GMAT (Subject) / Quantitative - Algebra (Lesson)

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• Addition of Powers For example,  x3 + x5 can be rewritten as x3(1+x2). Extracting a common factor out of an addition or subtraction of powers will solve the GMAT problem at hand 99% of the time. Before continuing, try to figure out on your own how we turned x3 + x5 into x3(1+x2).
• The product of two negative numbers is 160. If the lesser of the two numbers is 4 less than twice the greater, what is the greater number? we can derive two equations xy = 160 x = 2y-4 2y-4 * y = 160 (y-2)*y = 80 FACTORIZATION = 80 is to be written as a product of two negative numbers, one that is 2 less than the other. Trying simple factorization of 80 qucikly leads to the value of y.  (y-10)*(y+8) = 80 Therefore, y=-8 is the only negative solution in the answer choices!
• Mark an Anna together were allocated n boxes of cookies to sell for a club project. Mark sold 10 boxes less than n and Anna sold 2 boxes less than n. If Mark and Anna have each sold at least one box of cookies, but together they have sold less than n boxes, what is the value of n? Mark sold n-10 boxes and Anna sold n-2 boxes. Because each person sold a least one box, it follows that n-10>1 and n-2>1 => n>11 Together, they sold less than n boxes, so (n-10)+(n-2) < n, such that n>11 and n>12 Therefore, n=11
• If 5-6/x = x, then x has how many possible values? 5-6/x = x   l*x 5x - 6 = x2 0 = x2-5x+6  0= (x-3) * (x-2)  x = 3 or 2  => two values 
• Jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume, respectively. If these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 5 percent sulfuric acid, approximately how many liters of the 2 percent solution will be required? Search for the two equations A + B = 60 => A = 60-B 0.02A+0.12B = 0.05*60  0.02 A + 0.12B = 3 0.02A + 0.12 ( 60-B) = 3  Solution: A = 42
• √(80+25)^2-8000 = √(80+25)2-8000 = Try to ballpark the figure - the answer choices are quite distant from each other. √(105*100-8000 Estimate! 1052 is around 105·100=10500, so 1052-8000 is approx. 10500-8000 = 2500. So you're looking for a root of a number that is a bit greater than the root of 2500. Find the root of 2500 = 50 - the clostest root is 55
• If x and y are integers, and 4^x·5^8=2^3x·5^x+y what is the value of y? --> 4x·58=23x·5x+y 2 and 5 are the common factors of the bases in both sides of this equation. --> (22)x·58=23x·5x+y On the left-hand side, we have a power of a power - multiply the exponents: --> 22x·58=23x·5x+y 1. ) Now the bases are the same on both sides, so we can equate the exponents: --> 2x=3x.Isolate X - move the 2x to the other side of the equation by subtracting 2x from both sides: --> 2x-2x = 3x-2x --> x = 0 2.) As for the Y second equation: --> 8=x+y. Plug in x=0: y=0+8 = 8
• Which of the following expressions is equal to 50 ∙ 5^a-2 ? Apply the reverse rule for dividing powers to transform 5a-2  into 5a / 52: --->   50 ∙ (5a / 52)= --->   50 ∙ (5a / 25) Next, reduce 25 with the 50 in the numerator leaving 2/1: --->   2 ∙ 5a
• Fractional exponen of roots ∛x2 = x2/3 = x(1/3)·2 = (x1/3)2 = (∛x)2
• If x=√64, and y=x+8, y may be which of the following? I) 0 II) 8 III) 16 ONLY CONSIDER Y as 8, not as -8 =Y ONL> POSITIVE! However, as a mathematical convention, a root sign always denotes the positive root. Thus, √9 denotes only  3, and not -3. That's why in this case √64 denotes positive 8 only. In other words, both positive and negative roots must be considered if we placed the square root ourselves, as part of solving an equation, for example.
• (√x)/(3√x)=? Correct.   (√x)/(3√x) = -->    x1/2 / x1/3 = -->    x(1/2) - (1/3) = -->    x(3-2)/6 = -->    x1/6 
• Proof: 3^5=3·81 Try to manipulate the expressions in each statements according to the rules of powers and roots. I) 83 = (23)3 = 29 II) 3·81 = 3·92 = 3·(32)2 =  3·34 = 31+4 = 35
• Multiplication / Division Roots As long as the root is equal, multiplying and dividing roots with different bases involves combining the two bases under the same root. Example (multiplying roots with different base, same root): √2·√8 = √(2·8) = √16 = 4 Example (dividing powers with different base, same root): √75 / √3 = √(75 / 3) = √25 = 5 Note that the above examples show a neat way to avoid difficult roots - √75 is not something that can be worked out without a calculator, but combining and reducing the two bases under the same root allows you to reduce the problem to a manageable √25 = 5. Reversed: √50 = √(25·2) = √25·√2 = 5·√2.
• Roots Addition / Subtraction As we've seen in our Powers, this is a big no-no. Adding √2 and √3 does NOT magically reach a result of √(2+3) = √5.  The same goes for subtraction: √7 - √4 ≠ √3In fact, there's not much you can do if faced with a situation involving adding or subtracting roots with different bases, except beware of the above traps, which will definitely be laid by our friends at GMAC
• Roots: Arithmetic operations involving equal Roots with different bases: Arithmetic operations involving equal Roots with different bases: 1) Multiplication / Division: Combine the two bases under the same root. √2·√8 = √(2·8) = √16 = 4 √75 / √3 = √(75 / 3) = √25 = 5 Note: split complex bases into building blocks under the same roots. Look for building blocks with easily calculable roots, such as Perfect squares. √50 = √(25·2) = √25·√2 = 5·√2 Combining bases under different roots is an illogical concept. For example: √25 / ∛5 ≠ √(25/5) 2) Addition / Subtraction: Do not add / subtract the bases. Beware of traps: √2+√3 ≠ √5 √7 - √4 ≠ √3
• Rules to: Which of the following is NOT equal to 9654.321·10^15? A positive exponent shows that the decimal point is shifted that number of places to the right. A negative exponent shows that the decimal point is shifted that number of places to the left. The rule: Maintain the balance between the digit term and the exponential term. If one goes up (↑) by a magnitude of 10, the other must go down (↓) by the same magnitude, and vice versa. if the exponent goes up (from 13 to 15), the number must reduce - the digit moves left two places:--> 965432.1·1013 = 9654.321·1015
• "in terms of k" phrases Look for Different units in the same problem (e.g. Dollars vs. Cents). PLUG IN: use this technique when variables are around. The surefire sign to Plug In is variables in the answer choices. "in terms of k" phrases!
• If a, b, and c are three positive integers and 3a = 4b + 2c, then what is the least possible value of 3a+2b+c ? PLUGGING IN! If b and c = 1, then:--> 3a = 4b + 2c --> 3a = 4+2 --> 3a = 6 --> a = 2 And thus, 3a+2b+c = 3·2+2·1+1 = 6+2+1 = 9 Alternative method: Since 3a = 4b + 2c, then 3a+2b+c = (4b + 2c) + 2b + c = 6b+3c. Now you can plug in the smallest possible value for b=c=, so that the required expression comes down to  6b+3c = 6·1+3·1 = 9.
• If in k+1 years from now John will be m years old, then how old was John t-1 years ago? In the question above, should we plug in for m? for k? for t? for all 3? Start by plugging in for one of them, and see whether the others can be determined from it, or they deserve their own numbers. You can start anywhere but you want your life to be easy. If you start by plugging in for k, you'll see that m and t are not limited by k: they can be any numbers you choose. If k=3, and m=10, then in k+1=3+1=4 years from now, John will be 10. he's 6 years old now. If we plug in t=5, we can find how old John was t-1 years ago: if t=5, then t-1=4. So 6 year old John was a 2 year old toddler, 4 years ago. "how old was John t-1 years ago?" The answer to that is 2 - that's your goal. Now go to the answer choices - they include all three variables, so you need to plug in all three eliminate answer choice which do not equal 2. k+t-m = 3+5-10 = 8-10=-2. Not equal to 2. POE this answer choice.
• Plugging In Work Order Plugging in is always a three step method: 1) Decide on a number (or numbers) 2) Work the question with the number(s), until you find your goal. When you're done, you should be able to phrase a summary sentence: "If k=..., then the right answer should equal ...". 3) Go to the answer choices, replace the variables with your chosen number(s), and eliminate ACs which do not match your goal.
• PLUGGING-IN : good numbers => small, positive integers.  Dozens of eggs -> 24, 36, etc. Number that's divisible by 15 => multiple of 15 such as 30 Fractions => a multiple of the denominators (simply multiply the bottoms of the fractions) Percents, choose 100 or multiples of it
• Basics of Must be questions The basic steps for Must Be Questions: Try to figure out the issue of the problem. That's how you'll know what to Plug In, especially, while plugging in more than once. Here are a few rules of thumb to get you going;If the question contains multiplication and divisibility, try to plug in positive vs. negative numbers.If the question contains exponents and roots, try to plug in integers vs. fractions.Plug In good numbers first, then POE.If several answers remain, ask yourself for each answer choice: Is it always true, for any number? Plug In again using DOZEN F, then POE. Keep Plugging In until only one choice remains.Remember the DOZEN F: Different (e.g. even vs. odds, prime vs. multiple, fractions vs. integers, positive vs. negative, etc.)OneZeroEqual numbers for different variablesNegativesFractions
• If -2<x<2, which of the following must be non-negative? x+2/x-1 etc.. Your target is non-negative, and the issue here is negative vs. non-negative. Plug in numbers that fit -2<x<2, then POE; Plug in x=0.5 to POE B. Plug in x=-1.5 to POE A and C. Plug in x=1.5 to POE E.
• If a is an integer such that a2/123 is odd, which of the following must be an odd integer? Only 1 makes equation ODD
• If x,y and z are positive integers, and 30x=35y=42z, then which of the following must be divisible by 3? I) x II) y III) z the one that cannot be divided by 3. Y

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