Inequalities

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Inequalities

Basic Principles to deal with inequalities

Adding the same expression to each side of an equality results in an equivalent inequality, that is,

                             if       a < b           then
                                 a + c < b + c
Ex:                                  2 < 10
                                  2 + 5 < 10 + 5
Subtracting the same expression from each side of an inequality results in an equivalent inequality, that is

                             if       a > b           then
                                 a - c > b - c
Ex:                                 -2 > -10
                                  -2 -5 > -10 - 5
Multiplying or dividing each side of an inequality by the same positive expression results in an equivalent inequality, that is

                             if       a > b           then
                                    ac > bc          when c > 0
Ex:                                 -2 > -10
                                -2 x 5 > -10 x 5
Multiplying or dividing each side of an inequality by the same negative expression results in a reversal inequality, that is

                             if       a > b           then
                                    ac < bc         when c < 0
Ex:                                 -2 > -10
                               -2 x -5 < -10 x -5
If both sides of an equality have the same sign, inverting both sides of the inequality results in a reversal inequality, that is,
                              if 0 < a < b           then
                                  0 < 1/a < 1/b
Ex:                             0 < 2 < 4   then   0 < 1/4 < 1/2
                              if a < b < 0           then
                                  1/b < 1/a < 0
Ex:                             -4 < -2 < 0   then   -1/2 < -1/4 < 0
If two inequalities are of the same type, i.e., both greater of both less, adding the respective sides results in the same type of inequality, that is,
                              if      a < b     and c < d    then
                                a + c < b + d
Ex:                                  2 < 5       and -5 < -2 then
                             2 + (-5) < 5 + (-2)
Note that            2 - (-5) = 5 - (-2)       (Just to remind you that this rule applies to adding only. Also realize that it has some interesting implications you will see in practices.)


		Problem Solving Questions If u > t, r > q, s > t and t > r, which of the following must be true? I. u > s II. s > q III. u > r A. I only B. II only C. III only D. I and II E. II and III If b < 2 and 2x - 3b = 0, which of the following must be true? A. x > -3 B. x < 2 C. x = 3 D. x < 3 E. x > 3 If 0 < x < 4 and y < 12, which of the following CANNOT be the value of xy? A. -2 B. 0 C. 6 D. 24 E. 48 If the quotient a/b is positive, which of the following must be true? A. a > 0 B. b > 0 C. ab > 0 D. a - b > 0 E. a + b > 0 Bill’s school is 10 miles from his home. He travels 4 miles from school to football practice, and then 2 miles to a friend’s house. If he is then x miles from home, what is the range of possible values for x ? A. 2 < x < 10 B. 4 < x < 10 C. 4 < x < 12 D. 4 < x < 16 E. 6 < x < 16 Which of the following describes all values of x for which 1 - x² > 0? A. x > 1 B. x < -1 C. 0 < x < 1 D. x < 1 or x > 1 E. -1 < x < 1 If it is true that x > -2 and x < 7, which of the following must be true? A. x > 2 B. x > -7 C. x < 2 D. -7 < x < 2 E. None of the above

	Answers: D. In combination, you get u > t > r > q and s > t > r > q but no relation between u and s. I and II are correct. D. From 2x - 3b = 0, then b = 2x/3 Given b < 2, then (from above), 2x/3 < 2 ==> x < 3 E. If x = 4, and y = 12 ==> xy = 48 If x = 4, and y < 12 ==> xy < 48 C a/b > 0 ==> both a and b > 0 or both a and b < 0 ==> ab > 0 D. The maximum distance is under the condition that he travels further and further away from school in the same direction, that is 10 + 4 + 2 = 16. In contrary the minimum one is under the condition that he travels closer and closer from school in the same direction, that is, 10 - 4 - 2 = 4 E. 1 - x² > 0 ==> x² - 1 < 0 ==> (x - 1)(x + 1) < 0 ==> -1 < x < 1 B. In combination, -2 < x < 7. A, C and D cannot be always true with all of x above.


	Data Sufficiency Questions Is it true that a > b 1. 2a > 2b 2. a +c > b + c A certain company currently has now many employees? 1. If 3 additional employee are hired by the company and all of the present employees remain, there will be at least 20 employees in the company 2. If no additional employees are hired by the company and 3 of the present employees resign, there will be fewer than 15 employees in the company. If x is equal to one of the numbers 1/4, 3/8 or 2/5 what is the value of x ? 1. 1/4 < x < 1/2 2. 1/3 < x < 3/5 If a, b and c are integers, is a - b + c greater than a + b - c ? 1. b is negative 2. c is positive If x and y are positive, is x/y greater than 1 ? 1. xy > 1 2. x - y > 0 If xy < 3, is x < 1 ? 1. y > 3 2. x < 3 If x + y + z > 0, is z > 1 1. z > x + y + 1 2. x + y + 1 < 0 Is x greater than 1.8 ? 1. x > 1.7 2. x > 1.9 Is z less than 0 ? 1. xy > 0 and yz < 0 2. x > 0 Is xy > 5 ? 1. 1 < x < 3 and 2 < y < 4 2. x + y = 5 Is x a negative number ? 1. 9x > 10x 2. x + 3 is positive. Is the value of n closer to 50 than to 75 ? 1. 75 -n > n - 50 2. n > 60 Is xy < 6 ? 1. x < 3 and y < 2. 2. 1/2 < x < 2/3 and y² < 64

		Answers: (1) 2a > 2b ==> a > b OK (2) a + c > b + c ==> a > b OK Both 1 and 2 are OK. Choose D. Let x be the current number of employees. (1) x + 3 > 20. You never know what exactly x is. (2) x - 3 < 15. The same implication as that of (1). However, (1) gives x > 17 and (2) gives x < 18. The only positive number that match both conditions is 17. Choose C. (1) is not sufficient since both 3/8 and 2/5 comply with the condition. (2) is also not sufficient with the same reason as that of (1). Even though you combine (1) and (2), you still find the two answers, i.e., 3/8 and 2/5. Choose E. Assuming a - b + c > a + b - c then 2c > 2b ==> c > b You now need both (1) the fact that b < 0 and (2) the fact that c > 0 to conclude that c > b. Choose C. (1) implies various possible values of xy. Try, for example x = 1/10, y = 11 or vice versa you get x/y < 1 and xy > 1 respectively. (2) x- y > 0 ==> x > y ==> x/y > 1. (Remember that x and y > 0) In such a problem, the method of substitution works best. (1) If y > 3, given xy > 3, the only possible x must be less than 1 == x <1 OK. (2) Leave it. x < 3, then x can be greater than 1 or less than 1. Choose A. Apply rule no. 6 on our review. (1) x + y + z > 0 and z > x + y + 1. Adding the two inequalities ==> x + y + z +z > x + y + 1 ==> z > 0.5 Not OK (2) x + y + z > 0 and x + y + 1 < 0 ==> -x - y - 1 > 0. Adding both ==> z - 1 > 0 ==> z > 1 OK. Choose B. Quite obvious. Choose B. (1) implies (a) x > 0 and y > 0 OR (b) x < 0 and y < 0 and with the condition of yz < 0, suggesting z > 0 or z < 0 depending upon y. Not OK. (2) gives you nothing further. (1) and (2) imply that y > 0 and using the condition of yz < 0, then you get z < 0. Choose C. (1) is not sufficient. Choosing x = 1 and y = 2, and x = 3 and y = 4, you get xy = 2 and 12, which do not match the question. (2) Choosing x = 4, y = 1, and x = 3, y = 2, you get quite a similar result. Not OK (1) and (2) see (2) above. Choose E. (1) 9x > 10x ==> -x > 0 ==> x < 0. OK (2) is obviously inconclusive. For example both x = -2 and x = 5 comply with the condition of x + 3 is positive. Choose A. (1) 75 - n > n - 50 ==> 125 > 2n ==> 62.5 > n or n < 62.5, suggesting n is less than the average of 75 and 50. Then n must be closer to 50 than to 75. OK (2) is not sufficiency since n could be 61, 62.5 or 70. Choose A. (1) if x = -3 and y = -3 then xy > 6. Not OK. (2) y² < 64 ==> -8 < y < 8. Trying the maximum possible xy < 8 x 2/3 < 16/3. You can conclude that xy < 6. Choose B.


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	Last updated: Sep 18, 2001