Identifying Quant Questions

Are the two distinct lines, l1 and l2, parallel? (1) l1 is perpendicular to l3. (2) l2 intersects l3 forming a right angle.

Q#: 33 Type: D/S yes/no Geometry question without diagram Issue: 1) Geometry: parallel lines WTS: 1) draw all the different options with the statement enable and see if they meet the criteria answer: c Elaboration: The issue here is whether either of the statements or both of them together define a set of parallel lines. Remember that parallel lines are straight lines with the same slope that never meet. Alone, the statements do not relate l1 and l2 to each other and are therefore insufficient. However, taken together, they produce two distinct parallel lines: l1 is perpendicular to l3 and therefore by definition creates a right angle between them. The intersection of l2 and l3 are also at a right angle. Therefore both lines intersect with l3 at a 90° angle. They have the same slope and therefore can never meet.

What Type of Question is this? How to solve it? Solve it? M is the set of all consecutive odd integers between A and B. If B>A>1, and both A and B are even integers, what is the average (arithmetic mean) of M? (A−B)/2 (A−B)/2 + 1 (A+B−1)/2 (A+B)/2 (A+B+1)/2

Q#: 1 Type: Problem solving Issue: Consecutive integers: Average of a Set = Average of First and Last Terms WTS: Plug in constant instead of variable A and B. Because there are variable in the answers. Answer: d Way:

What is the reciprocal of (3−√8)? 1/(√8 − 3) √8 − 3 3/(√8) 3 + √8 3·√8

Q#:34 Type: Quant Problem Solving Issue: Fractions: reciprocal WTS: 1) Ballparking 2) Remember that a the product of a number and it's reciprocal is 1 Answer: d Elaboration: Answer choices A and B can be eliminated immediately for being negative (√8 < √9=3). The product of a negative number with positive (3−√8) cannot be equal to one. Check C: 3/(√8) × (3−√8) = 9/(√8) − 3. This is not equal to one, so eliminate C. Checking D you get (3+√8) × (3−√8). That should ring a bell. The numbers fit in the quadratic formula (a−b)(a+b)=a2−b2, so (3+√8) × (3−√8) = 32−(√8)2 = 9−8 = 1. The product of (3−√8) and (3+√8) is equal to 1, therefore they are each other's reciprocal.

What Type of Question is this? What is the issue? How to solve it? Solve it? If m is a positive integer, which of the following must be a multiple of 6? (2m+3)(2m+4) (2m+2)(2m)(2m+1) (m+2)(m+3) (6m+1)(m+6) m(m+2)(m+3)

Q#: 10 Type: Problem solving- must be true question Issue: Sequences: Consecutive Integers - the Product of n Consecutive Integers is Divisible by n! WTS: Since the product of consecutive numbers in n! where n is the number is numbers. We must find three consecutive numbers as 3! is 6. Normally we would plug in and POE answers that are not always true. Answer: b elaboration: This is a MUST be true Q. Plug in numbers and try to eliminate four answer choices that are not always true. An answer that yields a value that is NOT a multiple of 6 for a single value of m cannot qualify as MUST be true for all possible values of m - eliminate it. Since the question must have a right answer, the last answer standing (the last one that has not been eliminated) must be the right answer. Plug in a good number such as m=2 into all answer choices: (2·2+2)(2·2)(2·2+1) = 6·4·5 is a multiple of 6, so this answer choice cannot be eliminated for m=2. Since all other answer choices are eliminated for the same plug-plug in, this is the right answer choice. Alternative explanation: Rearrange answer choice B as (2m)(2m+1)(2m+2) and see that it actually consists 3 consecutive integers. Recall that the product of n consecutive integers must be divisible by n!.Therefore, this expression must be divisible by 3!=3·2·1=6.

What Type of Question is this? What is the issue? How to solve it? Solve it? If a is a positive even integer, and b is a positive odd integer, and b>a, then the number of even integers between a and b, inclusive, must be (b−a)/2 (b−a−1)/2 (b−a+1)/2 (b−a+1)/2 + 1 b−a

Q#: 11 Type: Problem solving- Must be, with variables in the answer choices Issue: Sequences: Consecutive Integers - Counting Consecutive Multiples within a Range WTS: There are no real numbers in the answers -First we choose good number, then define a goal and finally plug-in, and POE. If necessary we repeat the process with different numbers. answer: c elaboration: When there are no real numbers in the question and the answers contain variables or expressions, plug in and eliminate! Remember - when plugging in numbers, check all answer choices. Plug in b=7 and a=2. The number of even integers in the range is 3: {2, 4, 6}. That's your Goal. Plug in b=7 and a=2 into the answer choices and eliminate. (b-a+1)/2 = (7-2+1)/2 = 6/2 = 3. All other answer choices are eliminated for this plug in, so this is the right answer choice.

What Type of Question is this? What is the issue? How to solve it? Solve it? A and B are two multiples of 14, and Q is the set of consecutive integers between A and B, inclusive. If Q contains 9 multiples of 14, how many multiples of 7 are there in Q? 20 19 18 17 16

Q#: 12 Type: Problem solving Issue: Sequences: Consecutive Integers - Counting Consecutive Multiples within a Range WTS: Use the formula where you: a) subtract the relevant (divisible) extremes b) divide by the regular interval c) add 1 to the result Answer: d Elaboration: To find the multiples of 7 between A and B, you must subtract the relevant extremes, divide by 7 and add 1. For that, you need to find the relevant extremes, or their difference B-A. To make things clearer, you can actually plug in numbers: a good number for A would be 14. Count 9 multiples of 14 to find B. Remember that A and B are also multiples of 14: A=14 - 28 - 42 - 56 - 70 - 84 - 98 - 112 - 126=B. Then use the method above to find the number of multiples of 7 between 14 and 126 inclusive: Subtract the relevant extremes: B-A=126-14=112. Divide by 7: 112/7=16 Add one: 16+1=17.

What Type of Question is this? What is the issue? How to solve it? Solve it? A and B are two multiples of 14, and Q is the set of consecutive integers between A and B, inclusive. If Q contains 9 multiples of 14, how many multiples of 7 are there in Q? 20 19 18 17 16

Q#: 17 Type: Problem Solving Issue: Sequences: Consecutive Integers - Counting Consecutive Multiples within a Range WTS: remember the formula: 1. subtract the relevant extremes 2. divide by the interval 3. add one answer: d Elaboration: To find the multiples of 7 between A and B, you must subtract the relevant extremes, divide by 7 and add 1. For that, you need to find the relevant extremes, or their difference B-A. To make things clearer, you can actually plug in numbers: a good number for A would be 14. Count 9 multiples of 14 to find B. Remember that A and B are also multiples of 14: A=14 - 28 - 42 - 56 - 70 - 84 - 98 - 112 - 126=B. Then use the method above to find the number of multiples of 7 between 14 and 126 inclusive: Subtract the relevant extremes: B-A=126-14=112. Divide by 7: 112/7=16 Add one: 16+1=17. Alternative method: Each multiple of 14 is a multiple of 7, so there are 9 multiples of 7 we already know of. Now, in the interval between any two consecutive multiples of 14 there is an additional multiple of 7. For example, the interval between 14 and 28 holds 21 as an additional multiple of 7 which we need to count. In the next interval (28 to 42 to the next multiple of 14), there's another hidden multiple of 7: 35. Therefore, to find the total number of multiples of 7, we need to take the known 9 multiples of 14, and count the number of intervals between them - each such interval holds one more multiple of 7. The secret is that the number of intervals is 1 less than the number of multiples: between 2 multiples there is only 1 interval, between 3 multiples we count 2 intervals, etc. Thus, between 9 multiples we'll have 8 intervals, and 9+8=17.

What Type of Question is this? What is the issue? How to solve it? Solve it? In a sequence of N consecutive integers, are there exactly two multiples of 8? (1) N=18 (2) The first integer in the sequence is not a multiple of 8.

Q#: 18 Type: D/S Yes/No Question Issue: Consecutive Integers - Counting Consecutive Multiples within a Range WTS: Understand the variables the affect the number of multiples of a given number within a set of consecutive integers. Answer: e Elaboration: This is a Yes/No DS 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 whether there are exactly two multiples of 8 included in a set of N consecutive integers. Plug in values for the integers in the set, and count the number of multiples of 8. Then ask yourself "is this always true, for any number? Think of DOZEN F numbers to show opposite answers. Stat. (1): Plug in a good number for the integers in the set, such as the integers between 1 and 18, inclusive. In this case, there are exactly two multiples of 8 in the set (8 and 16), and the answer is "Yes". However, is this always true, for any number? Plug in a DOZEN F case, such as the integers between 0 and 17, inclusive. Now there are three multiples of 8 (0, 8, and 16), and the answer is "No". That's a "maybe", so Stat.(1)->Maybe->IS->BCE. stat. (2): Alone, this statement allows plugging in all integers between 1 and 18, inclusive, yielding two multiples of 8 (8 and 16) and an answer of "Yes". However, since the number of terms isn't determined in stat. (2), the set could include, for example, only 3 integers (such as 1, 2, 3), yielding zero multiples of 8 and an answer of "No". That's a "Maybe", so Stat.(2)->Maybe->IS->CE. stat. (1)+(2): combined, the two statements limit your plug ins to sets of 18 consecutive integers beginning with an integer that is not a multiple of 8. Plug in the integers between 1 and 18 to get two multiples of 8 and an answer of "Yes"; now Think of a DOZEN F case to get a "No" answer, such as the set of integers between -1 and 16, inclusive. This set satisfies the terms set in both statements, yet yields three multiples of 8 (0, 8 and 16) and an answer of "No". Still "Maybe", so Stat.(1+2)->Maybe->IS->E.

What Type of Question is this? What is the issue? How to solve it? Solve it? In a sequence of N consecutive positive integers, are there exactly four multiples of 5? (1) N=16 (2) The greatest number in the sequence is a multiple of 5.

Q#: 19 Type: D/S Yes/No Issue: Sequences: consecutive integers- counting consecutive multiples within a range WTS: Try starting the sequence of numbers at different starting points and with different lengths within the constraints given by statements one and two to see what the possibilities are in terms of the number of multiples Answer: C Elaboration: This is a Yes/No DS 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 the number of multiples of 5 in a range of consecutive integers. Plug in different sets of numbers that satisfy the statements and show the answer could be "Yes" or "No". Stat. (1): Plug in different numbers for set N: If set N includes all the integers from 1 to 16, inclusive, then there are only 3 multiples of 5 in set N: (5, 10, and 15) and the answer is "No". Can you find a "Yes", i.e. a set of 16 consecutive integers with 4 multiples of 5? Move the entire set 4 integers forward (i.e. the set of integers from 5 to 20), so that set N includes another multiple of 5 (20) without losing the first one (5). The new set includes exactly 4 multiples of five (5, 10, 15, 20), yielding an answer of "Yes". Therefore, the answer is "Maybe", and Stat.(1)->Maybe->IS->BCE. Stat. (2): This statement allows the set of 16 consecutive integers 5 to 20, yielding an answer of "Yes". However, Without setting a number of terms for set N, this statement allows smaller sets such as {3, 4, 5}, which also ends with a multiple of 5, but contains only one multiple of five and an answer of "No". Therefore, the answer is "Maybe", and Stat.(2)->Maybe->IS->CE. Stat. (1+2): Any set of 16 consecutive integers which ends with a multiple of 5 must always begin with another multiple (try it: 5-20, 10-25, etc.). Such a set will always include exactly four multiples of 5 - the first, the last, and two in between. Thus, the answer is always "Yes", and Stat.(1+2)->S->C.

What Type of Question is this? What is the issue? How to solve it? Solve it? If M is the set of all consecutive multiples of 9 between 100 and 500, what is the median of M? 300 301 301.5 302.5 306

Q#: 2 Type: Problem solving Issue: Consecutive integers: Average of a Set = Average of First and Last Terms WTS: Since this is a set of integers with a set interval one only has to find the first and last integer of the set. so the idea is finding the fist and last number divisible by 9 in the range between 100 and 500. namely 108 and 495. Answer: 301.5 elaboration:

What Type of Question is this? What is the issue? How to solve it? Solve it? What is the sum of the terms in a certain sequence of consecutive integers? (1) The least term in the sequence is 8. (2) The average of (arithmetic mean) terms is 16.

Q#: 20 Type: D/S "What is the value of..." Issue: Sequences: consecutive integers- Calculating the Sum of Consecutive Integers WTS: 1) Remember that the average in sequences of consecutive integer is equal to the sum of the first and last numbers of the set divided by two 2) to find the sum of a set of consecutive integers you only need the average and the number of terms Answer: c Elaboration: This is a "what is the value of..." DS question. In this type of question, a statement will be sufficient only if it leads to a single value of the variable (or expression) you're asked about. The issue is finding the sum of a series of consecutive integers. Recall the method for finding the sum of a series of consecutive integers: you need the average of the set, and the number of terms. If the statement(s) provide this information, then it is sufficient. Stat. (1): the value of the least number of the set does not tell you how many terms are there, nor their average. Stat.(1)->IS->BCE. Stat. (2): The average of the terms in the set is a useful piece of information, but you're still missing the number of terms in the set. If the set includes 3 terms or 30 terms, the sum of the set will be completely different. Stat.(2)->IS->CE. Stat. (1)+(2): Recall that the average of a set of consecutive integers is also the average of the first and last terms. Therefore, if the least term is 8 and the greatest term is x, then (8 + x)/2 = 16, according to Stat. (2). From here it is possible to find the largest term x, and the number of terms can also be found - simply count the number of consecutive integers between 8 and x. You have what you need to find the sum of the terms in the set, so Stat.(1+2)->S->C.

What Type of Question is this? What is the issue? How to solve it? Solve it? What is the sum of the terms in a sequence of consecutive integers? (1) Exactly half of the terms in the sequence are non-negative. (2) There are 16 terms in the sequence.

Q#: 21 Type: D/S "What is the value of" Issue: Sequences- Consecutive integers- the sum of consecutive integers WTS: 1) understand that to find the sum of the the set of consecutive integers you need the average of the set of integers and the number of the set of integers. 2) to find the average of a set of consecutive of integers you need the first and last elements of the set answer: c Elaboration: This is a "what is the value of..." DS question. In this type of question, a statement will be sufficient only if it leads to a single value of the variable (or expression) you're asked about. The issue is the sum of a set of consecutive integers. Recall that the sum of a a set of consecutive integers is the product of the set's average and the number of terms in the set (Average × number of terms). These are the two pieces of the puzzle needed for a statment(s) to be sufficient. Stat. (1): half of the terms in the sequence are non negative, or ≥0. Since the set includes consecutive terms, this means that the other half are negative. Plug in a set of numbers to simulate the problem: Try {-2, -1, 0, 1}, which yield a sum of -2-1+0+1=-2. Now plug in a different set, such as {-3, -2, -1, 0, 1, 2}, yielding a different sum of -3-2-1+0+1+2=-3. No single value of the sum of the set can be determined, so Stat.(1)->IS->BCE. Stat. (2): The fact that the set includes 16 terms does not alone tell you what these terms are, so a single value for the sum of terms cannot be determined. Stat.(2)->IS->CE. Stat. (1)+(2): Combined, the two statements fix the terms of the set to a single possible group consecutive integers: the integers between -8 and 7, where half (-8 to -1) are negative and the other half (0 to 7) are non-negative. Knowing this enables you to figure out the sum of the set using the average and the number of terms. Stat.(1+2)->S->C.

What Type of Question is this? What is the issue? How to solve it? Solve it? The sum of all consecutive odd integers from −21 to 31, inclusive, is 130 135 150 156 235

Q#: 22 Type: Quant problem solving Issue: Sequences: consecutive Integers- The sum of consecutive integers WTS: 1) use the formula to find the average of sets of consecutive integers 2) Use the formula to find the number of elements in a set of consecutive integers 3) notice the the question only asks for consecutive odd integers!!! Answer: b Elaboration: According to the formula, we require both the average of the set and the number of terms in the set to calculate its sum: Sum of terms = Average of terms · Number of terms The number of terms is calculated by subtracting the smallest term from the greatest term, dividing by 2 (since it's a series of odd integers) and adding one: 31-(-21) = 52 ---> 52/2 + 1 = 27 As for the set's average, average the first and last terms to find the "middle" of the set: (-21 + 31) / 2 = 10/2 = 5 Plug in the sum and the number of terms into the formula: 5×27 = 135

What Type of Question is this? What is the issue? How to solve it? Solve it? In a phase III clinical trial of Dosaxin, a new drug, patients received a progressively growing dosage of Dosaxin for a few days. On the first day, each patient received 15 milligrams of Dosaxin. On each of the following days, the daily dosage was m milligrams greater than the dosage received the day before, reaching a dosage of 43 milligram on the last day of the trial. For how many days did the trial last, if each patient received a total amount of 145 milligrams of Dosaxin during the whole trial ? 4 5 6 7 9

Q#: 23 Type: Quant word problem Issue: Sequences: Consecutive integers: the sum of consecutive integers WTS: 1) Use the formula for finding the average of sets consecutive of consecutive integers using the first and last elements 2) Use the formula to find the sum of a set of consecutive integers. 3) plug in and solve Answer: b Elaboration: Sum of a list of numbers with constant increments can be calculated using: Sum = Average x Number of numbers, where average = (largest value + smallest value) / 2. Use the above to calculate the number of days i.e. Number of numbers. Given that Sum = Average x Number of days: 145 = [(43 + 15) / 2] x Number of days 145 = (58 / 2) x Number of days 145 = 29 x Number of days 145 / 29 = Number of days 5 = Number of days. Hence, this is the correct answer.

What Type of Question is this? What is the issue? How to solve it? Solve it? If the sum of all consecutive integers from A to B (A<B), inclusive, is S and the average (arithmetic mean) of these integers is M, then, in terms of A, B, and S, what is the value of M? 2S/(A+B+1) 2S/(A+B) (B−A)/2 S/(B−A+1) (B−A)/(2S)

Q#: 24 Type: Quant Word Problem Issue: Sequences: consecutive integers: the sum of consecutive integers WTS : 1) Use the formula for finding the average of sets consecutive of consecutive integers using the first and last elements 2) Use the formula to find the sum of a set of consecutive integers. Either plug in or solve algebraically Answer: d Elaboration: Remember the method to find the sum of a set of consecutive integers within a certain range: 1) Find the average of the set: average the first and last terms to find the "middle" of the set. 2) Count the number of terms in the set: subtract the extremes, and add one. 3) Multiply the "middle" number (the average of the set) by the number of terms. The sum of consecutive integers = Average x (largest - smallest + 1) --> S = M x (B - A + 1) --> S / (B - A + 1) = M. Hence, this is the correct answer. Alternative method: Plug in good number for A and B such as A=1 and B=3. The sum S is then 1+2+3=6, and the average M=2. Your Goal is M=2 - eliminate all answer choice that do not match it for the numbers above.

What Type of Question is this? What is the issue? How to solve it? Solve it? If k is a positive integer and the greatest common divisor of k and 45 is 15, then the greatest common divisor of k and 900 may be any of the following EXCEPT 15 45 60 150 300

Q#: 25 Type: Quant Problem solving Issue: Integers: GCD- Greatest common divisor WTS: if the GCD of k and 45 and 15 it means that K is not divisible by 45 therefore 45 cannot be the GCD of k and 900 answer: b Elaboration: Remember that the GCD (Greatest Common Divisor) of two numbers is the greatest number by which both numbers are divisible. According to the question stem, the GCD (greatest common divisor) of k and 45 is 15. That means that the greatest number both k and 45 are divisible by is 15. If k would have been divisible by 45, the GCD of k and 45 would have been 45 itself. Hence, k is not divisible by 45. If k is not divisible by 45, then the GCD of k and any other number (in this case 900) cannot be 45.

What Type of Question is this? What is the issue? How to solve it? Solve it? If n is a positive integer, which of the following must be divisible by 3? n·(n2−1) (n−1)·n·(n+3) n·(n+1)3 (n−2)·(n+1)·(n+3) n2·(n−3)

Q#: 26 Type: "Must be" problem solving with variables in the answer choices Issue: Integers: divisibility by 3 WTS: Since there are variables in the answer choices we should plug in simple numbers. Since this is a must be questions we only have to fid one exception to the rule to cross out a particular answer. answer: a elaboration: This is a MUST be true question. For every answer choice, ask yourself, "Is this always true, for any number?" Plug in a positive integer for n and eliminate answer options that do not give an integer divisible by 3. Plug in n = 2 i.e. n (n² - 1) = 2 (2² - 1) = --> 2 (4 - 1) = --> 2 (3) = 6 Since 6 is divisible by 3, this answer choice cannot be eliminated for this plug in. All other answer choices can be eliminated by plugging in 2 or 1, so this is the correct answer.

In a sequence of 12 numbers, each term, except for the first one, is 12^11 less than the previous term. If the greatest term in the sequence is 12^12, what is the smallest term in the sequence? −12^11 0 12^11 11·12^11 12^12

Q#: 29 Type: Quant problem solving question Issue: sequences: arithmetic sequences WTS: 1) Remember the formula fro arithmetic sequences answer: c Elaboration: A_12 = 12^12 + (12-1) × -(12^11) Expand the brackets: --> A_12 = 12^12 - 12·12^11 + 1·12^11 Since 12·12^11 = 12^12, this simplifies to --> A_12 = 12^12 - 12^12 + 1·12^11 --> A_12 = 12^11

What Type of Question is this? What is the issue? How to solve it? Solve it? Q is the set of all integers between A and B, inclusive. The average (arithmetic mean) of Q is m. If A<B, and B=190, what is the value of A? (1) Q contains 40 terms that are greater than m. (2) m=150

Q#: 3 Type: Value DS question Issue: 1) Consecutive integers: Average of a Set = Average of First and Last Terms, 2) Averages WTS: Since the average of a set of consecutive integers can also be easily calculated as the average of its first and last terms. Missing Piece: The value of A to find m WTS: Since the average of a set of consecutive integers can also be easily calculated as the average of its first and last terms. Answer: B Elaboration: Stat. (1): Since m is the middle of the set, it follows that there are 40 members smaller than m. This statement wants to lead you to the conclusion that there are 81 members in set Q: 40 members greater than m, 40 members smaller than m, and m itself. Thus, A can be calculated by counting down 81 consecutive integers from B=190 (A in this case would be 110, but you don't really have to know that). But is that the only possible value for A? The problem lies with the assumption that m itself is a member of set Q, but that fact is not stated anywhere. In fact, there is even no guarantee that m is an integer. If set Q has an even number of members (i.e. only 80 terms), then m, as the median of the set, is calculated by averaging the pair of middle terms; m would still have 40 members greater and 40 members smaller than it, but will not itself be a member of set Q. However, since in this case set Q has only 80 members, this will determine a different value for A (i.e. A=111). No single value can be determined for A, so Stat.(1)->IS->BCE. Stat. (2): If m is an integer, it follows that m is also a member of set Q. Since the average is also the median, then number of terms smaller than m is equal to the number of terms greater than m, until B=190. It is therefore possible to count down the same number of consecutive integers from 150 and find a single value for A. Stat.(2)->S->B. Alternative explanation for stat. (2): since the average of a set of consecutive integers is the average of the first and last members of the set, (A + B)/2 = m. Since B=190 and m=150, this leads to a single equation with A: (A+190)/2 = 150. From here you can determine a single value for A.

Martha has 5 dogs and 4 cats. She takes each animal separately for a daily walk. She can take the animals for a walk to the park or to the barber shop. How many different options for a walk with a single animal does Martha have? 8 9 10 18 20

Q#: 35 Type: Quant problem solving Issue: Combinations WTS: Write Se boxes for all the variables add boxes that have an OR relationship and multiply boxes that have an AND relationship Answer: d

The Coen family consists of a father, a mother, two children and a dog. A photographer is about to take the family's picture. How many different arrangements (of standing in a row) does the photographer have, if it is known that the father insists on standing by his woman, as the song says? 12 24 48 60 120

Q#: 36 Type: Quant problem solving Issue: Combinations from a single source WTS: 1) treat the the father mother object as a single object 2) figure out the combinations of the father and mother object answer: c Elaboration: Calculating the number of different arrangements while treating the father and mother as one object yields 4! choices: 1st 2nd 3rd 4th 4 x 3 x 2 x 1 But, since the big "father+mother" object actually has 2! arrangements possible (father to the right of mother, or to her left), multiply the result by 2!: 4!×2! = 24×2 = 48 different family arrangements.

A "Sven" number is defined as a five-digit number which goes according to the following rules: the leftmost digit is even, any digit to the right of an even digit must be an odd digit, and any digit to the right of an odd digit can only be one of the digits 1 or 7. How many different 5-digit numbers are "Sven" numbers? 20 80 160 220 250

Q#: 37 Type: Quant Problem Solving Issue: Combinations WTS: 1) break up each digit into separate SeBoxes answer: c Elaboration: Calculate the number of possible choices for each of the digits separately, using SeBoxes. Then multiply accordingly. Remember that a "Sven" number is still a 5 digit number. This means that the first digit cannot be zero - for example, 04352 is not a 5-digit number. First digit - even: 4 choices (2, 4, 6, 8. Since you need 5 digit numbers, the first digit cannot be zero 0). Second digit - odd: 5 choices (1, 3, 5, 7, 9). Third (to the right of an odd digit): 2 options (1 or 7) Since 1 and 7 are both odd, the fourth and fifth digits will be to the right of an odd digit, and therefore also have 2 choices each - 1 or 7. Use SeBoxes to mark the number of choices for each digit, and multiply the boxes, since you need ALL digits to form each number. That's a total of 160 options: 1st 2nd 3rd 4th 5th 4 x 5 x 2 x 2 x 2 = 160

Ed throws a blue die and a red die. How many combinations are there in which the two dice do not yield the same result and the sum of the results is not 9? 10 24 26 30 36

Q#: 38 Type: Quant problem solving Issue: Combinations with forbidden choices WTS: 1) calculate the total number of choices and subtract the "forbidden" choices Answer: c Elaboration: In this problem, the limitations are phrased very clearly, leading to a "Forbidden combinations" approach: [Total combinations - "Forbidden" combinations] = Good combinations Calculate the "Forbidden" combinations, then subtract them from the Total combinations to get the Good combinations (the wanted choices). There aren't that many Forbidden choice, so you can just count them: Start with the results in which the dice are the same: 1-1, 2-2, 3-3, 4-4, 5-5 and 6-6 - total of 6 "Forbidden combinations" where the dice yield the same result. None of these equals 9, so count the "Forbidden combinations": Results that sum up to 9 are 3-6, 6-3, 4-5, and 5-4 - a total of 4 Combinations. So there are 10 "Forbidden" choices. Calculate the total number of possible results for rolling two dice using SeBoxes. With six choices for the Die I AND six choices for Die II, the total number of combinations is: Die I Die II 6 x 6 So the number of good choices are 36-10 = 26 results.

How many 4 digit numbers are there, if it is known that the first digit is even, the second is odd, the third is prime, the fourth (units digit) is divisible by 3, and the digit 2 can be used only once? 20 150 225 300 320

Q#: 39 Type: Quant Problem solving question Issue: Combinations with forbidden choices WTS: Calculate the total number of choices and subtract the forbidden choices Answer: d Elaboration: Because of the limitation on the usage of "2", use the "forbidden combinations" method - calculate all combinations and then subtract those you cannot use. [Total combinations - "Forbidden" combinations] = Good combinations For the first digit, remember that question asks for numbers (not codes), so zero is not an option: zero is indeed even, but a number cannot begin with zero. Find the total number of combinations, disregarding the restrictions on "2", by breaking down the problem set-by step using SeBoxes: The options for the digits are: First digit: 4 options ( 2, 4, 6, 8). Second digit: 5 options (1, 3, 5, 7, 9). Third digit: 4 options ( 2, 3, 5, 7). Fourth digit: 4 options (0, 3, 6, 9). That gives us a total of 320 options: 4 x 5 x 4 x 4 = 320 Now, define the "Forbidden" combinations: the phrasing "the digit 2 can be used only once" means that the Good combinations include those numbers where the digit "2" does not appear at all, or appears once. Therefore, the "Forbidden" combinations include the remainder: those scenarios in which the digit "2" appears more than once. Since "2" can only appear in the two even digits, the "Forbidden" combinations include those numbers where both of the even digits are "2". We have to reduce the number of options in which 2 appears twice. To do that we can assume that 2 does appear twice and check the other places: First digit: 1 option (2) Second digit: 5 options (1, 3, 5, 7, 9). Third digit: 1 option (2). Fourth digit: 4 options (0, 3, 6, 9). That's a total of 20 "forbidden options", in which 2 appears twice: Now we subtract the "Forbidden" choices from the total choices to find the good options - numbers in which the digit 2 appears only once, or does not appear at all: 1 x 5 x 1 x 4 = 20 320-20=300 options.

What Type of Question is this? What is the issue? How to solve it? Solve it? If x, y and z are consecutive integers, and x<y<z, which of the following is NOT always true? y·z is even. The average (arithmetic mean) of x, y and z is y. y^2−x·z is odd. x·y<y·z x·y·z is even.

Q#: 4 Type: Problem solving- Must be with variable in the answers Issue: 1) Sequences: Consecutive Integers - the Product of n Consecutive Integers is Divisible by n! 2) The mean in a set of consecutive integers is equal to median WTS: Use the properties of consecutive integers and plug numbers into the answers to solve a) The product of n consecutive numbers is divisible by n! b) The mean in a a set of consecutive intgers is equal to the median Answer: d

If a rectangular box has two faces with an area of 30, two faces of area 60, and two faces of area 72, what is its volume? 60 90 162 300 360

Q#: 41 Type: Geometry question without diagram Issue: Solid geometry:rectangular box WTS: 1) write into 3 equations with 3 variable and solve answer: e Elaboration: Remember, the volume of a rectangular solid is Length x Width x Height. Form the equations of the three different faces of the box, and isolate the 3 required dimensions. Based on the question assume that: Length x Width = 30 --> Length = 30/Width Width x Height = 72 --> Height = 72/Width Length x Height = 60 --> (30/Width) x (72/Width) = 60 Based on this (30 x 72)/60 = Width² --> 36 = Width² --> 6 = Width. Plug in width=6 into the above equations. The volume is Length x Width x Height or (30/6) x 6 x 12 = 5 x 6 x 12 = 360. Hence, this is the correct answer.

An electrical supplies store sells 140 power stabilizers every week. If each power stabilizer costs the store $16, what is the minimum selling price per unit that will ensure a weekly profit of at least $5600 from sales of power stabilizers? 24 26 40 56 60

Q#: 43 Type: Quant word problem Issue: profit WTS: translate the problem into an equation and solve answer: d elaboration: Remember: profit = revenues - expenses Find the minimum profit per unit. Since the question asked for the minimum selling price (revenue), don't forget to add the cost (expenses) of the stabilizer to the store. Divide the required profit by the number of units to find the minimum dollar profit-per-unit. 5600 / 140 = 560 / 14 = 40. Therefore, in order to make a $40 profit per unit, the store must charge 40+16=$56 per unit. Alternative method: numbers in the answer choice and a specific question call for reverse Plugging in. Assume that the selling price is $56. Take away the cost of $16 to get a profit of $40 per unit. Multiply by 140 to reach the minimum profit of $5600.

If |5x-5|>x-1, which of the following must be true? x>1 x<1 x>5 x<5 None of the above

Q#: 44 Type: Quant problem solving question Issue: 1) absolute value 2) Inequality WTS: 1) Split the inequality into two separate inequalities to represent the negative and positive options for the absolute value Answer: e Elaboration: 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. For each answer choice, ask yourself "is this always true? for any number?". Plug in numbers for x to find those that fit |5x−5|>x-1. When you have found a number that fits the inequality, POE all answer choice 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. Using POE, you can eliminate all other answer choices. Note: The inequality |5x-5|>x-1 is true for any x≠1.

How many three digit numbers that do not contain the digit 2 are there? 100 200 512 648 729

Q#: 47 Type: Quant Problem solving question Issue: combinations: choosing form a single source with repetition WTS: 1) Make SeBoxes for each digit and multiply Answer: d Elaboration: This problem presents a case of choosing from a single source - DIGITS. work the problem in a step-by-step mode. Consider whether problem includes repetition or not - whether the items chosen can be chosen more than once or not. Also, note the digit(s) that cannot appear. A three digit number requires three SeBoxes. Since the problem doesn't state that the digits cannot repeat themselves, this is a problems with repetition, and the SeBoxes' size should remain the same. Now, consider the constraints presented by the problem: Since you cannot use the digit 2, you have only 9 options for each box (the digits 0,1,3,4,5,6,7,8 or 9). Also, since the problem asks for NUMBERS, the first digit cannot be 0, and therefore has only 8 options. Multiply the numbers in the boxes, because you need all three digits to form the number. The number of three digits numbers without the digit "2" is therefore: I II III 8 x 9 x 9 = 648

In the Land of Oz only one or two-letter words are used. The local language has 66 different letters. The parliament decided to forbid the use of the seventh letter. How many words have the people of Oz lost because of the prohibition? 65 66 67 131 132

Q#: 48 Type: Quant Problem solving question Issue: Combination: choosing form a single source with repetition WTS: make 4 SeBox sets for all the forbidden word options a) single letter word b) 2 letter word starting with the forbidden letter but ending with a different letter c) 2 letter word ending with the forbidden letter but starting with a different letter d) 2 letter word starting and ending with the forbidden letter answer: e Elaboration: The number of lost words is the difference between the number of words originally and the number of words after the prohibition. Remember to consider the two scenarios presented by the question: 1 letter words, and 2 letter words. Calculate the total number of options for each scenario before and after the prohibition, then subtract the two scenarios before from the two scenarios after. Within the two letter scenario: break the problem into two SeBOxes (1st letter, 2nd letter). Remember that the problem does not state that each word has to contain two different letters - thus, this is a case with repetition, and the SeBox size remains the same. Before the prohibition there were 66 one-letter words. There were also 662 two-letter words: 1st 2nd 66 X 66 = 66^2 After the prohibition there are 65 letters, so there are 65 one-letter words, and 652 two-letter words: 1st 2nd 65 X 65 = 65^2 The lost words are the difference between the number of words before and after the prohibition: (66+66^2)-(65+65^2) = 66-65+(66^2-65^2) = 1+(66^2-65^2) Expand the second part using the third recycled quad: a2-b2 = (a-b)(a+b): 66^2-65^2 = (66-65)×(66+65) = 1×131 = 131. So the total number of forbidden words is 1+131=132.

How many combinations options are there for a code consisting of 2 even digits followed by 2 odd digits, if no digit can be repeated? 200 250 400 625 1000

Q#: 49 Type: Quant problem solving question Issue: Combinations: choosing form a single source without repetition WTS: Make SeBoxes for each option and don't forget to subtract used up options as you go answer: c Elaboration: This problem presents a case of Single source - DIGITS. Break the problem in a step-by-step method using SeBoxes. There are 4 digits, so you need 4 SeBoxes. Since no digit can be repeated, there's no repetition and the SeBoxes' size changes from one digit to the next. There are 5 even digits and 5 odd digits. Once you used an even digit, you have only 4 options for the second even digit. The same goes for the odd digits, therefore the number of combinations for such a code is: even1 even2 odd1 odd2 5 x 4 x 5 x 4 = 400

What Type of Question is this? What is the issue? How to solve it? Solve it? If n and m are positive integers, and 3 is not a factor of m, then m may be which of the following? (n−1)·n·(n+1) (n−3)·(n−1)·(n+1) (n−2)·n·(n+2) (n−1)·n·(n+2) (n−3)·(n+1)·(n+2)

Q#: 5 Type: Problem solving - may be Issue: 1) Sequences: Consecutive Integers - the Product of n Consecutive Integers is Divisible by n! WTS: Since the question asks which of the following may be a value for m, all you need is on example that works you need to understand the concept of consecutive integers. The answers are representations of different orders of consecutive integer. One must see if these orders of consecutive integers can potentially skip multiples of 3. One may plug in simple numbers where the smallest number is one. and then try it with an even number. Answer: d Elaboration: Based on the information given in the question, m must be a positive integer that is not divisible by 3. Since the problem asks which of the following may be a value for m, there are four answer choices that may NOT be the value of m - because they will be divisible by 3 no matter what the value of n is. All you need is a single example of a viable m to prove that an answer choice may be m. Plug in a positive integer for n and see which answer choice gives a value of m which meets the aforementioned conditions. Plug in 2 for n i.e. (n−1)·n·(n+2) = (2-1)(2)(2+2) = (1)(2)(4) = 8. 8 is a positive integer not divisible by 3, so m could be 8. Hence, this is the correct answer. Alternative explanation, for the algebraically minded: Since the question asks "which of the following may be the value for m", it follows that there are four answer choices which, for some reason, may NOT be the value of m. Notice that answer choice A is the product of 3 consecutive integers. Recall that "the product of n consecutive integers is divisible by n!"; Therefore, the product of 3 consecutive integers is divisible by 3! = 6, and hence will definitely be divisible by 3. Therefore, choice A cannot be the value of m. Answer choices B, and C can be eliminated for a similar reason. They are the products of 3 integers that differ by exactly 2, and so they must be divisible by 3 as well. As for answer choice E: n-3 acts in the same way as n in terms of divisibility by 3 - either both are divisible by 3 (3 and 6), or both are not divisible by 3 (4 and 7). Therefore, n-3 may serve the same function as n, and E can also be treated as a product of 3 consecutive integers n·(n+1)·(n+2). This last bit is a little difficult to see - a difficulty which is easily bypassed by Plugging In!

How many 5 letter combinations can be made from the letters of the word VERMONT if the first letter has to be a vowel and the last letter has to be a consonant, and each letter can be used only once? 21 42 120 600 720

Q#: 50 Type: Quant problem solving question Issue: Combination: choosing from a single source without repetition WTS: Make Se Boxes for each of the letter and don't forget to subtract options that have been used up answer: d Elaboration: This problem presents a case of choosing from a Single source - letter Break the problem in a step-by-step method using SeBoxes. There are 5 items in the required combinations, so you need 5 SeBoxes. Since you can use each letter only once, there's no repetition and the SeBoxes' size changes. Finally, since changing the order of the chosen letters yields a different word, the order of choice matters. When there are restrictions on certain terms, start the filling the SeBoxes from those "problematic" terms: The first letter can only be a vowel - 2 choices The fifth letter can only be a consonant - 5 choices Then move to the other terms: The second through fourth letters can be anything, but since we already used two letters, there are only 5 choices left for the second letter (2 out of 7 have already been used) and then 4 and 3 choices respectively for the third and fourth letters (the number decreases with each consecutive letter, because there is no repetition). The SeBoxes should therefore look like this: 1st 2nd 3rd 4th 5th 2 x 5 x 4 x 3 x 5 2×5×4×3×5 = 600 combinations

Which of the following equations describes a line that is perpendicular to y=x+6? y = x − 6 y = −1 − x y = 6x y = 6x+1 y = 1/x + 6

Q#: 51 Type: Quant Geometry problem solving without diagram Issue: Geometry: coordinate geometry: slope WTS: 1) remember that in the y = ax +b for a is the slope of the line Answer: b Elaboration: The equation of a line is y=mx + b, where m is the slope of the line and b is the y-intercept. Also, the slope of a line perpendicular to a line with slope m is -1/m. A line perpendicular to the line y=x+6 must have a slope of -1 because the slope of y=x+6 is 1. The slope of y=-1-x i.e. y=-x-1 is -1. Hence, this is the correct answer.

M and N are among the 5 runners in a race, and there can be no tie. How many possible results are there where M is ahead of N? 10 25 35 60 80

Q#: 54 Type: Quant problem solving question Issue: Combinations: permutation: choosing from a single source without repetition- order matters WTS: use the permutations formula answer: d Elaboration: This problem requires common sense much more than strict formulas. Calculate the total number of options to arrange 5 terms in a row. Order matters here, since each order yields a different result in the race. There are 5! = 120 ways of ordering 5 runners. Now think - if there is no possibility for a tie, in how many of these 120 arrangements is M ahead of N? Considering the fact that there is no possibility of a tie, M will be either ahead or behind N. Intuitively, there is no real reason why there would be more results with M ahead of N than behind. Thus, the number of combinations where M is ahead of N should be a straightforward half of the total number of combinations - the total number of ways of ordering 5 runners in a race. There are 5! = 120 ways of ordering 5 runners, and half of that is simply 120/2=60. If you want to analyze this by numbers, do the following analysis: If N is last, then there are 4 places where M is ahead of N (1st, 2nd, 3rd and 4th places), and none where he is behind. If N is in fourth place, there are 3 places where M is ahead (1st, 2nd, 3rd) and one where he is behind. If N is in third place, there are exactly two places ahead and two behind. If N is in 2nd place, there is only one place ahead and 3 places behind. Note that this situation is the exact symmetrical opposite of the situation where n is in fourth place. If N is first, there are no places where M is ahead, and four places behind - the exact symmetrical opposite of the situation where N is last. If you go down the above list and sum the number of cases either way (ahead/behind), you will find that the number of places where M is ahead (4+3+2+1+0) is equal to the number of cases where M is behind N (0+1+2+3+4). This proves that there is no real bias either way, and thus the number of possible results where M is ahead of N should be exactly half of the total number of results, or 120/2=60.

A photographer is to take group photographs of a class of students for the school magazine such that each photograph should have five students. If there are four girls and four boys in the class and each photograph must not have two girls or two boys standing together, how many different photographs can the photographer take? 80 288 4^4 576 288^2

Q#: 55 Type: Quant problem solving question Issue: Combinations:choosing from a single source without repetition- order matters WTS: draw SeBoxes for each of the potential spots and go one spot at a time and figure out how many options you have left answer: d Elaboration: This fearsome looking question requires some thought. First, how can we make sure that we won't have two boys or two girls standing together? If we start with a boy on the left side of the row, then we cannot place another boy next to him - we must place a girl. Since that girl cannot have another girl standing next to her, we must alternate the third place with a boy again, etc. It follows that each group photo of five must alternate boy/girl/boy. However, there are two scenarios here, depending on whether we start with a boy (B-G-B-G-B) OR a girl (G-B-G-B-G). Calculate each scenario separately, then add the results, as they have an OR relationship. Within each scenario: The problem presents a case of choosing from Two sources - Boys and Girls. Keep in mind that the options from each source cannot be repeated, as no boy can logically appear twice in the same photo. Let's start with the photographs with girls and boys standing in the order GBGBG. Since there's no repetition, the number of options for each successive child from the same gender decreases: for example, if there are 4 choices for the first girl, there are only 3 choices for the next girl (third place), and two choices for last girl (5th place). The same goes for boys (2nd and 4th places). G B G B G 4 x 4 x 3 x 3 x 2 = 288 B G B G B 4 x 4 x 3 x 3 x 2 = 288 Now, do the same for photographs with girls and boys standing in the order BGBGB. Note that intuitively, this scenario should have the same number of permutations, as we have an identical 4 boys and 4 girls to choose from. Since there is an OR relationship here between either of the two types of photographs that can be taken, add the combinations. 288 + 288 = 576. Hence, this is the correct answer.

How many different arrangements are possible to place seven different books on a shelf if all three math books must be placed next to each other? 120 240 360 540 720

Q#: 56 Type: Quantitative problem solving question Issue: Combinations: choosing form a single a single source - number of ways to arrange k items - k! WTS: 1) group the three math book as a single item, and then find the internal combinations as well as the external combinations and multiply answer: e Elaboration: In questions in which objects must be adjacent, count them as one object at first. Here, treat the 3 math books as one - It's as if you have 4 books plus one "big" book. The question is then reduced to a relatively simple scenario of arranging 5 books (4 books plus 1 "big" math book) so you have 5! options to arrange them. However, there's another step: the three math books are not REALLY one book, and they can be rearranged between themselves. So now, since you can arrange the three math books in k!=3! different arrangements, multiply by the number of internal arrangements within the "big" book (3!): --> 5!×3! = 120×6 = 720

There are 10 points on a circle. A hexagon can be formed by linking 6 of the 10 points. How many such hexagons are possible? 60 120 200 210 600

Q#: 59 Type: Quant problem solving question Issue: Combinations: from the same source- no order WTS: throw the variable into the formula and solve Answer: d Elaboration: This problem presents a case of choosing from a Single source - points on a circle. You need to pick 6 out of 10 points. Does the order of picking them matter here? For a certain hexagon it doesn't matter if you chose vertex x before or after vertex y - in both cases, x and y will be vertexes of the hexagon. Therefore, the order of choosing the vertexes doesn't matter. You can either use the formula, in which case n=10 and k=6: C_(n,k)= (n!)/((n-k)! x k!)= 10!/((10-6)!x6!)= 10!/(4!x6!) = 210 or you can break down the problem using SeBoxes. There are 10 possible points to choose from for the first Vertex. You cannot choose the same vertex twice (no repetition), thus the number in each box decreases by 1. Since the order of choosing doesn't matter, divide the product of the boxes by the number of possible arrangements of the 6 chosen vertexes = 6!:

A and B are two multiples of 36, and Q is the set of consecutive integers between A and B, inclusive. If Q contains 9 multiples of 9, how many multiples of 4 are there in Q? 18 19 20 21 22

Q#: 71 Type: Quant Problem solving question Issue: Counting consecutive multiples within a range WTS: 1) Find the relevant ranges 2) Subtract them from each other 3)divide by x 4) add one Answer: b Elaboration: To find the multiples of 4 between A and B, you must subtract the relevant extremes, divide by 4 and add 1. For that, you need to find the relevant extremes, or their difference B-A. Note that since A and B are multiples of 36, they are also multiples of 9 and 4, so A and B serve as the relevant extremes for all cases. The question states that the number of multiples of 9 between A and B is 9. Use this information and the method of finding the number of multiples of 9 to find the range B-A: subtracting the extremes in the range (B-A), dividing by 9 and adding 1 should yield a result of 9. --> (B-A)/9 + 1 = 9 --> (B-A)/9 = 8 Therefore,(B-A) = 8×9 To find the multiples of 4 are in that range? Do the same, but now divide by 4: (B-A)/4 + 1 = --> (8×9)/4 + 1 = --> 9×2 + 1 = 19 Alternative method: Plug in a good number for A such as 36. Count 9 multiples of 9 to find B. Remember that A and B are also multiples of 9: A=36 - 45 - 54 - 63 - 72 - 81 - 90 - 99 - 108=B. Then use the same method to find the number of multiples of 4 between 36 and 108 inclusive: Subtract the relevant extremes: B-A=108-36=72. Divide by 4: 72/4=18 Add one: 18+1=19.

A jar contains 40 marbles, of which 15 are orange and 25 are beige. if 18 of the marbles were randomly removed, what is the probability that the next marble to be removed at random is beige? (1) Of the marbles removed, the number of beige marbles is twice that of orange marbles. (2) Of the first 16 marbles removed, 10 are beige.

Q#: 72 Type: Quant D/S what is the value of Issues: 1) Probability: the probability formula WTS: understand that you must know the exact number of beige and orange marbles Answer: a Elaboration: This is a "what is the value of..." DS question. In this type of question, a statement will be sufficient only if it leads to a single value of the variable (or expression) that you're asked about. The issue here is the probability of drawing a beige marble after some of the marbles were taken out. In order to answer this question you have to know how many beige marbles remained in the jar after removing 18 marbles. Note that you already know the total number of marbles after the exclusion. According to Stat. (1), If a total of 18 marbles were removed in which x are orange are and 2x are beige, then: --> x+2x = 18 --> 3x = 18 --> x = 6 Thus the number of beige marbles removed, is 2x = 12, and after the removal there are 25-12 = 13 beige marbles in the jar and 15-6 = 9 orange marbles in the jar. The probability of drawing a beige marble in this situation is thus 13/22 (the number of beige marbles out of the total number of marbles remaining). Stat.(1)->S->AD. According to Stat. (2), There is no telling what is the color of the marbles drawn after the 16th marble: The two additional marbles could be either beige or orange. Therefore, you cannot know how many beige marbles remained after the exclusion, and thus you cannot calculate that number out of the remaining 22 marbles. Stat.(2)->IS->A.

If n is an integer between 1 and 96 (inclusive), what is the probability that n×(n+1)×(n+2) is divisible by 8? 1/4 1/2 5/8 3/4 7/8

Q#: 73 Type: Quant problem solving question Issue: 1) Probability: more than one scenario WTS: 1) Test out a few scenarios, find the rules and solve answer: c Elaboration: Between 1 and 96 half of the numbers are even and half are odd, thus the probability that n is even (which will make n+1 odd and n+2 even) is 1/2. This is the probability of the first scenario. As for the second scenario: The probability that a number in a range is a multiple of 8 is the number of multiples of 8 in the range divided by the total number of items in the range. The denominator, the number of items in the range is the difference between the extremes plus 1: 96-1+1 = 96 numbers. Now to the numerator, the number of multiples of 8 in the range: When counting the multiples of x within a given range (in this case x = 8): 1) Find the relevant extremes - the nearest multiples of 8 within the specified range: The effective range is actually 8 to 96 inclusive. 2) Subtract the relevant extremes, and divide by 8: 96 - 8 = 88. 88/8 = 11 3) Add one: 11 + 1 =12 Thus the probability that n+1 is a multiple of 8 is 12/96 = 1/8. Finally, the total probability is 1/2 + 1/8 = 4/8 + 1/8 = 5/8

What Type of Question is this? What is the issue? How to solve it? Solve it? Which of the following functions satisfies the condition f(x) = f(-x) for all values of x? f(x) = x^3 f(x) = x+1 f(x) = (x/2) + 1 f(x) = (x/2) - 1 f(x) = x^2 + 1

Q#: 8 Type: Problem solving- With variable in the answer Issue: 1) Positive and negative 2) the effect of powers on sign 3) functions 4) integer "must be true" questions WTS: plug in a simple answer like 1 and POE the answers answer: e elaboration: Note the phrasing of the question: "for all values of x..". Only the right function will satisfy the condition for every possible value you choose; the other four answer choices may satisfy the condition for some values, but not for others. Don't mess with the algebra - plug in easy numbers (for instance, x=1) into each of the functions and POE. Keep plugging in and eliminating answer choices that do not meet the condition, until you are left with a single answer choice. If x=1, --> f(x) = f(1) = 12+1 = 2 --> f(-x) = f(-1) = (-1)2+1 = 2 For x=1, f(x) = f(-x). Since no other answer choice satisfies this condition for the same plug in, this is the right answer by POE. Alternative method: Note that squaring x eliminates the sign (positive or negative), which this is why this function will always have equal values for x and -x.

What Type of Question is this? What is the issue? How to solve it? Solve it? If x is a positive integer greater than 2, the product of x consecutive positive integers must be divisible by which of the following? I. x - 1 II. 2x III. x! I only II only II and III only I and III only I, II and III only

Q#: 9 Type: Problem solving- Must be. With variables in the answer fields Issue: Sequences: Consecutive Integers - the Product of n Consecutive Integers is Divisible by n! WTS: 1) The product on n consecutive integers is divisible by n! as such it will be divisible by any of the factors of n! 2) Since there are variables in the answers we should plug in good numbers answer: e elaboration: Plug in x = 3. In this case, 3 x 4 x 5 = 60. 60 is divisible by x - 1 = 3 - 1 = 2. 60 is divisible by 2x = 2 (3) = 6. 60 is divisible by x! = 3! = 3 x 2 x 1 = 6. Plug in another value of x to be sure because the question says 'must be'. This time try an even integer i.e. x = 4. In this case, 2 x 3 x 4 x 5 = 120. 120 is divisible by x - 1 = 4 - 1 = 3. 120 is divisible by 2x = 2 (4) = 8. 120 is divisible by x! = 4! = 4 x 3 x 2 x 1 = 24. At this point, you can choose answer choice E with a high degree of certainty that all of the statement must be true. If you have the time, take a look at the algebra and prove why the statements must be true: The product of x consecutive integers is divisible by x!, so III is definitely true. Now, x! is 1⋅2⋅3⋅...(x-1)⋅x , so (x-1) and 2 are included in x! (that is, provided that x is greater than 2). Hence, this is the correct answer.