus 16 equals 25 or a C squared equals nine. Take the square root herbal sides the square cancels with with the square root, which is why we took the square. Irritable asides. We get three. So the length of A C is three and calling B is four. It's calling. B is larger.
102. Quantitative Comparisons Problem 83: since X is the radius of circle C. The area of circle C is pie X squared and the perimeter of circle C is two pi X. This perimeter is equal to the perimeter of the square s. Therefore, the site of S is pie X over two. So the area of the square is pie X over two squared, which is pi square X squared over four. Now pies approximately three we get for column a three X Square and for call him be we get nine times X squared over four, which is two and 1/4 X squared. Hence column A, which is 3/4 larger than call him be, is larger.
103. Quantitative Comparisons Problem 84: since consecutive angles of a parallelogram are supplementary, we get X plus five plus 20 minus y equals 1 80 Adding up the like terms we get X minus y plus 25 is equal 1 80 Subtracting 25 gives us X minus y equals 1 55 and finally adding white to both sides. We get X equals 1 55 plus why this equation says that 155 units has to be added to why to make it as large as acts hence X is greater than why and call him a is greater than calling B.
104. Quantitative Comparisons Problem 85: notice that our formula into the Asterix means to take the square root of end. We can rewrite this slightly by saying, If you raise something to the Asterix, it will equal the square root of that something. So whatever is plugged in here will pop out here and turning to call him a notice that we have into the fourth power inside the parentheses. So we will get into the fourth power here, which will pop out here, and the square root of into the fourth power is in square, so we now have in squared to the Asterix and then apply the same formula again, which will give us the square root of whatever's inside. The parentheses, which is in squared in the square root of n squared, is in. Technically, it's the absolute value of N because it will always be positive. But we're told that in is positive, so weaken drop the absolute value. It's unnecessary. Hence column A is equal to call and be
105. Quantitative Comparisons Problem 86: Let's solve the equation in order to determine the possible values for acts. Now the factors. A 10 or 10 times one and there some is 11 and their differences. Nine neither, which is seven. But other factors of 10 or five and two and there's some five plus two is seven. So the equation factors into and X and an ex of five and a two a negative and a negative setting. These factors equals zero we get X minus five equals zero and X minus two equals zero or X equals five and X equals two. If X is five, then call and be is 25 which is greater than call him a And if X is equal to two and calling B. S four, which is smaller than column A. This is a double case. Therefore, the answer is D.
106. Quantitative Comparisons Problem 87: notice that the right hand side of the equation look similar to call and be so. See if we can manipulate the equation and make it look exactly like Holland B. Squaring both sides. Well, cancer, the radical on the left. So we have X minus y, and on the right, we have a perfect square. Try no meal, which is X radical X squared minus twice the product plus one squared, which reduces the X minus two radical acts plus one canceling the X from both sides. Give. And now I'm multiplying through by negative one. We get why equals to radical X minus one. And sure enough, we were able to turn call. We were able to turn the expression in to call them be so call him A is equal to column B, and the answer is C.
107. Quantitative Comparisons Problem 88: we have vertical angles here on the top, so this angle is also acts. Now Recall that the angle some of the triangles 1 80 so x plus z plus wine equals 1 80 I also recall that a straight angle has 180 degrees. So why, plus V equals 1 80? Now there's notice that each of these expressions equal 1 80 Therefore, they must equal each other, canceling the wife from both sides. We get X Plus Z is equal to be which says that column may is equal to column B and the answer c.
108. Quantitative Comparisons Problem 89: Let's solve this problem by substitution. If X equals Y equals two, then the equation is satisfied because two times two equals four and call him a becomes four. Now four is greater than three. So for this case, call him a is larger enough X equals y equals negative, too. Then again, it satisfies the equation because negative, too. Times negative, too, is for and now call him a becomes negative. Two plus negative, too, which is negative. Four and negative four is less than three. So we have a double case and the answer is D.
109. Quantitative Comparisons Problem 90: let's determine the possible values for end by solving the equation. Now 16 divides into 256 16 times, so left side becomes 64 times 16 and the right side becomes four to the end plus one, because when you multiply bases, you add exponents. For example, X to the a times X to the B gives us X to the A. Plus B, now 64 is four to the board to the third power, and 16 is four square, so the left side is four to the fifth power. Therefore, five must equal in plus one. Since the basis are the same. Subtracting one we get in is equal to four. Therefore, calling a is larger than call him be.
110. Quantitative Comparisons Problem 91: recall that the slope of the line is the rise of the run, so M equals Delta y over Delta X. We're Delta spends for the change the change in the why or the change in the X calculating the slope between the 0.3 negative two and the origin 00 We get negative 2/3 now. Kathleen the slope between these two points we get No, Since the pairs of points are in the same line, their slopes have to be the same. Solving this equation for age by clearing fractions we get H plus two is equal to negative two and subtracting, too. We get H equals negative four and negative four is larger than negative five because on the number line negative four is to the right of negative five. It's called Mbia's larger.
111. Quantitative Comparisons Problem 92: let r be the radius of the smaller circle, then the circumference of the smaller circle is two pi r. So column B, which is twice the circumference of the smaller circle, becomes four pi r. Now, since the radius of the smaller circles are this, distances are as well and then notice that this distance here is the radius of the larger circle which now is to our so this conference of the largest circle is two pi times the radius, which is to our, which is four pi r, which is the same as calling be. Therefore, the answer is C.
112. Quantitative Comparisons Problem 93: Let's clear the fractions by both point. Both Collins by 14. The LCD. So call him a becomes seven. Route to and call him be becomes five times two or 10. And since everything's positive, weaken square. Both Collins, without affecting the inequality and then call him a becomes 49 times two and call them be becomes 100. So we get 98 versus 100. Since 100 is bigger than 98 call them be is larger.
113. Quantitative Comparisons Problem 94: let the radius of the circle be acts. So this distances acts and this distances acts. Now the area of a triangle is 1/2 the base times the height, which we're told is a no in a right triangle, the base and the height can be considered the legs. So because X and HSX so we get 1/2 X squared equals eight. We'll find by two we get X squared, equals 16 and then taking the square root, we get X equals four. Now the area of a circle is pie R squared, and we've already set the radius of the circle to be acts. So we get pie X squared, which is pi times foreswear because we determined access for which gives us 16 pie. Now 16 pie is smaller than 64 pi, therefore column B is larger.
114. Quantitative Comparisons Problem 95: Let's solve the equation to determine the possible values for n subtracting the export on the bottom from the exploded on the top. It gives his 20 minus two in now. Since the basis of the same, namely five, the exponents must be equal, so 12 equals 20 minus two. In subtracting 20 from both sides, we get negative. Eight equals negative to end, dividing both sides by negative too. We get four is equal to him. Hence the columns are equal and the answer is C.
115. Quantitative Comparisons Problem 96: Let's solve this promise. Substitution. Choosing X to be negative. One. Call him a becomes. Now. The absolute value of negative one is positive One and a positive number divided by a negative number is negative. Negative one. In this case, call him a is smaller than calling B. Now choose X to be positive one. And that suit value positive one is one and 1/1 is one. Now call him A Is larger than calling be So we have a double case and the answer is D.
116. Quantitative Comparisons Problem 97: performing the operations in this equation gives 12 times X equals 10 times. Why now? Column A is X over y. So let's create that term out of this equation. Dividing both sides of the equation by wide gives us 12 x over y equals 10. Now divide both sides by 12 and we get X over. Why equals 10 12 which reduces to 56 versus 4/5 and to determine which ones larger we could cross. Multiply five times five is 25 and six times four is 24 since 25 is greater than 24 56 is greater than 4/5 hence COLUMN A is greater than colon be.
117. Quantitative Comparisons Problem 98: There is not enough information here to the side because different sized circles can have different results, for example, and draw two different circles for column A for the 1st 1 choose this to be three and three and this court to before here the area the circle is nine pie. If we change, take the same chord of length four, but on a bigger, bigger circle where this is five, then the area is 25 pi. So here the area is smaller than call him being here. It's larger than calling B. This is a double case, so the answer is D.
118. Quantitative Comparisons Problem 99: since two is greater than one, it's square root is still greater than one, and since 1/2 is less than when it's square is less than one, therefore column A is larger than calling B
119. Quantitative Comparisons Problem 100: Since both columns are positive, we can square both columns to eliminate the radicals, and that won't change the direction of the inequality. So call him be becomes X plus y and call him a is a perfect square. Try no meal plus twice the product plus radical y squared and cancelling the Radicals we get X plus two x to radical X radical wide plus. Why now? Just noticed that column A is bigger than call him be by this amount, and this is a positive number. Therefore COLUMN A Is bigger than column B.
120. Quantitative Comparisons Problem 101: since X is greater than one, we can multiply both columns by axe. It won't change the direction of the inequality. No to stripping the X and column. May we get X minus one versus X Times X minus one Now again, since X is greater than one X minus, one is greater than zero, so we can divide both columns by X minus one. And it won't affect the direction that inequality canceling we get X for column B and one for column A and again, we're given that X is greater than one. Therefore column B is greater than column a.
121. Quantitative Comparisons Problem 102: adding, like terms and column B gives us four plus two acts and performing the multiplication column May by foiling, we get four plus two x plus another two acts plus X Square. Adding up like turns, we get four plus four X plus X squared. Now subtracting four from each column and two acts. Gives us two X plus X squared and call him A and gives us zero and column B now notes that were given that X is greater than zero. Therefore, this expression is greater than zero hence column A is greater than column B.
122. Quantitative Comparisons Problem 103: Let's solve this problem by substitution, letting X equal. Why equal to then it will solve. Satisfy the inequalities in the product of X and Y will be two times two or four and the sum of X and y. We'll also before. So in this case, the columns are equal. Now let's choose X equal. Why equal three, Then column A is three times three or nine, and Colin B. Is three plus three or six now call him a is larger. This is a double case there for the answer is D.
123. Quantitative Comparisons Problem 104: Let's solve the given inequalities to find the possible ranges for L and M dividing the first inequality by two. We get that l is greater than three and dividing the second inequality by three we get em is less than three. Since L is always bigger than three and M is always smaller than three. L is greater than him. It's calling a is larger.
124. Quantitative Comparisons Problem 105: Let's subtract X and Y from both columns, so call him a becomes nine x and column B becomes nine. Why now? Divide both columns by nine and we get X versus Why? But we're given that X is greater than why, hence colonnades larger.
125. Quantitative Comparisons Problem 106: Since P is greater than zero, we can divide both columns by P, which will cancel it out now. Each expression that it remains. In column A and column B Art is a difference of squares, so call him a becomes P squared minus one squared or P squared minus one and column B becomes P. Squared minus four. Subtracting piece square from both columns gives a negative one versus negative four. Negative one is greater than negative for because of negative one is to the right of negative four on the number line, hence call him a is larger.