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Inequalities are some of the more algebraically complex concepts. Inequalities can be defined as statements that show values that are greater or lesser than other values. The open end of the inequality sign(<) will always face the larger value, for instance, 5 > 3. Inequalities are used to relate known values as well as variables. For example, you could have a > 3, or x < y.
Inequalities can be manipulated algebraically or combined similarly to equations if all values in the inequality are known to be greater than zero. If you want to manipulate an inequality as you would an equation, you must make sure that all values are positive. This is because if you are multiplying or dividing an inequality by a value less than zero, then you must flip the inequality sign. That is, > becomes < and < becomes >.
If all values for the inequality are defined as greater than zero, prepare to manipulate algebraically. This means that you must not assume that a variable such as x is positive when it might be negative. Recognize that inequalities are not the same as equations in considering sufficiency for data sufficiency because there are possibly multiple values. For complex inequalities, you can consider alternative tactics, especially Modeling and Estimation, in the problem-solving part of the quantitative section.
$$\frac{1}{4}(2x^{2}+8)>34$$
$$-\frac{1}{2}(18+y)<-5y$$
Add 18 and 9y to each side, you are basically just moving the pieces of the inequality from one side to the other side of the inequality \(\rightarrow -18 + 18 + 9y < – 9y + 9y + 18\)
\(\rightarrow 9y < 18\)
Divide the inequality by \(\rightarrow 9 y < 2\).
A compound inequality is an inequality with multiple inequality signs. Basically, it has a value in the middle, which is greater than one value but less than the other value. e.g a > b > c.
Split a single compound inequality into as many simple inequalities as necessary, one for each inequality sign.
Seek for the sought value in each individual inequality.
Recombine compound inequality if that is helpful in solving the problem, or evaluating data sufficiency.
$$3<\frac{1}{2}(18+y)<-4y$$
Attempt these questions only if you are certain of the steps to complete the combination. It can be a little confusing if you are not very familiar with this kind of manipulation, especially when dealing with subtraction. The addition is much easier, but again, you have to be completely certain if you are going to combine systems under inequalities.
Seek methods of manipulation to facilitate the addition of two inequalities because it means you can just move variables on either side of each other and get the inequality facing the same direction and be able to add.
Beware of combining through subtraction because you have to be very technically specific on how to accomplish this.
Firstly, stack the inequalities vertically, aligning variables and ensuring the same direction for the inequality signs.
Then carefully add inequalities and make sure to keep the inequality signs facing the same direction.
Finally, follow the standard rules of inequality manipulation to simplify and solve for the individual variables you seek.
Simplify: \(5x +y > 45\) and \(2x – 6 < 2y\)
\(5x + y > 45\)
\(2y > 2x – 6\)
Firstly, stack the inequalities vertically, aligning variables and ensuring that the signs face opposing directions.
Then carefully subtract the inequalities by applying the sign of the subtracted from inequality(top inequality) for the resulting inequality.
Finally, you can simplify following the rules of inequality manipulation.
What is y greater than if 5x + y > 45 and 2x – 6 < 2y?
\(5x + y > 45\)
\(2x – 6 < 2y\)
\(2(5x + y > 45) \rightarrow 10x + 2y > 90\)
\(5(2x – 6 < 2y)\rightarrow 10x – 30 < 10y\)
\(2y — 30 > 90 – 10y \rightarrow 2y + 30 > 90 – 10y\)
If \(x < -1 < 0 < y < z < 1\), which of the following statements must be true?
A. I only
B. II only
C. III only
D. I and II
E. II and III
Set up your scratch pad by listing A through E vertically. Additionally, you can write down Roman numerals. They could help with elimination as only a limited number are presented for these types of problems. Draw a line on top to write what you seek:
A. I ×
B. II ×
C. III ×
D. I & II ×
E. II & III ✓
The first thing we need to do is separate the individual relevant variables from the inequality.
(a) \(x < -1\); (b) \(0 < y\); (c) \(y < z\); (d) \(z < 1\).
It is not necessary to continue with the second part of the inequality since obviously 0 > -1.
Generally, with algebraic manipulation, even when dealing with inequalities, you may want to try the algebraic manipulation first because that could be more efficient.
For I, \(xy < xz\), we know that x is common to both sides. So we can divide the inequality by \(x\rightarrow xy < xz÷x\). We just have to know if x is positive or negative. We know from (a) that \(x < -1\), which means that the inequality sign is going to flip \(\rightarrow {y} > z\). But we already know from (c) that \(y < z\). This means the Roman numeral I is not true, and we can eliminate any answer choice that includes I. That leaves us with choices BCE.
Given the structure of this question, we have to evaluate II and III individually because they are the only options left, and both combinations are present in the choices we have left.
We know that x < 0 and z > 0. So no matter the situation, \(xz\) has to be less than 0. II will have to be included, and that allows us to eliminate anything that does not include it. Therefore, it leaves us with the choice BE.
We also know from this evaluation that \(xz\) is negative. If we multiply that by another positive, the result will still be negative. \(\text(xz).\text(y) < 0\). III is also true. II alone is eliminated, and our correct answer, therefore, is choice E.
That is the technical approach, we can also take the plugging-in alternative.
A. I ×
B. II ×
C. III ×
D. I & II ×
E. II & III ✓
Since we have all the variables x, y, and z in our answer choices, it means we have a plugging-in value opportunity. The best way to do this is to pick easy values for x, y, and z and see what happens.
Let’s take \(x = -2; y = ¼; z = ½\).
For I, \(xy < xz: xy = -2 × ¼ = -½ ; xz = -2 × ½ = -1\). And we know that \(-1 < -½\). This proves that the Roman numeral I is incorrect. Therefore, A and D are eliminated.
We then have the same values for x and z.
\(xz < 0, -2 × ½ = -1 < 0\), so our choice must include II.
For \(xyz < 0\), \(y = ¼\) and \(xz = -1\). So xz.\(y = -1 × ¼ = -¼ < 0\). III is also correct, and this leads us to choice C as our correct answer.
Is \(a + b > 10\)?
When we combine the two, we need to subtract to confirm that a + b is not > 10
$$2a+3c>5$$
$$(c-2b<25)3$$
2a +3c > 5
-6b + 3c < 75
2a + 6b > -70
This statement alone does not definitively tell us whether a + b is or is not greater than 10. We can confidently say that this information is not sufficient and select choice E.
This statement alone does not definitively tell us whether a + b is or is not greater than 10. We can confidently say that this information is not sufficient and select choice E.
The best way to test and improve your understanding of GMAT Math Concepts as you prepare for the final exam is to go ahead and do as much practice as you can on your own. There are quite a lot of study resources available on our website that you can take advantage of. Get a study package that suits your needs and start preparing for your GMAT exams as soon as today. You’ll gain confidence to sit for your exams as you handle real exam questions. Best wishes.
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