Good Questions 2

Answer the following Non-GMAT question: 1/2 - 11/14 = ? -1/7 -2/7 -4/7 -7/4 -7/2

-2/7

In the equation x2+k·x+1=0, x is a variable and k is a constant. If the quadratic equation has two distinct real roots, which of the following could be true? I) k=0 II) k=−1 III) k=−3 I only III only I and II only II and III only I and III only

Close enough - you took 2 minutes and 31 seconds to answer this question. Correct. For a quadratic equation to have real two solutions, its discriminant (b2-4ac) has to be positive. Find the discriminant and see what values of k will make it positive. --> x2+k·x+1=0 --> b2-4ac=k2-4·1·1>0 --> k2>4 --> k>2 or k<-2 Thus k can equal -3 alternative explanation: Plug k=-3 into the equation and see that it satisfies the other requirements presented, i.e. that the equation will have two possible solutions. Do so by checking that the discriminant is positive when k=-3: If k=-3, then x2-3·x+1=0 Check the discriminant: b2-4ac = (-3)2-4·1·1 = 9-4 = 5, which is greater than zero. Thus, the equation will have two solutions when k=-3, so k=-3 is a possible value.

Answer the following Non-GMAT question: What is the value of 10/3 divided by 21/50 ? 63/5 500/63 63/500 5/7 7/5

Correct. Dividing by a fraction is actually multiplying by the reciprocal of the fraction: ÷ (a/b) = × (b/a)

If 3-5x < 1 / 2432, what is the smallest possible integer value of x? 2 3 4 5 6

Correct. Express both sides of the inequality in terms of a common base (3 in this case). Once the bases are the same, compare powers with each other. First, convert the negative power of 3 into a positive one i.e. 3-5x = 1 / 35x 1 / 35x < 1 / 2432 Cross multiply to get --> 2432 < 35x Express everything as a power of 3: 243 = 9 x 27 = 3 x 3 x 3 x 3 x 3 = 35 So 2432 < 35x can be expressed as (35)2 < 35x --> 310 < 35x Since the bases are the same, the powers can be compared directly ---> 10 < 5x ---> 2 < x Hence, the smallest possible integer value of x is 3.

If k=3−(4−x)², then k is greatest when x= −16 −13 0 3 4

Correct. Look at the equation k=3−(4−x)2. k is 3 minus something squared. That square cannot be negative, but if it's positive, 3 will be reduced. So you must find a value of x for which (4−x)2 is the smallest possible value, so that the complete equation comes as close to 3 as possible. Thus to not reduce 3 (and k), the square should be 0. This will happen if 4-x=0 --> x=4.

Hansel is 7 years older than Gretel. If the product of their ages (in years) is 60, then how old is Gretel ? 5 6 7 12 14

Correct. Numbers in the answer choices and a specific question ("How many years old?") call for Plugging In The Answers. You may feel like writing down one equation or more. This is just your algebraic urge, which is another stop sign for Reverse PI problems. Assume the amount in the answer choice is the age of Gretel and then follow the story in the problem. If everything fits - stop. Pick it. Otherwise - POE and move on, until you find an answer that works. Start with answer choice C: Assume Gretel is 7 years old. Then, Hansel must be 14 years old. The product of their ages (in years) is bigger than 60, so you can POE C. You need an answer that would lead to a smaller product, therefore their ages must be smaller. POE D and E. Now, plug in A or B to check which is correct. The correct answer is A. If Gretel is 5, Hansel is 12 and their age's product is 60.

If the equation x2−6x=k has at least one solution, which of the following must be true? k > 9 k < −9 k = 9 k ≤ 9 k ≥ −9

Correct. The "must be true" in this question refers to the condition in which a quadratic equation has at least one solution. For a quadratic equation to have at least one solution, its discriminant (b2-4ac) has to be non-negative. First, rearrange the equation in the form of ax2+bx+c=0: x2−6x=k --> x2−6x-k=0 Now find the discriminant: --> b2-4ac=(-6)2-4·1·(-k)=36+4k The discriminant has to be non negative, thus --> 36+4k≥0 --> 4k≥-36 --> k≥-9

√(80+25)² - 8000 = 25 45 55 452 552

Correct. The first part of this monster square root is actually recycled quadratic I: (a+b)2 = a2+2ab+b2. Expand the quadratic, and work from there. Alternatively, try to ballpark the figure - the answer choices are quite distant from each other. √(80+25)² - 8000 = √80²+2*25*80+25² -8000 = √80² + 50*80 + 25² - 100*80 = = √80² - 4000 + 25² = √80² - 2*80*25 + 25² = √(80-25)² = √55² = 55 Alternative method: 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: √2500 = √25*100 = √5² * 10² = √50² = 50 The only answer choice that is slightly greater than 50 is answer choice C - 55.

Is a·b>c·d? (1) a>c (2) b>d

Correct. This is a DS Yes\No 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. Try to plug in a, b, c, and d so that a·b is greater than c·d (yielding an answer of "Yes"), and different values so that it is smaller than or equal to c·d (yielding an answer of "No"). Stat.(1): alone, this tells you nothing about b and d. Plug in b=1000 and d=-1000 and vice versa to get opposite results of "Yes" and "No". No definite answer, so Stat.(1)->Maybe->IS->BCE. Stat.(2): alone, this tells you nothing about a and c. Plug in a=1000 and c=-1000 and vice versa to get opposite results of "Yes" and "No". No definite answer, so Stat.(2)->Maybe->IS->CE. Stat.(1+2): combined, the statements look sufficient, since each variable on the right is greater than one counterpart on the left-hand the answer is "Yes". However, is this always true, for any number? Plug in some DOZEN F numbers to try to reach an answer of "No": a=b=0 and c=d=-1. These numbers satisfy the statements, as 0>-1. However, a·b=0 and c·d=(-1)(-1)=1 so a·b is smaller than c·d, yielding an answer of "No". No definite answer, so Stat.(1+2)->Maybe->IS->E.

If |m+5|−5 = m, which of the following must be true? m = 0 m = −10 m is a multiple of 5. m ≥ −5 |m+5| > 0

You grossly underestimated the time this question took you. You actually solved it in 3 minutes and 20 seconds. Correct. The two scenario approach is confusing for this equation. Instead, work with the "must be true" element - For each answer choice, ask yourself: "must this be true?" (For example, "Must m = 0?). Eliminate answer choices by finding values of m that are supported by the equation in the question stem, but are not supported by the answer choice, thus proving that the answer choice is not always true. The key is not to plug in blindly, but to think of numbers that are useful for POE: Finding out that m can be numbers other than 0 or -10 immediately POEs A and B. Simply plug in m=1 into the equation in the question stem: |1+5|−5 = 1, so m COULD also equal 1. Thus, A and B are not MUST be true, as m does not HAVE to equal 0 or -10. m=1 is also a good example to use because it POEs C as well, as 1 is not a multiple of 5. E is POEd by the only value that can show that it is untrue: an absolute value can never be negative, so the only way to disprove E is to show that the absolute value could equal zero, (or that m could be -5). Thus, the problem can be unraveled with only two plug ins: 1 and -5. Since answer choice D is not eliminated for the same plug ins (both 1 and -5 are ≥ −5), this is the only remaining answer choice, and must be correct answer.

If |x| ≥ x+2, which of the following must be true? x ≥ 1 x ≥ 2 x ≤ -1 x ≤ -4 -1 < x < 1

x ≤ -1 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. In addition, this is a "must be" question, where plugging in is the standard tool to try to fail and eliminate each answer choice. Plug in numbers for x to find those that fit |x| ≥ x+2. When you have found a number that fits the inequality, POE all answer choices that are not true for this plug in. Keep plugging in until you have eliminated four answer choices. Whatever's left must be the right answer, i.e. must always be true. A possible elimination scenario follows... Try plugging in x = -5: |-5| ≥ -5 + 2 ---> 5 ≥ -3 -5 works for the inequality in the question stem, but since -5 ≤ -1, you cannot yet eliminate this answer choice. However, x = -5 does eliminate answer choices A, B, and E. Try plugging in x = -2: |-2| ≥ -2 + 2 ---> 2 ≥ 0 -2 works for the inequality in the question stem, but since -2 ≤ -1, you still cannot eliminate this answer choice. However, x = -2 eliminates answer choice D. By process of elimination, this is the only choice left standing. Therefore, it is correct.

If y<x<0, which of the following CANNOT be true? x·y > 2 x−y > 2 x/y > 2 x+y > −2 y/x < 2

x/y > 2 Correct. Both y and x in this question are smaller than 0, i.e., negative. Plug in appropriate numbers (remember, y<x) and POE anything that CAN be true. y and x are both negative so their quotient is positive. Also, y is smaller (read: more negative) than x (for example, x=-2 and y=-3), so x/y must be a fraction between zero and 1. Hence, x/y CANNOT be greater than 2.