Sequences: Overview

Q2...What is the volume of a cube whose edges are 0.1 meters long? 0.0001 cubic meter 0.001 cubic meter 0.006 cubic meter 0.01 cubic meter 0.3 cubic meter

B Correct. The volume of a cube is given in the formula Volume=s3 = 0.13 = (1/10)3 = 13/103 = 1/1000 = 0.001

To sum up:Average of a Set = Average of First and Last Terms

For any set of integers with a constant difference between any two consecutive terms (such as consecutive integers): The Average of the set = Average of any equidistant pair of terms around the median. From this insight, remember this simpler rule: The Average of a Set of consecutive integers = Average of the first and last terms of the set.

A whale goes on a feeding frenzy that lasts for 9 hours. For the first hour he catches and eats x kilos of plankton. In every hour after the first, it consumes 3 kilos of plankton more than it consumed in the previous hour. If by the end of the frenzy the whale will have consumed a whopping accumulated total 450 kilos of plankton, how many kilos did he consume on the sixth hour? A seemingly tough question, requiring several steps.

Fortunately, this question, and others like it, can be solved with the use of the properties of arithmetic sequences. An arithmetic sequence has several properties which are tested in GMAT problems. One of them is the average property: in a set of integers with a constant difference between them (including consecutive integers, where the difference is 1): Average of the set = Median of the set. Take a look at this simple set of consecutive integers: {1, 2, 3}. The median of the set is easily recognizable - since the set has an odd number of terms, the median is the number in the middle, or 2. The Average of the set is total / # of items = (1+2+3) / 3 = 6/3 = 2. This stays true even if the set includes an even number of consecutive integers, as in the set {1, 2, 3, 4}. since there is an even number of members in the set, the median is calculated as the average of the two middle terms: (2+3) / 2 = 5 / 2 = 2.5. Surprisingly enough, the average of the set is also (1+2+3+4) / 4 = 10 / 4 = 2.5. This example shows the most common occurrence of this property, which is sets of consecutive integers. However, this property is not confined to consecutive integers alone; In any set of integers which constitute an arithmetic sequence - any set of integers with a constant difference between each consecutive terms (be it a difference of 1 (1,2,3), two (2, 4 ,6) or 5 (2, 7, 12)) - the average = the median. Think of the average of a set of consecutive numbers as the exact "middle" of the set. Okay. Interesting concept. Can I see that whale question again? Your wish is our command. A whale goes on a feeding frenzy that lasts for 9 hours. For the first hour he catches and eats x kilos of plankton. In every hour after the first, it consumes 3 kilos of plankton more than it consumed in the previous hour. If by the end of the frenzy the whale will have consumed a whopping accumulated total of 450 kilos of plankton, how many kilos did he consume on the sixth hour? 38 47 50 53 62 D Correct. The question describes an arithmetic sequence with a difference of 3: in the first hour our whale consumes x kilos, in the second (x+3), in the third (x+6), etc. Adding these together will give a total of 450, from which we can find x, but that is not an easy calculation. By the time you're done with that, you might easily forget that the question does not ask for x, but rather for the consumption in the sixth hour, which is actually x+15. Instead, recall the average property of arithmetic sequences: Average = Median. Since the question kindly provides the total kilos of Plankton (450) and the number of hours (9), the average hourly consumption of Plankton can be easily calculated: 450 / 9 = 50. Therefore, the Median of our set of consecutive integers is also 50. Since the set has an odd number of members, the median is the number in the middle, or the 5th hour. If the whale consumes 50 kilos of Plankton in the 5th hour, he will consume 50+3 = 53 kilos in the sixth hour. Quick and easy - with the right approach.

To sum up:

1) Every GMAT question is solvable in a reasonable amount of time. If the solution doesn't present itself, don't panic! Begin with a similar, simple case first, take the question a few steps further, and look for some underlying repeating pattern. 2) In questions involving units digits of a product or power, focus only on the units digits of the factors.

In a certain series, each term is m greater than the previous term. If the 17th term is 560 and the 14th term is 500, what is the first term? 220 240 260 290 305

B Close enough - you took 1 minutes and 53 seconds to answer this question. Correct. In this arithmetic set, the difference between any two consecutive terms is m. Since there are three terms from a14 to a17 we can conclude that: --> a17 = a14+3·m --> 560 = 500+3·m --> m=20 We now can conclude that: --> a1 = a14-13·m --> a1 = 500-13·20 --> a1 = 240

If in a certain sequence of consecutive multiples of 50, the median is 625, and the greatest term is 950, how many terms that are smaller than 625 are there in the sequence? 6 7 8 12 13

B Correct. Since the median is 625, which is not a multiple of 50, then the two middle terms in the set (on both sides of 625) are 600 and 650. The same number of integers lie below and above the median, thus we can calculate the number of terms above 625 to know the number of terms below 625: --> ((950-650)/50)+1 = (300/50)+1 = 6+1 = 7 terms.

The sum of all consecutive odd integers from −21 to 31, inclusive, is 130 135 150 156 235

B Correct. The sum of odd integers from -21 to 21 is zero because each negative value cancels out its positive counterpart. Based on this, add odd integers from 22 to 31 to get the sum of consecutive integers from -21 to 31. 23 + 25 + 27 + 29 + 31 = 135. Hence, this is the correct answer.

Can you think of a series of 3 consecutive integers in which none of the integers is a multiple of 3? Let's look at an example 1, 2, 3 - includes 3. 2, 3, 4 - has 3. 3, 4, 5 - has 3 4, 5, 6 - doesn't include 3, but includes 6, which is a multiple of 3. And so on. Any set of 3 consecutive integers must contain one multiple of 3. One result of this insight is that the product of 3 consecutive integers must be divisible by 3. If x, y and z are consecutive integers, then one of them is divisible by 3, and therefore x·y·z will also be divisible by 3. Note that x, y and z must also include AT LEAST one even integer (as the above numerical example shows), so the product x·y·z must be divisible by 2 AND 3, so it must be divisible by 6. We can state that the product of 3 consecutive integers must be divisible by 3! = 3·2·1.

There's a general rule here that's worth remembering: the product of n consecutive integers is divisible by n!. For example: 4 consecutive integers are divisible by 4! = 24 5 consecutive integers are divisible by 5! = 120 etc.

How many multiples of 7 are there between 15 to 77, inclusive? The process of answering is as follows: 1) Find the extremes - the nearest multiples of 7 within the specified range. 15 is not a multiple of 7 - the nearest multiple of 7 within the range is 21. 77 is a multiple of 7, so we leave it as the higher extreme of our range. 2) Subtract the extremes, and divide by 7. (77 - 21) / 7 = 56 / 7 = 8 3) Add one. 8+1 = 9.

To sum up: Counting the number of integers within a given range - subtract the extremes, and add one. General case: Counting the multiples of x within a given range: 1) Find the relevant extremes - the nearest multiples of x within the specified range. 2) Subtract the relevant extremes, and divide by x. 3) Add one.

The GMAT quantitative section is not really about testing your math skills. Your MBA won't require you to know the solution of a quadratic equation or the formula for the volume of a cylinder. The GMAT checks whether you demonstrate cognitive processes that support decision making. One of these is the ability to recognize a recurring pattern in a seemingly meaningless body of data. For this reason, some GMAT problems will seem unsolvable at first, at least under the 2 minutes per question time limit. The following is a classic example: What is the unit's digit of 137? On the surface, there are no higher math concepts involved. This question is solvable by the simple, brute force method of multiplying 13 by itself 7 times. However, doing so in under two minutes is another story. There must be a trick, some easy way of quickly solving the question, or it wouldn't appear on the GMAT.

We will solve this question in a bit, but this lesson is not about unit's digits, or power multiplication. Keep this simple, optimistic concept in mind when taking the GMAT: "Every GMAT question is solvable in a reasonable amount of time. If the solution doesn't present itself, don't panic! Begin with a similar, simple case first, take the question a few steps further, and look for some underlying repeating pattern."

If Μ(m) is defined by the equation M(m)=(m−1)·M(m−1) for all positive m>1, and if M(4)=24, then M(3)= 6 8 12 72 96

B You underestimated the time this question took you. You actually solved it in 3 minutes and 40 seconds. Correct. The function in the question presents a sequence in which each term is dependent on the preceding term, i.e M(4) is dependent on M(3), etc. Start with what you know - the value of M(4). Since the problem presents the value of M(4), Plug in m=4 into the RULE: --> M(4) = (4−1)·M(4−1)= 3·M(3) = 24 Now Isolate the desired M(3): --> 3·M(3) = 24 --> M(3) = 24/3 = 8

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

D Correct. Variables in the question stem and in the answer choices? plug in and eliminate! In this case you can go with A=2, B=8 , for example. If A=2 and B=8, then the odd integers between them are 3, 5 and 7, whose average is (3+5+7)/3=15/3=5. That's your Goal - plug in A=2 and B=8 into the answer choices and eliminate any answer choice that does not match your Goal. --> (A+B)/2 = (2+8)/2 = 10/2 = 5. This is the right answer choice because all other answer choices are eliminated for the same Plug in.

The total number of plums that grow during each year on a certain plum tree is equal to the number of plums that grew during the previous year, less the age of the tree in years (rounded down to the nearest integer). During its 3rd year, the plum tree grew 50 plums. If this trend continues, how many plums will it grow during its 6th year? 43 41 38 35 32

D Incorrect. This is a sequences question. The question stem can be translated into a rule: An = An-1 - n, where n is the age of the tree in years. Keep in mind that the tree is 0 years old during its first year. Thus, A2 = 50. 35 is the answer you get when you assume that the tree is 3 years old (instead of 2 years old) during its 3rd year. C Correct. 4th Year: A3 = A2 - 3 = 50 - 3 = 47 5th Year: A4 = A3 - 4 = 47 - 4 = 43 6th Year: A5 = A4 - 5 = 43 - 5 = 38

Q3...F and E are midpoints of AB and AD respectively. If the side of square ABCD is 8, what is the area of triangle CEF? 8√2 9√3 18 16√2 24

E Correct. Don't hassle with needless calculations, be efficient - ballpark! Draw the figure and indicate every length you know. If the shaded region is hard to estimate, try estimating the un-shaded regions instead. Use the fact that E and F are midpoints to calculate the area of the unshaded right triangles. The area of a right triangle whose legs are a and b is (a·b)/2. The area of the shaded region is the area of the large square (82=64) minus the areas of the unshaded triangles, 64−16−16−8=24, and the correct answer is E. Sometimes relying on the figure can quickly bring you to the exact answer.

Which of the following functions satisfies the condition f(x) = f(-x) for all values of x? f(x) = x3 f(x) = x+1 f(x) = (x/2) + 1 f(x) = (x/2) - 1 f(x) = x2 + 1

E Correct. 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.

A computer program assigns consecutive numbers to the days of the week. Sunday is 1, Monday is 2, and so ... and Saturday is 7. Every day, the computer program calculates the value of parameter D according to the following definition: D is the number of the day times the value of D on the previous day. If the computer calculated D to be 10 on Wednesday, what would be the value of D calculated on the following Saturday? 210 1200 1250 2100 8400

E Incorrect. Translate this confusing question into a RULE of the sequence: Dn =Dn-1 × n Where Dn is the value of parameter D on day n. According to the question D4 = 10. Plug this number into the RULE to find all consecutive Values of D on Thursday=5, Friday=6 and Saturday=7. D Correct. Dn = Dn-1 × n --> D5 = D4 × 5 = 10 × 5= 50 --> D6 = D5 × 6 = 50 × 6= 300 --> D7= D6 × 7 = 300 × 7= 2100

A sequence is a set of numbers that follow a certain RULE. In fact, sequences can be viewed as a subset of functions questions. Examples: 1) An = n+2 The nth term in this sequence is determined by the RULE of n+2. The value of a term in this sequence isdetermined solely by its place in the sequence n. The first term (n=1) is A1=1+2=3. The second term (n=2). is A2=2+2=4.

2) An = An-1+4 The nth term in this sequence is determined by the RULE of An-1+4. The value of a term An in this sequence is determined by the value of the preceding term An-1. Note that in this case, it is necessary to know the value of one term in order to determine the value of any other term. If the problem states that the first term A1 is 3, then the second term A2=A1+4=3+4=7. The third term will then be A3=A2+4=7+4=11.

If the dimensions of a rectangular box are a by b by c inches, and a<b<c, the volume of the box in cubic inches is how many times as large as the area in square inches of largest face of the box? a b c b·c a/(b·c)

A Correct. Find the volume of the box and the area of its largest face in order to find the ratio between the volume and the side area. Each face of the box is actually a rectangle. The area of a face of a rectangular box is the product of two of the box's dimensions. The volume of a rectangular box is given in the formula: Volume=w·l·h. Make the algebra go away. When you have variables in the answer choices - plug in! Use numbers that are comfortable to work with; in this case 2<3<5 would suffice. The volume is w·l·h=2*3*5=30 and the largest face is 3*5=15. 30 is twice as large as 15, so the answer you're looking for is 30/15=2. Now plug in the same numbers (2,3,5) where applicable in the answer choices until you get "2". Make sure to check all answer choices. In this case, only answer A equals 2.

Every time Cole tries to teach his old Collie dog, Cachaca, a new trick, it takes 3 minutes more than half the time it took to teach it the previous trick. If it took Cole m minutes to teach Cachaca the latest trick, and m>5, how long did it take him to teach Cachaca the trick before the previous trick? 4m−18 4m+9 2m+9 2m+12 m−6

A Correct. Translate this confusing question into a RULE of the sequence: An = An-1 / 2 + 3 Where An is the number of minutes it takes to teach trick n. The question states that An = m, and asks for An-2 (the trick before the previous trick, or two tricks back), effectively requiring you to work backwards - find An-1 from An, and An-2 from An-1. It seem as if it might therefore be better to make the preliminary step of reversing your RULE to isolate each term in terms of the one following him. However, that is working the problem using algebra with abstract variables An-1 is long and prone to careless errors. Note that the answer choices also include expressions with m. Remember the old cry: Variables in the answer choices? Plug in numbers and POE! Since the problem requires m>5, plug in m=6. Teaching the current trick took m=6 minutes, which is 3 more than half the time it took to teach the previous one. 6 is 3 more than 3 minutes; this is half the time of the previous trick, so the previous trick took 3·2 = 6 minutes again. Using the same rationale, the trick before the previous trick also took 6 minutes. That's your GOAL. Plug in m=6 into the answer choices and eliminate any answer choices that do not match your goal of 6 minutes. 4·6 - 18 = 24-18 = 6. Answer choice A is the only one that matches your goal - all others do not even come close. Since all other answer choices are eliminated for this plug in, A is the correct answer.

If the height of a right cone C is 3 and the diameter of the base of C is 8, what is the distance from the apex of C to any point on the circumference of the base of C? 3√2 5 5√2 √(67) √(73)

B Correct. This is actually a 2D question in 3D disguise. Even though there is no figure in this question, try to draw one yourself to better understand the question. Look for recycled right triangles. A right cone is a cone where the apex (the tip of the cone) is directly above the center of the base. So if you draw a line from the center of the circle to the tip of the cone, you'd have a perfectly vertical line. What we need to find is the length of AB, which is the hypotenuse of right triangle ABO. The legs of triangle ABO are AO, the height of the cone, which equals 3, and BO, which is the radius of the cone's base=4 (the diameter is 8). Use recycled right triangle 3:4:5 to get AB=5.

If z=1/y, and the result of rounding z to the nearest unit is 2, what is the greatest possible value of y? 1/3 2/5 1/2 2/3 1

B Incorrect. The question asks for the greatest value of y. Luckily, there are only five possible answers to that question - the answer choices. Numbers in the answer choices provide another clue to the fact that it's time to Reverse Plug in. Since the question asks for the greatest possible value of y, begin with the greatest answer choice, which is E: Assume that y is 1, and work the problem accordingly. If the y=1 satisfies the conditions set in the question stem, choose E; if it doesn't, POE and move on to the next greatest answer choice (D). Keep going until one answer choice satisfies the conditions in the question stem. If y=2/5, then z=1/(2/5) = 5/2. Rounding 5/2 = 2.5 to the nearest unit reaches a result of 3, not 2. POe this answer choice and move on. C Incorrect. This answer choice does indeed provide a possible value of y; but is it the greatest possible value of y? D Correct. Plug in E: If y=1, then z=1/1=1. Rounding 1 to the nearest unit will reach a result of 1, which is incompatible with the question's requirements that z rounded to the nearest unit equals 2. Therefore, POE E, and move on to D. Plug in D: if y=2/3, then z=1/(2/3) = 3/2 = 1.5. Rounding 1.5 to the nearest unit reaches the required result of 2. since all the other answer choices are numbers smaller than 2/3, there's no need to check the other answer choices: D this is the right answer choice.

In a certain series, each term (except for the first term) is one greater than twice the previous term. If the fifth term is equal to the third term, then the first term is which of the following? −2 −1 0 1 2

B Well done. Translate the description of the sequence into a RULE: "...each term (except for the first term) is one greater than twice the previous term." An = 1 + 2·An-1 , n>1 Usually, you would plug in the information given in the question, "...fifth term is equal to the third term...". However, this statement does not fit easily into the rule of the sequence. Don't waste precious seconds writing more equations. Instead, focus on the issue of the question, "...what is the first term?", and plug in the answer choices. Start with answer choice C and proceed according to the rule: If A1=0, then A2=1+2·0=1, and A3=3, A4=7, A5=15. A3≠A5, therefore eliminate C. Go on plugging in the answers until one of them agrees with the A3=A5 condition. You may already notice that if the first term were larger than 0, then the next terms would be some sort of increasing series, so that there would be no chance that A3=A5, therefore try plugging in the answer choices smaller than 0. The correct answer is B, for if A1=−1, then A2=1+2·(−1)=−1, A3=1+2·(−1)=−1, and so on. If A1=−1 then all the terms of the series are −1, and in particular the fifth term is equal to the third term.

A........Wait - can you explain that bit again? Why can I disregard all digits except for the units digit? B.........Got it. But even 37 is a very big number. How does this help us?

B Yes, 3^7 is also a big number. But once again, we only need to know it's unit digit - not the value of the power. Start multiplying 3 by itself and look for a pattern: 3^1 = 3 3^2 = 9 3^3 = 32·3 = 9·3 = 27 3^4 = 33·3 = 27·3 = ? 3^4 is getting slightly uncomfortable to calculate. Focus on the units digits only: 7 times 3 is 21, so 34 = 27·3 must end with a 1! Using the same concept, what is the next step? what is the units digit of 3^5? 1 3 5 I'm not sure... B Correct! Focus on the units digit: 34 = ...1 35 = 34·3 =...1·3 = ...3 In other words: if 34 ends with a 1, multiplying 34 by another 3 must yield a number with a units digit of 1·3=3. That number is 35. Once we get the hang of focusing only on the units digit, finding the units digit of 37 is easy: 31 = 3 32 = 9 33 = 32·3 = 9·3 = 27 34 = 33·3 = 27·3 = ...1 35 = 34·3 = ..1·3 = ...3 36 = 35·3 = ..3·3 = ...9 37 = 36·3 = ..9·3 = ...7 The units digit of 137 is the same as the units digit of 37: 7. The correct answer is D.

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

B You grossly underestimated the time this question took you. You actually solved it in 6 minutes and 23 seconds. Correct. Counting the multiples of x within a given range: 1) Find the relevant extremes - the nearest multiples of x within the specified range. 2) Subtract the relevant extremes, and divide by x. 3) Add one. 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.

In a sequence of 12 numbers, each term, except for the first one, is 1211 less than the previous term. If the greatest term in the sequence is 1212, what is the smallest term in the sequence? −12^11 0 12^11 11·12^11 12^12

C You underestimated the time this question took you. You actually solved it in 2 minutes and 29 seconds. Correct. Remember the formula for an arithmetic sequence: An = A1+(n-1)d Thus, A12 = A1+(12-1)d Where d is the difference between each pairs of consecutive terms. Since each term is 12^11 less than the previous term, d=-12^11 A12 = 12^12 + (12-1) × -(1211) Expand the brackets: --> A12 = 12^12 - 12·12^11 + 1·12^11 Since 12·12^11 = 12^12, this simplifies to --> A12 = 12^12 - 12^12 + 1·12^11 --> A12 = 12^11

(Office Excel) Sequence Set: Q1...A circular region centered at C is inscribed in equilateral triangle ABC above. If the area of ΔABC is 12, then what is the area of the shaded region? 6-3π 6-π√3 12-π√3 12-3π 3+π√3

B You slightly underestimated the time this question took you. You actually solved it in 3 minutes and 46 seconds. Correct. "Shaded region" problems are a perfect opportunity to ballpark. Your given value is the area of ABC=12. Your target area is the area of the small shaded region. Run on each answer using elimination criteria, from the faster elimination to the slower. Start by checking if the expression is negative - An area can't be negative. Next check if the format is correct - no need to figure out what is subtracted from what, just to recognize the fact that you expect the format of " something - ∏·something". Next ballpark for estimated size and POE answers that are not in the ballpark. Here all that can be said about the target area is that it's terribly small. Each small triangle has area 12/4=3. The shaded region seems less than a third of that, i.e., less than 1. POE any answer that is not in the right ballpark.

Q5...Two equilateral triangles, each of area (√3), and three squares were combined to form the figure above. If this figure were to be cut out of paper and folded to form a prism of triangular base, what would be the volume of the prism? 2 3 2√3 4√3 6

C Close enough - you took 1 minutes and 23 seconds to answer this question. Correct. Note that the prism has a triangular base, i.e. the shape must be folded so that the triangles form the top and bottom. Below is a figure of the prism: Remember: Volume of a prism with a triangular base=(base area)(height) If s is the length of the side of the triangle, then Area of an equilateral triangle=s2√3/4 Use the area formula of an equilateral triangle to find the side s: √3 = s2√3/4 /×4 --> 4√3=s2√3 /:√3 --> 4=s2 --> s=2 Since the side of the triangles is also the side of the square (see figure), it is also the height of the resulting prism. Therefore, the volume of a prism with a triangular base=(base area)(length of the side)=√3x2=2√3 Hence the answer is correct.

Q2...If a circle is circumscribed around a quadrilateral, as shown above, what is the value of x? 105 110 115 120 125

C Correct. Angle DAB lies on arc DCB and angle DCB lies on arc DAB. Together, both angles lie on an arc that is equal to the entire circumference of the circle. Recall that the measure of an arc is double the measure of the inscribed angle that defines it. Therefore, the measure of the inscribed angle is half of the measure of the arc, and thus their sum is half the central angle of an entire circle = 360/2 = 180. --> x + 65 = 180 --> x = 115

Before the summer holidays, Bill got 100 mathematics exercises for homework. During the first week of his vacation, Bill worked out exercises number 1 to 5, 35 to 40, and 76 to 100. If Bill finished working out all his homework on the second week of the vacation, how many exercises did he complete on the second week? 66 65 64 36 34

C Correct. Counting the number of integers within a given range - subtract the extremes, and add one. General case: Counting the multiples of x within a given range: 1) Find the relevant extremes - the nearest multiples of x within the specified range. 2) Subtract the relevant extremes, and divide by x. 3) Add one. In this case, the number of items in a list of consecutive integers, is: largest value - smallest value + 1. Subtract the exercises finished on different days from the total number of exercises. 100 - [(100-76+1) + (40-35+1) + (5-1+1)] = 100 - (25 + 6 + 5) = 100 - 36 = 64. Hence, this is the correct answer.

If a, b, and c are three positive integers and 3a = 8b + 2c, then what is the least possible value of 5a+b+c ? 4 20 23 32 48

C Correct. Having trouble with the logic here? Don't fret, plug in and see what happens. Since you need the least value of a, start with plugging in the smallest possible value for b and c --> 1. If b and c = 1, then: --> 3a = 8b + 2c --> 3a = 8+2 --> 3a = 10 This isn't possible, since a has to be an integer. Our original assumption of b=c=1 must be wrong. Where should you go from here? Evidently, one of them is greater than 1. Since you're looking for the least possible value of a, enlarge c (and not b), because the coefficient of c (2) is smaller, and will result in the least increase in the value of a. Plug in b=1, c=2: --> 3a = 8b + 2c --> 3a = 8+4 --> 3a = 12 --> a = 4 And thus, 5a+b+c = 20+1+2 = 23.

Home Excel(Seq Set) Q1...Arcs AC, AB and BC are centered at B, C and A respectively, as shown above. If the area of equilateral triangle ABC is √3, what is the area of the shaded region? 2/3Π - 3√3 2/3Π - √3 2Π - 3√3 2Π - 3 2Π - √3

C Correct. Shaded region problems call for Ballparking and eliminating irrelevant answer choices. Use Ballparking and avoid unnecessary calculations. The shaded area is the target value that you seek. In your ballparking calculations you use the given value - in our case it's √3, the area of the triangle. Run on each answer using elimination criteria, from the faster elimination to the slower. Start by checking if the expression is negative - An area can't be negative. Next check if the format is correct - in this question the shaded area is the result of "subtraction with pi", don't waste time on trying to figure out what is subtracted from what. Just note that you expect a subtraction and that one of the members of the expression should include ∏. Next ballpark for estimated size and POE answers that are not in the ballpark. To better guesstimate the area of this slim shady (that is, shaded) region, "fold" the shaded regions onto the triangle and determine what is the ratio of the shaded area to the area of the triangle. After folding the shaded regions onto the triangle it is easier to guesstimate the area of the shaded region: The shaded regions add up to approximately half the area of the triangle. If you're not sure you see that, subdivide the triangle connecting the center of the circle with vertices A, B, C as shown it this figure: Therefore, your target is half of the triangle= √3/2≈0.9 POE any answer choice that isn't in the right ballpark.: This answer: 2π - 3√3 = 2⋅3 - 3⋅1.7 = 6 - 5.1 = 0.9 so this one is possible. After checking all four other answer choices, this is the only one in the ballpark. Click on the arrow on the bottom of the box above to scroll back to the other answers and see why each was eliminated.

Tommy spent six lucky days in Las Vegas. On his first day he won a net amount of only $20, but on each of the following days, the daily net amount he won grew by d dollars. If Tommy won a total net amount of $1620 during his stay in Las Vegas, how much did he win on the last day? 330 500 520 540 620

C Correct. 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 amount Tommy won on the last day i.e. largest value. Given that Sum = Average x Number of days 1620 = [(x + 20) / 2] x 6 Where x is the amount won on last day. 1620 = [(x + 20) / 2] x 6 --> 1620 = [(x + 20) / 2] x 6 --> 1620 / 6 = [(x + 20) / 2] --> 2 x 1620 /6 = (x + 20) --> 540 = (x + 20) --> 540 - 20 = x --> 520 = x Hence, this is the correct answer.

The average height of the four Torres towers is 800 feet. If the four Torres towers have a median height of 900 feet, what is the greatest possible height of the shortest building of the four? 100 feet 300 feet 500 feet 600 feet 800 feet

C Correct. The average is given in the formula: . Since the average height is 800 feet, it follows that the total height of the four Torres towers is 4×800 = 3200 feet. Out of this, the question asks for the greatest possible height of the shortest building; this will happen when the other 3 buildings are at their minimum possible height. The only other piece of data given by the question is the median of 900 feet. Since the set consists of an even number of towers, the median is calculated as the average height of the two middle towers, which must equal 900 feet. This serves as a minimum requirement for the height of these two towers - set the other towers at their minimum possible height, which still satisfies this requirement, to find the greatest height of the shortest tower. Take the minimal case: the two middle towers are 900 feet tall (satisfying the median requirement), and the tallest tower is also at the minimum height of 900 feet. At this minimal case, the shortest tower is 3200-3·900 = 3200-2700 = 500.

In a certain sequence the term an is given by the formula an=an−1+2n-1 for all integer n>1. If a7=49, what is a1? −1 0 1 2 3

C Correct. The issue of this question is finding a rule for moving backwards from the 7th term to the 1st term. Isolate an-1: an = an-1 + 2n - 1 --> an-1 = an − 2n + 1 Plug n into this relation to find the terms preceding the 7th term. Plug a7=49, and n=7 into an-1 = an − 2n + 1 so: a6=49−2×7+1=36 (Plugging in a7=49 and n=7) a5=36−2×6+1=25 (Plugging in a6=36 and n=6) Do you see the pattern? Each term an equals n2. Hence, a1 is 1. If you missed the pattern, proceed to calculate a4, a3, a2, and a1: a4=25−2×5+1=16 (Plugging in a5=25 and n=5) a3=16−2×4+1=9 (Plugging in a4=16 and n=4) a2=9−2×3+1=4 (Plugging in a3=9 and n=3) a1=4−2×2+1=1 (Plugging in a2=4 and n=2)

Josh spends a total of $5.50 buying N items in the convenience store. If each of the items is either a 5 cents single bubblegum, or a 50 cents bubblegum pack, then N may be which of the following? 99 100 101 109 140

C Correct. The question asks which of the following may be N - the number of items bought by Josh. This means that if you can find a single example, i.e a single configuration of 5 and 50 cent bubblegum which fits one of the answer choices, that's enough to choose that answer choice as the right one. Begin with the extremes to see what's going on. If all the items in N are 50 cent packs, Josh's $5.50 will only buy him N = 5.5/0.5 = 11 packs. However, if all the items in N are 5 cent singles, Josh's $5.50 will buy him N = 5.5/0.05 = 550/5 = 110 singles. Since none of these is in the answer choices, the truth must lie somewhere between 11 and 110: a mixture of singles and packs. Since the answer choices are closer to 110 than they are to 11, begin with the initial standpoint of "all 5 cent singles", and start trading singles for packs to reduce the total number of items. A 50 cent pack is worth ten 5 cent singles. Think: what happens to the TOTAL number of items when you trade ten singles for one pack? Trading 10 singles for one pack only decreases the Total number of items by 9, as the big pack is added. As the following table shows, 100 singles and 1 pack will satisfy the cost requirement of $5.50, and the total number of items is N=101. 5 cent Single bubblegum 50 cent pack total number of packs 110 0 110 110-10=100 0+1=1 100+1=101 100 × 5¢ + 1 × 50¢ = 500 + 50 cents = $5.50. Since you've found a single configuration of N=101 items which reaches $5.50, the number of items may be 101, and therefore this is the right answer choice.

As with other Functions questions, Plugging in is the way to solve most sequence questions. If a question gives you the RULE for the sequence and asks for the value of a term, simply plug in the necessary values and follow the RULE. In a certain series, each term (except for the first term) is one less than twice the previous term. If the third term is -7, then the first term is which of the following? -3 -2 -1 0 4

C Correct. Translating the description of the sequence in an equation leads to An=2·An-1 − 1. Since the question supplies the third term and asks for the first, it effectively requires you to work backwards - find the preceding term from the current term. Manipulating the formula into isolating An-1 can prove useful: 2An-1=An+1 An-1=(An+1) / 2 Now, plugging in A3=-7 will provide A2: A2 = (A3+1) / 2 = (-7+1) / 2 = -6 / 2 = -3. And therefore A1 = (A2+1) / 2 = (-3+1) / 2 = -2 / 2 = -1. Alternative method: Plug in the answer choices and work backwards. The question asks which of the following is the first term. Begin with C - assume that the first term is -1. Then the second term is 1 less than twice -1, or -3. The third term will again be one less then twice -3, or -7, which satisfies the condition state in the question.

Q3...If AB=3, BC=6, CD=7 and DE=9, what are the coordinates of point E in the figure above? (−8,8) (4,−4) (5,−5) (2,−2) (−4,4)

C Correct. You could waste time finding both the x and the y coordinates of point E, but you don't have to go that far. Focus on the x coordinate first, then glance at the answer choices - can you eliminate a few? Since parallel lines CB = 6 and DE = 9 begin from the same line CD, the x coordinate of point E is larger by 9-6=3 than that of point B (in other words, point E is 3 coordinates to the right of B). Since BA is parallel to CD, the x coordinate of point B is equal to that of point A = 2. Thus, the x coordinate of E is 2+3 = 5. Only answer C has an x coordinate of 5, so this is the right answer.

A sequence of numbers is defined by the formula an = 2^n, for any positive integer n. What is the difference between the mth term and the (m−1)th term, in terms of m? 2 2m m2 2^m 2^m−1

C Incorrect. From a mathematical point of view, this question deals with a sequence defined by a formula, and with powers. From a GMAT point of view, this question is a Plugging In question. The Stop Sign for Plugging In is the variable m in the question and in the answer choices. Plug in a good number for m, and solve. B Incorrect. E Correct. Plug in a good number for such as m=3. Work the problem using m=3 to find your Goal: The 3rd term a3=2n=23 The 2nd term a2=2n=22 Therefore, the difference is 23-22=4. That's your Goal. Now, plug m=3 into the answer choices and eliminate any answer choice which does not match your Goal. Only E is not eliminated, so E is the correct answer.

In a certain series, each term except for the first term is one less than twice the previous term. If the fifth term is equal to the second term, then the first term is which of the following? 2 1 0 −1 −2

Correct. Define the RULE of the sequence: An = 2×An-1 - 1 Each term is defined as a function of the preceding term. The question presents an equation between the second term and the fifth term, and seems to require a recursive exercise - find A5 in terms of A4, then find A4 in terms of A3, then find A3 in terms of A2, so in the end you'll have A5 in terms of A2 and a single equation. However, you probably got tired just reading the above paragraph, let alone working out the problem this way. The problem is that you are missing any real numbers in the question, so you can't plug numbers into the rule. Can you bypass this hurdle somehow? The problem presents numbers in the answer choices and a specific question - what is the value of the first term. These, along with the basic algebraic urge to start jotting down equations, are the basic and advanced stop signs of Reverse Plugging In. Begin with the middle answer choice: assume that A1 = 0. Then A2 = 2×A1 - 1 = 2×0 - 1 = -1. --> A3 = 2×A2 - 1 = 2×(-1) - 1 = -3 --> A4 = ... You can stop right there, as you can see where this is going - each term is going to be more negative than the previous one. A5 is going to be a much more negative number than A2, rather than its equal, as the question requires. POE C. In which direction should you go now? At this point, it's not very clear, so just pick a direction and GO. In any case, keep your eyes open to see if A5 is getting further away from A2 when Reverse PI other answer choices. The correct answer choice is B: assume that A1 = 1. Then A2 = 2×A1 - 1 = 2×1 - 1 = 1 (the same as A1) --> A3 = 2×A2 - 1 = 2×1 - 1 = 1 (begin to see a pattern here?) --> A4 = ... Once again, you can stop right there, as you can see where this is going - each term is equal to the preceding term - they're all 1. This way, A5 is going to be equal to A2, and B is the right answer choice.

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)

D Correct. 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!

The edges of rectangular box B are made out of wire. If the width of B is 4, the height of B is 8 and the length of B is 6, what is the total length of the wire? 24 36 48 72 96

D Correct. Remember, every rectangular solid has 12 edges. These 12 edges are made up of 4 lengths, 4 widths and 4 heights. The total length of wire on all edges is the sum of all 12 edges. Length of wire = 4(Length) + 4 (Width) + 4(Height) = 4 (Length + Width + Height) = 4 (6 + 4 + 8) = 4 (18) = 72. Hence, this is the correct answer.

What is the sum of the integers between 30 and 70, inclusive? Sure, you can start adding those integers one by one, but you won't necessarily enjoy it very much. You also do not have time to do this sort of long exercise on the GMAT. Follow these 3 simple steps: 1) Find the average of the set: average the first and last terms to find the "middle" of the set. (30+70) / 2 = 100 / 2 = 50 2) Count the number of terms in the set: subtract the extremes, and add one. 70-30, plus 1 = 40 +1 = 41. 3) Multiply the "middle" number (the average of the set) by the number of terms to find the sum: 50·41 = 50·(40+1) = 2000+50 = 2050.

Excellent. Remember: To find the sum of a set of consecutive integers within a certain range, follow these 3 steps: 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.

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. y2−x·z is odd. x·y<y·z x·y·z is even.

D Correct. Use the properties of consecutive integers to prove that there are four answer choices that MUST be true. Alternatively, find a single example of x, y and z which satisfies the question stem's conditions, yet shows that an answer choice is NOT true, thus proving that it is NOT ALWAYS true. 3 consecutive integers can have two possible cases: even - odd - even (such as 2, 3, 4) or odd - even - odd (such as 1, 2, 3) - check both options. Plug in odd and even values for x, y and z to eliminate answer choices. In order to check if this is always true, plug in even and odd values of x. If x = 2, y = 3 and z = 4, then 2·3 < 3·4 or 6 < 12. If x = 1, y = 2 and z = 3, then 1·2 < 2·3 or 2 < 6. So far so good, but is it always true, for any number? If both x and y are negative, their product will be positive. For example, If x = -2, y = -1 and z = 0, then xy = 2 and yz = 0. Based on this, xy is not less than yz. Hence, this answer choice is is NOT always true, and is thus the correct answer choice.

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)

D Very good! You took 2 minutes and 15 seconds to answer this question. Correct. 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.

If the terms of a certain sequence are defined by the equation An = An-1 + n for all n, what is the value of An in terms of An+1? An+1 − 1 An+1 + n + 1 An+1 − n An+1 − n − 1 An+1 − n + 1

D You slightly underestimated the time this question took you. You actually solved it in 3 minutes and 53 seconds. Correct. Since the question asks for An+1, take the RULE in the question one step further. Simply add 1 to any n in the rule: If An = An-1 + n Then An+1 = An + (n + 1). Now isolate An. An = An+1 - (n + 1) = An+1 - n - 1 Alternative Explanation: You can also try plugging in a few "n"s. Plugging in for A1 will work - you just need to be clear on which n you are plugging in, since the value of a term is also dependant on n. Let's say that A1=5 in that case, A2 = A1+2 = 5+2=7. (replace n with everywhere) A3 = A2+3 = 7+3=10. Now the question asks for An in terms of An+1. Define which n you're focusing on, and go: if n=2, we want the value of A2 = 7 (that's our goal), and we're plugging in A3=10 and n=2: A 10-1=9 does not match your goal of 7. Eliminate. B 10+2+1 = 13, not 7. Eliminate. C 10-2=8, not 7. Eliminate. D 10-2-1 = 7 -Good - don't eliminate it yet. E 10-2 1 = 9, not 7. Eliminate. And D is the only answer.

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

E Correct. Remember, the product of n consecutive integers is divisible by n!. For example: The product of 3 consecutive integers: 1 x 2 x 3 = 6 is divisible by 3! = 6. If you do not remember this rule, get a feel of the question plugging in good numbers. 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.

What is the sum of the odd integers from 45 to 65, inclusive? 495 550 555 600 605

E Correct. To find the sum of a set of consecutive integers within a certain range, follow these 3 steps: 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, since both extremes are included. 3) Multiply the "middle" number (the average of the set) by the number of terms. Remember that the difference between consecutive odd integers is 2. Therefore: Calculate the number of integers in the list by: [(largest integer - smallest integer) / 2] + 1. For example: To find the sum of odd integers from 7 to 17, do the following: 1) Average: (7+17)/2 = 24/2 = 12 2) Number of Terms: [(17-7)/2] + 1 = 10/2 + 1 = 5 + 1 = 6 3) Average × Number of Terms: 6 × 12 =72 Check: (7 + 9 + 11 + 13 + 15 + 17 = 72) To find the sum of the odd numbers from 45 to 65 inclusive, do the following: 1) Average: (45+65)/2 = 110/2 = 55 2) Number of Terms: [(65-45)/2] + 1 = 20/2 + 1 = 10 + 1 = 11 3) Average × Number of Terms: 55 × 11 = 605 Hence, this is the correct answer.

Q4...The shaded region above is formed by the intersection of squares ABCD and GECF. If the area of the shaded region is equal to the area of GECF, and BE=2, what is the length of EC? 2 / (1-√2) 2√2 - 2 2√2 4 2 + 2√2

E Correct. Trade in complicated calculations for quick Ballparking. Don't mess with uncomfortable numbers. Be efficient - guesstimate! Guesstimate the length of EC using the figure, which is drawn to scale. EC is about three times longer than BE. Therefore your target is ~6. Remember √2≈1.4, for ballparking sake. POE any answer choice that is not in the right ballpark.

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

E Incorrect. Counting the multiples of x within a given range: 1) Find the relevant extremes - the nearest multiples of x within the specified range. 2) Subtract the relevant extremes, and divide by x. 3) Add one. D Correct. 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.

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

E Incorrect. 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. C Correct. (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.

For every positive integer n, the nth term of a sequence is the sum of three consecutive integers starting at n. What is the sum of terms 1 through 99 of this series? 5,250 10,098 14,850 15,147 15,150

E You grossly underestimated the time this question took you. You actually solved it in 6 minutes and 16 seconds. Incorrect. To better understand what this strange sequence looks like, write down the first few terms. Plug in n=1,2,3 in the definition of the sequence. 1st term, i.e. n=1, this term is 1+2+3=6 2st term, i.e. n=2, this term is 2+3+4=9 3st term, i.e. n=3, this term is 3+4+5=12 Each term is greater than the previous term by 3, and our series is actually: 6, 9, 12...300. The last number (99th term) is reached by finding the pattern again: The first term is 6 (3*2), the second term is 9 (3*3), the third term is (3*4)...the 99th term is 300 (3*100). In words: the nth term is 3*(n+1). The question deals with summing up all these terms up to the 99th term, and the issue of the question is efficiently summing up consecutive integers. Recall that the sum of a sequence of consecutive integers is the average of the sequence times the number of terms. The average of such a sequence of consecutive integers is the average of the first and last terms. C Incorrect A Incorrect D Correct. Average of sequence is average of first and last terms, or (6+300)/2 = 306/2 = 153. Number of terms is 99, so the sum of the series is 153*99 = 15,147 (subtract 153 from 153*100 = 15,300)

We've seen that the average of the set of consecutive integers is considered the exact "middle" of the set (as it is equal to the median of the set). this important insight brings us to another property of a set of consecutive integers. The Average of a Set of consecutive integers = Average of any equidistant pair of integers around the median.

For example, take a set of consecutive integers {1, 2, 3, 4, 5}. "3" is the median, and therefore the average. Take a look at the pair "2" and "4" (one step away from the average 3): 2 is one down from 3, 4 is one up from 3. Adding the two will cancel the difference and leave "6". The average of 2 and 4 is 6/2 = 3 = the average of the group. The same happens if we take another step away from the median - the average of 1 and 5 is still 3, which is the median and the average of the entire set. The use of this insight depends on what data is provided by the question. In many cases, GMAT questions will describe a set of consecutive integers in terms of its first and last terms - all the integers between 22 and 55, for example. Since the average of the entire group is equal to the average of any pair of terms around the middle, we can use the first and last term provided by the above statement to calculate the average of the entire set. In other words, the average of the set of all integers between 22 and 55 is simply the average of the first and last terms of the set: the average of 22 and 55 = (22+55) / 2 = 77 / 2 = 38.5. This will also be the median of the set, according to what we've already learned of the properties of the average of a set of consecutive integers. Remember that this insight holds true for any set of integers with a constant difference between any two consecutive terms (arithmetic sequence) - not just for consecutive integers. For example, the multiples of 3 between 22 and 55 constitute a set of integers with a constant difference of 3 between any two terms, and thus the properties of consecutive integers apply in this case as well. If the question asks about the average of a set of all multiples of 3 between 22 and 55 (inclusive), simply find the first and last terms of the set - the first and last multiples of three within that range. Averaging the two multiples will give you the average of the entire set: 22 isn't a multiple of 3 - the first multiple of 3 within the specified range is 24. 55 isn't a multiple of 3 either - the last multiple of 3 within the specified range is 54. Therefore, the average of the set of multiples of 3 between 22 and 55 = the average of the first and last multiple of 3 withing the range = (24+54) / 2 = 78 / 2 = 39.

An arithmetic sequence is an example of a sequence with a constant difference between every two terms. A classic example would be a set of consecutive integers {1, 2, 3, 4}, in which the difference between any two terms is 1. If the difference between the terms is labeled d, then the following is the formula for the nth term in the sequence: An = A1+(n-1)d Translated into words, read: the nth term can be calculated by taking the 1st term and adding (n-1) "jumps" of d. Thus the 2nd term, A2, will be A1 + (2-1)d = A1 + d; the third term will be A1+2d, or two jumps away from the first term, etc.

In the following arithmetic sequence: 7, 13, 19, 25,... what is the place of the term with a value of 127? 17 18 19 20 21 E Correct. Arithmetic sequence: a sequence with a constant difference between every two terms, e.g., {1, 2, 3, 4} where the difference is 1. The formula for the nth term is (the difference is labeled d): An = A1+(n-1)d In the set sequence presented in the question, each term is 6 more than the previous term, beginning with 7. thus, a1 = 7 and d=6. Use the formula to find n when an = 127. An = A1+(n-1)d --> 127 = 7+(n-1)*6 --> 127 = 7+6n-6 --> 6n = 126 --> n = 21

Back to our question: What is the unit's digit of 137? A 1 B 3 C 5 D 7 E 9 We've established that we're not going to start multiplying 13 by itself 7 times - that way madness lies.

Instead, notice that the question asks for the units digit of 137 - not the value of the power. That alone should indicate that you're probably not supposed to work out the value of this huge number. Focus on the units digit: 3. We can disregard all digits from the tens digit onwards, as multiplying those by any number must yield a units digit of 0, and contribute nothing to the units digit of the end result. The unit's digit of 13^7 must therefore be the same as the units digit of 3^7. A.....Wait - can you explain that bit again? Why can I disregard all digits except for the units digit? B.....Got it. But even 37 is a very big number. How does this help us? B Let's multiply 13 by itself to see what we mean. Put one over the other: 1 3 x 1 3 You've done this sort of math in primary school. Multiply the units digit of the bottom number by the digits of the top number, and write the result below the line. 1 3 x 1 3 3 9 Now, multiply the tens digit of the bottom number by the digits of the top number, and write the result below the line, displaced by one place to the left. 1 3 × 1 3 3 9 1 3 Finally, add the two numbers below the line to get the end result: 1 3 x 1 3 3 9 1 3 1 6 9 Notice that the units digit of the end result 169 is determined solely by multiplying the units digit of 13 and 13: 3×3 = 9. The rest can be disregarded, as they only affect the digits in the tens position and upwards. If we were to multiply 169 by 13 again, the units digit of the product will still be determined solely by the units digit of 13 and 169 = 3×9=27, or units digit of 7. (Try it!)

3) An=An-1 - n + 2 The nth term in this sequence is determined by the RULE of An=An-1 - n + 2. The value of a term An in this sequence is determined both by the value of the preceding term An-1 AND by its place n in the sequence - both are needed to calculate the value of An. Again, it is necessary to know the value of one term in the sequence in order to determine the value of any other term. If the problem states that the first term A1 is 3, then the second term A2=A1-n+2=3-2+2=3. The third term will then be A3=A2-n+2=3-3+2=2.

We'll discuss the question types testing sequences later. For now, just remember these three possible general sequence types: 1) Place dependent - The value of a term An is determined solely by its place in the sequence n. e.g. An = n+2 2) Term dependent - The value of a term An is determined by the value of the some other term, such as the preceding term An-1. e.g. An = An-1+4 3) Combination - The value of a term An is determined both by the value of the preceding term An-1 AND by its place n in the sequence - both are needed to calculate the value of An. e.g. An=An-1 - n + 2

Pretty cool. Where does the whole discussion of the pattern come into place?

Well, what if the question read as follows: What is the units digit of 1327? Focusing on the units digit alone is helpful, but there's no way we'll continue calculating unit digits until the 27th power...there has to be an easier way. If you've been looking for it, you may have noticed a repeating pattern in the units digit of 3 above: 31 = 3 32 = 9 33 = ...7 34 = ...1 35 = ...3 And from here on, the pattern repeats itself: {3, 9, 7, 1, 3, 9, 7, 1...} Therefore, another way of phrasing the question "what is the units digit of 1327" is this: what is the 27th term in the above quadruple repeating pattern? Once we notice this repeating pattern, we can use it for easily working out the problem. Since our pattern has 4 repeating terms, the terms repeat themselves in place-multiples of 4: The term in the 4th place is 1; the term in the 8th place will also be 1; so will the 12th. In fact, in every power of 13 that is a multiple of 4, the units digit will be 1. Therefore, the units digit of 13^24 will also be "1". From here, finding the units digit of 1327 is a simple matter of moving down the pattern another 3 steps: 13^24 = ...1 13^25 = ..3 13^26 = ..9 13^27 = ..7

How many integers are there between 115 and 120, inclusive? The intuitive way to calculate this would be to subtract 115 from 120 and come up with "5", and that would, of course, be wrong. There are actually 6 integers between 115 and 120 - 115, 116, 117, 118, 119, and 120. It is very easy to forget to include the extremes when confronting this kind of seemingly easy question. In this case, simply subtracting 115 from 120 takes away the first extreme (115) which is subtracted along with all integers up to and including 114. The rule to remember here is this - remember that if both extremes should be counted, add one. How many integers are there between 21 and 87, inclusive? 87-21, plus 1, equals 66+1 = 67.

What if the question requires counting multiples of an integer within a range, rather than just counting consecutive integers? No problem! Same rule applies - subtract the extremes, divide by the integer, and add one at the end.