GMAT - Error Log

A tank containing water started to leak. Did the tank contain more than 30 gallons of water when it started to leak? (Note: 1 gallon = 128 ounces) (1) The water leaked from the tank at a constant rate of 6.4 ounces per minute. (2) The tank became empty less than 12 hours after it started to leak.

(1 alone) Does not tell you anything about the size of the tank. OUT (2 alone) Does not tell you anything about the rate of leakage. OUT (1 & 2) Lets do some simple math to begin. 30 gallons = 3840 oz 6.4 oz/min = 384 oz/hr Ah, nice! 3840 and 384 are easy numbers to work with. It should be pretty clear that... 10 hours to drain => 30 gallons less than 10 hours to drain => less than 30 gallons more than 10 hours to drain => more than 30 gallons However, (2) does not help us out in this category, as the time could be either 1 or 11. OUT (1 or 2) OUT (Neither Sufficient) CORRECT

K is a set of numbers such that (i) if x is in K, then -x is in k, and (ii) if each of x and y is in K, then xy is in K. Is 12 in K? 1. 2 is in K. 2. 3 is in K.

(1) 2 is in K --> according to (i) -2 is n K --> according to (ii) -2*2=-4 is in K --> according to (i) -(-4)=4 is in K and so on. Thus we know that 2, -2, -4, 4, 8, -8, 16, -16, ... are in K, so basically powers of 2 and their negative pairs. Is 12 in K? We don't know. Not sufficient. (2) 3 is in K --> according to (i) -3 is n K --> according to (ii) -3*3=-9 is in K --> according to (i) -(-9)=9 is in K and so on. Thus we know that 3, -3, -9, 9, 27, -27, 81, -81, ... are in K, so basically powers of 3 and their negative pairs. Is 12 in K? We don't know. Not sufficient. (1)+(2) From (1) 4 is in K and from (2) 3 is in K, hence according to (ii) 4*3=12 must also be in K. Sufficient. Answer: C.

If m, n, and p are integers, is m + n odd? (1) m = p^2 + 4p + 4 (2) n = p^2 + 2m + 1

(1) INSUFFICIENT: Given that m = p^2 + 4p + 4, If p is even: m = (even)^2 + 4(even) + 4 m = even + even + even m = even If p is odd: m = (odd)^2 + 4(odd) + 4 m = odd + even + even m = odd Thus we don't know whether m is even or odd. Additionally, we know nothing about n. (2) INSUFFICIENT: Given that n = p^2 + 2m + 1 If p is even: n = (even)^2 + 2(even or odd) + 1 n = even + even + odd n = odd If p is odd: n = (odd)^2 + 2(even or odd) + 1 n = odd + even + odd n = even Thus we don't know whether n is even or odd. Additionally, we know nothing about m. (1) AND (2) SUFFICIENT: If p is even, then m will be even and n will be odd. If p is odd, then m will be odd and n will be even. In either scenario, m + n will be odd. The correct answer is C.

[y] denotes the greatest integer less than or equal to y.Is d < 1? 1) d= y-[y] 2) [d] =0

(1) if y is an integer then [y]=y. if y is not an integer, say 3.2 then [y]=3. 3.2-3 = .2 and will always be less than 1. In other words the distance between a non integer y and its [y] will always be less than one - Sufficient (2) It is given that [d]=o, which is equivalent to 0<d<1. Plug in numbers to see that this is also sufficient Answer is D

If the positive integer x is rounded to the nearest ten, will the result be greater than x ? (1) If x is divided by 10, the remainder is even. (2) If x is divided by 5, the remainder is odd.

(1): INSUFFICIENT. If a positive integer is divided by 10, the remainder is the same as the integer's units digit. This statement thus implies that the units digit of x is 0, 2, 4, 6, or 8. If the units digit of x is 6 or 8, then rounding to the nearest ten will result in an increase; otherwise, it will not. The statement is insufficient. (2): INSUFFICIENT. If a positive integer is divided by 5, the remainder is: 0, if the integer's units digit is 0 or 5 1, if the integer's units digit is 1 or 6 2, if the integer's units digit is 2 or 7 3, if the integer's units digit is 3 or 8 4, if the integer's units digit is 4 or 9 According to this statement, then, the integer's units digit must be 1, 6, 3, or 8. If the units digit of x is 6 or 8, then rounding to the nearest ten will result in an increase; otherwise, it will not. The statement is insufficient. (1) AND (2): SUFFICIENT. If both statements are true, the units digit of the integer must be 6 or 8 (the only values common to the two statements). In either case, rounding to the nearest ten will result in an increase. The two statements together are sufficient. The correct answer is C.

A corporation that had $115.19 billion in profits for the year paid out $230.10 million in employee benefits. Approximately what percent of the profits were the employee benefits? (1 billion = 10^9)

(230.10 x 10^6/ 115.19 x 10^9) * 100 ≈ 0.2%

Formula for percent change

(new-old)/old) * 100

List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in S is 7 more than the least integer in T, how much greater is the average (arithmetic mean) of the integers in S than the average of the integers in T?

1) Based on given, choose sample numbers: S:{9,11, 13, 15, 17, 19...) and T:{2,4,6,8,10}. Since for any evenly spaced set median=mean=the average of the first and the last terms. The average of S is 18 and the average of T is 6. 18-6=12 2) For any evenly spaced set median=mean=the average of the first and the last terms. So the mean of S will be the average of the first and the last terms: mean=(x+x+9*2)/2=x+9, where x is the first term; The mean of T will simply be the median or the third term: mean=(x-7)+2*2=x-3; The difference will be (x+9)-(x-3)=12. 3) The formula for Sum of n numbers in an evenly spaced progression is denoted by: (n/2)*(2a₁+(n-1)d), where a₁= First term of the series Mean is expressed as Sum/n So, ([(2a+(10-1)*2)]/2)-([(2(a-7)+(5-1)*2)]/2) = 12

If (t-8) is a factor of t^2-kt-48, then k=?

1) Using factor method: (t-8)(t+6), therefore -kt=-2t, so k=2 OR 2) If (t-8) is a factor of the expression then t=8 is a solution. Plugging it back into the equation yields 8^2-8k-48=0. Solve for k=2

What is the largest integer n such that 1/2ⁿ>0.01

1/2ⁿ>0.01 is equivalent to 2ⁿ<100. Using trial and error find that 2⁶=64 and 2⁷=128 so 6 is the largest integer

If M is a positive integer, then M^3 has how many digits? 1) M has 3 digits 2) M^2 has 5 digits

1: take minimum and max. value for 3 digit integers min value if m = 100, M^3 = 1,000,000, which has 7 digits. min value if m = 999, M^3 = do not know exactly but should have 9 digits (since 1000^3 has 10 digits, 999 is slight lower than 1000. so 999^3 should have 9 digits). not suff... 2: again min and max. values for m^2 that has 5 digits. min value of m^2 = 10,000.00, m = 100 and m^3 = 1,000,000, which is the same as above in 1. min value if m^2 = 99,999, m = 316 approx. again do not know exactly the value of m^3 but it should have either 9 or 8 digits. again not suff. togather also do not know. so E.

MGMAT - CAT2 If x is an integer, then x(x - 1)(x - k) must be evenly divisible by three when k is any of the following values EXCEPT

Alternatively, use the rule that the product of three consecutive integers will always be a multiple of three. The rule applies because any three consecutive integers will always include one multiple of three. So, if (x), (x - k) and (x - 1) are consecutive integers, then their product must be divisible by three. Note that (x) and (x - 1) are consecutive, so the three terms would be consecutive if (x - k) is either the lowest of the three, or the greatest of the three: (x - k), (x - 1), and (x) are consecutive when (x - k) = (x - 2), or k = 2 (x - 1), (x), and (x - k) are consecutive when (x - k) = (x + 1), or k = -1 Note that the difference between k = -1 and k = 2 is 3. Every third consecutive integer would serve the same purpose in the product x(x - 1)(x - k): periodically serving as the multiple of three in the list of consecutive integers. Thus, k = -4 and k = 5 would also give us a product that is always divisible by three. The correct answer is B.

A square wooden plaque has a square brass inlay in the center, leaving a wooden strip of uniform width around the brass square. If the ratio of the brass area to the wooden area is 25 to 39, which of the following could be the width, in inches, of the wooden strip. I. 1 II. 3 III. 4 A. I only B. II only C. III only D. I and III only E. I, II and III

Area of brass square/ area of wooden strip = 25 /39 lets say length of the wooden plaque= l and length of the square brass = x then x^2 / (l^2 - x^2) = 25/39 =>39x^2 = 25l^2 - 25x^2 =>64x^2 = 25l^2 =>8x = 5l Width of wooden strip should be l-x =>x = 5l/8 so l -x = l = 5l/8 = 3l/8 Now 3l/8 could be any value depending on the value of l so answer is E.

Set A consists of five different numbers; set B consists of four different numbers, each of which is in set A. Is the standard deviation of set A less than the standard deviation of set B ? (1) Set A contains five consecutive integers. (2) The average (arithmetic mean) of set A is equal to the average (arithmetic mean) of set B.

Before beginning the problem, note the close relationship between the two sets. Specifically, set B is formed by removing one of the numbers in set A; alternatively, set A may be formed by adding a fifth number to set B. Therefore, the problem can be considered in terms of what happens when that single number is added or taken away. (1): INSUFFICIENT. According to this statement, set B is formed from set A by removing one of five consecutive integers. If the middle integer of set A -- which, by symmetry, is equal to the mean of the set -- is removed, the standard deviation will increase; in that case, the standard deviation of set B will be larger than that of set A. On the other hand, if one of the extremes of set A -- that is, the largest or smallest number in A -- is removed, then the standard deviation will decrease; in that case, the standard deviation of set B will be smaller than that of set A. Alternatively, it may be easier to view the situation in terms of adding a new number to set B to create set A. To use a concrete example, if set B consists of the integers 1, 2, 4, and 5, and the integer 3 is added, then the resulting set A (which consists of consecutive integers, as required) will have a smaller standard deviation, as the mean of the original set B is 3. On the other hand, if set B consists of the integers 1, 2, 3, and 4, and the integer 5 is added, then the resulting set A (which, again, consists of consecutive integers 1 through 5) will have a larger standard deviation, as the new number is an extreme of the distribution. (2): SUFFICIENT. If removing a certain value does not change the mean of set A, then the value that has been removed must equal the mean itself. (Remember, removing a value below the mean of a set will raise the mean; removing a value above the mean will lower the mean.) If the mean itself is removed from the set, the standard deviation must increase; therefore, the standard deviation of set B must be larger than that of set A. Alternatively, consider the situation starting from set B instead. If adding a fifth number does not change the mean of the set, then that fifth number must be equal to the original mean of set B. Adding the mean value will always reduce the standard deviation of a set, provided the standard deviation is not 0 (a fact that is guaranteed here, as the sets contain numbers that are different from one another); therefore, the standard deviation of the new set A must be smaller than that of set B. The correct answer is B.

Is q > t ? (1) qp^2 < tp^2 (2) qp^3 > tp^3

Both of these statements give us information about the relationship of q to t, but that relationship is muddled by the presence of p in the expressions. Generally, it is not a good idea to divide an equation or inequality by a variable since dividing by zero is illegal, and a variable might be equal to zero. However, note that p cannot be zero here, since if it were, the expressions in the statements would all be zero and thus equal. Therefore, p is definitely not zero, and we are free to divide each statement by p to simplify. (1) SUFFICIENT: Since p^2 must always be positive (it has an even exponent), we can divide both sides by p^2, certain that the inequality sign will not flip, and determine that q < t. We can definitively answer the question of whether q is greater than t: "no." Remember, on data sufficiency GMAT questions, a definite "no" answer is sufficient, just as a definite "yes" answer is sufficient. A statement will be insufficient only when the answer is "maybe" or "it cannot be determined." (2) INSUFFICIENT: Since p3 can be positive or negative (it has an odd exponent), we cannot determine which way the inequality sign will face when we divide both sides by p3. If p3 is positive, then the inequality sign will not flip. If p3 is negative, then the inequality sign will flip. The correct answer is A.

The sum of Doris and Fred's ages is y. Doris is 12 years older than Fred. How many years old will Fred be y years from now, in terms of y.

D+F = Y BUT F+12 = D THUS 2F +12 = Y F NOW = Y-12/2 , F Y YEARS FROM NOW = (Y-12)/2 +Y = 3y/2-6

Line l is defined by the equation y - 5x = 4 and line w is defined by the equation 10y + 2x + 20 = 0. If line k does not intersect line l, what is the degree measure of the angle formed by line k and line w? (Assume that all lines lie in one coordinate plane.)

First, let's rewrite both equations in the standard form of the equation of a line: Equation of line l: y = 5x + 4 Equation of line w: y = -(1/5)x - 2 Note that the slope of line w, -1/5, is the negative reciprocal of the slope of line l. Therefore, we can conclude that line w is perpendicular to line l. Next, since line k does not intersect line l, lines k and l must be parallel. Since line w is perpendicular to line l, it must also be perpendicular to line k. Therefore, lines k and w must form a right angle, and its degree measure is equal to 90 degrees. The correct answer is D.

A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women?

First, we must calculate the total number of possible teams (let's call this t). Then, we must calculate how many of these possible teams have exactly 2 women (let's call this w). The probability that a randomly selected team will have exactly 2 women can be expressed as w/t. To calculate the number of possible teams, we can use the Anagram Grid method. Since there are 8 employees, 4 of whom will be on the team (represented with a Y) and 4 of whom will not (represented with an N), we can arrange the following anagram grid: A - Y B - Y C - Y D - Y E - N F - N G - N H - N To make the calculation easier, we can use the following shortcut: t = (8!)/(4!)(4!). The (8!) in the numerator comes from the fact that there are 8 total employees to choose from. The first (4!) in the denominator comes from the fact that 4 employees will be on the team, and the other (4!) comes from the fact that 4 employees will not be on the team. Simplifying yields: t = 8!/4!4! t = (8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)/(4 × 3 × 2 × 1) × (4 × 3 × 2 × 1) t = 70 So, there are 70 possible teams of 4 employees. Next, we can use a similar method to determine w, the number of possible teams with exactly 2 women. We note that in order to have exactly 2 women on the team, there must also be 2 men on the team of 4. If we calculate the number of ways that 2 out of 5 women can be selected, and the number of ways that 2 out of 3 men can be selected, we can then multiply the two to get the total number of teams consisting of 2 men and 2 women. Let's start with the women: A - Y B - Y C - N D - N E - N 5!/2!3! = 10 So, the number of ways that 2 women can be selected is 10. Now the men: A - Y B - Y C - N 3!/2!1! = 3 Thus, the number of ways that 2 men can be selected is 3. Now we can multiply to get the total number of teams with exactly 2 women (and thus exactly 2 men): w = (10)(3) = 30. Since there are 30 possible teams with exactly 2 women, and 70 possible teams overall, w/t = 30/70 = 3/7. The correct answer is D.

Given that n* denotes the product of all the integers from 1 to n, inclusive so, 6*+2=6!+2 and 6*+6=6!+6. Now, notice that we can factor out 2 our of 6!+2 so it cannot be a prime number, we can factor out 3 our of 6!+3 so it cannot be a prime number, we can factor out 4 our of 6!+4 so it cannot be a prime number, ... The same way for all numbers between 6*+2=6!+2 and 6*+6=6!+6, inclusive. Which means that there are no primes in this range.

Given that n* denotes the product of all the integers from 1 to n, inclusive so, 6*+2=6!+2 and 6*+6=6!+6. Now, notice that we can factor out 2 our of 6!+2 so it cannot be a prime number, we can factor out 3 our of 6!+3 so it cannot be a prime number, we can factor out 4 our of 6!+4 so it cannot be a prime number, ... The same way for all numbers between 6*+2=6!+2 and 6*+6=6!+6, inclusive. Which means that there are no primes in this range.

If 4<[(7-x)/3], which of the following must be true? I. 5<x II. |x+3|>2 III. -(x+5) is positive A) II only B) III only C) I and II only D) II and III only E) I, II and III

I. 5<x --> not true as x<-5 . II. |x+3|>2, this inequality holds true for 2 cases, (for 2 ranges): 1. when x+3>2 , so when x>-1 or 2. when x+3<-2, so when x<-5 . We are given that second range is true, so this inequality holds true. Or another way: ANY x from the range (-5.1, -6, -7, ...) will make |x+3|>2 true, so as x<-5 , then |x+3|>2 is always true. III. -(x+5)>0 --> x<-5 --> true. Answer: D.

MGMAT - CAT 2 7 teams compete in a track competition. If there are 20 events in the competition, no event ends in a tie, and no team wins more than 3 events, what is the minimum possible number of teams that won at least one event?

In order to get the smallest possible number of teams as winners (of at least one event), we want to have as many teams as possible not win any events. How can we accomplish this? Since there are 20 events, there are going to be 20 events won. We want as few teams as possible to be the winners of those 20 events. To accomplish this, we will make each "winning" team win as many events as possible. We are told that no team wins more than 3 events. Thus, the maximum number of events that a team wins is 3. Team A B C D E F G # of wins 3 3 3 3 3 3 2 The chart shows that even when we award teams 3 wins each, the final team (G) still wins 2 events. No team ends up without a win. The correct answer is E.

Running at their respective constant rates, Machine X takes 2 days longer to produce w widgets than Machine Y. At these rates, if two machines together produce 5/4w widgets in 3 days, how many days would it take Machine X alone to produce 2w widgets? A) 4 B) 6 C) 8 D) 10 E) 12

Let T total number of days taken by machine X to produce W widgets. Then Work done by x is 1 day is W/T Work done by machine y in 1 day = W/(T - 2). Combined work done in 3 days = 3( W/T + W/(t-2) ) = 3W( 2T-2 ) / (T- 2) which is equal to 5W/4 So equating the two equations we get 12( 2T - 2 ) = 5T( T - 2) 24T - 12 = 5T2 - 10T 5T2 - 32T + 12 = 0 (T-6) (5T+2) = 0. Which gives T = 6. So no. of days to produce 2W widgets = 2T = 12. Ans: E

A certain clock marks every hour by striking a number of times equal to the hour, and the time require for a stroke is exactly equal to the time interval between strokes. At 6:00 the time lapse between the beginning of the first stoke and the end of the last stroke is 22 seconds. At 12:00, how many seconds elapse between the beginning of the first stroke and the end of the last stroke?

Let x = time required for each stroke = time interval between strokes Therefore when the clock strikes 6:00pm, the time requires for the six strokes = 6x (for the strokes) + 5x (for the interval between the strokes) = 11x = 22 secs => x = 2 seconds Therefore at 12:00, time between the beginning of the first stroke and end of the last = 23x (12x+11x) = 46 seconds or option (D).

At 10:00 a.m., Peter begins traveling on a certain bike path from Riverdale at a constant rate of 10 mph. If, at 2:00 p.m., John begins traveling from Riverdale on the same path at a constant rate of 15 mph, at what time will he catch up to Peter?

Let's use the Rate/Time/Distance chart below to organize the information in this problem. Since John left four hours later than Peter, his time can be represented as t - 4. When Peter and John meet, their distances will be equal: 10t = 15(t - 4) or t = 12. If Peter will have been cycling 12 hours when they meet, they will meet at 10:00 p.m. The correct answer is E.

If Mel saved more than $10 by purchasing a sweater at a 15 percent discount, what is the smallest amount the original price of the sweater could be, to the nearest dollar?

Letting P be the original price of the sweater in dollars, the given information can be expressed as (0.15)P>10. Solving for P gives P>66²/₃

154. A survey of employers found that during 1993 employment costs rose 3.5 percent, where employment costs consist of salary costs and fringe-benefits costs. If salary costs rose 3 percent and fringe-benefit costs rose 5.5 percent during 1993, then fringe-benefit costs represented what percent of employment costs at the beginning of 1993? a. 16.5% b. 20% c. 35% d. 55% e. 65%

Note that one element increased 3% and another one 5.5% and weighted average increase is closer to smaller increment so clearly that has higher weight. Weightage of fringe benefits is (3.5-3.0)/(5.5-3.0)=20% In my view easiest method would be, eliminating the percentage sign Let 3.5 be 35 5.5 be 55 and 3 be 30 then first eq: S+F=E 130S+155F=135E 5s=20F therefore, s=4F and then E=5F F/E=F/5F=20% Therefore, Answer is B The amount by which employment costs rose is equal to 0.035(salary costs + fringe benefit costs); On the other hand the amount by which employment costs rose is equal to 0.03*salary costs + 0.055*fringe benefit costs; So, 35(S+F)=30S+55F --> S=4F --> F/S=1/4 --> F/(S+F)=1/(1+4)=1/5=0.2.

If x+y+z > 0, is z > 1 ? 1) z > x + y + 1 2) x + y + 1 < 0

Note that: You can only add inequalities when their signs are in the same direction: If a>b and c>d (signs in same direction: > and >) --> a+c>b+d . Example: 3<4 and 2<5 --> 3+2<4+5 . You can only apply subtraction when their signs are in the opposite directions: If a>b and c<d (signs in opposite direction: > and <) a-c>b-d--> (take the sign of the inequality you subtract from). Example: 3<4 and 5>1--> 3-5> 4-1 . Back to the original question: If x+y+z > 0, is z > 1? (1) z > x + y + 1 --> as the signs are in the same direction we can add two inequalities: (x+y+z)+z>x+y+1 --> 2z>1 --> z>1/2 , so z may or may not be more than 1. Not sufficient. (2) x + y + 1 < 0 --> as the signs are in the opposite direction we can subtract two inequalities: (x+y+z)-(x+y+1)>0--> z-1>0--> z>1. Sufficient. Answer: B.

When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35.

Positive integer n is divided by 5, the remainder is 1 -->n = 5q+1 Positive integer n is divided by 7, the remainder is 3 -->n = 7p+3 Equate the two equations: 5q+1=7p+3 Simplify 5q = 7p+2 Plug in numbers, starting from a small number (0,1,2...) until you find the q and p that makes the equation true. You find that q=6 and p=4 makes the statement true. Plug into one of the original equation to see that n=31 n+k = 35x, so 31+k = 35x, say x is 1, then k must be 4, which is the smallest positive integer.

MGMAT - CAT2 Reiko drove from point A to point B at a constant speed, and then returned to A along the same route at a different constant speed. Did Reiko travel from A to B at a speed greater than 40 miles per hour? (1) Reiko's average speed for the entire round trip, excluding the time spent at point B, was 80 miles per hour. (2) It took Reiko 20 more minutes to drive from A to B than to make the return trip.

Say, speed for A to B = a miles per hour and speed for B to A = b miles per hour Also assume that distance between A and B = d miles Hence, total time taken in the trip = (d/a + d/b) = d(a + b)/ab And, total distance covered = 2d Statement 1: Average speed = 2d/[d(a + b)/ab] = 2ab/(a + b) = 80 Hence, 2ab = 80a + 80b ---> (2ab - 80a) = 80b ---> a = 80b/(2b - 80) For a ≤ 40 --> 80b/(2b - 80) ≤ 40 --> 80b ≤ 80b - 3200 --> 0 ≤ -3200 -----> This is not possible Hence, a must be greater than 40 Sufficient Statement 2: d/a = (d/b + 20/60) As we don't know anything about b and d, we cannot comment on a. Not sufficient

Town T has 20,000 residents, 60% of whom are female. what percentage of the residents were born in town T? 1. The number of female residents who were born in town t is twice the number of male residents who were NOT born in town T 2. The number of female residents who were NOT born in Town T is twice the number of female residents who were born in Town T

Set up a chart! From the passage, we know T has 20,000 residents of which, 60% or (60/100)*20,000 = 12,000 are female, and 8,000 (20,000-12,000) are male. We're asked what percentage of residents were born in town T. This would be found by (# of residents born in town T)/(# of residents in town T) * 100%. Since we already know the denominator, what the question is really asking is the number of residents born in town T. From statement (1), we're told: # of female residents born in town T = 2*(# of male residents not born in town T) We do now have the value for both LHS and RHS, and so statement (1) is not sufficient. Note: The question stem only gives us the total number of female and male residents, and never broke them down into those that were born in town T, and those that were born outside town T. From statement (2), we're told: # of female residents not born in town T = 2*(# of female residents born in town T) Let's say the number of female residents who were born in town T is x, then the number of female residents who were not born in town T would be (12,000-x). We can now equate both LHS and RHS and solve for the number of female residents who were born in town T. However, this still does not answer the question as we're lacking a figure representing the number of male residents born in town T. So statement (2) alone is not sufficient. Using both (1) and (2), we can now use the value we found in (2), to calculate the number of male residents who were not born in town T, and therefore find the number of male residents who were born in town T. This would allow us to answer the question, and so C is the answer.

MGMAT - CAT2 a, b, c, and d are positive consecutive integers and a < b < c < d. If the product of b, c, and d is twice that of a, b, and c, then bc =

Since the product of b, c, and d is equal to twice that of a, b, and c, we can set up an equation and discover something about the relationship between d and a: bcd = 2abc d(bc) = 2a(bc) Note that bc appears on both sides of the equation. It is multiplied by d on the left side, and by 2a on the right side. Since the left side of the equation must equal the right side of the equation, we can divide out bc (which we know doesn't equal zero, so we're allowed to divide) and we get the following: d = 2a Since a, b, c, and d are consecutive integers, d must be 3 more than a, or: d = a + 3 We can combine both equations and solve for a: 2a = a + 3 a = 3 If a = 3, we know that b = 4, c = 5, and d = 6. Therefore, bc = (4)(5) = 20. The correct answer is D.

MGMAT-CAT1 Is x > y? (1) ax > ay (2) a3 > 0

Since the question itself is worded in a very straightforward way and need not be simplified, we can proceed with the analysis of each statement. (1) INSUFFICIENT: While we know that ax > ay, we cannot determine whether x > y without knowing the sign of a. Recall that if a is positive, we can divide both sides of the inequality by a and keep the inequality sign the same, yielding x > y. However, if a is negative, when we divide both sides of the inequality by a, we need to reverse the sign, resulting in x < y. Therefore, when a is positive, x > y, but when a is negative, x < y. (2) INSUFFICIENT: If a3 > 0, it must be that a > 0. However, we know nothing about x and y. (1) AND (2) SUFFICIENT: Since we know that a is positive, we can divide both sides of the inequality ax > ay by a and keep the sign of the inequality the same, yielding x > y. The correct answer is C.

When a cylindrical tank is filled with water at a rate of 22 cubic meters per hour, the level of water in the tank rises at a rate of 0.7 meters per hour. Which of the following best approximates the radius of the tank in meters?

Since water is filling the tank at a rate of 22 cubic meters per hour, after one hour there will be a "cylinder of water" in the tank (smaller than the tank itself) with a volume of 22 m³. Since the water level rises at 0.7 meters/hour, this "cylinder of water" will have a height of 0.7 meters. The radius of this "cylinder of water" will be the same as the radius of the cylindrical tank. Volumecylinder = πr²h 22 =π r²(0.7) (use π ≈ 22/7 ) 22 ≈ 22/7 (7/10) r² r2 ≈ 10 r ≈√10

If the standard deviation of Set Y is 4, what are the greatest and least values that are within one standard deviation of the mean? (1) The median of Set Y is 5. (2) The mean of Set Y is 6.

Standard deviation is a measure of how far data points in a distribution fall from the mean. The problem tells us the standard deviation. In order to figure out the values of the greatest and least values that fall within one deviation, then, the only information we need is the mean. We can rephrase the question, "What is the mean of Set Y?" (1) INSUFFICIENT: This tells us nothing about the mean. (2) SUFFICIENT: If we know that the mean is 6, the greatest value that is one standard deviation from the mean is 10 and the least value that is one standard deviation from the mean is 2. The correct answer is B.

Craig sells major appliances. For each appliance he sells, Craig receives a commission of $50 plus 10 percent of the selling price. During one particular week Craig sold 6 appliances for selling prices totaling $3620. What was the total of Craig's commissions for that week?

TC=(50+.10S)A where TC=Total Commission, S=Selling Price, and A=number of appliances sold With the given, the equation can be written as: TC=50A+.10SA, where SA=3620 So TC=50*6+.10*3620 = 300+362 = 662

A certain right triangle has sides of length x, y, and z, where x < y < z. If the area of this triangular region is 1, which of the following indicates all of the possible values of y ?

The area of the triangle is xy/2 =1 (x<y<z means that hypotenuse is z) -->x=2/y . As x<y, then 2/y<y -->2<y² -->y>√2. Also note that max value of is not limited at all. For example y can be 1000000 and in this case xy/2 --> (x*1000000)/2 = 1 --> x=2/1000000 Think this way: We know x < y and (1/2)xy = 1 (z is the hypotenuse) So xy = 2 If x = y, then x = y = sqrt(2) But since x < y, x is less than sqrt(2) and y is greater than sqrt(2). Makes sense. Good thinking.

Two trains, X and Y, started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; Train Y, traveling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had train X traveled when it met train Y?

The concept used in these questions is Relative Speed. If two people walk in opposite directions (either towards each other or away from each other), their speed relative to each other is the sum of their speeds. e.g. If you are walking away from me at a speed of 2 miles/hr and I am walking away from you at a speed of 1 mile/hr, together we are creating a distance of 3 miles in 1 hr between us so our relative speed is 2 + 1 = 3 miles/hr On the other hand, when two people walk in the same direction, their relative speed is the difference between their speeds. e.g. if you are walking away from me at 1 mile/hr and I am walking towards you at 2 miles/hr, my speed relative to you is 2-1 = 1 mile/hr. Time taken to meet = Total distance traveled/Relative speed Speed of train X = 100/5 = 20 miles/hr Speed of train Y = 100/3 miles/hr Relative Speed = 20 + 100/3 = 160/3 miles/hr Distance between them = 100 miles Time taken to meet = 100/(160/3) hr = 15/8 hrs In this time, train X would have traveled 20 * (15/8) = 37.5 miles Faster Alternate Approach using Ratios : Time taken by train X : Time taken by train Y = 5:3 Then, Speed of train X:Speed of train Y = 3:5 Since they start simultaneously, they travel for same time. So the ratio of their distance covered should be same as ratio of their speeds. Distance covered by train X : Distance covered by train Y = 3:5 3/8 *100 = 37.5 miles (Distance covered by train X)

In pentagon PQRST, PQ=3, QR=2, RS=4,and ST=5. Which of the lengths 5, 10 and 15 could be the value of PT? A 5 only B 10 only C 5 and 10 only D 10 and 15 only E 5, 10 and 15

The length of any side of a triangle must smaller than the sum of the other two sides. The same for pentagon: the length of any side of a pentagon must be smaller than the sum of the other four sides. PQ+QR+RS+ST=3+2+4+5=14, so the length of the fifths side can not be more than 14. Answer: C (5 and 10 only). As mentioned by Bunuel, I will give another way to solve this problem using Triangle property. Join P & R In triangle PQR, the known two sides are 3 & 2 units so third side (Let x) must be between 1<x<5 i.e. it can take any value 2,3 or 4 In other words the maximum & minimum values possible are 4 & 2 respectively Join T & R In triangle PQR, the known two sides are 4 & 5 units so third side (Let y) must be between 1<x<9 i.e. it can take any value 2,3,4,5,6,7 or 8 In other words the maximum & minimum values possible are 8 & 2 respectively We want to know the values possible for PT In triangle PRT, the range of two sides is 1<x<5 & 1<x<9 units so third side (PT) must be between (2-2)< PT <(8+4) or 0< PT<12 i.e. it can take any value 1,2,3,4,5,6,7....10,11 (assuming sides can take only integer values) Thus both 5 & 10 are possible Thus Answer C Hope it helps.

The table above shows the cancellation fee schedule that a travel agency uses to determine the fee charged to a tourist who cancels a trip prior to departure. If a tourist canceled a trip with a package price of $1,700 and a departure date of September 4, on what day was the trip canceled? (1) The cancellation fee was $595 (2) If the trip had been canceled one day later, the cancellation fee would have been $255 more

The table above shows the cancellation fee schedule that a travel agency uses to determine the fee charged to a tourist who cancels a trip prior to departure. If a tourist canceled a trip with a package price of $1,700 and a departure date of September 4, on what day was the trip canceled? (1) The cancellation fee was $595 --> cancellation fee is 595/1700=35% of a package price, so the trip was canceled from 31 to 45 days prior to 4th of September. Not sufficient. (2) If the trip had been canceled one day later, the cancellation fee would have been $255 more --> $255 is 15% of of a package price, so the trip was canceled either 31 or 16 days prior to 4th of September (15% change is from 35% to 50% and from 50% to 65%). Not sufficient. (1)+(2) The trip was canceled 31 days prior to 4th of September. Sufficient.

One week, a certain truck rental lot had a total of 20 trucks, all of which were on the lot Monday morning. If 50% of the trucks that were rented out during the week were returned to the lot on or before Saturday morning of that week, and if there were at least 12 trucks on the lot that Saturday morning, what is the greatest number of different trucks that could have been rented out during the week?

There are numerous ways in which you can solve this question. Brute force method if the relation between rented and non-rented trucks in not very clear: Monday morning - 20 trucks Saturday morning - at least 12 trucks 50% trucks rented in the week were returned. maximum no of trucks rented out = ? I want to maximize the no. of trucks rented so I say - If 20 trucks were rented (i.e. all of them), then we should have 50% i.e. 10 of them back. But we have more; we have at least 12. So the no. of trucks rented out must be less than 20 (because they cannot be more than 20). What about 18? If 18 trucks are rented out, 2 remain in the lot through the week. Out of 18, 9 are returned so total 11 are in the lot. But we need at least 12 in the lot. Let's go further down and try 16. 4 trucks do not leave the lot. Out of 16, 8 come back so we have 12 trucks in the lot. (As we keep reducing the number of trucks rented out, the total number of trucks in the lot of Saturday morning keeps increasing. We need to maximize the number of trucks rented out which will be at the minimum possible value of total number of trucks in the lot.) Therefore, 16 trucks must have been rented out. Algebraic approach: As we increase the number of trucks rented, the total number of trucks in the lot on Saturday morning decreases since out of the rented trucks only 50% come back (while all non-rented trucks stay in the lot). (e.g If none of the 20 trucks are rented, the lot will have 20 trucks on Saturday. If 18 trucks are not rented, the lot will have 19 (18 + 1 rented comes back) trucks on Saturday morning.) So maximize the number of trucks rented, we should try to minimize the number of trucks in the lot on Saturday morning i.e. make it 12. N - Not rented trucks; R - Rented trucks N + R = 20 N + R/2 = 12 R = 16

If a and b are both single-digit positive integers, is a + b a multiple of 3? (1) The two-digit number "ab" (where a is in the tens place and b is in the ones place) is a multiple of 3. (2) a - 2b is a multiple of 3.

There is no obvious way to rephrase this question. There are too many possibilities for a and b that would yield an "a + b" which is a multiple of 3. (1) SUFFICIENT: The two-digit number "ab" can be represented by the expression 10a + b. Since 10a + b is a multiple of 3, 10a + b = 3k, where k is some integer. This can be rewritten as 9a + (a + b) = 3k (we are being asked about a + b). If we solve for the expression a + b, we get: a + b = 3k - 9a. 3k - 9a can be factored to 3(k - 3a). Since both k and a are integers, 3(k - 3a) must be a multiple of 3. Therefore a + b is also a multiple of 3.

When one line and another line are negatively recipricoal in slope (x, -1/x)

They are perpendicular to each other

When two lines do not intersect

They must be parallel

Patrick is cleaning his house in anticipation of the arrival of guests. He needs to vacuum the floors, fold the laundry, and put away the dishes after the dishwasher completes its cycle. If the dishwasher is currently running and has 55 minutes remaining in its cycle, can Patrick complete all of the tasks before his guests arrive in exactly 1 hour? (1) Vacuuming the floors and folding the laundry will take Patrick 36 minutes. (2) Putting away the dishes will take Patrick 7 minutes.

This is a logic problem, best approached by thinking about the real-world situation. (1) INSUFFICIENT: This statement tells us nothing about the time required to put away the dishes in the dishwasher. (2) SUFFICIENT: At first, this statement may seem insufficient, since it tells us nothing about how long Patrick will spend vacuuming and folding laundry. However, we do know that the dishwasher will finish in 55 minutes, and that Patrick must wait until the dishwasher completes its cycle before spending 7 minutes to put away the dishes. So, at the very least, we know that Patrick cannot complete his tasks in less than 55 + 7 = 62 minutes. Thus, the answer to the question "Can Patrick complete all of the tasks before his guests arrive in exactly 1 hour?" is definitely "No," regardless of how long the other tasks would take. Remember that a definite "no" answer is sufficient, just as a definite "yes" answer would be. "Maybe" or "It depends" are the only insufficient answers to yes/no data sufficiency questions. The correct answer is B.

If |x| · y+ 9 > 0, and x and y are integers, is x < 6? (1) y is negative (2) |y| < 1

We are told that |x| · y+ 9 > 0, which means that |x| · y > -9. The question asks whether x < 6. A statement counts as sufficient if it enables us to answer the question with "definitely yes" or "definitely no"; a statement that only enables us to say "maybe" counts as insufficient. (1) INSUFFICIENT: We know that |x| · y > -9 and that y is a negative integer. Suppose y = -1. Then |x| · (-1) > -9, which means |x| < 9 (since dividing by a negative number reverses the direction of the inequality). Thus x could be less than 6 (for example, x could equal 2), but does not have to be less than 6 (for example, x could equal 7). (2) INSUFFICIENT: Since the question stem tells us that y is an integer, the statement |y| ≤ 1 implies that y equals -1, 0, or 1. Substituting these values for y into the expression |x| · y > -9, we see that x could be less than 6, greater than 6, or even equal to 6. This is particularly obvious if y = 0; in that case, x could be any integer at all. (You can test this by picking actual numbers.) (1) AND (2) INSUFFICIENT: If y is negative and |y| ≤ 1, then y must equal -1. We have already determined from our analysis of statement (1) that a value of y = -1 is consistent both with x being less than 6 and with x not being less than 6. The correct answer is E.

Is x a multiple of 4? (1) x + 2 is divisible by 2 (2) 6 is a factor of 3x

We can rephrase the question as "is x divisible by 2 twice?" since a number that is divisible by four must have two 2's in its prime make-up (prime box). (1) INSUFFICIENT: if x + 2 is divisible by 2, then x itself must be divisible by 2, but not necessarily 4. (RULE: for x + y to be divisible by y, x itself must be divisible by y) We could also plug numbers here to see that the statement is insufficient. According to the statement, x could be 2 (2 + 2 = 4, which is divisible by 2) and x could be 4 (4 + 2 = 6, which is divisible by 2). One of these values is a multiple of 4 and one is not. (2) INSUFFICIENT: If 6 is a factor of 3x, than x must be divisible by 2. For 6 to be a factor of a number, that number's prime make-up (prime box) must consist of a 3 and a 2, the prime components of six. 3x naturally has a 3 in its prime box because of the 3 coefficient. x therefore must be the source of the 2 and so x is divisible by 2. This does not, however, guarantee us that x is divisible by 4. Again we plug numbers here to illustrate the lack of sufficiency. According to the statement x could be 2 (3 × 2 is divisible by 6) and x could be 4 (3 × 4 is divisible by 6). (1) AND (2) INSUFFICIENT: Statements 1 and 2 lead to the same conclusion, namely that x is even, but not necessarily divisible by 4. The correct answer is E.

If a, b, and c are positive integers such that a < b < c, is a% of b% of c an integer? (1) b=(a/100)⁻¹ (2) c = 100^b

We can rephrase the question as follows: Is abc/100² an integer? (1) INSUFFICIENT: We can simplify the given equation and substitute into our rephrased question: b=100/a ab=100 c100/100² c/100 The variable c is an integer, but we have no idea whether c is a multiple of 100. c/100 might be an integer, but it also might not be. (2) SUFFICIENT: We can substitute this value into our rephrased question: Is (ab100^b)/100² = ab100^(b-2) an integer? Because a and b are positive integers and a < b, we know that the value of b must be greater than or equal to 2. This is because b can't be zero because it is a positive integer. b can't be 1 because a is less than b and is a positive integer. But b can be greater than 2. Therefore, b−2 ≥ 0. If this is the case, then the minimum possible value of 100^(b-2) is 100⁰ = 1 and every other possible value is also an integer (100¹, 100², and so on). As a result, the product of the integers a, b, and 100^(b-2) must also be an integer. The correct answer is B.

Mira is making telescopes, each consisting of 2 lenses, 1 tube, and 1 eyepiece. Lenses can be purchased only in packs of 50, tubes only in packs of 10, and eyepieces only in packs of 30. However, half of the lenses in each pack are not usable for telescopes. If all parts are used only for the telescopes, what is the minimum number of lenses Mira must purchase to make a set of telescopes with no leftover components other than the unusable lenses?

We can use algebra or test the answers; both methods are shown below. Test the Answers Because the question asks for the minimum number of lenses, start with the smallest answer. Notice that the correct answer must be a multiple of 50, since the lenses must be purchased in packs of 50. Eliminate answer A. (B) If Mira buys 150 lenses, then 75 are usable to build telescopes. She would then need 37.5 tubes and 37.5 eyepieces - impossible. Eliminate answer B. (C) If Mira buys 300 lenses, then 150 are usable and she would need 75 tubes and 75 eyepieces. Tubes can be purchased in packs of 10, so Mira would have to buy 8 packs, for a total of 80 tubes. She'd have 5 left over, violating the terms of the question. Eliminate answer C. Note: only try one more answer. If D is incorrect, choose E and move on! (D) If Mira buys 600 lenses, then 300 are usable and she would need 150 tubes and 150 eyepieces. Tubes can be purchased in packs of 10, so Mira would be able to buy exactly 150 tubes. Eyepieces can be purchased in packs of 30, so Mira also would be able to buy exactly 150 eyepieces. All conditions of the question are satisfied; this is the correct answer. The correct answer is D. Algebra The project requires a lens to tube to eyepiece ratio of 2:1:1. However, because half of the lenses cannot be used, Mira must actually purchase twice as many as needed, for a ratio of 4:1:1. Create variable expressions to represent the purchased quantity of each supply. Use L, T, and E to represent the number of packs of lenses, tubes, and eyepieces, respectively, and express the overall quantities by multiplying the number of packs of each supply by the number of items per pack: lenses = 50L tubes = 10T eyepieces = 30E Because Mira needs 4 times as many lenses as tubes or eyepieces, relate the above quantities as follows: 50L = 4(10T) = 4(30E) 50L = 40T = 120E 5L = 4T = 12E From this equation, draw two separate ratios: 5L = 4T and 5L = 12E Because these ratios are in their most simplified form, T must be a multiple of 5, E must be a multiple of 5, and L must be a multiple of 4 and of 12. Therefore, the smallest possible value for L is 12. If Mira purchases 12 packs of lenses, she'll have 12 × 50 = 600 lenses. The correct answer is D.

If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2 ? (1) More than 1/2 of the 10 employees are women. (2) The probability that both representatives selected will be men is less than 1/10.

What is the probability of choosing 2 women out of 10 people w/10*(w-1)/9 and this should be <1/2 . So we have --> w(w-1)>45 this is true only when w>7. (w # of women <= 10 ). So the question asks is w>7 (1) More than 1/2 of the 10 employees are women --> w>5 not sufficient. (2) The probability that both representatives selected will be men is less than 1/10 -->(10-w)/10*(10-w-1)/9 < 1/10 --> (10-w)(9-w)<9 --> w>6, not sufficient (1)+(2) w>5 and w>6 : w can be 7, answer NO or more than 7, answer YES. Not sufficient. Answer E. You can use Combinations, to solve as well: 2Cw # of selections of 2 women out of employees; 2C10 total # of selections of 2 representatives out of 10 employees.

13. A company that ships boxes to a total of 12 distribution centers uses color coding to identify each center. If either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors, what is the minimum number of colors needed for the coding? (Assume that the order of the colors in a pair does not matter.) (A) 4 (B) 5 (C) 6 (D) 12 (E) 24

it's much easier just to try the answers than to attempt setting up an equation. the last two choices are WAY too big - (d) you clearly don't need a separate color for each distribution center, and (e) um, what? are you kidding me? i think (d) and (e) are probably in there for students who don't really understand the question stem; admittedly, this is not the world's easiest question to understand upon first reading. if you try 4, you'll get 4 (single colors) + 4!/2!2! (pairs of colors), which is 4 + 6 = 10. this is not enough. at this point you can just realize that the addition of a fifth color will add significantly more than 2 extra possibilities - you get the fifth color itself, as well as its combination with EVERY one of the existing colors - so it's easy to tell that 5 colors will be good enough without actually computing the total.

Name each place value: 1,342,365.1427

millions, hundred thousands, ten thousands, thousands, hundreds, tens, ones, tenths, hundredths, thousandths, ten thousandths

The population of the bacteria culture doubles every 2 minutes. Approx how many minutes will it take for the population to grow from 1000 to 500,000 bacteria ?

min 0 - 1,000 min 2 - 2,000=1,000*2^1 min 4 - 4000=1000*2^2; min 6 - 8000=1000*2^3 Be careful! It doubles every 2 minutes. So, 1,000*2^x=500,000 2^x=500. It asks you to approx. You know that 2^9=512, so x=9. Because the pop doubles every 2 minutes, the result=9*2=18 minutes.

Is xy < 6? (1) x < 3 and y < 2. (2)1/2 < x < 2/3 and y^2 <64.

(1) if x and y are large negative numbers than xy>6 but if they are both small positive or one is negative then xy<6. Statement is insufficient (2) Simplify y to be -8<y<8, taking both the largest x and y, we get that xy = (2/3)*8=16/3 which is always less than 6. Statement is sufficient Answer is B

Formula for Sum of n numbers in an evenly spaced progression

(n/2)*(2a₁+(n-1)d) where a₁ is the first term of the series

When to use a Venn Diagram vs. a Table for overlapping sets?

...

When you have a percent discount (example 10%), and a price P, what is the formula for the discount amount?

.10P

Today Rose is twice as old as Sam and Sam is 3 years younger than Tina. If Rose, Sam, and Tina are all alive 4 years from today, which of the following must be true on that day. I. Rose is twice as old as Sam II. Sam is 3 years younger than Tina III. Rose is older than Tina

As per the question: Tina = X yrs Sam= (X-3) yrs Rose = 2*Sam = 2(X-3) After 4 yrs Tina = X+4 yrs Sam = [(X-3)+4] yrs Rose = [2(X-3)+4] yrs Now,I understand and agree option 1 is not correct in 4+ yr scenario. Option 2 is also fine,infact that is the most evident option. III in not always true. In your algebraic approach in 4 years Rose is 2x-2 years old and Tina is x+4 years old. How is 2x-2 always more than is x+4? 2x-2>x+4 holds true for x>6, so if Tina is now 7 years old or older.

MGMAT - CAT2 Quentin's income is 60% less than Rex's income, and Sam's income is 25% less than Quentin's income. If Rex gave 60% of his income to Sam and 40% of his income to Quentin, Quentin's new income would be what fraction of Sam's new income?

Everything ultimately depends on Rex's income, and since the information in the problem involves percents, we should assign a value of $100 to Rex's income. 60% less than $100 is $40, so that's Quentin's income. Sam's income, 25% less than $40, is $30. 60% of Rex's income is $60, and 40% of his income is $40. So Sam gains $60 more, for a total of $90. Quentin gains $40, for a total of $80. We're looking what fraction Quentin's new income ($80) is of Sam's new income ($90): 80/90 = 8/9.

MGMAT-CAT1 In a certain game, a large bag is filled with blue, green, purple and red chips worth 1, 5, x and 11 points each, respectively. The purple chips are worth more than the green chips, but less than the red chips. A certain number of chips are then selected from the bag. If the product of the point values of the selected chips is 88,000, how many purple chips were selected?

Factor 88,000 = 11 × 5³ × 2⁶ Purple must be a composed of 2s however cannot be 2 because 5 < x < 11 So purple value = 8. And there are two 8s. Answer is 2.

If S is the infinite sequence S1 = 6, S2 = 12, ..., Sn = Sn-1 + 6,..., what is the sum of all terms in the set {S13, S14, ..., S28}?

For sequence S, any value Sn equals 6n. Therefore, the problem can be restated as determining the sum of all multiples of 6 between 78 (S13) and 168 (S28), inclusive. The direct but time-consuming approach would be to manually add the terms: 78 + 84 = 162; 162 + 90 = 252; and so forth. The solution can be found more efficiently by identifying the median of the set and multiplying by the number of terms. Because this set includes an even number of terms, the median equals the average of the two 'middle' terms, S20 and S21, or (120 + 126)/2 = 123. Given that there are 16 terms in the set, the answer is 16(123) = 1,968. The correct answer is D.

If x is an integer and y=3x+2, which of the following CANNOT be a divisor of y?

If y=3x+2...we know that y is not divisible by 3 because 3x will always be divisible by 3. So if you add +2, the resulting number will be 5,8,11, etc, never divisible by 3. Since we know that if a number is not divisible by 3 then its not divisible by 6, thats your answer

When 10 is divided by the positive integer n, the remainder is n-4. Which of the following could be the value of n? A. 3 B. 4 C. 7 D. 8 E. 12

It says that the remainder when you divide 10 by n is n-4 This basically can be translated into the following statement algebraically: 10=kn+(n-4) This is simplified as follows: 14=kn+n Further simplifying: 14=n(k+1) So n must be a factor of 14, which is 7

A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probablity that they will have exactly 2 girls and 2 boys? A) 3/8 B) 1/4 C) 3/16 D) 1/8 E) 1/16

Probability of having a girl or a boy is 1/2. Because the question asks for 2 boys and 2 girls,the number of possible combinations is 4!/(2! * 2!) = 6 Consider 2 girls and 2 boys as same. For each combination,probability is 1/2*1/2*1/2*1/2 = 1/16 therefore for all the combinations value will be 6*1/16 = 3/8. Hence the ans is A the solution posted above this one works. if you don't think of that, you can also just list out all the possibilities for two boys and two girls: bbgg bgbg bggb gbbg gbgb ggbb that's six different possibilities. the total number of ways in which the two children can be born is 2 x 2 x 2 x 2 (from the slot method) = 16, so the probability that you want is 6/16 = 3/8.

Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b|

St1: a-b <0. This means left hand side term will always be -ve and right hand side term will always be +ve.: SUFF St2: 1 < |a-b| we can't say anything about the sign of (a-b) so either side can be either positive or negative. : INSUFF

if x and y are non zero integers, is x^y<y^x? 1. x=y^2 2. y>2

Statement 1 : x=y^2 therefore we need to find out whether x^y<y^x or y^(2y)<y^(y^2) now compare 2y and y^2, we don't know whether y is positive, or negative, or between 0 and 1, any of the above (L.H.S. or R.H.S.) could be greater than other. Insufficient Statement 2: y>2, but what is x? insufficient Combining two statements: if y>2, y^2 will surely be greater than 2y. hence our answer is YES, y^(2y)<y^(y^2) C

A certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the computer science department. If none of the candidates is eligible for a position in both departments, how many different sets of 3 candidates are there to fill the 3 positions? A. 42 B. 70 C. 140 D. 165 E. 315

This can be a tricky question in terms of wording. As we have to find out the different sets of both, we need to multiply each of the combinations. Therefore 7C1 * 10C2 = 315

A certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11% from 1996 and revenues from truck sales in 1997 were up 7 percent from 1996. If total revenues from car sales and truck sales in 1997 were up 1 % from 1996, what is the ratio of revenue from car sales in 1996 to revenue from truck sales in 1996? 1:2 4:5 1:1 3:2 5:3

This is a weighted average question. Average of -11% and +7% is +1%. Using w1/w2 = (A2 - Aavg)/(Aavg - A1), we get w1/w2 = (7 - 1)/(1- (-11)) = 6/12 Revenue from Car:Revenue from Trucks = 1:2 T+C=R 89C+107T=101R 12C=6T C/T=6/12=1:2

When graphing linear inequalities when to shade above the line and when to shade below the line?

When y≥ shade above the line When y≤ shade below the line Alternatively you can use test points.

X^0

1

When given x=y in a formula where y=2x+3, what can you do to solve the formula?

If x=y then x=2x+3, solve for x

How to tell if a statement is insufficient in a y/n DS question?

If you reach at both a Y and a N

By what percent is 25 greater than 15?

We are asked by what percent 25 is greater than 15. 25 is 10 more than 15. Now we need to determine the percentage that 10, the difference, represents out of 15, the original number: 10/15 = 2/3 = 0.6666 = 66 2/3% The correct answer is E.

(a+b)(a-b)

a²-b²

At a garage sale, all of the prices of the items sold were different. If the price of a radio sold at the garage sale was both the 15th highest price and the 20th lowest price among the prices of the items sold, how many items were sold at the garage sale?

the highest price top: 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,....... we have 14 items with higher price than price of radio the lowest price top: 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,....... we have 19 items with lower price than price of radio the total number of items = 14 (higher price) + 1 (radio) + 19 (lower price) = 34

Relationship between the three sides of a triangle

the third side is less than the sum of the two other sides and less than the difference of the two other sides

(x+y)²

x²+2xy+y²

(x-y)²

x²-2xy+y²

√2 √3 √121 √169 √225 √400 √625

≈1.41 ≈1.7 ≈11 ≈13 ≈15 ≈20 ≈25

If d > 0 and 0 < 1 - c/d < 1, which of the following must be true? I. c > 0 II. c/d < 1 III. c^2 + d^2 > 1 A. I only B. II only C. I and II only D. II and III only E. I, II and III

0 < 1 - c/d < 1 --> add to all three parts of this inequality --> -1 --> -1<-c/d<0 multiply by -1 and as multiplying by negative flip signs --> . 0<c/d<1 I. c>0 --> as c/d>0 and d>0 , Always true. II. c/d<1 --> directly given as true. III. c²+d²>1 --> if c=1 and d=2, then YES, but if c=0.1 and d=0.2, then No, hence this one is not always true. Answer: C (I and II only).