Inequalities

If b<4, and 1 ≤ a < 5, the value of b/a CANNOT be 4 1 0 −5 −20

Correct. If b is smaller than 4 and a is at least 1, b/a has to be smaller than 4/1. Also, all other answer choices can eliminated by plugging in values which satisfy the inequalities, and for which b/a CAN equal the answer choices.

If a/b>1, then which of the following must be true? Correct. In "must be true" questions plug in numbers and eliminate answers. Remember to plug in wisely - for each answer choice, think of DOZEN F numbers which satisfy the inequality a/b>1, but for which the answer choice is NOT true. Plugging in a=2, b=1 eliminates answer choice B, C and E. These numbers satisfy answer choice A and A and D, so they cannot be eliminated for this plug in. But Are A and D always true, for any number? Plug in a=-2, b=-1 to eliminate A and leave yourself with the one right answer choice - D. Why is this answer always true? If a/b>1, then |a|>|b|. Hence, b/a has to be less than 1.

1. b<a 2. a<b 3. a^2<b^2 4. b/a<1 5. 1<a−b

Which of the following describes all the values of x for which x·(8−4x)<0? Correct. Don't let evil algebraic expressions throw you. Eliminate answer choices by plugging in values for x. Prove the incorrect answer choices are false by finding a value of x which satisfies the question stem, but is outside of the given range they offer. Alternatively, figure out when x·(8−4x) is negative, and eliminate any answer choice which doesn't describe the entire range of possibilities. By way of elimination, this is the correct answer: Answer choices C and D describe values not actually satisfying the question stem. Answer choice A is wrong since x may also be greater than 2, not only a negative number. Answer choice E is wrong since x may also be a negative number. This answer choice covers both options, and is therefore correct.

1. x < 0 2. x < 0 or x > 2 3. 0 < x < 1/2 4. 0 < x < 2 5. x > 2

Some GMAT problems require you to solve a set of simultaneous inequalities. We propose two possible approaches: Adding the inequalities, and plugging in the extremes. For example: if 2<x<10 and 5<-y<7, what can be said about x+y? The -y gets in the way of reaching the desired result of x+y. Multiply the second inequality by -1. Don't forget to flip the sign: 5<-y<7 /·(-1) --> -5>y>-7 Now, line the inequalities up so that the sign goes in the same direction: 2<x<10 -7<y<-5 Finally, add the inequalities: 2-7<x+y<10-5 --> -5<x+y<5

Adding the inequalities 1) First, line up the inequalities so that the sign goes in the same direction. 2) Add the inequalities, much as you would a set of simultaneous equations. Always add! There are rules for subtracting inequalities, but they involve specific cases and scenarios depending on whether the variables are positive or negative. It is better to manipulate inequalities by multiplying by -1 until they reach the state where adding will get you where you need to go.

If 8^-x < 1 / 32^3, what is the smallest integer value of x? 4 5 6 8 9

Correct. First, convert the negative power of 8 into a positive one i.e. 8^-x = 1 / 8x 1 / 8^x < 1 / 32^3 Cross multiply to put 8x out of the bottom of the fraction: --> 1 / 8^x < 1 / 32^3 /⋅8^x --> 1 < 8^x / 32^3 /⋅32^3 --> 32^3 < 8^x Express everything as a power of 2: 8 = 2 x 2 x 2 = 2^3 32 = 2 x 2 x 2 x 2 x 2 = 2^5 32^3 < 8^x can be expressed as (2^5)3 < (2^3)x --> 2^15 < 2^3x Now that the bases are the same, the powers can be compared directly ---> 15 < 3x ---> 5 < x Hence, the smallest possible integer value of x is 6.

If x and y are positive integers, and 4x2=3y, then which of the following must be a multiple of 9? I) x2 II) y2 III) xy

Correct. I) Since x is divisible by 3, x2 must be divisible by 3·3 = 9. II) If x2 is divisible by 9, then 3y = 4x2 must also be divisible by 9. Write this as a fraction: 3y/9 = {some integer}. Reduce by 3 to get y/3 = {some integer}. It follows that y is divisible by 3, and therefore y2 must be divisible by 32 = 9. III) Since each of the integers x and y must be divisible by 3, xy must be divisible by 3·3 = 9.

Which of the following describes all the values of y for which 1/y2 < 1/10? y > √10 or y < −√10 y < √10 y < −√10 −√10 < y < √10 y > √10

Correct. Manipulate the inequality as you would an equation. It's ok to multiply by y2 - since an even power is non-negative, no need to flip the sign. Remember that an inequality involving even powers has two possible cases: x2≤a: -√a≤x≤√a x2≥a: x≥√a or x≤-√a 1/y2 < 1/10 Multiply by y2: ---> (1/y2)*y2 < (1/10)*y2 ---> y2/y2 < y2/10 Reduce the y2 on the left side of the equation: ---> 1 < y2/10 multiply by 10 ---> 10 < y2 This is an inequality of the form x2≥a, sox≥√a or x≤-√a. In this case, this means y > √10 or y < −√10. If you're not sure, plug in values of y that satisfy the inequality, such as y=±4. All other answer choices are eliminated for these plug ins, so this is the right answer choice.

If |3s + 7| > 5, what is the range of values for s? 2/3 < s < 4 -2/3 < s or s < -4 -4 < s < -2/3 s > -2/3 s > -4

Correct. Solve absolute values of the number case by considering two possible scenarios: First scenario - copy the inequality without the absolute value brackets and solve: 3s + 7 > 5 --> 3s > 5 - 7 --> 3s > -2 Divide both sides by 3 to isolate s 3s/3 > -2/3 --> s > -2/3 Second scenario - remove the absolute value brackets. Put a negative sign around the other side of the inequality, AND flip the sign 3s + 7 < -5 --> 3s < -5 - 7 --> 3s < -12 Divide both sides by 3 to isolate s 3s/3 < -12/3 --> s < -4 Finally, combine the two scenarios into one range for s: -2/3 < s or s < -4 Hence, this is the correct answer.

How many different values of positive integer x, for which |x+8|<x, are there?

Correct. The issue of this question is an inequality combined with an absolute value. Note that this inequality is of the variable form, i.e. variables on both sides. Thus, the two-scenario approach will not necessarily work. Instead, plug in numbers to find out what the inequality really means. Plug in x=2 --> |x+8| = |2+8| = |10| = 10 is NOT smaller than x=2. Therefore, x cannot equal 2. Try other positive values of x: Plug in x=3 --> |x+8| = |3+8| = |11| = 11 is NOT smaller than x=3. Therefore, x cannot equal 3. Plug in x=0.5 --> |x+8| = |0.5+8| = |8.5| = 8.5 is NOT smaller than x=0.5. Therefore, x cannot equal 0.5. Notice a pattern? For every positive value of x, the left side of the equation will always be greater by than the right side. Thus |x+8| will never be smaller than x, and there are no values of x which satisfy the inequality.

If |m/5| > |m+10/10| , which of the following could be true? 2<m<10 m>10 0<m<2 m=5 5<m<7

Correct. The issue of this question is an inequality combined with an absolute value. Note that this inequality is of the variable form, i.e. variables on both sides. thus, the two-scenario approach will not necessarily work. Instead, plug in numbers to find out what the inequality really means. Notice that this is a "could be true" question. In this type of question, if just one plug in fits the inequality, you can say that the answer could be true, and therefore that it is the correct answer. Since the question is "could be" and not "must be" true, it is sufficient to plug in once. For example, plug in m=20. You get |4|>|3|, which is true, and thus it could be that m>10, and this is the correct choice. Also, by POE, this is the only choice left.

if x≠0 or 2, is 3/(x-2) >0 ? (1) 1/(x-2) >0 (2) -2/x-2 < 0

Correct. This is a DS Yes/No question. The issue is whether 3/(x-2) is positive. Answering a definite "Yes" or a definite "No" means Sufficient. If the answer is sometimes "Yes" and sometimes "No", it means Maybe, which means Insufficient. A fraction will be positive if the numerator and the denominator are either both positive or both negative. Since 3 is positive, 3/(x-2) is positive when x-2 is positive. Thus, the real issue in the question is actually the answer to the question "Is (x-2)>0" ? According to Stat. (1), 1/(x-2) >0 Since the numerator is positive (1), the fraction on the left will be positive if the denominator is also positive. Thus, X-2 > 0, and the answer is "Yes". Stat.(1)->S->AD. According to Stat. (2), Since the numerator is negative, the fraction on the left is negative when the denominator is positive. Thus, X-2 > 0 and the answer is 'yes'. Stat.(2)->S->D.

Is a<b? (1) a/b<1 (2) a·b>0

Correct. This is a Yes/No DS question. Answering a definite "Yes" or a definite "No" means Sufficient. If the answer is sometimes "Yes" and sometimes "No", it means Maybe which means Insufficient. The issue is Inequalities. Plug in numbers that satisfy the statements and show that a<b and the answer is "Yes". Then ask yourself "is this always true, for any number?" Think of DOZEN F numbers to reach an answer of "No" and prove the statement(s) insufficient. Stat. (1): Plug in a=2 and b=3, so that a/b = 2/3 <1. a<b, so the answer is "Yes". But is it always "Yes" for any number? Note that the question did not specify that a and b must cannot be negative. Plug in negative values for a and b, e.g. a=-2 and b=-3. a/b = -2/-3 = 2/3<1, so these values also satisfy the statement. However, in this case a=-2 > b=-3, so the answer is "No". That's a "Maybe", so Stat.(1)->Maybe->IS->BCE. Stat. (2): a·b>0 means that a and b share the same sign - either both positive or both negative. However, it doesn't tell you anything about their relative value - Plug in a=2 and b=3, and the answer is "Yes"; plug in a=3 and b=2 and the answer is "No". No definite answer, so Stat.(2)->Maybe->IS->CE. Stat.(1+2): the combination of the statements still allows the values we used for stat. (1): Plug in a=2 and b=3, so that a/b = 2/3 <1. a<b, so the answer is "Yes". However, plug in negative values for a and b, e.g. a=-2 and b=-3. a/b = -2/-3 = 2/3<1, so these values also satisfy the statement. However, in this case a=-2 > b=-3, so the answer is "No". That's a "Maybe", so Stat.(1+2)->IS->E.

Is 5<m<7? (1) |m|>5 (2) |m|<7

Correct. This is a Yes/No DS question. Answering a definite "Yes" or a definite "No" means Sufficient. If the answer is sometimes "Yes" and sometimes "No", it means Maybe which means Insufficient. The issue is inequalities involving absolute values. For each statement, try to show that m COULD be within the specified range ("Yes"), but could also be outside the range ("No"). Stat. (1): according to the rules of inequalities involving absolute value, either m>5 or m<-5. m>5 allows values of m=6, yielding an answer of "Yes", but also allows values of m=8, yielding an answer of "No". That's a "Maybe", so Stat.(1)->Maybe->IS->BCE. Stat. (2): according to the rules of inequalities involving absolute value, either m<7 or m>-7. m<7 allows values of m=6, yielding an answer of "Yes", but also allows values of m=4, yielding an answer of "No". That's a "Maybe", so Stat.(2)->Maybe->IS->CE. Stat. (1+2): combining both inequalities with their various scenarios is a little complicated, so try to find values of m that satisfy both and still give opposite results. m=6 satisfies both inequalities (|6|=6 is greater than 5, but less than 7) and yields an answer of "Yes". However, can you think of a DOZEN F value for m that is outside the specified range? Plug in m=-6, which also satisfies both statements (|-6| = |6| = 6), but is not between 5 and 7, yielding an answer of "No". That's still a "Maybe", so Stat.(1+2)->Maybe->IS->E.

Which of the following describes all the values of m for which m4 > 16? m > 4 −2 < m < 2 m > 2 m > 2, or m < −2 m > 4, or m < −4

Correct. When x2 ≤ a, then -√a ≤ x ≤ √a. When x2 ≥ a, then x ≥ √a or x ≤ -√a. In this question you are dealing with the second scenario: If m4 > 16, then m2 < -4 or m2 > 4 (m2 is the square root of m4, ±4 are the roots of 16). Since you are looking for the values of m (and not of m2), continue to handle the two inequalities involving m2. First scenario: m2 < -4 Stop. Is that even possible? m2 is never negative. One inequality down, one to go: Second scenario: If m2 > 4 means m < -2 or m > 2. That's all, folks! Alternative explanation: The question asks which of the following describes all of the values of m for which the inequality is true. This means that there are four answer choices here that do NOT describe all of the values of m. Plug in values of m which satisfy the inequality in the question stem, and eliminate the answer choices which do not allow these values. Keep plugging in values until you have eliminated four answer choices, and then the last one must be the correct one. m2>4 allows m to equal 3, since 32=9 will be greater than 4. Eliminate answer choices A, B and E, since they do not allow m to equal 3. C and D are close, with the main difference that D allows negative values of m, which C does not. Does the original inequality support negative values? Since anything squared is always positive, m can be negative as well: if m=-3, then (-3)2 will still equal 9, which is greater than 4. Thus, m=-3 is a valid value for m, and we can eliminate C because it does not allow m to be negative. D is the only remaining answer, and must be correct.

Some GMAT problems present inequalities involving absolute value. Dealing with these questions is a multi step process: First, isolate the absolute value so that it stands alone on one side of the inequality. Next, differentiate between two different cases: the number case, and the variable case. In this lesson we will deal with the number case. In this case, the other side is a number, i.e., the inequality has the form |something| < Number or |something| > Number We'll deal with the variable case (where the other side of the inequality includes a variable) later. For now, just remember this: Inequalities involving absolute values - the number case Solve inequalities with an absolute value on one side and a number on the other by considering two scenarios: 1) First scenario - copy the inequality without the absolute value brackets and solve. 2) Second scenario - remove the absolute value brackets. Put a negative sign around the other side of the inequality, AND flip the sign. Solve.

Example: If |x-6|<2, what is the range of values for x? Solve absolute values of the number case by considering two possible scenarios: First scenario - copy the inequality without the absolute value brackets and solve: x-6 < 2 --> x < 2+6 = 8 Second scenario - remove the absolute value brackets. Put a negative sign around the other side of the inequality, AND flip the sign x-6 > -2 --> x > -2+6 = 4 Finally, combine the two scenarios into one range for x: 4<x<8.

When dealing with inequalities involving even exponents, you must consider two scenarios. We also differentiate between two cases of quadratic inequalities, exemplified by x2≤a and x2≥a. Consider this example of the first case: x2≤4 If this were an equation (x2=4), it would have two solutions: x=2 or x=-2. Since this is an inequality, we consider the following scenarios:

First scenario: if x is positive, then x≤2. Second scenario: if x is negative, then x≥-2. Notice that the inequality sign flips. Combine the two scenarios into a range -2≤x≤2. To ensure that you considered the right scenarios, plug a number within the range (zero is a good idea) into the original inequality: 02 = 0 ≤ 4. Since it satisfies the original inequality, this is the right answer.

Which of the following describes all the values of y for which y < y2? 1 < y −1 < y < 0 y < −1 1/y < 1 0 < y < 1

Incorrect. Do not be tempted to divide by y and reach 1 < y - multiplying or dividing by an unknowns is forbidden, unless you know the unknown's sign. Which you don't. Don't get dragged into different scenarios of a positive or negative y - plug in numbers and try to eliminate 4 answer choices. The last one left standing is the right answer choice. Eliminate this answer choice by plugging in y=2. 2 < 22, so y=2 satisfies the inequality in the question stem, but this answer choice does not allow y=2. Correct. Plug in a good number for y: y=2 satisfies the inequality, as 2 < 22. Eliminate answer choices B, C and E, as they do not allow a value of y=2. A and D are not eliminated for this plug in: A 1 < 2 D 1/2 < 1 To decide between A and D, think of DOZEN F values for y. Try a negative y such as y=-2. y=-2 satisfies the inequality, as -2 < (-2)2. Plug in y=-2 into A and D: A is eliminated, because -2 is not greater than 1. D is the only answer choice not eliminated, so it must be the right answer choice.

If -13 < 7a + 1 < 29 and 19 < 2 - b < 23, what is the maximum possible integer value of a + b? -23 -18 -14 -13 -12 Correct. 7a + 1 and -b get in the way of reaching the desired result of a + b. Subtract 1 from the first inequality to isolate a. -13 - 1 < 7a + 1 - 1 < 29 - 1 --> -14 < 7a < 28 Now, divide by 7, to isolate a: -14/7 < 7a/7 < 28/7 --> -2 < a < 4 Subtract 2 from the second inequality: 19 - 2 < 2 - b - 2 < 23 - 2 --> 17 < -b < 21 Multiply the second inequality by -1, to isolate b. Don't forget to flip the sign: 17/(-1) < -b/(-1) < 21 / (-1) --> -17 > b > -21 Now, line the inequalities up so that the sign goes in the same direction: -2 < a < 4 -21 < b < -17 Finally, add the inequalities: -23 < a + b < -13 Since we're looking for the integer value of the sum, which must be smaller than -13, go down one integer to -14.

Incorrect. If you've done your calculations correctly, you must have reached the conclusion that a+b must be smaller than -13, which means that the sum cannot equal -13. Look for a smaller answer. First, line up the inequalities so that the signs go in the same direction. Then, add the inequalities, much as you would a set of simultaneous equations. Remember - in simultaneous inequalities, always add - manipulate inequalities by multiplying by -1 until they reach the state where adding will get you where you need to go Incorrect. Incorrect. If you've done your calculations correctly, you must have reached the conclusion that a+b must be smaller than -13. Don't be hasty, though - is -12 really smaller than -13?

What is the value of /a/divide by a ? (1) a < /a/ (2) blah blah blah... Read the question stem and statement (1) alone, and decide sufficient/insufficient.

Incorrect. Plug in numbers that fit the statement to get a grip on the question. Correct. Plug in good numbers that fit statement (1). For instance, plug in a=10. Start by checking if the number fits the statement. In this case - it doesn't. Pick another number that works. For instance, a=-10. In fact, no positive number will fit into a. (a=0 does not fit the statement either) Now answer the question ▐-10▐ /10 = 10/-10=-1. In truth ▐ a▐/a = -1 for any negative number. Therefore, statement (1)->Sufficient.

Is |a|>8? (1) a > −8 (2) −a < −8 Correct. Stat. (1): a>-8 allows all values of a greater than -8, positive or negative. a could be 9, yielding an answer of "Yes" (since |9|>8. However, a could also be 7, yielding an answer of "No", since |7| is not greater than 8. Stat.(1)->Maybe->IS->BCE. Stat. (2): Isolate a by multiplying both sides of the inequality by -1. Don't forget to flip the sign. The new inequality is a > 8. Therefore, |a| > |8|, and the answer is a definite "Yes". Stat.(2)->Yes->S->B.

Incorrect. This is a Yes/No DS question. Answering a definite "Yes" or a definite "No" means Sufficient. If the answer is sometimes "Yes" and sometimes "No", it means Maybe which means Insufficient. The issue is inequalities + absolute value. Plug in numbers that satisfy the statements and show that |a|>8 and the answer is "Yes". Then ask yourself "is this always true, for any number? Think of DOZEN F numbers to reach an answer of "No" and prove the statement(s) insufficient. Stat. (1): a>-8 allows all values of a greater than -8, positive or negative. a could be 9, yielding an answer of "Yes" (since |9|>8. However, a could also be 7, yielding an answer of "No", since |7| is not greater than 8. Stat.(1)->Maybe->IS->BCE.

Is y*a>y*b ? (1) a>b (2) y^2 * a> y^2 * b Correct. Start by plugging in good numbers that fit statement (1), such as a=3 and b=2. What about y? It is not limited by statement (1) so you can plug in positive and negative numbers such as y=2 or y=-2. For y=2 the answer to the question stem is Yes, and for y=-2 it is No. Therefore statement (1)->Maybe->IS->BCE. Go on to consider statement (2). Try to economize the process by using the same numbers, a=3 and b=2. For y, you can use y=2 or y=-2, as y2 equals 4 for both, which satisfies the statement. If y=2, the answer is "yes", but if y=-2, the answer is "no". Therefore statement (2)->Maybe->IS->CE. Combine the statements. This means that you should plug in numbers that comply with both statements. Actually, the numbers you have used so far (unless you have chosen y=0) fit both statements. So there is no "always" answer to the question stem. Therefore, statements (1)+(2)->Maybe->IS->E.

Incorrect. This is a Yes/No DS question. Answering a definite "Yes" or a definite "No" means Sufficient. If the answer is sometimes "Yes" and sometimes "No", it means Maybe which means Insufficient. The issue is inequalities. Plug in numbers that satisfy the statements and show that ya > yb and the answer is "Yes". Then ask yourself "is this always true, for any number?" Think of DOZEN F numbers to reach an answer of "No" and prove the statement(s) insufficient. Go on to consider statement (2). Try to economize the process by using the same numbers, a=3 and b=2. For y, you can use y=2 or y=-2, as y2 equals 4 for both, which satisfies the statement. If y=2, the answer is "yes", but if y=-2, the answer is "no". Therefore statement (2)->Maybe->IS.

Plugging in the extremes Another useful technique for dealing with several inequalities is to focus on their highest and lowest possible values. This is especially effective in cases where adding inequalities doesn't help: If 2≤x≤10 and 5<y<12, which of the following CANNOT be the value of x/y? A 1/6 B 2/5 C 5/6 D 1 E 3/2 To sum up: Two possible approaches for dealing with simultaneous inequalities: Adding the inequalities 1) First, line up the inequalities so that the sign goes in the same direction. 2) Add the inequalities, much as you would a set of simultaneous equations. Always add! It is better to manipulate the inequalities by multiplying by -1 until they reach the state where adding will get you where you need to go. Don't forget to flip the sign. Plugging in the extremes Focus on the highest and lowest possible values of the inequalities. Plug in those to reach the desired result proposed by the question. This is especially effective in cases where adding inequalities doesn't help.

Let's see: One of these answer choices must be outside the range of possible values for the fraction x/y. focus on the extremes to find the HIGHEST and LOWEST value for x/y. The highest value for x is 10 The lowest value for x is 2. The highest value for y is less than 12. (Since y can't be 12, we call this less than 12, or just under 12) The lowest value for y is more than 5. Now, consider our fraction x/y: The highest value for a fraction occurs when the numerator (x) is the highest, and the denominator (y) is the lowest. Therefore, the highest possible value of the fraction is 10 / more than 5: a little less than 2. The lowest value for a fraction occurs when the numerator (x) is the lowest), and the denominator (y) is the highest. Therefore, the lowest possible value for the fraction x/y is 2 / less than 12: a little more than 1/6. Since y can't be 12, x/y can't be 1/6, so A is the right answer.

Take a look at the following problem: ax > ay. Is x > y? a. Yes. b. No. c. I don't know. The lesson learned here is this: Do not multiply or divide an inequality by an unknown! If the problem stated that a>3, then you could surmise that a>0 and act accordingly. The same goes for a case where the problem states that a<-2 - then you know that a is negative and flip the sign of the inequality. But if you find yourself multiplying or dividing an inequality by a complete unknown - don't.

Precisely the answer we're looking for. We've already seen the single, greatest difference between an equation and an inequality: When you multiply or divide an inequality by a negative number, flip the direction of the sign. In the transition from ax > ay to x > y, you're dividing by a. However, unless you know a's sign, you cannot be sure what to do with the inequality sign: if a > 0 - leave the sign as is to get x > y. However, if a < 0 - then we must flip the sign to get x < y.

☺To live well mean to act☺well. To live well mean to act To live well means to act Living well mean to act Living well mean acting To live well mean acting

Right! This answer choice corrects the original Subject Verb Agreement mistake by changing the plural verb mean to the singular verb means to match the singular subject To live.

If a<b, is a2>a·b? (1) a·b<0 (2) a+b=0

You slightly overestimated the time this question took you. You actually solved it in 3 minutes and 17 seconds. Correct. This is a DS Yes/No question. The issue is whether a2 is larger than a·b. Answering a definite "Yes" or a definite "No" means Sufficient. If the answer is sometimes "Yes" and sometimes "No", it means Maybe, which means Insufficient. According to Stat. (1), --> a·b<0, thus a and b have opposite signs (one is negative and the other positive). From the question itself you know that: --> a<b, thus a is negative and b is positive. Since a is negative, a2 is positive. a·b is negative, thus the answer is 'yes': a2>a·b Stat.(1)->Yes->S->AD. According to Stat. (2), --> a+b=0. This can happen when both a and b are 0, or when they have equal values but opposite signs. From the question itself you know that: --> a<b, thus a is negative and b is positive. Since a is negative, a2 is positive. a·b is negative, thus the answer is 'yes': a2>a·b Stat.(2)->Yes->S->D.

An economic crisis are dangerous not only from an economic point of view, but also from a political viewpoint, as it is often followed by a political crisis. 1. An economic crisis are dangerous not only from an economic point of view, but also from a political viewpoint, as it is often followed by a political crisis 2. Since economic crises are often followed by political crises, they are dangerous not only from an economic point of view, but also from a political viewpoint 3. Economic crises is dangerous not only from an economic point of view, but also from a political viewpoint, as they are often followed by a political crisis 4. Because an economic crisis are often followed by a political crisis, it is dangerous not only from an economic point of view, but also from a political viewpoint 5. Economic crises is often followed by a political crisis, so it is dangerous not only from an economic point of view, but also from a political viewpoint

With a sharp rise in ecological awareness through media campaigns and changes in the educational sector, UNDERINE STARTS FROM HERE, the importance of nature preservation has even made its way into the political field. 1. the importance of nature preservation has even made its way into the political field 2. nature preservation's importance has even made its way into the field of politics 3. preserving the importance of nature has even entered politics 4. nature preservation's importance has even entered the political field 5. the importance of nature preservation has even entered politics You slightly overestimated the time this question took you. You actually solved it in 1 minutes and 30 seconds. Impressive work! This answer choice is both grammatically correct and stylistically effective as it replaces the original wordy phrases made its way into and the political field with the more concise entered and politics.

Inequalities can appear in a variety of ways in the GMAT. Take a look at the following signs, and learn to recognize them: ≠ means not equal > means that the greater quantity is to the left < means that the lesser quantity is to the left ≥ means greater than or equal to ≤ means lesser than or equal to

You should solve inequalities the same way you solve for one-variable, linear equations. However, there is one exception: If you multiply or divide the inequality by a negative number - flip the sign. Take a look at the following example, 17<-5x+7≤37 First, isolate the variable as in one-variable linear equations, subtract 7 from each part of the inequality, 17-7<-5x+7-7≤37-7 10<-5x≤30 Divide each part of the inequality by (-5), 10/-5 < -5x/-5 ≤ 30/-5 Now flip the signs of the inequality, because you are dividing by a negative number, -2 > x > or = -6

If |b| ≤ b-1, how many different values are there to b? 0 2 4 6 8

You underestimated the time this question took you. You actually solved it in 3 minutes and 13 seconds. Correct. The issue of this question is an inequality combined with an absolute value. Note that this inequality is of the variable form, i.e. variables on both sides. thus, the two-scenario approach will not necessarily work. Instead, plug in numbers to find out what the inequality really means. Remember that if |x| is smaller than y, y has to be positive. Since |b| has to be non-negative, for |b| to be smaller or equal to b-1, b-1 also has to be non-negative, which means b is positive. Now, if b is positive, decreasing its value by 1 will yield a smaller result then |b|, and thus there are no values that fit b. For example, Let's say that b=2: is |2| ≤ 2-1=1?The answer is no, so b cannot equal 2. What if b=3? |3| is not smaller than 2, so b cannot equal 3. The same goes for 4, or 10.5, or 10,000. Plugging in numbers is a way to see that the absolute value of a number cannot be smaller than the same number reduced by 1. Therefore, there are no values of b for which the inequality is satisfied.

Which of the following describes all the values of x for which |x+2|<2? (x−2)2<4 (x+2)2<4 −2<x<0 (x+2)2<2 (x−2)2<2 Alternative method: Once you've solved the inequality in the question stem, eliminate four answer choices by plugging in values which satisfy the inequality , but do NOT satisfy the answer choices. -4 < x < 0 allows the value of x=-1, but A and E do not. POE A and E. -4 < x < 0 allows the value of x=-3, but C does not. POE C. Between B and D: a little tricky to see, but -4 < x < 0 allows the value of x=--0.5, but D does not. POE D. B is the only answer choice which cannot be eliminated, so it has to be the right one.

You underestimated the time this question took you. You actually solved it in 3 minutes and 58 seconds. Correct. Solving inequalities involving absolute values involves two scenarios: 1) First scenario - copy the inequality without the absolute value brackets and solve. 2) Second scenario - remove the absolute value brackets. Put a negative sign around the other side of the inequality, AND flip the sign. Solve. Solve the inequality in the question stem. First scenario: --> x+2 < 2 --> x < 0 Second scenario: --> x+2 > -2 --> x > -4 Thus -4 < x < 0 Now match to the solutions of the answers, until you find an answer that has the same solution as the inequality in the question. Remember - When dealing with inequalities involving even exponents you are faced with two cases: x2≤a: -√a≤x≤√a x2≥a: x≥√a or x≤-√a In the answer, you have the inequality (x+2)2<4: --> -2 < x+2 < 2 --> -4 < x < 0 That's the same solution!

If 0<b, and 0>ab, which of the following CANNOT be true? You grossly underestimated the time this question took you. You actually solved it in 4 minutes and 31 seconds. Correct. If b is positive and ab is negative, then a must be negative. Try to establish the sign of the expressions in the answer choices and POE according to the inequality. If you're not sure, plug in numbers that satisfy the question's premise, such as a=-2 and b=2. Remember, you are looking for what CANNOT be true - eliminate every answer for which you can find a single example that it CAN be true. (3b+5) is positive. (a-1) is negative. Their quotient must be negative so it definitely CANNOT be greater than 8.

a. (3b+5)/(1−a)>8 b. (a−1)/(3b+5)<8 c. (3b+5)/(a−1)<8 d. (3b+5)/(a−1)>8 e. (1−a)/(3b+5)>8

Given that 1 / x < - 2/3, which of the following cannot be the value of x? Correct. The question asks which of the following cannot be the value of x. This means that there are four answer choices that CAN be the value of x, i.e. satisfy the inequality that 1/x < -2/3. Plug in the answer choices to eliminate wrong answers and find the best one. Plug in x = -3 / 2 ( 1 / x) = (1 / (-3/2)) Multiply by the reciprocal of the bottom fraction --> 1 ⋅ (-2/3) = -2/3 Since -2/3 is NOT smaller than -2/3, this answer choice cannot be the value of x, and is therefore the right answer.

a. -3/2 b. -1 c. -3/4 d. -3/5 e. -1/2

Which of the following describes all the values of x for which x(x−1)(x−2)≥0? Incorrect. Don't let evil algebraic expressions throw you. Plug in and throw them! Since the problem asks for all the values of x which satisfy the inequality, try to eliminate answer choices by finding a value of x which satisfies the inequality x(x−1)(x−2)≥0, but which is NOT supported by the answer choices - thereby proving that the answer choices do not include the entire possible values of x. If x=1, x(x-1)(x-2)≥0, so the above answer does not lend us the full range of values of x. Correct. This is correct by way of elimination: Plugging in x=1 eliminates answer choice B, since 1 satisfies the inequality and is not included in the range defined by the answer choice. Since numbers greater than 1 (such as x=10) also satisfy the inequality, eliminate answer choices C and E. Answer choice D may be eliminated by plugging in x=1/2, which satisfies the inequality in the question stem while being outside of the range 1 ≤ x ≤ 2 or x ≥ 2.

a. 0 ≤ x ≤ 1 or x ≥ 2 b. x ≤ 0 or x ≥ 2 c. 0 ≤ x ≤ 1 d. 1 ≤ x ≤ 2 or x ≥ 2 e. 0 ≤ x ≤ 2

If 8≤(5/x)≤16, and 10≤y≤25, what is the least value of x/y? Correct. Remember the denominator dynamics: the least value of a fraction is achieved when the numerator is the smallest, and the denominator is the greatest. Thus, to find the least value of x/y, you need the smallest x and the largest y. Manipulate the inequalities to find the relevant values of x and y. The greatest value of denominator y is 25. As for x, Separate the two inequalities involving x: 8 ≤ 5/x and 5/x≤16 Multiply by x (x is positive, because a negative x will not be yield a fraction greater than 8): 8x ≤ 5 --> x ≤ 5/8 And 5 ≤ 16x 5/16 ≤ x Thus, the least possible value of x/y is (5/16) / 25 = --> 5/(16·25) = --> 1/(16·5) =1/80.

a. 1/2000 b. 1/400 c. 1/80 d. 1/40 e. 1/32

If m·k<0, then all the following MUST be true EXCEPT Correct. k-m<0 means k<m. This isn't necessarily true. For instance, k=3, m=-2 are values which satisfy the inequality, and for which k-m is not negative.

a. k/m < 0 b. k−m < 0 c. m2k2>0 d. m/k < 0 e. (k−m)/(m−k) = −1

Now consider the second case: x2≥4 First scenario: if x is positive, then x≥2 Second scenario: if x is negative, then x≤-2. As before, the sign flips when dealing with negatives. In this case, we're left with x≥2 or x≤-2. Again, plug in a number within those ranges (such as 3 or -3) to check whether it satisfies the original inequality x2≥4.

to sum up: Dealing with inequalities involving even exponents (two cases): x2≤a: -√a≤x≤√a x2≥a: x≥√a or x≤-√a For example: If x2 ≤ 9, then -√9≤x≤√9 ---> -3 ≤ x ≤ 3. If x2 ≥ 9, then x≥√9 or x≤-√9 ---> x≥3 or x≤-3