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Inequalities math

****In summary, if you know the signs of the variables, you should flip the inequality UNLESS x and y have diff signs

Dividing out variables

Test-taking trick

If ≠ 2, which of the following is equal to [b * (a^2 - 4)] / [ab - 2b]?

(a) ab (b) a (c) a+2 (d) a^2 (e) 2b

What are two routes/ways for solving this problem?

+ Route 1: simplify the expression in the question stem until it looks like one of the answer choices

+ Route 2: plug in numbers

--> Notice how you can use Route 2 if you are having problems with Route 1 (i.e. the expression is hard to simplify, as could happen with e.g. absolute values)

--> Note that there are some pairs of numbers that lead you to just 1 answer, and there are some pairs that lead to 2 Just try diff #'s if I end up with 2 answers.

Algebra tips

+ Sometimes I can get away with just solving for the combined equation, rather than each individual variable that goes into the combined equation.

e.g. What is x/y?

(1) (x+y)/y = 3 (2) y=4

--> Here, I can manipulate statement 1 to calculate the value of x/y. I don't need to solve for x, and then solve for y. The better strategy is to solve for x/y!! Be direct!

+ Sometimes I can rewrite the original DS question in order to come up with a new rearrangement that is FAR easier to test.

e.g. If xy ≠ 0 and sqrt (xy/3) = x, what is y?

(1) x/y = 1/3 (2) x = 3

--> Best way to work this problem is to first do the math on the original question, rather than plugging in statements 1 and 2.

Then, the equation turns into y = 3x. Thus, the rephrased question is "what is x?", so the answer is B

Random Exponents and square roots

+ Remember that sqrt (x^2) = abs (x)

+ Even exponents "hide" the sign of the base.

e.g. If x^2 = 25, then abs(x) = 5, and x = +5 or -5

e.g. a^2 - 5 = 12, then a^2 = 17, then a = +sqrt(17) or -sqrt(17)

+ Even exponents hide the sign of the base

+ Odd exponents keep the sign of the base. Also, equations that involve only odd exponents (like a cube root) only has 1 solution

Exponents - base of 0 or 1

0 raised to any power equals 0

1 raised to any power equals 1

If x = x^2, then you know that x is either 0 or 1!

Exponents - basic rules

+ When raising fractions to a power, you can distribute the exponent to both the numerator and denominator

e.g. (3/4)^2 = 3^2 / 4^2

+ Similarly, exponents can be distributed to a product

e.g. (3x)^4 = 3^4 * x^4

+ When two exponential terms have the same base, you can add/subtract the exponents upon multiplication/division, respectively

+ Anything raised to the 0 power equals 1! The only exception is 0^0, which is undefined

+ When you raise an exponential term to an exponent, multiply the exponents! e.g. (z^2)^3 = z^6

Exponents - negative bases

When dealing with negative bases, pay attention to PEMDAS.

--> If the negative sign is inside parentheses, the exponent does not distribute.

-2^4 ≠ (-2)^4

Any negative base will follow the same pattern as -1. That is, negative bases raised to an odd exponent will be negative, and any negative bases raised to an even exponent will be positive.

Exponent - factoring out common terms to get GCD

Normally you can't do much with two exponential terms that are added/subtracted to each other (we haven't yet seen an example of this in previous flash cards!).

BUT, if the exponential terms have the SAME BASE, then you can factor out a common term.

11^3 + 11^4 = 11^3 (1 + 11) = 11^3 × 12

Factoring out exponential terms with common bases can be especially helpful when you're trying to boil the expression down to its prime factors.

For example, notice how you can use this method to see all the prime factors for: 11^3 + 11^4

--> 11^3 + 11^4 = 11^3 × 12 = 11^3 × 2^2 × 3

Exponents - odd exponents

Equations with odd exponents only have 1 solution...NOT 2 as you get when, e.g., you square something

x^3 = -125

--> x = -5 (NOT 5)

y^5 = 243

--> y = 3 (NOT -3)

Equations with an odd exponent and an even exponent is very dangerous! It's likely that there are 2 solutions.

Exponents with same base

You must be careful when dealing with equations that have the same base (or could have the same base) on both sides of the equation.

If 0, 1, or -1 is the base (or could be the base), since the outcome of raising those bases to powers is not unique.

e.g. 0^2 = 0^5 = 0^29 = 0.

Thus, if 0^x = 0^y, then you can't claim that x=7

The same rule applies for 1 and -1.

Tricky exponents

What is (1/8) ^ (-4/3)?

Step 1: Because you have a negative exponent, you take the reciprocal of the base.

(8) ^ (4/3)

Step 2: Solve!

Answer: 16

Trap: Don't inverse the exponent when you're also inversing the base! That's wrong wrong wrong!

Roots - Simplifying roots

You can simplify roots by either combining or separating them in multiplication and division.

sqrt(25×16) = sqrt(25) × sqrt(16) = 5×4 = 20

sqrt(50) × sqrt(18) = sqrt(50×18) = sqrt(900) = 30

sqrt(144÷16) = sqrt(144) ÷ sqrt(16) = 12÷4 = 3

sqrt(72) ÷ sqrt(8) = sqrt (72÷8) = sqrt(9) = 3

But, you CAN'T combine or separate roots in addition and subtraction.

GMAT TRICK: The GMAT will try to trick you into splitting the sum or difference of 2 numbers inside a radical into 2 individual roots. Also, they'll try to trick you into combining the sum or difference of two roots inside one radical. Don't do it!

Remember that you can only separate or combine the product or quotient of 2 roots. You can't separate or combine the SUM or DIFFERENCE of 2 roots.

Wrong: sqrt(16+9) = sqrt(16) + sqrt(9) = 4+3 = 7

Right: sqrt(16+9) = sqrt(25) = 5

Memorize square and square roots

sqrt(2) = 1.4

sqrt(3) = 1.7

sqrt(5) = 2.25

sqrt(196) = 14

Quadratics - Disguised quadratics

The GMAT will try to trick you by disguising quadratics - that is, they'll put it in a form that doesn't look like ax^2 + bx + c = 0.

Example #1: Solve for x, given that 36/x = x-5

Example #2: Another example: 3x^2 = 6x.

** How do you solve this example??

DON'T JUST DIVIDE THIS OUT!! You MUST turn this into a quadratic. If you divide it out, you only get one solution, and you end up missing the other solution.

The right way to do it is:

3x^2 - 6x + 0 = 0

x(3x-6) = 0

x = 0, 2

-->When you divide it out, you COMPLETELY miss one of the solutions!!!! According to MGMAT, only divide when you're sure it doesn't equal 0.

Quadratic - Exception (taking the square root) / speed trick

Some quadratics are easier to solve when you do NOT set one side equal to 0 -- that is, when the other side of the quadratic is a perfect square. To solve, simply take the square root of both sides of the equation.

e.g. if (x+3)^2 = 25, what is x?

You COULD solve this problem by distributing out the LHS, setting RHS equal to 0, and factoring/solving. BUT, it's much easier to simply take the sqrt of both sides. Just be sure to consider the positive and the negative sqrt.

sqrt((x+3)^2) = sqrt (25) --> note that square-rooting the square of something is the same as taking the absolute value of that thing)

abs(x+3) = 5 --> x+3 = ±5 --> x = -3 ± 5 ---> x = 2, -8

Quadratics - zero in the denominator

Remember when dealing with quadratics and finding solutions that you CAN'T allow any division by 0. Thus, no solutions can yield a denominator of 0.

Example: (x^2 + x - 12) / (x - 2) = 0

When you factor this, you get:

[(x-3) * (x+4)] / (x-2) = 0

If the numerators are 0, then the whole expression is 0, so the solutions are x=3 and x= -4.

BUT if x=2, then the expression becomes undefined. Thus, x=2 is NOT a solution. x=2 is illegal, and x≠2.

Quadratics - Disguised quadratics

Never forget to completely set one side of a quadratic equation equal to 0. Otherwise, I'll miss solutions

How would I solve (x^2 + 6x + 9) / (x+3) = 7?

What's the methodology (don't need the answer)?

Step 1: multiply both sides by (x+3)

Step 2: subtract both sides by 7x+21

Step 3: solve --> eventually you get x = -3 or x = 4

Step 4: Discard -3 as a solution for x, since this value would make the denominator zero

Inequalities

+ Adding constants to both sides of an inequality --> this is allowed

+ Adding variables to both sides --> this is allowed

+ Multiplying/dividing by a positive number on both sides of an inequality --> this is allowed

+ Multiply/dividing by a negative number on both sides of an inequality --> This is NOT ALLOWED! If you do so, you MUST FLIP the inequality sign.

...this means that you can't multiply or divide an inequality by a variable, UNLESS you know the sign of the number that the variable stands for.

The reason is that you won't know whether you should flip the inequality sign or not.

Therefore, only if you know the variable is positive/negative, then you can multiply/divide accordingly.

Inequalities - adding them up (1/3)

You can combine 2 inequalities by adding them together. In order to add them together, you have to ensure that the inequalities are facing the same side.

Question: Is a+2b < c+2d?

(1) a < c

(2) d > b

The answer is (c) since you need both statements (1) and (2) to know that each term is less than the other corresponding term.

To actually solve this, you would add statements 1 and 2 together.

Step 1: flip statement 2 so you end up with a<c and b<d

Step 2: add statement 2 to statement 1:

(a<c) + (b<d) = a+b < c+d

Step 3: doing this twice yields a+2b < c+2d

Inequalities - adding them up (2/3)

+ You can ADD inequalities together - in fact this is a powerful method on the GMAT

+ HOWEVER< you can't SUBTRACT or DIVIDE two inequalities.

+ FURTHERMORE, you can only MULTIPLY inequalities together under certain conditions

You can only multiply inequalities together if BOTH sides of BOTH inequalities are positive.

Otherwise, DON'T DO IT!

Example: If m and n are both positive, is mn < 10?

(1) m<2 (2) n < 5

Since everything is positive, you can multiply the inequalities together and conclude that mn<10.

Answer is C!

Inequalities - adding them up (3/3)

Is mn < 10?

(1) m < 2

(2) n < 5

It is tempting to multiply these two statements together and say that mn < 10. However, m and n could be negative!!!!

Therefore, this is insufficient.

Only multiply inequalities together if both sides are positive.

Inequalities - more operations

Question:

If r > 0 and s > 0, is r/s < s/r ?

(1) r/3s = 1/4

(2) s = r+4

Since we know that both sides are positive, we can cross-multiply the inequality without flipping the sign

Plus, since we know that both r and s are positive, we can take the square root of both sides without changing the meaning of the inequality.

Thus you end up with: is r < s ?

**Key takeaway: if signs are the same, you can cross-multiply and take square roots of inequalities!

Fractions

You must know what happens when you raise a fraction to a power.

The results depends on the SIZE and the SIGN of the fraction.

*I will need to regenerate the following scenarios on the GMAT

EVEN EXPONENTS (such as 2)

(-3/2)^2 -> Result is bigger(-1/2)^2 -> Result is bigger

(1/2)^2 -> Result is smaller

(3/2)^2 -> Result is bigger

ODD EXPONENTS (such as 3)

(-3/2)^3 -> Result is smaller

(-1/2)^3 -> Result is bigger

(1/2)^3 -> Result is smaller

(3/2)^3 -> Result is bigger

Fractions

Any positive proper fraction raised to a power greater than 1 will result in a number smaller than the original fraction.

Any positive proper fraction raised to a power between 0 and 1 will result in a number larger than the original fraction.

Is (3/4)^(3/4) > (3/4)?

Yes! You shouldn't think about it as cubing, then taking the 4th root...

Test-taking trick - positive/negative analysis

Whenever you see 2 variables being multiplied together and you need to know if one of the variables is positive/negative, then simply make a table and do positive/negative analysis!

That's the only way I can prevent myself from getting confused and selecting the wrong answer!

Example: is d > 0?

(1) bc < 0

(2) cd > 0

--> Both statements by themselves are not sufficient. If you combine them, the only easy way to keep track of what happens when b/c/d is + or - is by making a table.

When you do that, you'll see that the answer is very clearly E!

Advanced - absolute values

You can either take an algebraic solve-the-formula approach, or you can take a positive/negative analysis approach.

If you have 2 variables in 2 or more absolute value expressions, then you should set up a positive/negatives table to solve those problems. The reason you should do this is because these problems are tough to solve with algebra.

But, if you have 1 (not 2) variable in 2 or more absolute value expressions, you can take an algebraic approach. You would simply test cases where the absolute values are either positive or negative.

Advanced - Quadratic formula

The solutions for a quadratic formula ax^2 + bx + c = 0 are:

x = [-b ± sqrt(b^2 - 4ac)] / 2a

If sqrt(b^2 - 4ac) > 0, there's 2 solutions.

If = 0, there's 1 solution.

If <0, there's NO solutions.

It's rare for the GMAT to bring up a problem that requires the quadratic formula!

Advanced - using conjugates to rationalize denominators

If you have a sqrt in the denominator, it's easy to simplify (just multiply it out)...but if you have an expression involving a sqrt in the denominator, it becomes tricky.

e.g. 4 / [3-sqrt(2)]

To simplify this expression, you have to multiply it out by the conjugate of the denominator

For a + sqrt(b), the conjugate is a - sqrt(b)

For a - sqrt(b), the conjugate is a + sqrt(b)

Advanced - formulas

Remember that the "domain" is the possible inputs to a formula. The "range" is the possible outputs.

Therefore, the range for f(x) = x^2 is f(x) >= 0, but the domain is all numbers (even 0).

Direct proportions

y = kx

y/x = k

Inverse proportions

y = k * (1/x)

xy = k

I need to know BOTH of these equations so I can work these problems on the GMAT in under 2 minutes!!!

Sometimes I'll have to pick numbers in order to solve the problem.

e.g. Current is inversely proportional to resistance. If a wire has 4 amps of current, but then the resistance is cut to 1/3 of the original value, then how many amps will flow through the wire?

Here, I must pick numbers to solve the problem - there's no other way! (Answer: 12)

Exponential growth

You can recognize an exponential growth question when it says something like "the ratio of values in any 2 consecutive years is constant", which implies exponential growth.

Also when it says something like "by what factor...", you can infer that it's an exponential growth question since you are dealing with a multiplier of some original quantity.

Remember that with exponential growth, you multiply something by the same constant every period of time (instead of adding, as in linear growth).

The way to express this growth as an equation is:

y = y0 * k^t--> k represents the constant multiplier for one period. Commonly, exponential growth multipliers take the form of percentage multipliers. e.g. if a quantity increases by 7% each period, k=1.07

Shortcut for exponential growth questions

If something increases 8-fold in an hour, then how much does it increase in 10 minutes?

What are the two ways I can solve this question???

(1) by plugging the data into an exponential growth equation, seeing that k^60 = 8, then calculating the increase in 10 minutes by saying k^10 = (8^(1/60)) ^ 10 and calculating it through to arrive at sqrt(2)

(2) by realizing that it doubles every 20 minutes. For something to double every 20 minutes, it would have to increase at a rate of sqrt(2) every 10 minutes -- this way, 2 periods of 10 minutes leads to an increase by a factor of sqrt(2) * sqrt(2)

Advanced Optimization / max-min problems

If you have a problem where you have to find the maximum (or minimum) possible value, then create a table that tests the extreme scenarios for x and y (that is, the situations where x and y are minimimzed/maximized)

e.g. If -7 <= a <= 6 and -7 <= b <= 8, then what is the max possible value of ab?

The proper way to solve this is to (a) find min and max for a and b, (b) multiply min(a) by min(b), then do min(a) by max(b), then do max(a) by min(b), then max(a) by max(b), and (c) see what you got!

Inequalities become particularly tricky when you're dealing with one of two scenarios:

(a) taking the reciprocal of an inequality

(b) squaring an inequality

Advanced - Reciprocals of inequalities (1/2)

Taking reciprocals of inequalities is similar to multiplying/dividing by negative numbers.

You need to consider the positive and negative cases of the variables involved.

The rule is that, if x < y, then:

(1) If x and y are positive, then 1/x > 1/y

e.g. If 3 < 5, then 1/3 > 1/5

(2) if x and y are negative, then 1/x > 1/y

e.g. if -4 < -2, then -1/4 > -1/2

Advanced - Reciprocals of inequalities (2/2) (continued)

(3) If x is negative and y is positive, then 1/x < 1/y --> DO NOT flip the inequality

e.g. if -6 < 7, then -1/6 < 1/7.

(4) If you don't know the signs of x and y, you can't take the reciprocal

Advanced - squaring inequalities (1/2)

As with reciprocals, you can't square both sides of an inequality unless you know the signs of both sides of the inequality.

However, the rules for squaring inequalities is different than those for reciprocating inequalities.

(1) if both sides are negative, then flip the inequality sign when you square

e.g. if x < -3, then you can say x^2 > 9, no problem.

BUT if x > -3, then you can't say x^2 > 9 (i.e. you can't square both sides) since you don't know if x is positive or negative. If x is negative then x^2 < 9...but if x is positive, then x^2 could be either greater than 9 or less than 9.

Advanced - squaring inequalities (2/2)

(2) If both sides are known to be positive, then don't flip the inequality sign when you square.

e.g. if x > 3, then you can say x^2 > 9 no problem -- after all, you know that both sides are positive.

BUT if x < 3, then you can't say x^2 < 9 (i.e. you can't square both sides) since you don't know if x is positive (in which case x^2 < 9) or if x is negative (in which case x^2 can be greater or less than 9)

(3) If one side is positive and the other is negative, you can't square both sides!

e.g. If you know that x<y with x negative and y positive, you can't say anything about x^2 vs. y^2. You get different solutions when plugging in -2 and 2, -2 and 3, or -2 and 1 for x and y, respectively.

(4) If the sides are unclear, then you can't square both sides! e.g. is x^2 > y^2, given that x>y? No! How about when x > 0? Definitely not. So answer is E.

Advanced - inequalities

When working inequality problems, I must be sure to write everything down. These problems become tricky when you multiply across and have to flip signs, deal with positive/negatives, etc.

e.g. 4/x < -1/3

How do you solve this??

Since positive/negatives matter when working with inequalities, create a table with both scenarios and solve accordingly

(1) Case 1: x > 0

x can't be positive since the left-hand side of the eqn. is less than -1/3. Therefore, x must be negative.

(2) Case 2: x < 0

You have to test this scenario anyways, but it's easy.

Multiple across and flip signs...12 > -x, so -12 < x. Since we also know that x < 0, the answer is -12 < x < 0.

You can't divide a variable out unless you're sure it's not 0.

i.e. it’s an illegal operation if the variable can possibly equal zero. Conversely, if you know FOR SURE that y is not zero, then you can divide by y.

About this deck

Author: Kunal K.

Created: 2014-04-05

Updated: 2014-07-05

Size: 39 flashcards

Views: 30

Created: 2014-04-05

Updated: 2014-07-05

Size: 39 flashcards

Views: 30

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