Data Sufficiecy Flashcards

Question

If x and y are positive, is x3 > y? (1) x^1/2 > y (2) x > y A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. D. EACH statement ALONE is sufficient. E. Statements (1) and (2) TOGETHER are NOT sufficient.

Answer 1

Good post? | x can be between 0 and 1 thus E

Answer 2

Originally Posted by missionGMAT Is x less than y? (1) x-y+1< x+1 and x-1 < x The target question asks "Is x < y?" Statement 1: x-y+1< y Since we also know that x < x+1, we can combine the two inequalities to get x < x+1 < y From this we can see that x < y, so statement 1 is sufficient Statement 2: x-y-1< y Now we also know that x-1 < x, but what can we do with this? We know that x and y are both greater than x-1, but we cannot say for certain whether or not y is greater than x. As such, statement 2 is not sufficient and the answer is A Cheers, Brent

Answer 3

Originally Posted by missionGMAT For a set of 3 numbers, assuming there is only one mode, does the mode equal the range? The median equals the range The largest number is twice the value of the smallest number Notice that if all 3 numbers are different, then we will have 3 different modes. So, if there is only 1 mode, then there are two possible cases: case a: 2 numbers are equal and the 3rd number is different case b: all 3 numbers are the same Statement 1: If the median equals the range, does the mode equal the range? Well, does the median equal the mode here? The answer is yes. Here's why: If we have case a, then the median must equal the mode (since it would be impossible for the middle-most number to be different from the other 2 values). So, the median = mode = range If we have case b, then the median must equal the mode, since all 3 numbers are equal. So, the median = mode = range Since we can be certain that the mode equals the range, statement 1 is sufficient. Statement 2: We can use counter-examples to show that this statement is not sufficient. The 3 numbers could be 3, 3, 6 in which case the mode equals the range The 3 numbers could be 3, 6, 6 in which case the mode does not equal the range So, the answer is A

Answer 4

The DS wants to know whetehr the percentage of directors can be known. let us consider total employee (count) - x director (count) - d manager (count) - x-d average salary of total employee = a we have to find - d/x-d = ? or d/x = ? or d/ x-d = ? from stem 1: (x-d)(a+5000) = ...? we just have the connection between agerage salary of managers to average salay of all employee. as no other information is given to fill-up the right hand side of equation. we even can not go further. INSUFFICIENT. A, D OUT. from stem 2: d(a+15000) = .... ? the explanation is reciprocal for d (as per stem 1). B OUT. LET US COMBINE -(x-d)(a-5000)+d(a+15000) = ax => xa - ad - 5000x - 5000d + ad + 15000d = ax => xa -ax - 5000 x + 15000d = 0 => 5000x= 15000 d => x = (15000/5000)d =>x/d =3 ( we got the ratio) c is the pick. alternatively, we can choose signs for average salaries of d,m. Thus C is the answer.

Answer 5

``` Originally Posted by missionGMAT S is a set of integers such that i) if a is in S, then –a is in S, and ii) if each of a and b is in S, then ab is in S. Is –4 in S? ``` (1) 1 is in S. (2) 2 is in S. Statement 1: All we can conclude is that 1 and -1 are in set S INSUFF Statement 2: If 2 is in set S, then -2 is in set S (by rule i). If 2 and -2 are in set S, then we can conclude is that -4 is in set S (by rule ii) SUFF So, the answer is B

Answer 6

Good post? | Draw it and you will see that A is parallel to C . (C)

Answer 7

06-30-2011, 08:27 AM #2 missionGMAT Within my grasp! ``` Join Date May 2011 Location Singapore Posts 110 Rep Power 2 ``` 1 out of 1 members found this post helpful. Good post? | Official Answer is A. Lets examine the question first. We have been given considerable data in the question itself: Let t be the time taken to drive first x miles and t' be the time taken to drive the rest of distance. Let total distance be Y. So from the given data, we can form the following equations: ``` x= 50t and Y-x = 60t' . We need to find t. ``` Statement 1: Trough this statement we get, t+t' = 10 and Y = 530. We had four variables in the equations from statement so we need atleast four equations to find the value. After combining the equations from statement 1, we get the desired result. Hence SUFFICIENT Statement 2: We cannot get two more equations, hence INSUFFICIENT.

Answer 8

Really A tough one !!! A is my pick. Critical point is - the question stem contains a LOT of information REQUIRED to solve the DS. Let us consider the stem- Guests at a recent party ate a total of fifteen hamburgers. Note - 15 hamburgers were eaten. Each guest who was neither a student nor a vegetarian ate exactly one hamburger. Note - Only NON student and NON-veg 1 burger each. Note - 2: there must be 15 guests who were neither Student nor Veg. No hamburger was eaten by any guest who was a student, a vegetarian, or both. Note -ONLY non student and non veg ate 1 burger each. If half of the guests were vegetarians, Note - 50% of the guests were veg. Note 2: 50% of the guests were NON-VEG. how many guests attended the party? Links - we have - 1. the number of burger eaten (15 nos.) 2. Veg - 50% 3. Non-veg - 50% 4. Non-Veg Non-students eaten burger . S, their % will be equal to 15 nos. We have to check whether we get sufficient information to find the % of non-veg & non-student. From St 1: (1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians. Veg guest - 50%. Ratio of Student (veg) and Non-student ( veg) - 2:3. So, Student (veg) - 20% and Non-student (Veg) - 30%. As this rate is half the rate for non-veg. for non veg Student : Non-student - 4:3 ~ 29% : 21% (SUFFICIENT). we get the % of Non veg non student - ~ 21%. St 1 alone is sufficient. Now for Statement 2 (2) 30% of the guests were vegetarian non-students. Total veg - 50% Veg non student - 30%. So, veg Student 20% Here we have no information of link / hints about NON - VEG STUDENT / NON VEG NON STUDENT. INSUFFICIENT ! We can not solve for % of non-student non-veg = 15 Please correct me if I am wrong. What is the source of this question ? Official Answer ?

Answer 9

Confusion between a and c There is no doubt in miy mind that the answer is C, lets take a real life example: Man: My bank account increased so much this month! Woman: How much was the ending balance? Man: Lets put it this way if it increased by 12% the ending balance would have been $504! Woman: Ok, that gives me your beginning balanace, but how much did it ACTUALLY increase? Man: 8% Woman: Ah, now I can figure it out. C People are mistaking finding ANY ending balance with finding THE ending balance. For instance if statement (2) was: If, during this period of time, the increase in the balance in the checking account had been 14%, then the balance in the account on May 30 would have been $513. The answer could not be D, there could not have been two ending balanaces, therefore, the answer would be E because we DO NOT know the ACTUAL increase.

Answer 10

Good post? | The answer is E If we knew the additional information: exact distance between (A;B) and the origin and (G;H) and the Origin; then we would have been able to determine the answer. If the distance between (A;B) and the origin and (G;H) and the Origin = 1/4 then the distances between the points (A;B) and (C;D) and the points (C;D) and (E;F)are equal, otherwise they aren't equal

Answer 11

Good post? | ``` (1) x is an integer. xn = x(-n)=> x2n = 1 = x0 2n = 0 => n = 0 If n = 0 then x^n – x^-n = 0 will be true for all values of x. INSUFFICIENT (2) n ≠ 0 xn = x(-n)=> x2n = 1 => x = 1 or x = -1 INSUFFICIENT Combining both statements, x = 1 or x = -1 INSUFFICIENT Hence E. ```

Answer 12

Originally Posted by shyamprasadrao IF X & Y are positive, is 3X > 7Y ? 1) X > Y + 4 2) -5X < -14Y SPOILER: Official Answer D. I guess its B Statement 1: We can use counter examples to show this is insufficient case a: x=10, y=1 gives us 3x is greater than 7y case b: x=10, y=5 gives us 3x is not greater than 7y Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT Statement 2: -5X < -14Y Multiply both sides by -1 to get: 5x > 14y Multiply both sides by 3/5 to get: 3x > 42y/5 Big point: If y is positive, then it must be true that 42y/5 > 7y. We can now add this inequality to the existing inequality to get: 3x > 42y/5 > 7y From here we can see that 3x > 7y, which means statement 2 IS SUFFICIENT and the answer is B

Answer 13

Good post? | The target question asks us whether W is positive. Statements 1 and 2 combined: We can use counter examples to show that the statements combined are not sufficient. Given the conditions in both statements, there are (at least) 2 cases: Case a: C=52 and W=-1, so W is negative Case b: C=52 and W=1, so W is positive Since we cannot answer the target question with certainty, the statements combined are not sufficient. So, the answer is E

Answer 14

Good post? | Okay, I'll post my version of a solution: Statement 1: x^3 + x is divisible by 4 When we factor the expression we see that x(x^2 + 1) is divisible by 4 For x(x^2 + 1) to be divisible by 4, there are essentially 3 possible cases to consider: case a: x is divisible by 4 (in which case x is divisible by 2) case b: x is divisible by 2, and x^2 + 1 is divisible by 2 (in which case x is divisible by 2) case c: x^2 + 1 is divisible by 4, and x is not divisible by 2 case b is impossible, for the following reason: If x is even (i.e., divisible by 2) then x^2 is also even, in which case x^2 + 1 is odd. If x^2 + 1 is odd, it cannot be divisible by 2 case c is impossible for the following reason: First, if x^2 + 1 is divisible by 4, then x^2 + 1 is even, in which case x^2 is odd. If x^2 is odd, then x must be odd. Now, if x must be odd, it is not possible for x^2 + 1 to be divisible by 4 We know this because, if x is odd, we can say that x = 2k+1 for some integer k (all odd numbers can be written as 2k+1) This means that: x^2 + 1 = (2k+1)^2 + 1 = 4k^2 + 4k + 2 = 4(k^2 + k) + 2 Since 4(k^2 + k) is definitely divisible by 4, we can see that 4(k^2 + k) + 2 is NOT divisible by 4, which means x^2 + 1 is NOT divisible by 4. If x^2 + 1 is NOT divisible by 4, we can eliminate case c, which means case a MUST be true, which means x MUST be divisible by 2. So, statement 1 is sufficient. Statement 2: 5x + 4 is divisible by 6 If 5x + 4 is divisible by 6, we know that 5x+4 must be divisible by 2, which means 5x+4 is even If 5x+4 is even, 5x must be even (since EVEN + EVEN = EVEN) If 5x is even, x must be even (since ODD x EVEN = EVEN) If x is even, then x is divisible by 2 So, statement 2 is sufficient, and the answer is D

Answer 15

Originally Posted by albema If x is a positive integer, is the remainder 0 when (3x + 1)/10? (1) x = 3n + 2, where n is a positive integer. (2) x > 4 SPOILER: E Statement 1: If x = 3n + 2, then 3x + 1 = 3(3n + 2) + 1 = 9n + 7 So, we can now reword the target question to be "Is the remainder 0 when (9n+7)/10?" Well, the remainder IS zero when n=7 but not when n=8 So, statement 1 is NOT SUFFICIENT Statement 2: If x=13, then the remainder does equal 0 when (3x + 1)/10 If x=14, then the remainder does not equal 0 when (3x + 1)/10 So, statement 2 is NOT SUFFICIENT Both statements combined At this point we know that the statements combined are not sufficient because, we already showed that statement 1 is not sufficient because n=7 and n=8 yielded different answers to our new target question. In both cases (when n=7 and when n=8), the resulting value of x is greater than 4, so statement 2's condition was already met. This means the answer is E

Answer 16

Originally Posted by dominicsavio Q1) Each of 25 balls is either red, blue, or white and has a number from 1 to 10. If one ball is selected at random, what is the probability that the ball is either white or has an even number on it? (1) Probability that the ball is both white and has an even number = 0 (2) Probability that the ball is white minus probability that the ball has an even number = 0.2 Note that we know NOTHING about the way this group of 25 balls was generated or selected. Therefore, we have to work completely off the givens from the two cases (except that all probabilities have to be a multiple of .04). Looking for: P(W or E) = P(W) + P(E) - P(E and W) = ? (1) "Probability that the ball is both white and has an even number = 0" So, P(W or E) = P(W) + P(E) INSUFFICIENT (2) "Probability that the ball is white minus probability that the ball has an even number = 0.2" P(W) - P(E) = .2 P(W or E) = P(E) + P(E) + .2 - P(E and W) P(W or E) = 2*P(E) + .2 - P(E and W) Still INSUFFICIENT...there's too much we don't know! (1) and (2) From (1), we know that P(W and E) = 0 From (2), we know that P(W) = P(E) + .2 P(W or E) = P(W) + P(E) - P(W or E) P(W or E) = 2*P(E) + .2 INSUFFICIENT, since P(E) can have multiple values. --> E Q4) Each employee in a certain taskforce is either a manager or a director. What percent of employees are directors? (1) Average salary of managers is $500 less than average salary of all employees (2) Average salary of directors is $1500 greater than the average salary of all employees Let D denote number of directors Let M denote number of managers What is D/(D+M) ? (1) "Average salary of managers is $500 less than average salary of all employees" Let s1 be the average salary of directors Let s2 be the average salary of managers ``` s2 = (s1*D+s2*M)/(D+M) - 500 (s2+500)(D+M) = s1*D+s2*M s2*D+s2*M + 500D + 500M = s1*D+s2*M s1*D = s2*D + 500(D+M) s1*D - s2*D = 500(D+M) D/(D+M) = 500/(s1-s2) ``` INSUFFICIENT (2) "Average salary of directors is $1500 greater than the average salary of all employees" This gives the same amount of into as part 1, so INSUFFICIENT. (1) and (2) From part 2, we know that s1 = 1500 + (s1*D+s2*M)/(D+M) (s1 - 1500) (D+M) = s1*D + s2*M s1*M - 1500 (D+M) = s2*M (s1 - s2)*M = 1500(D+M) (s1-s2) = 1500(D+M)/M Plugging this into above... ``` D/(D+M) = 500/(1500(D+M)/M) D/(D+M) = M/(3(D+M)) D = M/3 ``` Therefore, D/(D+M) = D/(D+3D) = 1/4 = 25% Answer is C

Answer 17

"All we know is that x is a prime number. We want enough information to determine which prime number x is. Our method, then, is to try to find more than one prime that fits with whatever information we're given. If we can, the information is insufficient; if we can't—if we can find only one prime that fits with the information—then the information is sufficient. (1) INSUFFICIENT: If x < 15, x could be 2, 3, 5, 7, 11, or 13. Eliminate (A) and (D). (2) INSUFFICIENT: If (x – 2) is a multiple of five, then x is 2 more than a multiple of 5. So the question is: Can we find more than one prime number that is 2 more than a multiple of five? Yes. Multiples of 5 are 0, 5, 10, 15, 20, and so on. Two more than 5 is 7—a prime number. While 2 more than 10 is 12, which isn't a prime number, 2 more than 15 is 17, which is prime. Eliminate (B). In combination: statement 1 narrowed down the possible values of x to 2, 3, 5, 7, 11, and 13. Remember that 0 is a multiple of 5 as well. So both 2 and 7 are 2 more than a multiple of 5, so we cannot find a single answer to the question using both statements. Choose (E)." I assumed the answer was B because i didn't know 0 is a multiple of 5. So 0 is a multiple of any integer? What's the technical rule for this? Thanks in advance

Answer 18

``` x – total number of sedans Question: x(x-1)/(20*19)>3/4 ? 1) x≥15 Case 1 x=15 15(15-1)/19*20>3/4 ? 15*14/19*20>3/4 ? 3*14/19*4>3/4 ? 4*3*14 > 4*19*3? NO case2 x=19 19*18/19*20>3/4 ? 9/10>3/4 ? 36>30 ? Yes A insufficient 2) Y –convertibles Y/20*(Y-1)/193/4 ? 4*3/19*5>3/4? 4*3*4>19*5*3? No! We know the answer when Y=1 then x=19 YES so INS. ``` Both:INS (already shown) IMO E

Answer 19

Good post? | | . . . (thousands)(hundreds)(tens)(units)(DECIMAL POINT)(tenths)(hundredths)(thousandths)(ten-thousandths)...

Answer 20

Originally Posted by dv_dheeraj A certain list consists of several different integers. Is the product of all the integers in the list positive? 1) The product of the greatest and smallest of the integers in the list is positive. 2) There is an even number of integers in the list Statement 1: case a: {-3, -1} product is positive case b: {-3, -2, -1} product is negative INSUFFICIENT Statement 2: case a: {-3, -1} product is positive case b: {-3, 1} product is negative INSUFFICIENT Statements 1 AND 2: Statement 1 tells us that the smallest and largest integers are either both positive or both negative This means that the integers in the set are either ALL positive or ALL negative Case a: the integers in the set are ALL positive This means the product must be positive Case b: the integers in the set are ALL negative Since there is an even number of integers, we can pair up every negative integer with another. Since the product of each pair will be positive, the entire product must be positive In both cases, the product must be positive --> SUFFICIENT C Cheers, Brent

Answer 21

First, how many people contributed altogether? n friends gave. Then n friends told n friends --> n^2 more people So, the total number of contributors = n + n^2 ``` Statement 1: we have a ratio: (first n people)/(all contributors) = 1/6 This means that n/(n + n^2) = 1/16 Cross multiply to get 16n = n + n^2 Subtract n from both sides: 15n = n^2 Divide both sides by n to get 15 = n SUFFICIENT ``` Statement 2: 500(total number of contributors) = 120,000 500(n + n^2) = 120,000 500n + 500n^2 = 120,000 Rearrange (since it's a quadratic equation) to get 500n^2 + 500n - 120,000 = 0 Divide both sides by 500 to get n^2 + n - 240 = 0 From here we can see that if we factor, we'll get something like (n+something)(n-something)=0 This means that one value of n will be positive and one value will be negative. Since n must be positive, we can assume that there is only one solution here SUFFICIENT Aside: If you want to factor, you get (n+16)(n-15)=0 This means that n=-16 or n=15 n must equal 15 So the answer is D

Answer 22

``` Answer:E we know: distance=time*rate or D=T*R Given, t=3600*1/2 So, Distance=1800*R The question asked: 1800*R>5280*6 or R>17.6 ?? ``` stm1=>R>16; R can be greater/less than 17.6; INSUFF stm2=>R16

Answer 23

Originally Posted by Masuraha I think the answer should be 'A'. From the first equation we can say (x/y) > (2/7) Given that x>0 and x/y > 2/7 so y has to be positive and hence greater than 0. Considering the second equation: dividing both side by y we get -1 0. If Y to 0. INSUFF 2. Is also insuff ; for example y = -3; x = 5 we also have -(-3)

Answer 24

Good post? | x=number of managers y=number of directors To answer the question it is sufficient for us to know the ratio x:y Now, let a=average salary of managers and b=average salary of directors So, Statement 1 : a = (ax+by)/(x+y) - 5000 => y(a-b)=-5000(x+y) => y(b-a)=5000(x+y) - (1) Statement 2 : b = (ax+by)/(x+y) + 15000 => x(b-a)=15000(x+y) - (2) Looking at the right hand sides of both the equations and assuming that neither x and y is 0, we can say that none of expressions on either side of each equation equals to zero. So, dividing (2) by (1), x/y = 3 which gives us the ratio x:y which is sufficient to calculate the % of directors. Hence both statements combined together is sufficient to solve the problem. Hence (C) Reply Reply With Quote

Answer 25

Good post? | f=30! let’s find out how many tens there are in 30! In order to find out the 10s we need to find out the multiples of 5. 5, 10, 15, 20 25, 30, note that 25 = 5^2. We have 7 5s (five 5s +two 5s in 25=seven 5s) and thus 7 10s (because there are 2s there in 30! to multiply them by 5s to get 10s) 1) d can be anything from 1 up to 7 2) obviously not suff, Both suff. d should equal 7 C

Answer 26

``` Originally Posted by Bobogee In the xy-plane, oes the line with equation y=3x+2 contain the point (r,s) 1. (3r+2-5)(4r+9-5) = 0 2. (4r-6-5)(3r+2-5) = 0 SPOILER: C Hi Bobogee, ``` You have posted the wrong question. It should read: In the xy-plane, does the line with equation y=3x+2 contain the point (r,s) 1. (3r+2-s)(4r+9-s) = 0 2. (4r-6-s)(3r+2-s) = 0 Now, if (r,s) is on the line defined by the equation y=3x+2, then (r,s) must satisfy the equation y=3x+2. For example: We know that the point (5, 17) is on the line y=3x+2, because when we plug x=5 and y=17 into the equation, we get 17 = 3(5)+2 and the equation holds true. So, we can reword the target question to be "Does s = 3r + 2?" 1. (3r+2-s)(4r+9-s) = 0 From this, we know that either (3r+2-s) = 0 or (4r+9-s) = 0 If (3r+2-s) = 0 then s = 3r+2, in which case the answer to our new target question is yes If (4r+9-s) = 0 then s = 4r+9, in which case the answer to our new target question is no Since we get two different answers to the target question, statement 1 is NOT SUFFICIENT 2. (4r-6-s)(3r+2-s) = 0 From this, we know that either (4r-6-s) = 0 or (3r+2-s) = 0 If (4r-6-s)) = 0 then s = 4r-6, in which case the answer to our new target question is no If (3r+2-s) = 0 then s = 3r+2, in which case the answer to our new target question is yes Since we get two different answers to the target question, statement 2 is NOT SUFFICIENT Statements 1&2 combined: Since (3r+2-s) is the only expression common to both statements, it must be true that 3r+2-s = 0, in which case y MUST equal 3r+2 As such the answer is C

Answer 27

Originally Posted by 800 If @, # and $ represent three different positive digits such that @+# = $, what is the value of # ? 1. $=4 2. @>1 Official Answer will be posted later. Statement 1: $=4 Since @+# = $ and since the 3 digits are all different, statement 1 yields only two possible case: case a) @=1, #=3, $=4 case b) @=3, #=1, $=4 Since # can equal either 1 or 3, statement 1 is INSUFFICIENT Statement 2: @>1 No info about #, so statement 2 is INSUFFICIENT Statements 1 & 2: Statement 1 yields only 2 possibilities: a) @=1, #=3 and b) @=3, #=1 Statement 2 rules out case a) This leaves us with only 1 possibility: @=3, #=1, so statements 1 & 2 combined are SUFFICIENT and the answer is C

Answer 28

Good post? | This is a nice question, that relies on our ability to find counter-examples in order to determine sufficiency. Statement 1: Given the conditions in statement 1, there are different sets of numbers that produce different answers to the target question. If the product of the greatest and smallest of the integers in the list is positive, then here are two cases: case a: the numbers are -3, -2, -1 in which case the product is negative case b: the numbers are -3, -1 in which case the product is positive Since we get two different answers to the target question, statement 1 is NOT SUFFICIENT Statement 2: Given the conditions in statement 2, here are two cases that produce different answers to the target question. case a: the numbers are -3, 2 in which case the product is negative case b: the numbers are -3, -2 in which case the product is positive Since we get two different answers to the target question, statement 2 is NOT SUFFICIENT Statements 1 AND 2: If the product of the greatest and smallest integers is positive, then there are two possible cases to consider: case a: the largest and smallest integers are both positive, in which case all of the integers are positive, in which case their product MUST BE POSITIVE case b: the largest and smallest numbers are both negative, in which case all of the numbers are negative. Now if all of the numbers are negative AND there is an even number of integers, then the product will still be positive. Why? Well, we know that the product of any 2 negative integers will be positive. If there is an even number of negative integers, then we can pair up each negative integer with another negative integer to create a positive product. As such, the product of all of the negative integers MUST BE POSITIVE. Since both cases yield the same answer to the target question (i.e., the product is positive), statements 1 & 2 combined ARE SUFFICIENT and the answer is C

Answer 29

Originally Posted by ranjeet_1975 Is x/3 + 3/x > 2? (1) x < 3 (2) x > 1 Official Answer is SPOILER: C Why A should not be the answer? as the equation comes to (x - 3)^2 > 0 Here's my solution: Statement 1 We can show this is insufficient through counter-example. If x < 3, then: a) x could equal 1, in which case x/3 + 3/x IS greater than 2 b) x could equal -1, in which case x/3 + 3/x IS NOT greater than 2 Since statement 1 yields 2 different answers to the target question, it is not sufficient Statement 2 If x > 1, then x must be positive. This is very useful, because if we know that x is positive, we can take our target question "Is x/3 + 3/x > 2?" and multiply both sides by 3x to get a new target question " Is x^2 + 9 > 6x?" If we take our new target question and subtract 6x from both sides, we get "Is x^2 - 6x + 9 > 0? Finally, if we factor the left hand side, we get "Is (x - 3)^2 > 0?" Well, (x - 3)^2 is almost always greater than zero. The only time it is not greater than zero is when x = 3. As such, statement 2 is not sufficient. Statements 1 & 2 combined From statement 2, we reworded the target question as Is (x - 3)^2 > 0? Statement 1 tell us that x cannot equal 3 This means that (x - 3)^2 must ALWAYS be greater than 0 As such, the two statements combined are sufficient, and the answer is SPOILER: C

Answer 30

Good post? | Ans C I Total cars = 4n + 23 II Total cars = 6n + 5 Solving both we can find the total cars

Answer 31

If you consider the statement 2 as not having the brackets, you cannot arrive at the answer using statement 2 alone. so consider stmt 1: ``` X^2 < (XY+X) ==> (X^2 - X) < XY ==> equation A consider stmt 2: XY/Y^2 - Y < 1 ==> XY/Y^2 < Y ==> XY < Y^3 ==> equation B ``` merge equation A and B ==> (X^2 - X) < XY < Y^3 ==> X^2 - X < Y^3 since X and Y can take any values > 1, we cannot determine whether X < Y. Hence both the statement together are not sufficient to answer this question. Hope this helps !!!

Answer 32

Originally Posted by aspire2011 Hi All, If x < -3 then x^2 > 9. eg if x < -3, say x=-5 then x^2 = 25 satisfies x^2 > 9. When x=-5, x^3=-25 x3 < -9 hence the correct answer should be 'D' (either statement is sufficient), but Official Guide answer is 'A' - any ideas why ? Thanks in advance for your time. A is the correct answer because. x^2 >9 means x>3 or x<-3

Answer 33

B C D 1. It has been solved in another thread. 2. P+C=2/3 of 60=40 option 1 says, C>2P => P+C>3P => 40>3P So, P12, even this gives us many answer. Combining, we get a unique solution as 13. So C is the answer. 3. (x^2+1)(x+5) if x is odd, both the factors will be even, hence the product will be even. option 1 says, x is odd, sufficient to solve. option 2 says, each prime factor of x^2 is greator than 7. All the factors of x^2 are factors of x. So each prime factor of x>7, so there is no 2 as the prime factor, and all other prime nos are odd. If all the factors are odd, number x will be odd as well. Its sufficient. So answer is D

Answer 34

B C B C 1: 1993 - 1000 1994 - (1+x/100)*1000 1995 - (1+(x+y+xy/100)/100)*1000 so, 2 helps us solve it, hence answer B 2. f=30! If we analyse 30! has factors(which can be a multiple of 10), 10 20 30 and 5 15 25 as we have enough multiples of 2 to make them 10. so in total 10^7 is a factor of f option 1 says 10^d is a factor of f, d can be 0,1,2.....7 => insufficient option 2 alone is insufficient, but with 1, d>6, so d=7, so the answer is C 3. 8! is a multiple of a^n if we analyse it, other than 2 all nos occur less than 6 times in the factorisation of 8! So n=6 gives us the answer. option 1 says a^n=64, 2^6=64, 4^3=64 so its insufficient. 4. My answer is E. Actually I am stumped.

Answer 35

Good post? | Originally Posted by marianha A person invested $5,000 for two years at a certain rate, compounded annually. If he invested $5,000 again two years later, did the interest rate increase? 1). After the first two years, the total amount became $5,800 2). To the nearest hundred, the amount of the interest for the second two years is $500 SPOILER: imo c Official Answer E i think it should be E because there is no data mentioned about the interval of compounding... not sure if there is some default assumption in such problems...

Answer 36

Let us consider, G-day scenario, St 1. is not enough. As sqrt deliver -/+ value. The UNIQUE value can not be obtained from st1. (A, D OUT) St 2. by inserting value of R (R=5) from st 2. we get - ``` (Y+3)(Y-1) - (Y-2)(Y-1) = R(Y-1) => (Y^2 - Y + 3Y +3) - (Y^2 - Y - 2Y +2) = 5Y - 5 => Y^2 + 2Y + 3 - Y^2 + 3Y - 2 = 5Y - 5 => 5Y - 1= 5Y - 5 => 0 = -4 ``` the equation does not exist. B out. Combining st 1 and st 2 -- no solution. E is the pick.

Answer 37

Imo A. 1. ineq is false for x<4. So insuff. So A.

Answer 38

We have to prove whether (2^x/y)^x(2^)(x^2) < y^x Since (2^)(x^2) >0, y^x >0 So from the two statements, we have to prove that y^x >0 1). x>y We can not conclude whether y<,= or > 0 INSUFFICIENT 2). y< 0 , we don't know whether y^x y<,= or > 0 INSUFFICIENT Because if x is odd (+ve or -ve number) then y^x 0, Combining, We don't know whether x is odd or even number. INSUFFICIENT Hence E.

Answer 39

``` Good post? | Imo E. Taking number line considering both stmts: z--x--0----------y x--y-----0--z ```

Answer 40

Thanks for your replies.... I am not sure if the below answers r correct or not but this is what i have as "Official Answer's" (source - got it from some other forum):- Answer to 1st question: E Answer to 2nd question: B (edited) think the ANS is E. Explanation :- 1) (x-1)^2 = 0 Insuff 2) x^2 - 1 > 0 x >1 or x

Answer 41

C Good post? | m>0, n>0 is (m+x)/(n+x) > m/n ``` st 1) m< 1 but x could be positve or negative and if x is negative, then i can get different numbers such that one of m+x is negative and n+x is positve nad the expression of LHS is negative which canot be greater than RHS. not sufficient st 2) x>0 .. as all expression are positive, we can multiple and reduce the equation to x *(n-m) > 0 x> 0 but we dont know if n-m > 0 not sufficent combing x>0, n-m > 0 .. sufficient ``` C

Answer 42

D ``` x > 0 , is x>3?? st 1) (x-1)^2 > 4 x-1 > 2(for other expression x will come as negative x > 3 sufficient st 2) (x-2)^2 > 9 x-2 > 3(for other expression x will come as negative x > 5 sufficient ``` D

Answer 43

This is my approach.. 3x > 7y to be proved... ``` given in 1: x > y+4 substitue this in the to be proved 3y+12 > 7y 12 > 4y 3 > y ``` ie y 7(2.9) 3x> 20.3 hence i concluded 3x min is 20.3 and 7y max is 20.3 hence A is sufficient second statement 5x > 14y divide by 2 2.5x > 7y so 3x > 7y Hence 'D' is the answer.

Answer 44

Good post? | From st 1, n^2 = integer - does not confirm that n is an integer. For example, sqrt(2) is not an integer, BUT (sqrt(2))^2 is integer. SO, 1 is not sufficient. A,D out. From st 2, sqrt(n)= integer. SO, n must be an integer. ENOUGH. B is the answer.

Answer 45

from the q -- if k is < -2 then that equation breaks. 1. k-12 is odd then k can be any odd value. 2. Prime numbers can never be negative. Since 2 (or 3 respectively) are the lowest prime numbers and k-12 is a prime number, k>=14. Along with -2 it gives a negative exponent ….if 2 is a prime, then k=14, if 3 is the first prime k=15. SUFFICIENT?

Answer 46

IMO E. The unique value of P is required. From statement 1, we can say that P is 30 OR its multiple ( 30,60,90, 120 ... ). Not enough for unique value of P. A & D out. From statement 2, we can say that P is 70 OR its multiple (70, 140, 210 ...) Not enough for unique value of P. B out. Taking to gather (st1 &st 2), P is 210 OR its multiple (210, 420, 630 ...) NOT enough for unique value of P. C out. The answer is E. What is Official Answer and OE ?

Answer 47

``` Good post? | To find the roots of quadratic equation we use discriminantformula: D=b^2-4xc Then we find roots 1)(–b+sqrt D)/2x 2)(–b-sqrt D)/2x Statement 1 ) b<0 It doesn’t give us any information about absolute value of sqrtD. Both are not sufficient as well ``` IMO E

Answer 48

Let’s assume that there are X male and Y female students in the class. 0.72X and 0.8Y have applied to college. We are asked what is X/(X+Y)? 1) We know X+Y but not the proportion of X and Y. INS. 2) We know tat (X+Y)*0.75=0.72X+0.8Y We know the ratio of x to y so Suff. ``` Just for fun we can find out the ratio. 0.75Y+0.75X=0.72X+0.8Y 0.05Y=0,03X Y=0.03X/0.05=3X/5 X/(X+Y)=X/(X+3X/5)=5X/8X=5/8 B ```

Answer 49

``` Good post? | Problem 1. If mv

<0 m is negative then p should be negative D Reply Reply With Quote ```

Answer 50

For 1. from stem 1. we got the number of male employee. NO hints about female employee. (NOT sufficient to determine ratio) from stem 2. let x & y is the number of male & female employee (respectively) so, 9.8 x + 9.1 y = 9.3 (x + y) => 9.8x + 9.1y = 9.3x + 9.3y => 9.8 x-9.3x= 9.3y - 9.1 y => 0.5 x = 0.2 x. We got the ratio of male to female employee, 5:2. (sufficient) the crutial point is RATIO not the NUMBER !

Answer 51

Good post? | In the present scenario, we have to equate the bases on both sides of the equation ; then only we can compare the powers. so if the RHS base is -2 then it is not equal to LHS base i.e 2. so here, you cannot even consider the 2nd case.

Answer 52

From the statements above, if A then B but the vice-versa many not be true. i.e if john is sun burned today , it may be because he has gone to beach the day before or also because of someother reason. Hence, IMO: we cant say from the data provided if john has gone to beach yesterday

Answer 53

I think the answer is that BOTH statements in itself are sufficient D. Here is my attempt: A states |x+3| = 14 Thus, x = 14-3 = 11 B states that (x+2)^2 = 169 => (x+2)(x+2) = 13*13 or x = 11 So, both statements in itself are sufficient

Answer 54

1 out of 1 members found this post helpful. Good post? |

Answer 55

Originally Posted by novel hope this helps you--- for eg 14.25 in this 4-units place.1-tens place 2-tenths place 5- hundreds place. as per statement i units digit is greater than 1 so it has to be atleast 2 which is greater than 1.5.Hence this is sufficient. As per statement 2 tenths digit is greater than 5 it can be 0.6,1.6,2.6...hence this is insufficient. The question is little tricky if we read it cursory. The answer is A. The main question stem confirmed that the number is positive. And St 1 says it is greater than 1. that must be ltleast 2. (sufficient) St 2 says the tenth digit is greater than 5. it does not says anything about the CRUTIAL unit digit. The crutial point is understanding the difference between tens & tenth !

Answer 56

Good post? | From 1 Tallest 1/3 n avg=6ft2.5 in , So total hight = 74.5 in X n/ 3 For the rest 2/3n avg=70 in, So total hight = 70 in X 2n/3 Total hight (74.5 n/3) + (70X2 n/3) in = 215.5 n/3 in SO, avg hight of n earthlings = (215.5n/3)/ n = Suffucient From 2, total hight is given NOT the value of n. INSUFFICIENT A is the answer

Answer 57

Originally Posted by atishree question stem: r + s > 2t stmt(1) t > s ``` add both the inequalities r + s + t > 2t + s => r + s - s > 2t - t (just to explain the middle step) => r > t (1) is suffiecient ``` ``` now, question stem: r + s > 2t r> s adding both the inequalities 2 r + s > 2t + s => 2r + s - s > 2t => 2r > 2t r>t (since 2 is a positive number, no change of sign) (2 is sufficient) ``` ans : D each stmt is sufficient You can plug in any values for solving such problems: r + s > 2t, is r > t r = 5 , s = 3 , Hence t = 4 Or r=3 , s = 5 and t = 4 (1) t > s From this stem can you say t > s ? Definitely no (from r=3 , s = 5 and t = 4) (2) r > s From this stem can you say t > s ? Definitely no (from r=3 , s = 5 and t = 4) Thus you can proceed. No need of going into the basics of algebra every time.

Answer 58

Agree with E. Is a>0? st1] x^2 - 2x + a >0 x * (x-2) > -a Here, if either x=0 or x=2 then the LHS is 0 and 'a' is positive. If (x-2)0 then 0 0 ax^2 > -1 If x=0 then 'a' could be any number- positive, negative or 0. If x^2 is non-zero i.e. positive then 'a' could be a negative number, making a*x^2 a negative fraction greater than -1 or 'a' could be any positive number. Insuff. Together, If x>2 or x could be a positive value and the answer to our main question is yes.

Answer 59

Good post? | (1) W + C > 50 - Insufficient (2) C > 48 - Insufficient Combining (1) and (2), when C = 49, then according to condition 1, W + 49 > 50. This means W > 1. when C = 55, then according to condition 1, W + 55 > 50. This means W >= -4. Not clear whether W is positive or negative. Hence E.

Answer 60

(n-x) + (n-y)+ (n-z)+ (n-k) = 4n -(x+y+z+k) Stmt(1) (x+y+z+k)/4 = n or x+y+z+k=4n stmt is sufficient. stmt(2) even though x,y,z,k are consecutive you get the sum in terms of x not n so INSUFFICIENT. IMO the ans should be A.

Answer 61

Good post? | 1. (3n/m) = Integer. Insufficient as m can be 6 or 9 and n can be (14,16,18..) and (15,18,21..) respectively. 2. (13n)/m = Integer. Since m is less than 13, thus n has to be a multiple of m. Sufficient. IMO B.

Answer 62

If x and y are positive, is x^3 > y? (1) sqrt(x)> y x and y are given as positive. But they need not be integers. statement 1 can be rewritten as x >y^2 take x and y values such that they satisfy condition 1 i-e let x=25 and y=3 Then check if x^3>y . It satisfies For x=0.2 and y=0.1 such that 0.2>0.01. But x^3=0.008 is less than y=0.1. Hence insufficient. Similarly for statement 2 also take one value with integers and another value with decimals. One set will show that x^3>y and another set of values will show that x^3

Answer 63

Good post? | D. (1) Every positive integer that is NOT a perfect square has an EVEN number of distinct factors, and every positive integer that IS a perfect square has an ODD number of distinct factors. So from statement (1) we know that the answer to the question is NO. Sufficient. (2) Every positive integer that is a perfect square has an odd number of odd factors, and so the sum of the factors will be odd. So if we know that the sum of the distinct factors of N is even, then we can be sure that N is not a perfect square and that the answer to the question is NO. Sufficient.

Answer 64

answer is 360.here how it works so in problem frankie should always stand behind Joey. a) if joey is in forst position then rest 5 positions of line can be arranged in 5!=120 ways b) if joey is in second place then for the first place any poeple out of remianing 4(apart from joey and frankie) can be there..i.e in 4C1 or in 4 ways and then the remianing 3,4,5,6 places can be arranged in 4! ways.so over all ways are =4 * 4!=96 c) if joey is in third place then first 2 places can be filled in 4c2 *2! and remaing 3 places(4,5,6) in 3! ways. so total ways=4c2*2!*3!=72 ways d) if joey is in fourth place then first 3 places can be filled in 4c3 *3! and remaing 2 places(5,6) in 2! ways. so total ways=4c3*3!*2!=48 ways e) if joey is in fifth place then first 4 places can be filled in 4c4 *4! and remaing 1 places(6) in 1 way. so total ways=4c4*4!*1=24 ways so total ways=360

Answer 65

``` Good post? | z(z-4)>5 A) if z=0, 0>5 thus wrong answer choice B) if z=0, 0>5 thus wrong answer choice C) if z=0, 0>5 thus wrong answer choice D) if z=0, 0>5 thus wrong answer choice E) if z=-2, 12>5 CORREC ```

Answer 66

C k^2 + k - 2 factorize = (k+2)(k-1) for the expression to be > 0, k should not lie between -2 and +1. so when k < 1, the expression can be both > 0 and < 0 and when k > -1, the expression can be again both > 0 and < 0 but when -1 < k < 1, the expression can be -ve only

Answer 67

Without going into details, we know both 1 and 2 are insufficient by own. Solving for together. Main Equations: A) RSQ + x + TSU = 180. B) QRS + UTS = 90 By 1) QRS is an isoceles triangle with angles RQS = RSQ. Therefore, RSQ = (180 - QRS)/ 2 By 2) Similarly for triangle STU, we get: TSU = (180 - UTS)/ 2 ``` Putting in main equation A (180 - QRS)/ 2 + x + (180 - UTS)/ 2 = 180 180 - (QRS + UTS)/2 + x = 180 x = (QRS + UTS)/ 2 x = 90/ 2 = 45. ``` SPOILER: C

Answer 68

I have solved it this way: Accordin to the first statement: r + 0.2r*n = 288 ( where n is number of hours after initial 4 hrs) so n = (288/r - 1)/0.2 if r = 100, it gives the maximum value for n, it comes to be around 9.4. This 9.4 + 4 will give number of working hours greater than 10 if r = 288 (for e.g) then n=0 and number of working hours = 4, a number less than 10, so statement1 is insufficient

Answer 69

``` Originally Posted by torpedo Is y – x positive? (1) y > 0 (2) x = 1 - y st 1) dont know about x .. insuffient st 2) x = 1-y if y=1 then no insufficient combining we know that y>0 but we dont know if y< 1 or y>=1 ``` E

Answer 70

Perimeter = 360 ==> l + w = 180 -----------------(1) Stmt 2: As far as data sufficiency is concerned, it doesnt really matter how you interpret the line - "difference between length and width". If you take it as l-w = 60, then l = 60+w. Substituting this in Equation 1, you get 60 + w + w = 180 2w = 120 or w=60. Then length becomes 120. If you interpret it as w-l = 60, then w = 60+l Substituting in equation 1, l+60+l = 180 2l = 120 or l = 60. Then w = 120. Depending on your interpretation, length can be 120 or 60. But the main point is length can be calculated. That's all we need to say Stmt 2 is sufficient. The problem of whether the larger value is always the length or not does not affect the outcome of our result. For example, I can say Stmt 2 gives me l=60 and w=120 (even though that doesnt meet your definition of a rectangle, since w>l). Then I have answered the question "what is the length?", which is what the question asks. Answer : D

Answer 71

In the xy plane, does the line with equation y=3x+2 contain point (r,s)? 1. (3r+2-s)(4r+9-s)=0 2. (4r-6-s)(3r+2-s)=0 Just giving a try ... 1. either (3r+2-s) =0 or (4r+9-s) =0 or both cant be equal to zero for obvious reasons if 3r+2-s=0 then s=3r+2 so the points are (r,3r+2) putting in the stem 3r+2=3r+2 the line contains the points but if 4r+9=s then points are (r,4r+9) putting in stem 4r+9=3r+2 so its not possible hence 1 alone cant solve 2. Similarily for 2 ... 2 cant solve alone. Now using both 1 and 2 using the two eq only one of the three can be zero (3r+2-s) or (4r+9-s) or(4r-6-s) and since 3r+2-s is in both the statements it is the term that is zero and we already know that 3r+2-s satisfies eq 1 hence the answer is that both statements together give the answer regards

Answer 72

Originally Posted by redpearl Official Guide Quant Review - 2nd Edition - DS-Question 46 What is the value of x^2 - y^2 ? (x square minus y square) (1) x - y = y + 2 (2) x - y = 1 / (x+y) This is an official question. Official Answer and answer explanation is available. My query is for second condition. How can we mutiply both sides by (x+y) unless it is specified that x+y is not equal to zero or x not equal to -y. Should we assume that denominator is not equal to zero in such GMAT questions ? Statement 2 tells us that x-y = 1/(x+y). From this, we can conclude that x+y does not equal zero. If x+y did equal zero, then 1/(x+y) would be undefined in which case it couldn't equal some other value. So, knowing that 1/(x+y) equals some other value, it's safe to conclude that x-y does not equal zero. Cheers, Brent - GMAT Prep Now

Answer 73

I Choose C. ``` Given that n = 3x + 2 t = 5y + 3 Derive nt = 15xy + 9x + 10y + 6 What is the remainder when nt is divided by 15 ? ``` Here if we know that x is a multiple of 5 & y is a multiple of 3 we know that the remainder when nt is divided by 15 is 6. st1] n - 2 is a multiple of 5 since n = 3x + 2 & (n - 2) is a multiple of 5 Substract 2 from both sides and we know that 3x is a multiple of 5; hence x is a multiple of 5. But we do not know whether y is a multiple of 3. Insuff. St2] t is a multiple of 3 5y + 3 is a multiple of 3; hence 5y has to be a multiple of 3; hence y is a multiple of 3. But no info about x. Insuff. Together we know that the remainder is 6.

Answer 74

The Official Answer is E, but the reason is that there is rule for preceding tiles but there is no pattern/rule is mentioned for succeeding tiles. After combining both the statements,u can come down to GY, GR,RY, RR...but not beyond this. Source: manhattangmat

Answer 75

Statement 1. If n is not even, then (n - 1) and (n + 1) are both even. Furthermore, either (n - 1) or (n + 1) is a multiple of 4. Therefore (n - 1)(n + 1) wll be a multiple of 8. We cannot, however, say what the remainder is when (n - 1)(n + 1) is divided by 24. For example, if n = 7, then (n - 1)(n + 1) = 48 and the remainder is 0. But if n = 9, then (n - 1)(n + 1) = 80 and the remainder is 8. Insufficient. Statement 2. If n is not a multiple of 3, then either (n - 1) or (n + 1) will be a multiple of 3. Therefore (n - 1)(n + 1) is a multiple of 3. We cannot, however, say what the remainder is when (n - 1)(n + 1) is divided by 24. For example, if n = 7, then (n - 1)(n + 1) = 48 and the remainder is 0. But if n = 8, then (n - 1)(n + 1) = 63 and the remainder is 15. Insufficient. Together. From statement 1 we know that (n - 1)(n + 1) is a multiple of 8. From statement 2 we know that (n - 1)(n + 1) is a multiple of 3. Together the statements tell us that (n - 1)(n + 1) is a multiple of 8*3 = 24, and therefore the remainder must by 0. Sufficient. The correct response is C.

Answer 76

Each alone is sufficient , D. i'm considering x !=0, otherwise the fraction can be indefinite.

Answer 77

Given that ab is not equal to zero, which means a is not equal to 0 and b is not equal to 0 and the line passes through the origin. ``` From statement 1 the slope of the line k is negative,which means the the line can passes can be only in 2nd and 4th quadrant passing through the origin. The point (a,b) can be in 2nd or 4th quadrant. If it is in 2nd quadrant then b is positive.If the point lies in 4th quadrant b is negative. Hence insufficient. ``` From statement B,a

Answer 78

Statement 1: Insufficient Statement 2: Insufficient Statement 1 and 2 x + y < 73/72 = 1 + 1/72 x + y could be 1 + 1/74, 1 + 1/76, 1 + 1/200 , 1, 1/10, 1/16, 20, and so on So we can't conclude that whether x+y would be greater than, less than, or equal to 1 but we say only that x+y < 1 + 1/72 Insufficient Ans: E

Answer 79

Good post? | At a certain store, each notepad costs x dollars and each marker costs y dollars, If $10 is enough to buy 5 notepads and 3 markers, Is $10 enough to buy 4 notepads and 4 markers Instead? (1) Each notepad costs 1less than $1. (2) $10 Is enough to buy 11 notepads. here we have the equation as 5x+3y =10 where x= notepad and y = marker each notepad costs less than $1 so assume max value of x as 99/100 in this case we get x= 0.99 and y= 2.01 approx respectively now 4*0.99 + 2.01*4 > 10 hence sufficent Statement 2 also states that x can be 10/11=0.909 hence sufficent Hence Official Answer is D Any suggestions

Answer 80

(1) says a < 0 Divide both sides of the equation a · |b| < a – b by a.|b| and you get 1 > 1/|b| - 1/a for b>0 -(i) 1 > 1/|b| + 1/a for b0 & b=0 => a · |b| < a – b true and for a a · |b| < a – b false So, (2) alone is not sufficient Using (1) and (2) also don't give consistent results, hence E

Answer 81

Is |a-c| < b ? st1] The sum of any two numbers among a, b and c is greater than the third number. This is one property of any given triangle and for this principle to work each of these 3 nos. must be greater than the difference between the other 2 nos. Hence |a-c| must be greater than b in the same way as |b-c| < a or |a-b|s the Official Answer?

Answer 82

Good post? | a) x^-1/3= 1/x^1/3<1... Here, we can use either a + or a - integer to satisfy the inequality, therefore x can either be greater or less than 1. The only thing off limits are decimals. Insuff. Together, all we have determined is that x can either be + or -... Seems that both are insuff.. Ans E.

Answer 83

hjafferi answers Good post? | Imo A

Answer 84

Good post? | IMO. A. We already know that each box in J is lighter than those in shelf K. The median box is the 45th, the heaviest of shelf J.

Answer 85

Originally Posted by lawrencl What fraction of this year's graduating students at a certain college are males? (1) Of this year's graduating students, 33% of the males and 20% of the females transferred from another college. (2) Of this year's graduating students, 25% transferred from another college. Let m = the number of male graduating students and f = the number of female graduating students. From statement 1 we know that .33m and .20f are transfers. We have no way to determine the relative values of m and f. Insufficient. From statement 2 we know that .25(m + f) are transfers, but again we have no way to determine the relative values of m and f. Insufficient. ``` Putting the statements together: .33m + .20f = .25(m + f) .33m + .20f = .25m + .25f .33m - .25m = .25f - .20f .08m = .05f 8m = 5f m/f = 5/8 ``` So we know that the ratio of males to females is 5 to 8, and therefore we can say that 5/13 of the graduating students are male. The two statements together are sufficient and the correct response is C.

Answer 86

Well I think the answer is also A Here is my attempt to the explanation k – m – p is ODD? A states k and m are even and p is odd ---------------------------------------------- using the subtraction rule for even and odd Even - Even = Even Even - Odd = Odd Odd - Odd = Even We know that k-m is even and p is odd so k-m-p is also odd B states that "k, m, and p are consecutive integers." ------------------------------------------------------------ Case 1: So k,m,p can be 1,2,3 for example and so we get Odd-Odd = Even using the substraction rule stated earlier Case 2: k, m, p can be 2,3,4 for example and so we get Odd-Even = Odd So Statement 2 is INSUFFICIENT and the answer is A

Answer 87

Good post? | Hi, As per the figure given in the Set1 . the angle x denotes angle BCD. can u let me know how we can use (1) to find that. yes if the question asks angle BCA then thats = 50 so (A) can be answer. Put your questions on MBAchase GMAT video forum to get videos of solutions. http://mbachase.com/phpbb3/viewforum.php?f=57

Answer 88

From Stem : Price Paperback = $ 8 Price Hardcover = $ 25 , Number paperback >10 , P = { 11, 12, ....} P and H are integers St1> 25* Number Hardcover >=150 Number Hardcover > =6 Then , H = { 6, 7, ....} Since question requires a unique value, INSUFFICIENT St2> 8P + 25H < 260 Since Greatest Common Factor of { 8, 25, 260} is different from 1, More than one solution , INSUFFICIENT From St1 and St2 > While P and H can only be integers Plug the minimmum values for H and P at 8P + 25H < 260, Eg. ( P, H ) at 8P + 25H < 260, ( 11, 6 ) , 238 < 260, VALID ( 11, 7 ) 263 < 260, WRONG ( 12, 7 ) 271 < 260, WRONG From this point, all other combinations will NOT comply the inequality. Then , there is a unique value for H = 6 answer = C

Answer 89

a. Insufficient, since the value of (x+y) cannot be found from x^2 + y^2. b. Insufficient, since value of (x+y) cannot be found from 2xy. Considering options a and b: (x+y)^2=(x^2 + y^2 + 2xy)=2+4=6. => x+y = 6^(1/2) But we do not know if x+y is positive or negetive. Thus D(IMO).

Answer 90

IMO (C) (1) - Product of 3 digits is 30 => digits are a non-repeatable combination of {5,3,2} or {6,5,1}. This does not tell whether n is greater (e.g. 651) or smaller (e.g. 235) than 550 (2) - Sum of 3 digits is 10 = > This does not tell whether n is greater (e.g. 721) or smaller (e.g. 235) than 550 Using (1) & (2), we conclude that n consists of 3 digits in {5,3,2} => The greatest possible value of n is 532 (which is less than 550), hence answer is (C)

Answer 91

Good post? | From the question: zy < xy < 0 Implies: 1. Since zy and xy are < 0 , then if y > 0, z < 0 and x < 0, and if y < 0, then x > 0 and z > 0. 2. If y > 0 - - divide by y, no change in sign, so z < x 3. If y < 0 - - divide by y, change the sign, so z > x (typo fixed) Statement 1: z < x From #2 above, we know that y > 0. Also we know that x < 0 and z < 0. So, |x-z| + |x| = x-z + (-x) = -z Since z < 0, then -z > 0 and -z = |z| SUFFICIENT Addendum: Take an example s.t. x,z0 we know that |x-z| = x-z. Statement 2: y > 0 From #2 we know that z < x and same answer as statement 1 Put your questions on MBAchase GMAT video forum to get videos of solutions.

Answer 92

Answer B? Given c>2 (y intercept) Is c/m >3? (x intercept is less than -3 , thefore eq becomes mx=-c or -c/m3) St 1: Pass through point -2,2 (brickinthewall, i think you have the coordinates swapped) ``` Therefore y=mx+c becomes 2=-2m+c or c=2+2m i.e 2(1+m)=c Substituting back in Given 2(1+m)>2 or m>0 ``` Doesnt really help to find out if c/m>3 -Insufficient St2: ``` m=1/2 therefore is c/m>3 becomes 2c>3 or c>1.5 which is true as we already know that c>2 therefore sufficient ``` Ans: B

Answer 93

IMO C Question: Is z-y = y-x => Is x+z = 2y This is a yes/no question 1 alone: (x + y + z)/3 < 4 (x + y +z) < 12 Insufficient 2 alone: Median of {x, y, z} = y Median of the set {x, y, z, 4}, should either be (x+y)/2 OR (y+4)/2 Since Median of the set {x, y, z, 4} < Median of {x, y, z} Either (y+4)/2 < y OR (x+y)/2 < y, which is evident from the information already provided (x < y < z) and is insufficient. Combining (1) and (2) Median of the set {x, y, z, 4} must be (y+4)/2 Therefore, we obtain the following two conditions: y> 4 (from 2) (x + y +z) < 12 (from 2) Minimum value of y = 5 And so x+z < 7 Thus we obtain (X+z) is always less than 2y Ans: No Sufficient (C) However, I may still be wrong, considering the fact that I am solving DS questions at the end of an extremely hectic day What is the Official Answer

Answer 94

Good post? | 1) Since y could be positive or negative, we don't know whether to flip the inequality sign when cross multiplying (3y * 1). If y is negative five then we are multiplying my -15, so we would need to flip the sign. But if y is positive 5 we need to keep the sign the same way. 2) Subtract p from each side. -x < -y. Multiply by negative one, flip the sign. x > y. SUFFICIENT. In statement two the guess work of positive/negative is taken out. B

Answer 95

Good post? | Did you mean (n+2)^n -- that is, "n plus two, raised to the nth power" -- in the second equation? If so, I think the answer is B, with the only value for n being 3. 125 is 5^3, and since natural numbers have unique factorization that is the only way it can be expressed as a product of natural numbers. n can't be negative because then the right hand side would be 1/something. This breaks down into two equations, n+2=5 and n=3. Therefore n=3 is the only solution. Any other thoughts?

Answer 96

1. Each of Mary's n friends donates $500 --> 500n 2. Then, each of the n friends gets n others to donate $500 --> n*n*500=500n^2 Total donations = 500(n^2+n) ``` Second statement: total = $120,000 500(n^2+n) = 120,000 n^2+n = 240 n^2+n-240 = 0 (n+ 16)(n- 15)=0 n can't be negative, so n=15. ``` First statement: If the first people donate 1/16 of the total amount, then the rest donate 15/16 First people: 500n (from above) = 1/16 * (total) --> total = (500n) * 16 Rest: 500n^2 = 15/16 * total --> total= (500n^2)*16/15 500n*16 = 500n^2 * 16/15 n = n^2/15 1 = n/15 15=n Either condition is sufficient. And no fractions in the answer.

Answer 97

from the q -- if k is < -2 then that quation breaks. 1. k-12 is odd then k can be any odd value. 2. Prime numbers can never be negative. but k-12 can be 2,3,5. So can be odd except 14. For all values it satisfy the Q step quation. Now combining as well k can be many posive odd value and one even value. For everything Q stem is valid. So E.

Answer 98

Good post? | | it should be E. Key is to take the numbers 0<1

Answer 99

A The number m is 9999....(x times)8 e.g 999998 or something like that. So sum of digits = 9*x +8 =116 So x=12 So m is 9999999999998 (12 times 9 followed by 8) m+2 = 10^13 So n=13

Answer 100

Good post? | | Any successive number set's median=mean. So its A.

Answer 101

Let's See The question is asking is distance between x-y is greater than distance of x from 0 minus distance of y from 0 the answer will be yes or no. When this kind of scenario possible? Draw a number line if x and y both positive and y 0 the answer is NO. Choose B HTH as I am tackling GMAT probs after 9 long months and almost an equal number of dings from B-Schools :-P

Answer 102

A. x=10a+b and y=10b+a then x-y=9a-9b which is divisable by 9 B. x= 53, y=35 then x-y =18 which is divisable by 9 x= 64, y=35 then x-y =29 which is NOT divisable by 9 So ans is A.

Answer 103

stmt (1) gives these possible values (considering x + 3 >= 20) i.e. 17, 18, 19... - insufficient stmt(2) gives these possible values (considering x - 3 <15) i.e. 15, 16 , 17 combining both there is only one possible solution i.e. 17 employees

Answer 104

i think you got h = 12 as wrong first of all.... See ...here ..OP< PQ sTEm 1 cordinates of O = (o,o) , P (6,8) plot this two points on graph... So what you get is ......right angled triangle .....with ....op = 10 ... h = 8 (along axis )... Now if we have pq > op then pq should be > 6 so we get ..min area = 1/2 * 8 * (6 + 7 ) > 48...so A is SUFF

Answer 105

I Choose C. Is a> 0 ? St1] a^3-a < 0 a (a^2 - 1) < 0 For this to hold good the following 2 conditions have to be met 1] a > 0 while a^2 -1 < 0 if a^2 - 1 < 0 then a < 1 while a>0; possible and our answer to the main question is yes. 2] a < 0 while a^2 - 1 > 0 Is a^2 -1 > 0 then a1 but we have assumed that a 0 1> a^2 this means that a is a positive or negative proper fraction. Insuff. Together we know 'a' has to be a positive fraction and our answer to the question is yes. Suff.

Answer 106

``` a) a^n = 64 8! = n(a^n) = n* 64 n = 630 So, a^630 = 64 Therefore, a = (64)^ (1/630) .... which will ultimately come down to square root of a number which in turn will have 2 values (one +ve and another -ve). ``` So, we cant have a definite answer b) n = 64 8! = 64 (a^64) a^64 = 630 Therefore, a = (630)^(1/64) ..which will ultimately come down to square root of a number which in turn will have 2 values (one +ve and another -ve). Here also we cant have a definite answer Combining both a) & b) a^n = 64 & n = 64 we'll have similar scenarios as above. Besides, n(a^n) = 64* 64 which certainly not equal to 8! (so violates the given condition) Thus, answer is E. Thanks and regards, Vivek

Answer 107

Re: Equations and Unknowns -I 1) P = 2A + L => pablo received the maximum amount. sufficient. ``` 2) P - L = 2A => P = L + 2A same as (1). sufficient. ``` hence answer is "each statement itself is sufficient" , D.

Answer 108

Good post? | I think it is C. 1] can have multiple values of s and hence this is not sufficient for the answer. 2] Insufficient because all this tells us that r = -s. [equation is: t-r=t-(-s)] Here if s is on the left side of 0, then r comes on the right side, which is incosistent with teh diagram shown. If s is at 0 ie s=0, then r =0. possible as question stem does not tell us anything definite abt r and s not being 0. Combining the two gives us a definite location of s. and we know that r = -s ==> r and s are equidistant from 0. Hence C

Answer 109

Official Answer is accurate in my opinion. 1- If X^2 is fraction, then y is fraction as well. 2- If X>Y then X should be 0.8 kind of number so X can be bigger that Y. Thus fraction.

Answer 110

In the xy plane, does the line with equation y=3x+2 contain point (r,s)? 1. (3r+2-s)(4r+9-s)=0 2. (4r-6-s)(3r+2-s)=0 Just giving a try ... 1. either (3r+2-s) =0 or (4r+9-s) =0 or both cant be equal to zero for obvious reasons if 3r+2-s=0 then s=3r+2 so the points are (r,3r+2) putting in the stem 3r+2=3r+2 the line contains the points but if 4r+9=s then points are (r,4r+9) putting in stem 4r+9=3r+2 so its not possible hence 1 alone cant solve 2. Similarily for 2 ... 2 cant solve alone. Now using both 1 and 2 using the two eq only one of the three can be zero (3r+2-s) or (4r+9-s) or(4r-6-s) and since 3r+2-s is in both the statements it is the term that is zero and we already know that 3r+2-s satisfies eq 1 hence the answer is that both statements together give the answer regards

Answer 111

Good post? | ds 21 What was Janet’s score on the fourth physics test she took? (1) Her score on the fourth test was 12 points higher than her average (arithmetic mean) score on the first three tests she took. (2) Her score on the fourth test raised her average (arithmetic mean) test score from 87 to 90. Can anyone help with this? (I) Her score on the fourth test was 12 points higher than her average score on the first three tests she took. If Janet scored 10 points on each of the first 3 tests, then the average for the first 3 tests would be 30/3 = 10. Therefore, she would have scored 10 + 12 = 22 points on the fourth test. BUT if Janet scored 15 points on each of the first three tests, the average of the first 3 tests would be 45/3 = 15, and she would have scored 15 + 12 = 27 points on the fourth test. Since we have no idea what the average of the first three tests is, this is INSUFFICIENT. (2) Her score on the fourth test raised her average (arithmetic mean) test score from 87 to 90. The formula for Average is: A = S/N. Where A is the average, S is the sum of all the terms, and N is the number of terms. Here, 87 = S/3 or Sum of Terms = 3 * 87 = 261. We're told that her fourth scored raised her average to 90. We can call her fourth score "X". Using the Average formula once again: 90 = new average 261 + x = sum of terms 4 = # of terms. From here, 90 = 261 +x/4 or 4 * 90 = 261 + X. So 360 = 261 + x, and x = 360-261 or X = 99 which is the score on the fourth physics test she took. SUFFICIENT. Of course, we don't have to solve this problem, we just need to recognize that B is sufficient. Hope that was clear.

Answer 112

Here are the solutions: 1. First Find out the sum for each class X= ( 54 * 70 ) = 3780 (This is a simple multiplication) Y = (73 * 40 ) + (73 *1) = 2993 (This way makes my life easier !!) Z = (30 * 76 ) + (76 *2) = 2432 The total is 9205 and the total number of students is 127 Now, I approximated that to 9000 / 125 = 72 Looking at the weights 32 is the least of the weights and 54 and 41 have the Avgs 70 and 73 respectively...hence my answer is B 2. Assume 5 digits : x, y , z, w, and s Let x be at the 10000th place, y at 100th, z at 100th, w at 10th and s at the units place. Hence the number will be 10000x+1000y+100z+10w+s The reflection as defined = 10000s+1000w+100z+10y+x The difference between the two = 9999x +990y+0-990w-9999s That is = 9999(x-s)+990 (y-w) That is = 99(101(x-s) + 10 (y-w)) Hence the difference will always be divisible by factors of 99 Here the choices have only 9 Hence the answer is 9 3. The answer has to be 53 , I don't know how 40 can be right. The ratio d/r = 5/6 and r/e = 8/ 3 hence the common ratio = d:r:e = 20: 24:9 Hence minimum is 53 4. The two trains are travelling toward each other. Let t be the number of hours travelled by each when they meet. The distance covered by both the trains when they meet will be: 30t + 40t = 120 We need to find the distance between them 1 hour before they meet that is in t-1 hours Therefore the distance they cover in t-1 hours= 30(t-1) + 40(t-1) = 30t+40t -70 = 120 - 70 =50 miles They cover 50 miles before they meet , hence the distance has to be 70 miles. The longer solution is: 30t +40t = 120 t= 12/7 Train 1 at 30 mph covers 360/7 miles in 12/7 hours Train 2 at 40 mph covers 480/7 miles in 12/7 hours in t-1 = 12/7-1 = 5/7 hours: Train 1 at 30 mph covers 150/7 miles in 5/7 hours Train 2 at 40 mph covers 200/7 miles in 5/7 hours Together they cover 350/7= 50 miles. Hence the difference is 120 - 50 = 70 miles Hope this helps!! Sunetra

Answer 113

(1) - Insuff -- no information on highest score in each class (2) - Insuff -- no information on highest and lowest scores (1) & (2) together still Insuff --- highest score in each class can be different So 'E'

Answer 114

Good post? | 981,495 a prime number? ? Use these Divisibility Tricks Divisibility tricks (help detect if a long number is prime): How do you know if a number is divisible by 3? Sum of the digits is divisible by 3 How do you know if a number is divisible by 4? Last two digits are divisible by 4 How do you know if a number is divisible by 5? Last digit is either a 0 or a 5 How do you know if a number is divisible by 6? Number is even and divisible by 3 How do you know if a number is divisible by 7? Number divides evenly by 7 (there is no shortcut) How do you know if a number is divisible by 9? Sum of the digits is divisible by 9 Is 47 a prime number? Yes Is 117 a prime number? No (divisible by 3, 9, etc.) Is 981,495 a prime number? No (divisible by 3, 5, etc.)

Answer 115

x^3-3*X^2+2X X(X-1)(X-2) So a number (number-1) (number-2) If number odd then odd*even*odd; May or may not be devisible Number even definitely divisible Now A -> X devisible by 4 so Yes B -> X even so Yes Hence D

Answer 116

B Let F be total tickets for F seminar G be total tickets for G seminar X be total tickets Therefore X = F + G Also from the question F/G=2 I.e F =2G Total percentage of tickets purchased = total tickets purchased / total tickets From statement 2 = (60%F + 80%G)/(F+G) Put F = 2G, we get answer as G will cancel itself in the ratio.

Answer 117

wow, Ian the finite set can be normally distributed, and it's not that unusual for such sets to be symmetric around their means (bell-shaped). The z-score and other stats are quite handy for navigating the distributed values. Yet it says here 82+ scores fall in the interval 98-100%. If we follow you then math is not useful (?), as the whole science of math is about making assumptions and basing the premises on such assumptions. The infinite sets can be tested for normally distributed properties - we need to move into calculus and apply differentials, derivatives to find delta of increasing/decreasing function(s) -> +- infinity here it provides normal distribution property and suggests the set is dispersed within 6 st.dev-s. By using st(2) we know that 10/500=2% which corresponds to one st.dev. Hence, we find 1 st.dev from the other 2 st.dev (82-72)/2=5. The question says below 16% will retest and this means 2%+14% will retest or 76-5-5=66 score (students earning less than 66 score will have to retest). The 2nd statement is Sufficient and the 1st is Not. However, I agree with Ian, GMAT won't ask normal distribution and other probability distributions because the questions involving stat parameters (t,f,z-score, chi square) may appear very ambiguous. I have already seen that GMAT only asks questions which are 100% accurate and devoid of ambiguity. This one is very difficult to prove as we are dealing with discrete values (the number of students) and continuous values (scores). The values corresponding to students and scores should be precisely estimated within their data sets not to look insensible.

Answer 118

``` grandh01 wrote: The only contents of a parcel are 25 photographs and 30 negatives. What is the total weight, in ounces, of the parcel’s contents? (1) The weight of each photograph is 3 times the weight of each negative. (2) The total weight of 1 of the photographs and 2 of the negatives is 1/3 ounce. ``` OA is C Let us assume that the weight of photograph = P, and the weight of negative = N We have to find the value of 25P + 30N (1) The weight of each photograph is 3 times the weight of each negative. P = 3N, but we don't know the values of P or N; NOT sufficient. (2) The total weight of 1 of the photographs and 2 of the negatives is 1/3 ounce. P + 2N = 1/3 or 3P + 6N = 1 25P + 30N = 5(5P + 6N) = 5(2P + 3P + 6N) = 5(2P + 1); NOT sufficient. Combining (1) and (2), we have 2 equations and 2 variables, so we can find the values of P and N, and hence the value of 25P + 30N; SUFFICIENT. The correct answer is C.

Answer 119

If x + y >0, is x > |y|? If x+y>0 then either x, y or both will be positive. (at least one will positive and greater than zero) (1) x > y If x >y then x is positive. Y can be either negative or positive If y>0 then x would have to be greater than absolute value of y for example x is 36 whereas y is 15. If yabs y (2) y <0 then x would be positive and absolute value of y cannot be less than x Therefore both equations are sufficient on their own. OA is D

Answer 120

If this were a real GMAT question (it's clearly not) it would ask "is x an even *integer*" to make it clear that the question is not only asking if x is even, but is also asking whether x need be an integer at all. If you know in advance that x is an integer, the answer is D here. If x need not be an integer, the answer is C, since many fractions will satisfy each individual statement. In any case, you just about got to the answer with your approach, and I thought your approach was interesting, so we can use it to finish answering the question: prat_agl wrote: Taking them together, if 3x is even and 5x is also even then their difference will also be even . i.e, 2x is even Yes, if 5x and 3x are even integers, then their difference, 2x, must also be an even integer. Notice we now know that 3x and 2x are both even integers, so their difference must also be an even integer, which means 3x - 2x = x is an even integer. So the two statements together are sufficient. _________________ Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details. private GMAT tutor in Toronto IMO D Rule is Even *Even = Even Odd * Even = Even Odd *Odd = Odd Since both 3 and 5 are odd and their multiplication is even therefore x old have to be even. Hence both statements are sufficient.

Answer 121

IMO The answer is A I Since both triangles have the same angles, they are similar. If ABC has 4 times the area of KLM, each side of ABC is twice the size of KLM. So the hyp of ABC = 2*10 = 20 So I is sufficient II Knowing one side and the hypt I can get the area of KLM sqrt(10^2-6^2) = 8. Thus, I can get the area of KLM = 24 Area of triangle ABC = 24*4 = 96 But I can't do much else with this.. So II is insufficient So answer is A

Answer 122

consider option 1 (The sum of these integers is positive but smaller than 20) when we take 2,4,6,8 (4 consecutive positive integers) the sum is 20. but we need sum to be less than 20 and positive. so the numbers we are left are (0,2,4,6) , (-2,0,2,4). if we take (-4,-2,0,2) the sum is not positive. with these two sets the product is 0, which means it is not positive. hence it can be answered with option 1 alone. consider option 2 (The product of the smallest two of these integers is positive) take (2,4,6,8) and (-4,-2,0,2). the product of smallest two of these sets is positive. but the option (2,4,6,8) product is positive. where as (-4,-2,0,2) product is 0 (not positive). so we are not able to answer with option 2. Hence, it is A p

Answer 123

Mon Aug 27, 2012 8:10 pmThankQuote mehaksal wrote: Jane gave Karen a 5 m head start in a 100 race and Jane was beaten by 0.25m. In how many meters more would Jane have overtaken Karen? One approach is to treat this as a problem about PROPORTIONS. When Karen finishes the 100-meter race, Jane is behind by .25 meters. Thus, the distance traveled by Jane = 100 - .25 = 99.75 meters. Since Jane give Karen a head start of 5 meters and finishes the race .25 meters behind Karen, Jane catches up by 5 - .25 = 4.75 meters. Thus, for every 99.75 meters that Jane travels, she catches up by 4.75 meters. To determine how much farther Jane must travel to catch up by another .25 meters, set up and solve the following proportion: 4.75 meters/99.75 meters = .25 meters/x meters x = 5.25. Thus, to catch up to Karen, Jane must travel another 5.25 meters.

Answer 124

E Went about it by plugging numbers in the equations Statement no 1 N^2-n is not divisible by 4 Plug in different numbers In case of 2 it is even but n^2-n yields 2 which is not a multiple of 4 Similarly plug in 4 and n^2-n yields 13 which is a multiple of 4 Therefore n can/cannot be even Whereas -3 is odd but n^2-n yields 12 which is a multiple of 4 Similarly 3 is odd but n^2-n yields 6 which is not a multiple of 4 Therefore n can/cannot be odd Hence statement 1 alone must not be sufficient as conflict Statement 2 Multiples of 3: 3,6,9,12..... Which are odd as well as even Therefore statement 2 is also insufficient Combining the two statements Plug in 3,6,9,12 in n^2-n we he get that none of the multiples of 3 (both even and odd) result in answers that are multiples of 4. Hence both statements together are insufficient. Answer must be E

Answer 125

B whenever martin has a restaurant bill with an amount between $10 and $99, he calculates the dollar amount of the tip as 2 times the tens digit of the amount of his bill. if the amount of martin's bill was between $10 and $99, was the tip calculates by martin on this bill greater than 15 percent of the amount of the bill? ``` Since the amount on the bill is a two digit number, it can be represented in the form 10A + B. Then the amount of the tip = 2A. Is 2A/(10A+B) > 15/100 ? i.e. 200A > 150A + 15B ? i.e. 50A > 15B ? i.e 10A>3B ? ``` We can now rephrase the question to - If the amount on the bill is 10A + B and the tip is 2A, Is 10A > 3B ? Quote: 1 the amount of the bill was between $15 and $50. If 10A + B = 16, then A = 1 and B = 6. 10*1 < 3*6 (10A < 3B) If 10A + B = 31, then A = 2 and B = 1. 10*3 > 3*1 (10A > 3B) Since we don't have a definite answer, statement I is insufficient to answer the question. Quote: 2 the tip calculated by martin was $8. Tip = 2A = 8. So the value of A is 4 10A = 40 is always greater than 3B(3B = 0 if B = 0 and 3B = 27 if B = 9). So, statement II is sufficient to answer the question Answer B

Answer 126

E Interesting! If a^2 * b^2 * c^3 = 4500. Is b+c = 7? Quote: (1) a, b and c are positive integers a^2 * b^2 * c^3 = 4500 Case 1: a^2 * b^2 * c^3 = 2^2 * 3^2 * 5^3 a = 2, b = 3 and c = 5. Is b+c=7 ? No! b+c = 3+5 = 8. Case 2: a^2 * b^2 * c^3 = 3^2 * 2^2 * 5^3 a = 3, b = 2 and c = 5. Is b+c=7 ? Yes! b+c = 2+5 = 7. I am sure that you might have considered the above cases BUT(and a big one) did you consider the case where the value of a or b is 1? Case 3: a^2 * b^2 * c^3 = 6^2 * 1^2 * 5^3 a = 6, b = 1 and c = 5. Is b+c=7 ? No! b+c = 1+5 = 6. Case 3: a^2 * b^2 * c^3 = 1^2 * 6^2 * 5^3 a = 1, b = 6 and c = 5. Is b+c=7 ? No! b+c = 6+5 = 11. Since we don't have a definite answer, statement I is insufficient to answer the question. Quote: (2) a > b So? Irrelevant. Since we don't have a definite answer, statement II is insufficient to answer the question. Quote: ](1) a, b and c are positive integers PLUS (2) a > b The value of a is greater than b in two cases, case 2 and case 3: Case 2: a^2 * b^2 * c^3 = 3^2 * 2^2 * 5^3 a = 3, b = 2 and c = 5.(a>b) Is b+c=7 ? Yes! b+c = 2+5 = 7. Case 3: a^2 * b^2 * c^3 = 6^2 * 1^2 * 5^3 a = 6, b = 1 and c = 5.(a>b) Is b+c=7 ? No! b+c = 1+5 = 6. Since we don't have a definite answer, [Statement I + Statement II]-combined isn't sufficient to answer the question. Answer E

Answer 127

D gmat25 wrote: A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n? (1) The probability that the two bulbs to be drawn will be defective is 1/15. (2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15. OA given is Op D, see in for both the Op'ns i made separate eq'ns and for Op A i got the solution but m not able to get the solution for Op B, so if someone can solve Op B that will be really helpful. Statement 1 gives something away: since P(both bulbs are defective) > 0, there must be at least 2 defective bulbs. Since the question stems indicates that nt work, since it will decrease all the numerators. Thus, n=3. Sufficient. The correct answer is D.

Answer 128

A shrey2287 wrote: 122. An=An-2+11, n>2. Is 633 in the sequence? 1) A1=39 2) A2=43 OA : A Since An=An-2+11, it represents an AP with d=11/2 = 5.5(A3=A1+11, A5=A3+11 and so on, So A2 = A1+5.5 & A3=A2+5.5), nth term of an A.P sequence is given by N=a+(n-1)d, here N = 633 (we want to find whether is in the series or not) Statement1: a=39, N=633, d=5.5 633=39(n-1)5.5 5.5(n-1)=594 (n-1)=108 n=109 So 633 is 109th term of the series if A1=39 Sufficient ``` Statement2: 43=39, N=633, d=5.5 633=43(n-1)5.5 5.5(n-1)=590 (n-1)=107.272727....... n=108.2727..... since n is a number, it can't be in decimal. Insufficient ``` Hence, A

Answer 129

E IMO E Plug in 1 and 3 in statement 1. Both values satisfy. So no concrete answer can be obtained. Therefore equation 1 is not satisfactory. Factories statement 2. Two factors 1 and 3. Again no single answer? Hence statement 2 is not satisfactory. Combing both equations no additional info can be obtained. Hence E must be the answer

Answer 130

A The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is the cost of 7 chocolates, 11 biscuits and 9 ice creams? (A) The cost of 5 chocolates, 7 biscuits and 3 ice creams is 217. (B) The cost of 4 chocolates, 1 biscuit and 3 ice creams is 141. We need to find whether linear combination of two equations can result in a third equation or not. By linear combination I mean simple arithmetic operations that can done on two equations, like addition/subtraction of two equations, multiplication/division of an equation by a constant. Please note that this the methodical solution which uses some higher level mathematical concept (namely linear independence). Solving proper GMAT question does not need such concepts. Given: Say cost of one chocolate = a, cost of one biscuit = b and cost of one ice-cream = c. Then (3a + 5b + 5c) = 195 => (3a + 5b + 5c - 195) = 0. We have to find (7a + 11b + 9c) = ?. Say, (7a + 11b + 9c) = x => (7a + 11b + 9c - x) = 0 Statement 1: (5a + 7b + 3c) = 217 => (5a + 7b + 3c - 217) = 0 Now the question is whether we can complete the required equation by linearly combining (3a + 5b + 5c) = 195 and (5a + 7b + 3c) = 217. If we can, then the following relation must hold for some constant m and n, m*(3a + 5b + 5c - 195) + n*(5a + 7b + 3c - 217) = (7a + 11b + 9c -x) As a cannot contribute in b, b in c and so on, m and n must follow the following relations, (the relations are obtained by equating the coefficients of a, b and c) 1. 3m + 5n = 7 2. 5m + 7n = 11 3. 5m + 3n = 9 If there exists a set of value for m and n for which all the three relations are satisfied, then we can easily find x. In fact x will be equal to (195m + 217m). For this case we can find such a set of value for m and n: m = 3/2 and n = 1/2 Sufficient. Statement 1: (4a + b + 3c) = 141 => (4a + b + 3c - 141) = 0 Applying same procedure as above, m and n must satisfy the following relations, 1. 3m + 4n = 7 2. 5m + n = 11 3. 5m + 3n = 9 Try to solve these three relations, you'll find there is no such (m, n) for which all the three relations are satisfied. Thus we cannot complete the required equation. Not sufficient. Correct answer is A.

Answer 131

E X can be equal to zero

Answer 132

D take present age as x. ``` considering option 1: age of carol 6 year`s ago = x-6 half her present age = x/2 so, x/2 = x-6 -> x = 12 can be answered with A alone ``` ``` considering option 2: age of carol 3 years from now = x + 3 age of carol 7 years ago = x -7 from the statement, x+3 = 3*(x-7) -> x =12 so it can be answered with B alone ``` hence, it is D

Answer 133

B IMO B Statement 1 plug in 1,2. This does not provide any concrete info. Statement 1 is not satisfactory In statement 2lug in the lowest value y can have I.e 1. This results in the expression value in excess of 1000. Hence statement 1 must be correct.

Answer 134

A Yeap answer is A i plugged in with numbers sqrt 36x : i tried with 4 16 36 and found that stmt 1 is true sqrt 3x+4: i plugged in 4 16 36 and also 20 stmt 2 not sufficient.

Answer 135

D Answer is 0

Answer 136

C PGMAT wrote: If x and y are integers, is X > Y? 1) x + y > 0 2) y^x < 0 Can some one explain this with examples? C 1. x=3, y=1 => x+y>0. Is X > Y ? YES x=1, y=2 => x+y>0. Is X > Y ? NO INSUFFICIENT 2. (-3)^-1 Y ? YES (-1)^-3 Y ? NO INSUFFICIENT Combining 1 and 2: x+y>0 => At least one of them has to be positive. If we consider both as positive, y^x < 0 cant be possible. So, one has to positive and one has to be negative. Consider: X -ve and Y to be +ve: y^x < 0 can't be possible. So, Y is negative and X is positive. => X>Y. (-2)^3 0 => Is X > Y ? YES SUFFICIENT

Answer 137

C wrote: Are x and y both positive? (1) 2x-2y=1 (2) x/y>1 c (1) 2x - 2y = 1 x and y both positive means that point (x, y) is in the first quadrant. 2x - 2y = 1 implies y = x - 1/2, and it's an equation of a line and the question asks whether this line is only in first quadrant, which is not possible; NOT sufficient. (2) x/y > 1 x and y have the same sign. But we don't know whether they are both positive or both negative; NOT sufficient. Combining (1) and (2), 2x - 2y = 1 implies x = y + 1/2 x/y > 1 implies (x - y)/y > 0 Substituting the value of x, 1/y > 0 implies y is positive and since x = y + 1/2, so x is also positive; SUFFICIENT. The correct answer is C.

Answer 138

IMO A Statement 1: even - even - odd is odd. Hence statement 1 is sufficient Statement 2: what if two integers are odd try putting in 5,6 and 7 result would be even. Also I try to put in -1,0,1 to check if the rule holds. In this case it does not. Hence statement 2 is insufficient

Answer 139

E wrote: Is quadrilateral RSTV a rectangle? (1) The measure of ∠RST is 90 degrees (2) The measure of ∠TVR is 90 degrees Each statement alone is clearly insufficient. The question really comes down to combining them. Based on the ordering of the letters, we know that RST and TVR are opposite angles. We certainly could draw a rectangle based on that information, but could we draw any other shape? So, we really need to answer: if the opposite angles in a quadrilateral are both 90 degrees, does the shape have to be a rectangle? The answer turns out to be no. It's tough to demonstrate that without drawing a diagram, but picture two right angle triangles with the same hypotenuse but different legs. We can "glue" the triangles together to form a quadrilateral and, because the legs are different lengths, only the opposite angles will both be 90 degrees. For example, if our triangles were: 5, 5root3, 10 (30/60/90 degree angles) and 6, 8, 10 (not 30/60/90 degree angles) We could glue them together on the 10 side to create a quadrilateral and only the two opposite angles would be 90 degrees (and the sides would be 5, 5root3, 6 and 8, clearly not a rectangle). So, even after combination our shape may or may not be a rectangle: choose E.

Answer 140

E grandh01 wrote: Product of xy>27? 1) 2y OA is E I thought the answer was C, because used together we can say no xy is not greater than 27 (1) 2 < x < 5 but we have no info about y; NOT sufficient. (1) 6 > y but no info on x; NOT sufficient. Combining (1) and (2), 2 < x < 5 and y < 6 If x = 3, y = 1, then xy = 3. Here xy < 27 If x = 4.9, y = 5.9, then xy = 28.9 > 27 No definite answer; NOT sufficient. The correct answer is E.

Answer 141

B Good post? | IMO B ac/bd = a/d * c/b > c/b therefore a/d > 1. since a,b,c,d are postive integers therefore the expression must be greater than zero. in statment 2 a > d therefore a/d must be greater than 1 Hence statement II is sufficient. Hence B

Answer 142

Good post? | IMO E too Set No 14 says that the Official Answer is C. but i like E best....

Answer 143

IMO D I got the answer. there was a misprint in the question. read the equation as Root {(2*root 63) + 2/(8 + (3*root 7))}.. the mistake is the positive sign. the question has mistakently been written as negative. the correct question would be with the positive sign here is how i solved it ....eventually realize that 3* root 7 = root 63 now let root 63 be A then the equation would read root {2*{A+1/(8+A)} ....taking 2 common root {2*{[A*(8+A)+1]/(8+A)} root {2*{[8A+A^2 +1]/(8+A) ``` now put A = 3*root 7 then 8A+A^2-1 = 8*3*root 7 + 9*7 +1 = 8*3*root 7 + 63 +1 = 8*3*root7 +64 take 8 common 8(8+ 3* root 7) put 3*root 7 = A then = 8(8+A) ``` put the expression back in the main equation we get root {2*{8*[8+A]/[8+A]}} = root 8*2 = root 16 = 4 hence D is the answer

Answer 144

A The answer is A The whole relation between x^2 and y^2 is a bit misleading. Question - Is x>y Statement 1 - x >|y| which means that x is positive and greater than y, irrespective of y being negative or positive. Sufficient Statement 2 - |x|>y. Just shows that mod x is greater than y. x can be either negative and less than y or positive and greater than y. Insufficient I don't actually see the relation between x^2 and y^2 playing a role in solving this question. Regards Anup

Answer 145

E Night reader wrote: Is q positive? (1) qp^2 is not negative (2) q^2 is positive (2) q^2 is positive q can be both negative and positive. Not Sufficient. (1) qp^2 is not negative When p = 0 then q can be both -ve or +ve. Not Sufficient. Combine both: No Use as when p^2 = 0 then p = 0 and we cannot say anything about q. IMO E

Answer 146

E 1) A cant send 41, for if A sent 41, A being second highest, the highest would have been greater than 41 and we would have > 75 so the highest is 41. we have 75-41=34 reps to be distributed between A and the other 4 countries. a way to do this would be arrange them in descending order as no countries have the same number of reps. say A =10, we can have the arrangements as 41+ 10+9+8+5+2=75 say A= 9, we can have the arrangements as 41+9+8+7+6+4=75 not sufficient as A can be greater or less than 10. 2) A<12 say A=11, we can have 30+11+10+9+8+7=75 say A=9, we can have 40+9+8+7+6+5=75 not sufficient as A can be greater or less than 10 taking both statements together we can have the same examples used for statement 1. not sufficient hence E

Answer 147

The question should probably say "distinct prime factors". Using both statements, it could be that k = (2)(3)(5)(7)(11), and p = (2)(3)(5)(7)(13), in which case m would have more than 5 prime factors (it would be divisible by the six primes 2, 3, 5, 7, 11 and 13). Or it could be that the prime factors of k and p are exactly identical, so say k = (2)(3)(5)(7)(11) and p = (2^2)(3)(5)(7)(11), in which case m has only five prime factors. All we can say for sure is that the number of distinct prime factors of m is somewhere between 5 and 10, inclusive. So the answer is E. _________________ Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details. private GMAT tutor in Toronto

Answer 148

D wrote: Is x^2 > 4x - 3y? (1) y > 0 and x > 4 (2) 4x - 3y = -1 Target question: Is x^2 > 4x - 3y? Statement 1: y > 0 and x > 4 To determine whether or not this statement is sufficient, it may be useful to first rephrase the target question. First move all all terms to one side to get: Is x^2 - 4x + 3y > 0? Factor the first 2 terms to get: Is x(x-4) + 3y > 0? At this point, we'll use the given information. If y>0, then 3y>0 (in other words 3y is positive) If x>4, then (x-4)>0 and x>0 (in other words, x-4 and x are both positive) If 3y, x-4 and x are all positive, then x(x-4) + 3y must be greater than 0 So, statement 1 is SUFFICIENT Statement 2: 4x - 3y = -1 This one is a little more straightforward. Here, we'll use the original target question: Is x^2 > 4x - 3y? If 4x - 3y = -1, we'll replace take the target question and replace 4x - 3y with -1 to get: Is x^2 > -1 Since x^2 must be greater than or equal to zero, it must be the case that x^2 > -1 As such, statement 2 is SUFFICIENT, and the answer is D Cheers, Brent

Answer 149

shrey2287 wrote: At a certain store, each notepad costs x dollars and each marker costs y dollars. If $10 is enough to buy 5 notepads and 3 markers, is $10 enough to buy 4 notepads and 4 markers instead? Since $10 is enough to buy 5 notepads at $x each and 3 markers at $y each, we get: 5x + 3y ≤ 10. The question stem asks whether $10 is enough to buy 4 notepads and 4 markers: 4x + 4y ≤ 10 2x + 2y ≤ 5 x+y ≤ 2.5. Question stem rephrased: Is x+y ≤ 2.5? Quote: (1) Each notepad costs less than $1. (2) $10 is enough to buy 11 notepads. It is important to recognize not only how a problem is restricted but also how it ISN'T. Neither statement restricts the cost of the MARKERS (y). Try EXTREME values. If x = .01 and y = .01, both statements are satisfied and 5x + 3y ≤ 10. In this case, x + y = .01 + .01 = .02, which is less than 2.5. If x = .01 and y = 3, both statements are satisfied and 5x + 3y ≤ 10. In this case, x + y = .01 + 3 = 3.01, which is NOT less than 2.5. Thus, the two statements combined are INSUFFICIENT. The correct answer is E. ``` _________________ Mitch Hunt GMAT Private Tutor and Instructor GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "Thank" icon. Contact me about long distance tutoring! ```