May 1- Practice Exam Review

Evenly Spaced Sets: *CRITICAL*

A set of numbers in which each number is a set distance, or increment, from the next. 1, 4, 7, 10 is an evenly spaced set with an increment of 3. (x+3, x+6, c+9) In any evenly spaced set: •The arithmetic mean (average) is equal to the median. The set comprising 1, 2, 3, 4, and 5 has a mean of 3 and a median of 3. •The mean and the median of the set are both equal to the average of the first and last terms in the set. For the set {1, 2, 3, 4, 5}, the first term is 1 and the last term is 5. The average of these two terms is 3, which also equals both the median and the mean of the entire set. •The sum of the elements in a set equals the mean multiplied by the number of items in the set. For the set {1, 2, 3, 4, 5}, the mean is the middle term, 3. There are five terms in the set, so the sum is 3*5 = 15.

EXAMPLE: HOW TO THINK ABOUT CONSECUTIVE INGETER PROBLEMS. * If n is a positive integer and r is the remainder when n^2 - 1 is divided by 8, what is the value of r? (1) n is odd. (2) n is not divisible by 8.

Think in baby steps - what do we know? (n^2)-1 = (N+1)(n-1) N= a positive integer (n-1), n, and (n+1) - three consecutive integers! *** KEY TO THE CASTLE: Recognize the difference of squares. Then NOTE that they are talking about consecutive integers. boom BEFORE going to the answer choices. Okay, NOW .. The q is asking us to take the first and third consecutive integers and multiply them together (n-1)(n+1) , then divide by 8 and see what the remainder, r, is 1. n is odd. Hmmm. If n is odd, what else can I figure out? Well, n-1 and n+1 must both be even! Right, (n-1), n, (n+1) - if n is odd 3 then n-1 is 2 and n+1 is 4. That means that it's divisible by 2. Actually, it must be divisible by 4, because each of the two numbers is separately divisible by 2. In other words, the product (n+1)(n-1) has at least two factors of 2 so it's divisible by 4. But the problem asks about 8. Maybe I'll just try a few things, just in case: So, we could have 0, 1, 2, as our three consecutive integers. Product of 0 and 2 is 0. 0/8=0 remainder 0. r=0 What about: 2,3,4 , well n=3 so (n-1)(n+1) = (2)(4)= 8. 8/8=1 with a remainder 0.... hmm that is interesting. What's my next possibility? 4, 5, 6. The product of 4 and 6 is 24 and 24/8 = 3 with a remainder of 0. For this particular case, r=0 yet again! All right. Is this a pattern or did I just get lucky that the first three all gave the same remainder? What have I been doing? I've been taking two consecutive even integers and multiplying them together, then dividing by 8. The two starting integers are each divisible by 2, as I figured out before oh, and here's the new thing! Every other even integer, by definition, is a multiple of 4, not just a multiple of 2. 0, 2, 4, 6, 8, 10, 12 starting with zero, every other one is a multiple of 4. So, whenever I multiply two consecutive even integers, one of them has to be a multiple of 4. What does that mean for the product? The product has to contain three 2s as factors, not just two 2s. The product will gain at least one 2 from the plain even integer and the product will gain at least two 2s from the multiple of 4 even integer. That means the product must be a multiple of 8! So, whenever I divide by 8, I'm going to have a remainder of zero. Bingo! Statement 1 is sufficient. Eliminate answers B, C, and E. Moving on to statement 2: (2) n is not divisible by 8. Hmm. The middle integer of my three consecutive integers is not divisible by 8. What are some possibilities? I can re-use my first possibility from statement 1: 0, 1, 2. When we tested this case before, the remainder, r, was zero. What's the next thing I can try? 1, 2, 3. (Note: make sure to use whatever's possible next, NOT simply what we used for the previous statement. The previous statement had different parameters.) So, the product of 1 and 3 is 3, and 3/8 is 0 remainder 3. r = 3. Great! I have (at least) two different possible values for r, so I know this statement is insufficient. Eliminate answer D. The correct answer is A.

Sum of n consecutive integers and divisibility There are two cases, depending upon whether n is odd or even: •If n is odd, the sum of the integers... •If n is even, the sum of the integers is...

•If n is odd, the sum of the integers is always divisible by n. Given 6+7+8, we have n = 3 consecutive integers. The sum of 6+7+8, therefore, is divisible by 3. •If n is even, the sum of the integers is never divisible by n. Given 6+7+8+9, we have n = 4 consecutive integers. The sum of 6+7+8+9, therefore, is not divisible by 4.

Evenly spaced sets: let's practice... •The mean and the median of the set are both equal to....

•The mean and the median of the set are both equal to the average of the first and last terms in the set. For the set {1, 2, 3, 4, 5}, the first term is 1 and the last term is 5. The average of these two terms is 3, which also equals both the median and the mean of the entire set.

In any evenly spaced set: •The sum of the elements in a set equals the mean multiplied by...

•The sum of the elements in a set equals the mean multiplied by the number of items in the set. For the set {1, 2, 3, 4, 5}, the mean is the middle term, 3. There are five terms in the set, so the sum is 3*5 = 15.

If n is an integer between 10 and 100, is the tens digit of n even? (1) The remainder when n is divided by 4 is equal to the remainder when n is divided by 5. (2) The only prime factor of n is 3. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient. EACH statement ALONE is sufficient. Statements (1) and (2) TOGETHER are NOT sufficient.

(1) SUFFICIENT: You can test cases or use an algebraic approach. Both methods are shown. Test Cases Investigate different possible values of n. The first case that jumps to mind is the case where n = (4)(5) = 20. In this case, the remainder is 0 when n is divided by either 4 or 5. This is true for 20, 40, 60, and 80. In all cases, the tens digit is even. Would other possible values work? Yes! For example, if n = 21, then the remainder is 1 when divided by either 4 or 5. If n = 22, the remainder is 2 when divided by either 4 or 5. This pattern holds up until n = 24, where the remainder is 0 when divided by 4, but the remainder is 4 when divided by 5. The pattern begins again at 40, 60, and 80. The statement is valid for values of n from 20 through 23, 40 through 43, 60 through 63, and 80 through 83. In every case, the number has an even tens digit, so the statement is sufficient. Algebra Let the remainder (in both cases) be designated as r. Then n is r greater than a multiple of 4, and n is r greater than a multiple of 5. Taking the two observations together reveals that n must be r greater than a multiple of both 4 and 5 - i.e., r greater than a multiple of 20. The remainder r can only be 0, 1, 2, or 3 (a remainder can never be equal to or greater than the divisor, so it can't be 4 or greater). Therefore, r is equal to a multiple of 20 plus 0, 1, 2, or 3. All of those values have an even tens digit. (2) SUFFICIENT: According to this statement, n is a power of 3. The only such powers between 10 and 100 are 33 = 27 and 34 = 81. Both of these values have an even tens digit. The correct answer is (D).

Key Takeaways for Solving Consecutive Integer Problems: (1) They can test you on a concept without naming that concept. How do you recognize it?

(1) They can test you on a concept without naming that concept. Be able to recognize a consecutive integer problem in disguise. The form n^2-1 is probably the most common indicator and n^3- n is another very common indicator because n3 " n = n(n^2 - 1 ) = n(n+1)(n-1).

Key Takeaways for Solving Consecutive Integer Problems: (2) Also be able to recognize some of the common useful pieces of information that might be given on consecutive integer problems.

(2) Also be able to recognize some of the common useful pieces of information that might be given on consecutive integer problems. Knowing something as seemingly simple as whether certain terms are even or odd can make a big difference, especially if the question deals with divisibility or remainders.

Key Takeaways for Solving Consecutive Integer Problems: (3) If you do the figuring out work while you're studying, you will save yourself a ton of time and mental effort on the test.

(3) If you do the figuring out work while you're studying, you will save yourself a ton of time and mental effort on the test. You may be able to figure out a decent proportion of this stuff during the test, but you'll go a lot further if you study how to recognize it in the first place.

Really Important Concept: (n - 1)(n)(n + 1) is a multiple of 3. ...what is the rule here?

(n - 1)(n)(n + 1) is a multiple of 3. Any three consecutive positive integers include exactly one multiple of 3 (if you don't remember this rule, try some real numbers and prove it to yourself). Of the three terms n - 1, n, and n + 1, one is a multiple of 3 but we have no way to determine which one.

Is positive integer n - 1 a multiple of 3? (1) n^3 - n is a multiple of 3 (2) n^3 + 2n^2+ n is a multiple of 3

AWESOME- REVIEW The question is in very simple form already; rephrasing the question isn't useful. The statements, however, can be rephrased. Statement (1) gives the formula n3 - n. We can first factor out an n to get n(n2 - 1). Next, we can factor (n2 - 1) to get (n + 1)(n - 1). So, n3 - n factors to n(n + 1)(n - 1). Notice that the three factored terms represent consecutive integers: n - 1, n, and n + 1. Now, let's rephrase statement (2). We first factor out an n to get n(n^2 + 2n + 1). Next, we can factor (n^2 + 2n + 1) to get (n + 1)(n + 1). So, n^3 + 2n^2 + n factors to n(n + 1)(n + 1). Notice that the factored terms represent two consecutive integers, with the larger of the two represented twice. (1) INSUFFICIENT: (n - 1)(n)(n + 1) is a multiple of 3. Any three consecutive positive integers include exactly one multiple of 3 (if you don't remember this rule, try some real numbers and prove it to yourself). Of the three terms n - 1, n, and n + 1, one is a multiple of 3 but we have no way to determine which one. (2) SUFFICIENT: n(n + 1)(n + 1) is a multiple of 3. This tells us that either n or n + 1 is a multiple of 3. The question asks whether the term n - 1 is a multiple of 3. Recall that n - 1, n, and n + 1 represent three consecutive integers and also recall that any three consecutive integers include exactly one multiple of 3. Note that we do not need to get this information from statement (1). If the multiple of 3 is either the n or the n + 1 term, then the n - 1 term cannot be a multiple of 3. The correct answer is B.

If x + 2y = z, what is the value of x? (1) 3y = 4.5 + 1.5z (2) y = 2 Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient. EACH statement ALONE is sufficient. Statements (1) and (2) TOGETHER are NOT sufficient.

C-> A trap here. Start with Statement 2. Why is that there? Clearly insufficient, huh? The test makers want us to think we need it for A... but do we? NO. Don't be the FOOL. If we solve the equation x + 2y = z in terms of x, we can rephrase the question. x = z - 2y The question becomes "What is z - 2y ?" (1) SUFFICIENT: We can manipulate this statement to solve for z - 2y: 3y = 4.5 + 1.5z divide both sides by 1.5 2y = 3 + z z - 2y = -3 (2) INSUFFICIENT: We cannot solve for the expression z - 2y, nor can we find the value of z separately. The correct answer is A.

Consecutive integers... Every other even integer, by definition, is a multiple of 4, not just a multiple of 2. So, whenever I multiply two consecutive even integers...

Consecutive integers... Every other even integer, by definition, is a multiple of 4, not just a multiple of 2. 0, 2, 4, 6, 8, 10, 12 starting with zero, every other one is a multiple of 4. So, whenever I multiply two consecutive even integers, one of them has to be a multiple of 4. What does that mean for the product? The product has to contain three 2s as factors, not just two 2s. The product will gain at least one 2 from the plain even integer and the product will gain at least two 2s from the multiple of 4 even integer. That means the product must be a multiple of 8! So, whenever I divide by 8, I'm going to have a remainder of zero.

List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in S is 7 more than the least integer in T, how much greater is the average (arithmetic mean) of the integers in S than the average of the integers in T ? A. 2 B. 7 C. 8 D. 12 E. 22

For any evenly spaced set median = mean = the average of the first and the last terms. So the mean of S will be the average of the first and the last terms: mean = (x + x + 9*2)/2 = x+9, where x is the first term; The mean of T will simply be the median or the third term: mean = (x - 7) + 2*2 = x - 3; The difference will be (x + 9) - (x - 3) = 12. Answer: D.

Evenly spaced sets: INCLUSIVE Given "a set of integers between 15 and 19 inclusive," we include the ....

Given "a set of integers between 15 and 19 inclusive," we include the numbers 15 and 19 in the set: {15,16,17,18,19}. Given "a set of integers between 15 and 19," we do not include 15 and 19 in the set: {16,17,18}.

Bag A contains red, white and blue marbles such that the red to white marble ratio is 1:3 and the white to blue marble ratio is 2:3. Bag B contains red and white marbles in the ratio of 1:4. Together, the two bags contain 30 white marbles. Q: How many red marbles could be in bag A? 1 3 4 6 8

Organization is critical: Organize Bag A, Bag B, Bag A+B - then look at WHAT ARE THEY ASKING ME? - Figure out how many White marbles there are. By using a chart.. you get to 6A+4B=30 If a=1; then 6A=6 If b=1; then 4B=4 --- Until you get to 6A+4B=30 Turns our A =3 B=3 (18)+(12)=30 Go back to A's ratios: 1 red to every 3 white or 1x+3x There are 6 white marbles - meaning x=2- 1*2=2 red marbles -- if there are 18 white marbles then 3x=18 x=6 or if x=6, then 1x=6 red marbles - There could be 6 red marbles or 2 red marbles We are told that bag B contains red and white marbles in the ration 1:4. This implies that WB, the number of white marbles in bag B, must be a multiple of 4. What can we say about WA, the number of white marbles in bag A? We are given two ratios involving the white marbles in bag A. The fact that the ratio of red to white marbles in bag A is 1:3 implies that WA must be a multiple of 3. The fact that the ratio of white to blue marbles in bag A is 2:3 implies that WA must be a multiple of 2. Since WA is both a multiple of 2 and a multiple of 3, it must be a multiple of 6. We are told that WA + WB = 30. We have already figured out that WA must be a multiple of 6 and that WB must be a multiple of 4. So all we need to do now is to test each candidate value of WA (i.e. 6, 12, 18, and 24) to see whether, when plugged into WA + WB = 30, it yields a value for WB that is a multiple of 4. It turns out that WA = 6 and WA = 18 are the only values that meet this criterion. Recall that the ratio of red to white marbles in bag A is 1:3. If there are 6 white marbles in bag A, there are 2 red marbles. If there are 18 white marbles in bag A, there are 6 red marbles. Thus, the number of red marbles in bag A is either 2 or 6. Only one answer choice matches either of these numbers. The correct answer is D.

Product of n consecutive integers and divisibility: The product of n consecutive integers is always divisible by ...

Product of n consecutive integers and divisibility: The product of n consecutive integers is always divisible by n! Given 6*7*8*9, we have n = 4 consecutive integers. The product of 6*7*8*9, therefore, is divisible by 4! = 4*3*2*1 = 24.

(n - 1)(n)(n + 1) is a multiple of 3. Q: Which one is a multiple of three?

RULE: Any three consecutive positive integers include exactly one multiple of 3 (if you don't remember this rule, try some real numbers and prove it to yourself). *Of the three terms n - 1, n, and n + 1, one is a multiple of 3 but we have no way to determine which one.

Recall that N^3+2n^2+n represent ...what? What is the rule here? When you see this... think WHAT?

Recall that n - 1, n, and n + 1 represent three consecutive integers and also recall that any three consecutive integers include exactly one multiple of 3.

The number m is the average (arithmetic mean) of the positive numbers a and b. If m is 75% more than a, then m must be a) 30% less than b b) less than b c) 50% less than b d) less than b e) 75% less than b

STRATEGY: -Review the problem and set it up. Average = Sum/# Note you have variables a & b... choose SMART #'s and plug & play - Pick smart #'s We can use smart numbers or try an algebraic approach; both methods are shown below. Smart Numbers Choose numbers that satisfy the given statements. It's easier to begin by picking an m and a that satisfy the second statement (m is 75% more than a); then b can be calculated fairly easily based on the relationship between m and a. If a = 100 then m = 175. According to the first statement, 175 is the mean of 100 and b; therefore b = 250. The question asks for a value that's in the answers, so assign a varaible; let's call it x: m is x% less than b m = b - (x%)b 175 = 250 - (x/100)(250) 175 = 250 - 2.5x 2.5x = 75 x = 75(2/5) = 30 Thus m is 30% less than b. Algebra If m is 75% more than a, then m = a + 0.75a = 1.75a. According to the first statement, m (or 1.75a) is the mean of a and b; therefore, 1.75a = (a + b)/2. Multiply by 2 to get rid of the fraction: 3.5a = a + b 2.5a = b The question asks for a value that's in the answers, so assign a varaible; let's call it x: m is x% less than b m = b - (x%)b substitute in to get rid of m and b: 1.75a = 2.5a - (x/100)(2.5a) multiply by 100: 175a = 250a - 2.5xa divide by a: 175 = 250 - 2.5x 2.5x = 75 x = 75(2/5) = 30 Thus m is 30% less than b. The correct answer is A.

If (243)^x(463)^y = n, where x and y are positive integers, what is the units digit of n? (1) x + y = 7 (2) x = 4

STRATEGY: Re-word to better understand the question. Take a deep breath and really understand - what is asking me? What do I know? How is it trying to trick me? Clearly 243 and 463 raised so the something is a huge number that the GMAT would never ask me to calculate. - DUH, so what might they secretly be asking? hmmm... two 3's as the units digit *HINT* *HINT*- think like CEO. We know from the question that x and y are integers and also that they are both greater than 0. Because we are only concerned with the units digit of n and because both bases end in 3 (243 and 463), we simply need to know x + y to figure out the units digit for n. Why? Because, to get the units digit, we are simply going to complete the operation 3^x × 3^y which, using our exponent rules, simplifies to 3^(x + y). CRITICAL.... So we can rephrase the question as "What is x + y?" (1) SUFFICIENT: This tells us that x + y = 7. Therefore, the units digit of the expression in the question will be the same as the units digit of 37. (2) INSUFFICIENT: This gives us no information about y. The correct answer is A.

Let's practice... In any evenly spaced set: •The arithmetic mean (average) is ....

The arithmetic mean (average) is equal to the median. The set comprising 1, 2, 3, 4, and 5 has a mean of 3 and a median of 3.

If x is a positive integer, is (x)(x + 2)(x + 4) divisible by 12? (1) x2 + 2x is a multiple of 3. (2) 3x is a multiple of 2.

This problem asks about the divisibility of (x)(x + 2)(x + 4). Though we may be tempted to distribute the terms in this algebraic expression, it is easier to analyze the divisibility of the product in its current factored form. Notice that each term is two more than the preceding term; this means that this expression is a product of three consecutive even or odd integers. (It is important to note that though each number is two more than the preceding term, this does not necessarily mean that each number is even: 1, 3, 5 and 2, 4, 6 both fit this form). Thus we are trying to determine if the product of three positive, consecutive odd or even integers is divisible by 12. For a product to be divisible by 12, it must be divisible by both 3 and 4. We can pick numbers for x to see whether there are any larger patterns we can use. If x = 1, our product is (1)(3)(5), which is divisible by 3. If x = 2, our product is (2)(4)(6), which is divisible by 3 and 4. If x = 3, our product is (3)(5)(7), which is divisible by 3. If x = 4, our product is (4)(6)(8), which is divisible by 3 and 4. For these four examples, the product of three consecutive odd or even integers was divisible by 3 in every case. Will that be true no matter what we try? Yes. Three consecutive even integers will always contain one integer that is a multiple of 3 because every 3rd even integer is a multiple of 3. For example, in the set {2, 4, 6, 8, 10, 12, ...}, we can see that every 3rd number is a multiple of 3. Similarly, three consecutive odd integers will always contain one integer that is a multiple of 3. On the other hand, the product was only divisible by 4 when x equaled 2 or 4 in our examples. In order for the product to be divisible by 2, the numbers must contain at least two factors of 2. The product of three consecutive odd numbers will be odd and therefore contain no 2's. However, the product of three consecutive even integers will contain at least three 2's and therefore be divisible by 4. Thus, we can rephrase the question "is x even?" (1) INSUFFICIENT: Here we are told about the divisibility of a sum. We can express this sum as a product by factoring out the common term: x2 + 2x = x(x + 2) We now have the product of two consecutive even or odd integers. Statement 1 tells us that this product is divisible by 3, so at least one of the terms must be divisible by 3. We can test numbers (remember that we are only allowed to pick numbers that make statement 1 true). If we let x = 3, then statement 1 is true. Is x even? No. Next, we want to try to prove insufficiency. We have a "no" answer already. What kind of number would we need to try in order to get a "yes" answer? An even number. If x = 6 then statement 1 is still true. Is x even? Yes. Overall, using statement 1, we can't tell whether x is even. (2) SUFFICIENT: In order for 3x to be a multiple of 2, the x term must contain at least one 2. We can see this by referring back to our odd and even rules: if (odd)(x) = even, then x must be an even number. The correct answer is B.

For positive integer k, is the expression (k + 2)(k^2 + 4k + 3) divisible by 4? (1) k is divisible by 8. (2) K+1/3 is an odd integer. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient. EACH statement ALONE is sufficient. Statements (1) and (2) TOGETHER are NOT sufficient.

[Q7 on Quant Pract test 05/01] STRATEGY: 700-800 level 1. Simplify the question. What is it really asking? 2. LOOK For patterns (k+1)(k+2)(k+3) 3. Note that when you're multiplying consecutive integers you need the # that it is divisible by to be in the problem or have enough prime factors - example: 4 must have 2 even integers (2*3*4) (6*7*8) ... both divisible by 4 but (5*6*7) is NOT The quadratic expression k^2 + 4k + 3 can be factored to yield (k + 1)(k + 3). Thus, the expression in the question stem can be restated as (k + 1)(k + 2)(k + 3), or the product of three consecutive integers. This product will be divisible by 4 if one of two conditions are met: If k is odd, both k + 1 and k + 3 must be even, and the product (k + 1)(k + 2)(k + 3) would be divisible by 2 twice. Therefore, if k is odd, our product must be divisible by 4. If k is even, both k + 1 and k + 3 must be odd, and the product (k + 1)(k + 2)(k + 3) would be divisible by 4 only if k + 2, the only even integer among the three, were itself divisible by 4. The question might therefore be rephrased "Is k odd, OR is k + 2 divisible by 4?" Note that a 'yes' to either of the conditions would suffice, but to answer 'no' to the question would require a 'no' to both conditions. (1) SUFFICIENT: If k is divisible by 8, it must be both even and divisible by 4. If k is divisible by 4, k + 2 cannot be divisible by 4. Therefore, statement (1) yields a definitive 'no' to both conditions in our rephrased question; k is not odd, and k + 2 is not divisible by 4. (2) INSUFFICIENT: If k + 1 is divisible by 3, k + 1 must be an odd integer, and k an even integer. However, we do not have sufficient information to determine whether k or k + 2 is divisible by 4. The correct answer is A.