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Please find my notes on Inequalities. Hope it would be a help to you. Please add if you have anything more

Linear Inequalities

An inequality looks like an equation except that the equality symbol = is replaced by one of the inequality symbols <, >, ≤ , or ≥. The solutions of inequalities are, in general, intervals of numbers on the real line instead of just a single number or a small set of numbers.

To solve an inequality that contains a variable means to find all values of the variable for which the inequality is true. Unlike an equation, an inequality usually has infinitely many solutions, which form an interval or a union of intervals on the real line.

Rules of Inequalities

Adding the same quantity to each side of an inequality gives an equivalent inequality

A≤B=>A+C≤B+C

Subtracting the same quantity from each side of an inequality gives an equivalent inequality

A≤B=>A-C≤B-C

Multiplying each side of an inequality by the same positive quantity gives an equivalent inequality

If C>0, then A≤B=>CA≤CB

Multiplying each side of an inequality by the same negative quantity reverses the direction of the inequality

If C<0, then A≤B=>CA≥CB

Taking reciprocals of each side of an inequality involving positive quantities reverses the direction of the inequality.

0<A≤B=>(1/A)≥(1/B)>0

Inequalities can be added

If A≤B and C≤D, then A+C≤B+D

Solving Inequalities

Linear inequalities involving variables may be solved by using the Rules of Inequalities to reduce them to a form like x≤a, , x≥a, x<a, or x>a. where x is our variable and a is a real number

Ex 1: Solve the inequality 3x-1≤5 Adding 1 to each side results in an equivalent inequality 3x≤6 Then multiply each side by (1/3) to get another equivalent inequality (1/3)(3x)≤(1/3)(6)

Since this is equivalent to the original inequality, you can say that any number x satisfies the original inequality if and only if it satisfies x≤2.

Ex2: Solve the inequality 3x<9x+4 First subtract 3x from each side to get an equivalent inequality 0<6x+4 Then subtract 4 from each side to get another equivalent inequality -4<6x Now multiply each side by (1/6) to get a third equivalent inequality -(2/3)<x

Since this is equivalent to the original inequality, you can say that any number x satisfies the original inequality if and only if it satisfies x>-(2/3)

Now let us move to solving Solving a Pair of Simultaneous Inequalities

Ex3: Solve the inequalities 4≤3x-2<13 ( Note that this is the same as saying 4≤3x-2 and 3x-2<13 )

First, add 2 to each side of each inequality to get 4≤3x and 3x<15 Then multiply each side of each inequality by (1/3) to get the solution (4/3)≤x and x<5

Ex4: Solve the inequalities 2x+1≤4x-3≤x+7 Subtract 2x and add 3 to each side of the first inequality to get 4≤4x => 1≤x (multiplying each side by (1/4))

Subtract x and add 3 to each side of the second inequality to get 3x≤10 => x≤((10)/3) (multiplying each side by (1/3))

Thus, the original pair of inequalities is equivalent to the pair 1≤x and x≤((10)/3)

Nonlinear Inequalities

Inequalities involving squares and higher powers of the variable are nonlinear inequalities

Solving Nonlinear Inequalities

a) If necessary, rewrite the inequality so that all nonzero terms appear on one side of the inequality sign b) If the nonzero side of the inequality involves quotients, bring them to a common denominator. c) Factor the nonzero side of the inequality. d) List the intervals determined by factorization e) Make a table or diagram of the signs of each factor on each interval. In the last row of the table, determine the sign of the product (or quotient) of these factors. f) Determine the solution set from the last row of the table. Be sure to check whether the inequality is satisfied by some or all of the endpoints of the intervals (this may happen if the inequality involves ≤ or ≥).

Ex5: Solve the inequality x²-5x+6≤0 All the nonzero terms are already on one side of the inequality sign, and there are no quotients, so the first two steps are taken care of. Now factor (x-3)(x-2)≤0

According to your knowledge of the sign of a product or quotient, you can solve this if you know exactly when each of the factors is positive or negative. Since x-3 is negative for x<3 and positive for x>3, and x-2 is negative for x<2 and positive for x>2, the problem is reduced to examining what happens in the four cases x<3, x>3, x<2 and x>2, and checking the values x=3 and x=2 as well.

So the intervals you need to examine are 1) x<2 2) 2<x<3 3) x>3 you can easily see that (2,3) is in the solution set and that x<2 and x>3 are not

So the final solution is 2≤x≤3

Absolute Value

|a|=a if a≥0 and –a if a<0

More generally, the distance from x to a on the number line is given by |x-a|.

Properties of Absolute Value

Let a, b, and c be real numbers, with c>0, and let n be an integer

|ab|=|a||b| |(a/b)|=((|a|)/(|b|)) if b≠0 |a^n|=|a|^n |x|=c is equivalent to x=±c |x|<c is equivalent to -c<x<c |x|>c is equivalent to (x<-c or x>c)

The Triangle Inequality

If a and b are any real numbers, then |a+b| ≤ |a|+|b|

Solving Equations and Inequalities Involving Absolute Value

Ex6: Solve the equation |2x-5|=3

This is equivalent to 2x-5=3 or 2x-5=-3 ( By property |x|=c is equivalent to x=±c ) 2x=8 or 2x=2 x=4 or x=1

Ex7: Solve the inequality (3/(|3x+7|))≥1 Since the absolute value of any number is nonnegative, |3x+7|≥0; as long as 3x+7 not equal to 0, you can multiply each side of the inequality by |3x+7| without affecting the inequality

3≥|3x+7|

-3≤3x+7 and 3x+7≤3 (By property |x|<c is equivalent to -c<x<c) -10≤3x and 3x≤-4 -((10)/3)≤x and x≤-(4/3) -((10)/3)≤x≤-(4/3)

Remember, though, that you had to insist on 3x+7 is not equal to 0. so the value x=-(7/3) must be excluded The solution is thus -((10)/3)≤x≤-(4/3), x not equal to -(7/3)

Some more important results on Inequalities

AM-GM inequality: For all positive real numbers a1,a2,a3,a4…. The following equality holds good

(a1+a2+a3+….an)/n ≥ (a1*a2*a3…an)^1/n ; equality occurs when a1=a2=a3=….=an

Weighted AM-GM inequality: For all positive real numbers a1,a2,a3,a4…. an. And for all non negative real numbers x1,x2….xn and have sum equal to1.

Inequality 1: The measure of an exterior of a triangle is greater than the measure of either remote interior angle.

Inequality 2: If one side of a triangle is longer than a second side, then the angle opposite the first side is larger than the angle opposite the second side.

In triangle ABC: If AC > AB, then Angle[ABC] > Angle[ACB]

Inequality 3: If one angle of a triangle is larger than a second angle, then the side opposite the first angle is longer than the side opposite the second angle.

In triangle ABC: If Angle[ABC] > Angle[ACB], then AB > AC

Inequality 4: The sum of the lengths of any two sides of a triangle is greater than the length of the third side

In a Triangle ABC: AB + BC > AC; AB + AC > BC; AC + BC > AB

Inequality 5: If two sides of one triangle are congruent to two sides of another triangle, but the included angle of the first triangle is larger than the included angle of the second then the third side of the first triangle is longer than the third side of the second triangle

In Triangle ABC and Triangle DEF; If AB =DE; BC = EF ;and Angle[ABC]>Angle[DEF], then AC > DF

Inequality 6: If two sides of one triangle are congruent to two sides of another triangle, but the third side of the first triangle is longer than the thirds side of the second, then the included angle of the first triangle is larger than the included angle of the second

In Triangle ABC and Triangle DEF; If BA = ED; BC = EF; AC > DF,then Angle[ABC]> Angle[DEF]

Inequality 7: Of all the plane figures with a given perimeter, the circle has the greatest area.

Inequality 8: Of all the plane figures with a given area, the circle has the least perimeter

Inequality 9: Of all the solids with a given surface area, the sphere has the greatest volume.

Inequality 10: Of all the solids with a given volume, the sphere has the least surface area.

Inequality 11: Of all the triangles with a common base and perimeter, the isosceles triangle has the greatest area.

Inequality 12: Of all the triangles with a common base and area, the isosceles triangle has the smallest perimeter

Inequality 13: Of all the triangles with a given area, the equilateral triangle has the least perimeter

Inequality 14: Of all the triangles with a given perimeter, the equilateral triangle has the greatest area.

Inequality 15: Of all the n-gons inscribed in a given circle, the regular n-gon has the greatest area.

Inequality 16: Of all the quadrilaterals with a given area, the square has the least perimeter.

Inequality 17: A quadrilateral with given sides has the greatest area when it can be inscribed in a circle.

Note: For this section, I will be including some example in latter time

Ex2: Solve the inequality 3x<9x+4 First subtract 3x from each side to get an equivalent inequality 0<6x+4 Then subtract 4 from each side to get another equivalent inequality -4<6x Now multiply each side by (1/6) to get a third equivalent inequality -(2/3)<x

Since this is equivalent to the original inequality, you can say that any number x satisfies the original inequality if and only if it satisfies x>-(2/3)

I have this odd feeling that I shouldn't subtract (3x) from both sides because I don't know whether (x) is positive or negative. This is because the direction of the signs change when multiplying or dividing by a negative. So I'd leave the equation in its original form for a DS question.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Ex2: Solve the inequality 3x<9x+4 First subtract 3x from each side to get an equivalent inequality 0<6x+4 Then subtract 4 from each side to get another equivalent inequality -4<6x Now multiply each side by (1/6) to get a third equivalent inequality -(2/3)<x

Since this is equivalent to the original inequality, you can say that any number x satisfies the original inequality if and only if it satisfies x>-(2/3)

I have this odd feeling that I shouldn't subtract (3x) from both sides because I don't know whether (x) is positive or negative. This is because the direction of the signs change when multiplying or dividing by a negative.

So I'd leave the equation in its original form for a DS question.

Any thoughts?

Adding or subtracting a variable on both sides of inequalities is always fine as there will not be a change in the sign of inequality. However, we cannot multiply or divide a variable on both sides as there is a possibility of change in the inequality.

Ex3: Solve the inequalities 4≤3x-2<13 ( Note that this is the same as saying 4≤3x-2 and 3x-2<13 )

First, add 2 to each side of each inequality to get 4≤3x and 3x<15 Then multiply each side of each inequality by (1/3) to get the solution (4/3)≤x and x<5

4≤3x-2 gives 6≤3x

No ?

jade3 wrote:

Ex4: Solve the inequalities 2x+1≤4x-3≤x+7 Subtract 2x and add 3 to each side of the first inequality to get 4≤4x => 1≤x (multiplying each side by (1/4))

you are right! I also identified the same typo error!

boo2gom wrote:

jade3 wrote:

Ex3: Solve the inequalities 4≤3x-2<13 ( Note that this is the same as saying 4≤3x-2 and 3x-2<13 )

First, add 2 to each side of each inequality to get 4≤3x and 3x<15 Then multiply each side of each inequality by (1/3) to get the solution (4/3)≤x and x<5

4≤3x-2 gives 6≤3x

No ?

jade3 wrote:

Ex4: Solve the inequalities 2x+1≤4x-3≤x+7 Subtract 2x and add 3 to each side of the first inequality to get 4≤4x => 1≤x (multiplying each side by (1/4))

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