GMAT Math Problems I got wrong

Does the integer k have a factor p such that 1 < p < k ?

(1) k > 4! (2) 13! + 2 ≤ k ≤ 13! + 13 (1) Since k > 4!, then k > 24, because 4! = (1)(2)(3)(4) = 24. However, k may or may not be a prime number. For example, if k = 27, then the factor p could be 3 or 9, but if k = 29, which is a prime number, then k would not have any factors between 1 and 29; NOT sufficient. (2) From this it can be concluded that k could be any of twelve integers: 13! + 2, 13! + 3, 13! + 4 .... 13! + 13, where 13! is the product of the integers from 1 to 13. Note that 2 is a factor of 13! + 2, since it is a factor of both 13! and 2. Similarly, 3 is a factor of 13! + 3; 4 is a factor of 13! + 4; and so on for all the values of k. Thus, for each number k from 13! + 2 to 13! + 13, there is a factor p such that 1 < p < k; SUFFICIENT.

What is the average (arithmetic mean) of x1, x2, x3, x4, x5, and x6?

1) The average of x1, x2, x3, x4 = 28 2) The average of x4, x5, x6 = 21 The issue is finding the average of six numbers. In averages questions, any two parts of the average formula (the Plumber's Butt) allow you to get the third part. Since you have the number of things (six numbers), the missing piece is the Total. Stat. (1) gives you the average of four of the numbers. Using the Plumber's Butt, you can calculate the total of x1, x2, x3, and x4, but that doesn't help in finding the total of all six numbers. Therefore, Stat. (1) → IS → BCE. Stat. (2) gives you the average of three of the numbers. Using the Plumber's Butt, you can calculate the total of x4, x5, x6, but that doesn't help in finding the total of all six numbers. Therefore, Stat. (2) → IS → CE. Since the value of x4 is unknown, the total of all six numbers cannot be calculated, and neither can the average. Therefore, Stat. (1)+(2) → IS → E.

Is rectangular box B a cube? (1) All six faces of B have the same area. (2) The cylinder of greatest volume that can be placed inside B occupies π/4 of the volume of B.

A: Statement (1) ALONE is sufficient, but Statement (2) alone is not sufficient to answer the question asked. The issue is finding out whether B is a cube. In a cube, the length (l), width (w), and height (h) are all equal. Stat. (1): The areas of B's faces are lw, lh, and wh. According to Stat. (1), they are all equal: If lw = lh, then w=h. if lw = wh, then l=h. Therefore, l=w=h, so the rectangular box B is definitely a cube. Stat.(1) → Yes → S → AD. Stat. (2) is slightly trickier. The largest cylinder that can fit in a box has the same height as the box, and its diameter is equal to either the box's height or width (whichever is smaller). The volume formula for a cylinder is V=πr^2h, while the volume formula for a box is V=lwh. Therefore, πr^2h = π/4 x lwh The variable h can be any number, and the ratio in Stat. (2) will still be maintained. So height can be 2 or 10. This means B can either be a cube or a rectangular box.

The three competitors in a race have to be randomly chosen from a group of five men and three women. How many different such trios contain at least one woman?

First calculate the number of total arrangements. You need three SeBoxes. Remember that the number decreases because there is no repetition (you cannot choose the same person twice) and that the order does not matter because there aren't any different roles: (8 x 7 x 6) / (3 x 2 x 1) = 56 Now, calculate the number of forbidden options—those in which there are no women, only three men. This is actually picking three competitors out of only five men, not ordered: (5 x 4 x 3) / (3 x 2 x 1) = 10 And lastly, subtract the number of "only men" trios from the "total number of trios" to find the number of "at least one woman" trios: Good combinations = 56 −10 = 46.

50 students in a certain class took a test. How many of them got a score that is less than 82? (1) The average score is 82. (2) The median score is 82.

For Stat. (1), it could be that all the students got an 82, and then no students got a score less than 82. Or it could be that one student got an 80, one got an 84, and the rest all got an 82, in which case one student got less than 82. The average in both cases is 82, but the number of students who got under 82 is not the same, so Stat.(1) → IS → BCE. For Stat. (2), you could Plug In the same possible numbers you used for Stat. (1), so Stat. (2) is insufficient as well. Therefore, Stat.(2) → IS → CE. For Stat. (1+2), the same two examples apply. Even the combination of statements does not yield sufficient information. Thus, Stat.(1+2) → IS → E.

If x and y are integers, and y = 1−x, then which of the following must be true?

If y is even then x is odd Remember that odd − odd = even and odd − even = odd.

A game at the state fair has a circular target with a radius of 10 centimeters on a square board measuring 30 centimeters on a side. Players win prizes if they throw two darts and hit the circular area on at least one of the two attempts. If Jim hits the board on both throws and is equally likely to hit any point on the board, what is the probability that Jim wins the game?

In this game you have to hit at least once to win. For "at least" questions, it is easier to calculate the probability of NOT HITTING the target on both shots, and then subtract the probability of this happening from 1: Odds of not hitting once = (9-π)/9 Odds of not hitting twice = (9-π)² / 81 Probability = 1 - forbidden probability 81/81 - (81 - 18π - π²)/81 Answer = (18π - π²) / 81

Which of the following describes all the values of y for which 1/y² <1/10?

Manipulate the inequality as you would an equation. It's OK to multiply by y². Since an even power is non-negative, there's no need to flip the sign. Start by multiplying by y: 1/y² x y² < 1/10 x y² 1 < y²/10 then 10 < y² y² > 10: −√10 > y OR y > √10

If m and k are positive integers, is m! + 8k a multiple of k? 1) k < m 2) m = 3k

Multiple of n ± Multiple of n = Multiple of n. Since 8k is a multiple of k, the issue here is whether m! is a multiple of k as well. Remember the definition of m!: m! = m x (m-1) x (m-2).... x 1 Stat. (1): since k<m, you know that m! includes k as a factor. For example, if k=2 and m=3, then m!=3⋅2⋅1 is a multiple of k. Therefore, both m! and 8k are multiples of k, and their sum must also be a multiple of k. Stat.(1) → Yes → S → AD. Stat. (2): m is equal to k × some integer and must be a multiple of k. Therefore, both m! and 8k are multiples of k, and their sum must also be a multiple of k

If in a certain sequence of consecutive multiples of 50 the median is 625 and the greatest term is 950, how many terms in the sequence are smaller than 625?

Since the median is 625 (which is not a multiple of 50), 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 find the number of terms below 625. To calculate the number of multiple of 50 between 650 to 950: Subtract the extremes: 950−650 = 300 Divide by 50: 300/50 = 6 Add 1: 6+1 = 7

A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?

Since there are 50 odd and 50 even balls in the box, the probability of selecting an odd ball at random is 50/100 = 1/2; the probability of selecting an even ball is the same. For the sum of the three numbers on the selected balls to be odd, either 1) the numbers must all be odd, or 2) exactly one of the numbers must be odd and the other two numbers must be even, which can occur in one of the three ways listed below. 1) Probability of selecting odd, odd, odd = 1/2 x 1/2 x 1/2 = 1/8 2) Probability of selecting odd, even, even = 1/2 x 1/2 x 1/2 = 1/8 Probability of selecting even, odd, even = 1/2 x 1/2 x 1/2 = 1/8 Probability of selecting even, even, odd = 1/2 x 1/2 x 1/2 = 1/8 Adding all four probabilities gives 4 x 1/8 = 1/2

A committee of six is chosen from eight men and five women so as to contain at least two men and at least three women. How many different committees could be formed?

Start with the first group: two men and four women. You have to pick two men out of eight and four women out of five. Order does not matter, and since you are picking humans, there is no repetition. C(8,2) = 28 C(5,4) = 5 5 x 28 = 140 ways of arranging 2 men and 4 women Ok, now for the committees with three men and three women, you have to pick three men out of eight and three women out of five with no order and no repetition: C(8,3) = 56 C(5,3) = 10 10 x 56 = 560 ways of arranging 3 men and 3 women This gives you a grand total of 140+560=700 committees with at least two men and three women.

John tossed a fair coin 3 times. What is the probability that the outcome was tails exactly twice?

With a fair coin, the probability of getting a heads and the probability of gettting a tails are both 1/2. Within each scenario, break down the triplets of tosses into separate single events (first toss, second toss, and third toss). Calculate the probability for each event one at a time, and then multiply the probabilities, since the events have an "and" relationship. There are three "good" scenarios here: Tails on toss 1 and 2, and heads on toss 3: Probability = 1/2 x 1/2 x 1/2 = 1/8 Tails on toss 1 and 3, and heads on toss 2: Probability = 1/2 x 1/2 x 1/2 = 1/8 Tails on toss 3 and 2, and heads on toss 1: Probability = 1/2 x 1/2 x 1/2 = 1/8 Add these probabilities up to find the answer: 1/8 + 1/8 + 1/8 = 3/8

If a < 1 and a ≠ 0, then which of the following must be true?

a^3 < a^2

A whale goes on a feeding frenzy that lasts for 9 hours. For the first hour, he catches and eats x kilograms of plankton. In every hour after the first, he consumes 3 kilograms of plankton more than he consumed in the previous hour. If, by the end of the frenzy, the whale consumes a whopping 450 kilograms of plankton, how many kilograms did he consume on the sixth hour?

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. 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.

0.02^4 / .002²

Convert to: (2 x 10^-2)^4 / (2 x 10^-3)² 2^4 x 10^-8 / 2² x 10^-6 4 x 10^-2 = .04

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?

(B-A / 9) + 1 = 9 (B-A/9) = 8 | B-A = 8 x 9 | B-A = 72 (72/4) + 1 = 18 + 1 = 19 We take the relevant extremes (in this case A and B are relevant extremes because 9 and 4 and both factors of 36) subtract them and divide by the multiple and add 1.

If r and s are positive integers is r/s an integer

1) Every factor of s is also a factor of r 2) Every prime factor of s is also a prime factor of r 1) The integer s is by definition a factor of itself. From this, every factor of s is also a factor of r. Therefore, r/s must be an integer; SUFFICIENT 2) From this example, if r = 18 and s = 6, then 6 has two prime factors (2 and 3), each of which are also factors of 18, and 18/6 is an integer (3). However if r=18 and s=8, then r has two prime factors (2 and 3) and s has a prime factor of 2, which satisfies the condition. Even though the prime factor of s is a prime factor of r, r/s (18/8) is not an integer. INSUFFICIENT

If a and n are integers, and a^3=360n, then n must be divisible by which of the following?

360 =36⋅10 = 2^3 x 3^2 x 5 The variable n must complete 360n's factors to triplets. So n must include at least the missing 3⋅5^2 to reach the minimum. 3 x 5^2 = 75 Thus, n must be divisible by 25.

Tammy climbed a mountain in two days. She spent a total of 14 hours climbing the mountain. On the second day, she walked at an average speed that was half a kilometer per hour faster but 2 hours less than what she walked on the first day. If the total distance she climbed during the two days is 52 kilometers, how many kilometers per hour did Tammy walk on the second day?

Assign each of the values in the question to the appropriate cell in the table, highlight the requested value, and go on to Plug In the answers. The numbers in each row must fit the Distance = Speed × Time formula. First day = v (speed) x t (time) = distance Second day = v + .5 (speed) x t-2 (time) = D Total = 14 hours 52 km (distance) Solve for time (t)+(t-2) = 14; 2t = 16 t = 8 t-2 = 6 Plug in C = 4 First day = 3.5 x 8 = 28 2nd day = 4 x 6 = 24 24 + 28 = 52 km. Therefore C is correct

John invested $100 in each of the funds A and B. After one year, the value of the money in fund A was $10 higher than the value of the money in fund B. After another year, the value of the money in fund A was $25 higher than the value of the money in fund B. If the value of the money in each fund increased by a fixed interest compounded annually, what was the annual interest of fund A?

Assume the amount in the answer choice is the annual interest of fund A and then follow the story in the problem. If everything fits, then stop—pick it. Otherwise, POE and move on until you find an answer that works. First, Plug In answer choice C (40%). If you assume that the annual interest of fund A is 40%, then after one year, the values of the two funds were A1 = 100 x 1.4 = 140 B1 = 140 - 10 = 130 Therefore, the annual interest of B was 30%. After another year, the values of the two funds were A2 = 140 x 1.4 = 196 B2 = 130 x 1.3 = 169; 196 - 169 = 27 (not 25) The difference is $27—a little bit too high. Eliminate answer choice C, and Reverse Plug In. If both interests are smaller, the difference will probably be smaller too, so go on to Reverse Plug In for answer choice B. A1= $100 × 1.3 = $130 B1= $130 − $10= $120 A2 = $130 x 1.3 = $169 B2 = $120 x 1.2 = $144; 169-144 = $25 (Bingo!)

Papayaya, a popular soft drink, contains only four ingredients. Soda water comprises 4/7 of Papayaya and natural lemon juice makes up 1/3 of Papayaya. The amounts of sugar and papaya puree in Papayaya are equal. Due to a malfunction, the mixing machine mixes double the regular amount of lemon juice and three times the regular amount of papaya puree. If no other changes were made to the amounts of the ingredients, what is the fractional portion of soda water in the drink that comes out of the malfunctioning machine?

Don't use algebra and fractions! The question revolves around the invisible unknown of the amount of Papayaya juice mixture. Plug In a number for the total amount of Papaya and calculate the quantities with real numbers instead of fractions. Since the question asks about fractions, multiply the denominators to find a good number to Plug In. Plug In 7×3 = 21 liters for the total amount of Papayaya. Then the regular amounts of soda and lemon are Soda = 4/7 x 21 = 12 Lemon = 1/3 x 21 = 7 Sugar = 1 and Puree = 1 The malfunctioning machine produces Lemon = 2 x 7 = 14 Puree = 3 x 1 = 3 The two other quantities remain the same, so the total is 14 + 12 + 3 + 1 = 30. Therefore, the fraction of soda in the beverage is 12/30 or 2/5.

Is quadrilateral ABCD a parallelogram? 1) All four internal angles of ABCD are equal. 2) AC divides ABCD into two congruent triangles

For Stat. (1), the sum of angles in a quadrilateral is 360°. Each of ABCD's angles measures 360/4 = 90. Therefore, ABCD is a rectangle, which is a type of a parallelogram. Stat.(1) → Yes → S → AD. Stat. (2) can be true for parallelograms, but this statement is also true for other kinds of quadrilaterals, such as kites (kites are not parallelograms). Hence, Stat.(2) → Maybe → IS → A.

Is m > n? (1) |m+n| < |m| + |n| (2) |m| > |n| + 1

For each statement, Plug In numbers in order to arrive at opposing answers. Stat. (1) tells you that the distance of the sum is smaller than the sum of distances. The only way this is true is if m and n have different signs so that adding the two will yield a sum that is closer to 0 than just adding their absolute values. Plug in values for m and n: if m = −2 and n = 5, then m < n and the answer is "No." if m = 2 and n = −5, then m > n and the answer is "Yes."

Among the members of the Malmo family, there are three times as many members who do not wear glasses as members who do. Half of the Malmo family members have blue eyes, and 4/5 of the family members who have blue eyes do not wear glasses. What percent of the Malmo family members wear glasses and do not have blue eyes?

For this problem, set up a table with the columns as "Glasses", "Not Glasses" and "Total". Then set up the rows as "Blue Eyes", "Not Blue Eyes" and "Total". We know that the total members that wears glasses is x and the members who do not wear glasses is 3x. The total is 100. Therefore, 4x = 100 and x = 25 and 3x = 75. We know that the total amount of family members that have blue eyes is 50 and 50 for those who do not have blue eyes. 4/5 of 50 is 40 and therefore 40 members have blue eyes and do not wear glasses. This means that 10 members with blue eyes wear glasses. Since we know that the total amount of members who wear glasses is 25 and 10 of those individuals have blue eyes, there must be 15 members who wear glasses and do not have blue eyes.

What is the lowest possible common multiple of two distinct integers, each greater than 125?

Note that when two numbers don't share any common factors, their LCM is simply their product. For example, the LCM for 7 and 5 (which share no common factors) is simply 35 (i.e., 35 is the smallest number divisible by both 5 and 7). If you notice the word "distinct" in the question stem, 126⋅127 seems like the quickest and easiest answer—the product of two next integers greater than 125. However, that would've been too easy, right? The GMAT will not present you with dumb questions. If you've just solved a question in 40 seconds, then you must have missed something. The general principle behind this question is this—the smallest possible LCM of two integers is reached when the two integers have the greatest number of factors in common (factors which are only counted once in the LCM). This should lead you to using 126 (the smallest possible integer), and a multiple of 126, so that the two integers will have the greatest number of common factors that are only counted once in the LCM. Thus, you can use 126 and 126⋅2, so that the common multiple will be the greater of the two. Since 2⋅126 is divisible by both itself and by 126, it is the LCM of 126 and itself. 126⋅2 = 252 Therefore, 252 is the correct answer.

If x⋅y≠0, what is the value of x/y? (1) − |x| = y (2) − |y| = x

Note that x⋅y≠0 means that neither one of the variables is 0. The issue of this question is absolute value, so plug in positive and negative numbers. First check that they satisfy the statement(s) and then calculate x/y. Stat. (1): Plug in for the variables: If you plug in x = 2 and y = 2, then −|x| = y is not true, so we can't use this plug-in. Similarly, −|x|=y is not true if y is positive, for example, if you use x = −2 and y = 2. If you use x = 2 and y= −2, then −|x| = y is true, so we can use this plug-in. In this case, x/y=−1. However, the equation is also true if x is negative, for instance if x=−2 and y=−2. In this case, x/y = 1. Therefore, BCE. Stat 2 is the same result -> CE Stat. (1+2): For Stat. (1) to be true, y must be negative, and similarly, for Stat. (2) to be true, x must be negative. For example, if x= −2 and y=−2, then both statements are satisfied and x/y=1 If you try to change only one of the variables in the equation above (keeping both negative), the equation would not be true. In other words, x=y. Therefore, x/y=1 always, so Stat.(1+2) → S → C.

Each cell of Type X divides into a certain number of X-cells every hour. Each Type Y-cell also divides into a constant number of Y-cells every hour, but not necessarily at the same rate as X. At a certain time, container A contained 10,000 X-cells and container B contained 10,000 Y-cells. After one hour, there were 30,000 more X-cells in container A than Y-cells in container B. After another hour, there were 330,000 more X-cells than Y-cells. What is the division rate per hour for the X-cells?

Numbers in the answer choices and a specific question call for Plugging In the answers. You may feel like writing down one equation or more. This is just your algebraic urge, which is another stop sign for Reverse PI problems. This question contains a wealth of information, but don't panic! Try out each answer choice in an orderly manner. Assume the amount in the answer choice is the number of cells to which one X-cell divides every hour, and then follow the story in the problem. Keep everything organized by writing down the number of cells in each stage of the question. Let's go over the process for this answer choice: Start with answer choice C. Assume every X-cell divides into six cells every hour. Then 10,000 X-cells would become 60,000 after one hour. It follows that after one hour there were 30,000 Y-cells in container B (since Y has 30,000 less). Since this means that the original 10,000 Y-cells divided into 30,000 cells after one hour, we find that a Y-cell divides into three every hour. Thus, after another hour, there were 6 × 60,000 = 360,000 X-cells and 3 × 30,000 = 90,000 Y-cells. Subtract the number of Y-cells from the number of X-cells: Difference = 360,000 − 90,000 = 270,000. The difference is 270,000, not 330,000 as required. This answer does not fit the problem. POE C. By now, you may see that the correct answer involves higher division rates. If you grasp that in real time, you can POE A and B. If not, just go on Reverse Plugging In! If an X-cell divides during one hour into seven, then after one hour there were 70,000 X-cells, and therefore 40,000 Y-cells (30,000 less). The number of initial Ys was 10,000, and there were 40,000 after one hour. Therefore, Y's division rate is 4. After another hour there were: 7 ×70,000 = 490,000 X-cells, and 4 × 40,000 = 160,000 Y-cells. The difference between the numbers of cells after the second hour is Difference = 490,000 − 160,000 = 330,000

A bag contains one chip labeled "1," two chips labeled "2," three chips labeled "3," and so on, all the way up to 100 chips being labeled "100." If one chip is randomly chosen, then what is the chance that it has an even number less than or equal to 50?

Probability = Wanted Outcomes / Total Outcomes You're trying to calculate the chance of picking an even number less than or equal to 50, so for the top of the fraction, you need to calculate the number of even chips from 2 to 50. Since there are two chips labeled "2," four chips labeled "4," and so on, there are a total of 2 + 4 + 6 +...+ 50 even chips labeled from 2 to 50. You can calculate this sum by multiplying its average by the number of terms in it. Since it's a sequence of consecutive even integers, its average is equal to the average of the extremes. 2+50/2 = 26 You can calculate the number of terms by subtracting the extremes, dividing by 2, and then adding 1. 50-2 = 48/2 = 24 + 1 = 25 Finally, multiply these numbers to find the sum (i.e., the number of outcomes in E). 26 x 25 = 650 Now, find the denominator. For this, you need to find the total number of chips, which is equal to 1+2+3+...+100. You can calculate it using the same method you used for the numerator. The average of this sequence is 100+1/2 = 50.5 The number of terms is 100-1 / 1 = 99 + 1 =100 50.5 x 100 = 5050 (sum) 650/5050 = 13/101

Paul, a painter, paints only flowers or cats in his notebook. The pictures are drawn in either regular pencil or charcoal (but not in both). The total number of pictures is 39 greater than the number of cat pictures drawn in pencil. How many charcoal flower pictures are in Paul's notebook?

Set up the 3 x 3 table: We know the total is x + 39 where x = number of cat pictures drawn in pencil. We are looking for the number of charcoal flower pictures. (1) The notebook contains 7 flower pictures drawn in pencil. Mark 7 in the pencil flower cell and add to x. The total amount of pencil pictures = x + 7. Therefore the total number of charcoal pictures equals x+39 - x+7 = 32. Not enough info to solve for charcoal flower pics. (2) The notebook contains 11 cat pictures drawn in charcoal. Mark 11 in the cats charcoal cell. This gives us a total of x+11 cat pictures. Subtract x+39 - x+11 = 28 which is the total number of flower pictures. Still not enough info. C) is the correct answer because we know the total flower pictures (28) and the flower pencil pictures (7), therefore the number of flower charcoal photos must be 21.

In a certain province in France, there are 15 cities. If a single road segment connects only two cities, how many road segments are required in order to connect the cities so that each city is connected to all other cities with a single road segment?

Set up the SeBoxes: There are fifteen possible cities for the first SeBox. The number decreases by 1, as you cannot repeat a city in the same connection. Multiply the boxes since this is an "and" relationship (each of the fifteen cities has fourteen other cities it can connect to): 15 x 14 Last question: Does the order of choosing matter? When paving roads in France (particularly), it doesn't matter if you connect Marseilles to Lyon or Lyon to Marseilles—it's the same road! Therefore, the order of selection of each pair doesn't matter, and to compensate for that, you need to divide by the number of possible internal arrangements—in this case for two cities: 2!. (15 x 14) / 2 = 15 x 7 = 105

What is the probability that an integer selected randomly from 101 to 550, inclusive, begins with 1, 2, or 3, and ends with 4, 5, or 6?

Start calculating the probability from the denominator: the number of possible outcomes is the number of integers in the range. Out of these, find the number of wanted outcomes - integers that have the required digits. Use SeBoxes to calculate the number of choices for such integers, then divide the "good" choices by the total choices possible to get the answer. The denominator is all the numbers in the given range. Subtract the extremes of the range and add 1 to get 550-101+1 = 450 numbers in the range. As for the number of wanted outcomes: break down the problem into hundreds, tens and units digits using SeBoxes: The hundreds digit can be 1, 2, or 3 = 3 possible choices The second digit: since the question does not state that the digits cannot be repeated, the second digit can be any digit 0-9 = 10 possible choices The third digit can be 4, 5, of 6 = 3 choices: 3 x 10 x 3 = 90 So the probability of selecting an integer which satisfies the terms of the question is 90/450 = 9/45 = 1/5

In a sequence of 19 numbers, after the first term, each term is 9 more than the previous term. If the last term in the sequence is 200, what is the first term?

The formula for the nth term of an arithmetic sequence, An, is given by the formula: An = A1 + (n-1)d where d is the difference between any two consecutive terms in the sequence. In the given sequence, d=9 and A19 = 200. Plug in d=9, n = 19, and An = 200 into the formula and solve for A1: 200 = A1 + (19-1)(9) 200 = A1 + 162, 200 - 162 = 38 38 = A1

Diagonal BD divides quadrilateral ABCD into two triangles. Are these triangles congruent? 1) AD || BC 2) AB = CD

The issue here is whether quadrilateral ABCD must be divided into two congruent triangles by diagonal BD. By finding even one quadrilateral that cannot be divided into two congruent triangles, you have proved that both statements are insufficient. If both statements are true, then the quadrilateral is either an isosceles trapezoid (a trapezoid whose legs are of equal length) or a parallelogram. However, an isosceles trapezoid meets conditions 1 and 2 but when divided by line BD, the resulting triangles are not congruent.

Are a and b both integers? (1) a^2 + b^2 is an integer (2) a/b is an integer

The issue is integers vs. fractions. Plug In good numbers that satisfy the statements and show that a and b are integers and show that the answer is "Yes." Then ask yourself, is this always true for any number? Plug in other numbers in order to reach an answer of "No" and prove the statement(s) insufficient. For Stat. (1), if you plug in a=b=2, then a^2 +b^2 is an integer, so these numbers satisfy the statement's requirements. In this case, a and b are both integers, so the answer is "Yes." However, if you plug in square roots, like a=b=√2, then a^2+b^2 is still an integer. With this second pair of plug-ins, a and b are not integers, giving an answer of "No." Therefore, Stat.(1) → Maybe → IS → BCE. For Stat. (2), if possible, use the same Plug-Ins you chose to prove Stat. (1) insufficient: both a=b=2 and a=b=√2 satisfy the equation because any number divided by itself is 1. The first case yields an answer of "Yes" and the second "No." Therefore, Stat.(2) → Maybe → IS → CE. For Stat. (1+2), since we have used the same set of numbers (a=b=2 and a=b=√2) to prove each of the statements insufficient, we can still use the same sets to reach both an answer of "Yes" and an answer of "No" for their combination, so Stat.(1+2) → Maybe → IS → E.

Sets S and T contain an equal number of terms, all of which are positive integers. The variable x is the median of S and y is the average (arithmetic mean) of T. Is x>y? (1) The sum of the terms in S is greater than the sum of the terms in T. (2) S consists of consecutive even numbers, and T consists of consecutive odd numbers.

The median is the middle number in a set of numbers, arranged in ascending or descending order. If the set has an odd number of terms, then the median is the middle number. If the set has an even number of terms, then the median is the average of the middle two elements. According to Stat. (1), since both sets contain an equal number of terms, the greater sum in S indicates a greater average than T. However, this tells you nothing about the comparison between the median of S and the average of T. An example where x > y is easy. Just choose two sets of consecutive integers with the same number of elements, but make sure the sum of S is greater than the sum of T. If S = {2,3,4} and T = {1,2,3}, then x = Median of S which is 3. y = Average of T which is 2. Therefore, x>y BUT... where does it say that the sets must be consecutive? Stat. (1) doesn't say that. Once we loosen that restriction, it is easy to construct a set S that is greater than T ={1,2,3} in sum but has a median less than 2. For example, the sum of S={1,1,1000} is definitely greater than the sum of T, but the median (the number in the middle) is 1, so x is smaller than y. Thus BCE According to Stat. (2), in sets of consecutive numbers, the average is also equal to the median. However, you still do not know which is greater: the average/median of S or the average/median of T. Therefore, Stat.(2) → IS → CE. According to Stat. (1+2), in both sets, the average is equal to the median. If both sets have the same number of elements, then the set with the greatest sum (S) also has the greatest average (which is also the median). Therefore, the answer is "Yes." Stat.(1+2) → S → C.

If x and y are positive, and x² + y² =100, then for which of the following is the value of x+y greatest?

The single equation with two unknowns does not provide a single solution for x and y. Plug the answers into the question, and see which value of x (and the corresponding value of y) yields the greatest value for the sum x+y. Since the question asks for the greatest value, there's no choice but to Plug In all five answer choices and see which is the greatest. x =10 x² = 100 y² = 0 x + y = 10 x = 8 x² = 64 y² = 100 - 64 = 36 y =6 8 + 6 = 14 x = 7 x² = 49 y² = 100 - 49 = 51 y = 7+ x + y = 14+

On the number line above, the segment from 0 to 1 has been divided into fifths, as indicated by the large tick marks, and also into sevenths, as indicated by the small tick marks. What is the least possible distance between any two of the tick marks?

The small tick marks are placed at 1/7, 2/7, 3/7, 4/7, 5/7, and 6/7, and the large tick marks are at 1/5, 2/5, 3/5, and 4/5. The least common denominator is 35, so the tick marks in ascending order are placed at 5/35, 7/35, 10/35, 14/35, 15/35, 20/35, 21/35... Smallest distance is 1/35

How many strings of six letters can be made using only the letters A, B, and C or only the letters D, E, and F ?

There are two scenarios presented by this problem: a six-letter combination made up of the letters A, B, C and a separate six-letter combination made up of the letters D, E, F. 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 a single source—letters. Since you need a six-letter combination, the three letters can repeat, and the SeBoxes' size remains unchanged. Set up the problem as n x n x n x.. n^k Let's start with the A, B, and C words. For each of the six places in the combination, you can use all three letters. When the number of choices for each place remains the same, use n^k where n is the number of choices and k is the number of places. Thus, you have 3^6 options. The same goes for the D, E, and F words. Since it's an "or" relationship, add the two results: 3^6 + 3^6 or 2 x 3^6

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 wife?

This problem presents a case of choosing from a single source—family members It also presents an added twist: the father and the mother have to stay together. When two or more objects have to be adjacent to each other, begin by considering them as one object at first. Then break down the problem in a step-by-step method using SeBoxes. There are four items (father + mother object, 2 children, and 1 dog), so you need four SeBoxes. Since we want everyone in the picture, there's no repetition, and the SeBoxes' size changes. Finally, since we're asked for the different ways to arrange the family in a row, the order of choice matters. After calculating the number of arrangements, remember to multiply by the number of internal arrangements in the big object. Calculating the number of different arrangements while treating the father and mother as one object yields 4! choices: 24 Finally, 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!: 24 x 2 = 48 arrangements

On a five-question multiple choice quiz, each question can be either correct or incorrect and is worth 20 points. If Brenda wants to score AT LEAST 60% on the quiz, in how many ways can she answer questions correctly and incorrectly?

Three correct: You need to calculate the number of ways to select three out of the five questions that are answered correctly. The order in which these are selected doesn't matter. Therefore, you need to use the combinations formula to calculate the number of ways to get three out of five questions correct. C(5,3) = 10 Four correct: Similarly, the number of ways you can select four out of five questions is C(5,4) = 5 Five correct: Lastly, the number of ways you can select five out of five questions is C(5,5) = 1 Total: Therefore, there are 10+5+1=16 ways that Brenda could get three, four, or five questions right on the quiz and score at least a 60%.

If t is a non-negative integer, is 21! divisible by 3^t? (1) t is the product of two distinct single-digit prime numbers that are each less than 7. (2) 0 < t < 9

To evaluate the question "is 21! divisible by 3^t?", follow the steps below: First break down 21! and 3t: 21! = 1 x 2 x 3 x 4 x 5 x 6 ... 21 3^t = 3 x 3 x 3...multiplied t times Understand what the question is asking: If 21! is divisible by 3^t, then (21! / 3^t) is an integer. For this to happen, all the 3's in the denominator must CANCEL with the 3's in the numerator. So the question is really asking, "does 21! have at least t 3's?" Count the 3's in 21!: Since we only care about how many 3's there are in 21!, non-multiples of 3 factors such as 2, 4, 5, or 7 are irrelevant. Focus only on the multiples of 3: 3 = one 3; 6 = one 3; 9 = two 3s; 12 = one 3; 15 = one 3; 18 = two 3s; 21 = one 3; therefore there are nine 3s We conclude that as long as t is 9 or less, then all of the 3's in the denominator will be reduced by 21! and give an integer result, so the answer to the question "Is 21! divisible by 3^t?" will be "Yes". 1) Could be 2 x 3 = 6 or 3 x 5 = 15. Answer: Maybe -> BCE 2) 0 < t < 9; t is less than 9 so we are good! B.

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?

Translate this confusing question into a rule of the sequence: An = (An-1 / 2) + 3 The variable 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 find An−2 from An−1. 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. The number 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. If you plug in m=6 into answer choice A, you get 4m−18=4(6)−18=24−18=6

Bill downloads the movie Revenge of the Avengers to his computer in 2.5 hours using a download manager that downloads from 3 sources marked A, B, and C. Each source provides download at a constant rate, but the rates of different sources are not necessarily identical. If the movie was downloaded from sources A and C alone, it would take 4 hours to complete the download. The next day, source B is available, but the other sources are inactive. How long will it take to download the trailer of the movie, a file that is 1/40 the size of the movie, from source B alone?

What we have here is a downloaded file. The size of that file is undetermined, and there is once again an invisible variable that calls for Plugging In. The number we Plug In as the size of the file should be divisible by the times mentioned in the question—2.5 and 4 hours. It should also be divisible by 40 so that the size of the trailer will also be an integer. Don't spend too much time on finding the right number to Plug In. Just multiply all those factors, 2.5 x 4 x 40 = 400 and that's your good number to Plug In. So assume you have is a 400 MB file downloaded from 3 sources. Now take it from there. After we Plug In 400 MB as the size of the movie file, we can calculate the download rate when sources A, B, and C are active: Calculate the rate of A+B+C = 400 MB / 2.5 Hour Rate = 160 MB/hr Then we have the download time from sources A and C alone: A+C = 400 MB / 4 hr = 100 MB/hr Now it's easy to calculate the download rate from source B alone: 160 MB - 100 MB = 60 MB Which brings us to the final step: The trailer is 1/40 the size of the movie. We have the download rate from source B, so we have all we need. Work = 400 / 40 = 10 MB Use this to calculate the time by dividing work by rate: 10 MB / 60MB/hr = 1 / 6 hr = 10 min

Which of the following CANNOT be true (for any real x)? 1) x²- x + 2 2) x² - 2x + 1 3) 2x² - x + 1

When can't a quadratic equation be true for any real x when it has no real solution? This happens when the discriminant b² − 4ac is negative. 1) Plug in 1 for a, -1 for b and 2 for c. Answer is -7, thus no real solution. 2) Plug in 1 for a, -2 for b, and 1 for c. Answer is 0, thus one solution. 3) Plug in 2 for a, -1 for b and 1 for c. Answer is -7, thus no real solution.