Problem Solving Flashcards

Question

``` Good post? | PS Questions (GMAT Guru's Erin and 800Bob Please Help) Part 2 9. 1 – [2 –(3 – [4 – 5] + 6) + 7] = (A) –2 (B) 0 (C) 1 (D) 2 (E) 16 ``` SPOILER: D ``` 11. If 0.497 mark has the value of one dollar, what is the value to the nearest dollar of 350 marks? (A) $174 (B) $176 (C) $524 (D) $696 (E) $704 ``` SPOILER: D ``` 12. A right cylindrical container with radius 2 meters and height 1 meter is filled to capacity with oil. How many empty right cylindrical cans, each with radius 1/2 meter and height 4 meters, can be filled to capacity with the oil in this container? (A) 1 (B) 2 (C) 4 (D) 8 (E) 16 ``` SPOILER: C ``` 16. A family made a down payment of $75 and borrowed the balance on a set of encyclopedias that cost $400. The balance with interest was paid in 23 monthly payments of $16 each and a final payment of $9. The amount of interest paid was what percent of the amount borrowed? (A) 6% (B) 12% (C) 14% (D) 16% (E) 20% ``` SPOILER: D ``` 12. An instructor scored a student’s test of 50 questions by subtracting 2 times the number of incorrect answers from the number of correct answers. If the student answered all of the questions and received a score of 38, how many questions did that student answer correctly? (A) 19 (B) 38 (C) 41 (D) 44 (E) 46 ``` SPOILER: D

Answer 1

1) D....Apply PEMDAS or BODMAS...ans:2 2) E....350/0.497 ~704 3) C....2^2*1=n*(1/2)^2*4...n=4 4) D ..amnt borrowed=400-75=325 Total amnt paid=75+23*16+9=452 amont paid as interest=452-400=52 ->52/325=16% 5) I think there is something missing in the question..it is not mentioned the marks awarded for the correct answer. If we assume it to be 1 than ans is 38+(12/2)=44...D

Answer 2

Good post? | | b (65*4)+(80*6)+(77*5)/15

Answer 3

IMO A. If the score for 5 students was 8 --> the total score for these 5 students is 40 (40 + 5y) / 10 = x solve for Y and it is answer A. 2x - 8

Answer 4

``` for me the answer is e correct=x incorrect=50-x x-2(x-50)=38 solve for x=46 correct ```

Answer 5

Good post? | Stmt 1: A = 37, B = 40 => median is the 38th element A has 18 students below 80(median) B has 20 students below 78 (median = (20 th element +21st element/2) => median of the combined set can be 80 or atleast > 78 as 38 students are below 80 Avg = 37*82 + 40*74/77 => avg < 78 (only if equal number of students are there, the avg will be 78) Median > avg Sufficient Stmt 2: No info on how many students are in A & B individually. Insufficient Ans A.

Answer 6

Good post? | i think i would agree with mastermind. as per what karmaholic says: __________________________________________________ ____________ Marks obtained in four tests = 160 (4*40) [40 = 80% of 50 and 50 is the average score] Total Marks = 600. So the student needs another 440 marks from remaining 8 tests. Now the max he can get in these tests is 60. __________________________________________________ _____________ now we need the min number of tests in which he has to score max ie 60 if no of tests with max score =2 then 2x60 + 6x50 = 420 .. option D incorrect no of tests with max score =3 then 3x60 + 5x50 = 430 .. option C incorrect no of tests with max score =5 then 4x60 + 4x50 = 440 .. correct option B

Answer 7

B D hi Vinay, "On the inside edge" means that the border of the land is untillable: (1000 x 20) + (1000 x 20) + (2000 x20) + (2000 x 20) = 120,000 ft squared Then there's another untillable part that cuts the field into two squares: 30 x 1000 = 30,000 ft. squared Total untillable land = 30,000 + 120,000 = 150,000 ft. squared Total land = 2000 x 1000 = 2,000,000 ft. sq % tillable land = (Total land - untillable land) / Total land = (2000,000 - 150,000) / 2000,000 x 100% = 92.5% Hence, 93% Btw, Vinay, are these questions from the 800score tests? Tina

Answer 8

Not sure if these are right, but here are my answers: ``` 1. = (5C3+5C3) / 10C3 = 20 / 120 = 1/6 => ANSWER D ``` 2. A-L = 12 letters 20 desirable possibilities [10 for forward 3-letter sequences (like ABC) 10 for backward 3-letter sequences (like FED)] 12*11*10 = total possibilities => 20/(12*11*10) = 1/66 => ANSWER B ``` 3. = 10P2 * 8P2 * 6P2 * 4P2 * 2P2 = 10*9*8*7*6*5*4*3*2 = 10! => ANSWER E ``` ``` 4. = x! / (n!)(x-n)! v (x+1)! / (n+1)! (x-n)! = 1 v (x+1) / (n+1) = n+1 v (x+1) = n v x => Answer depends on values of n and x => ANSWER D ``` 5. 12! / (x!)(12-x)! = 12! / (x+p)!(12-x-p)! => x(12-x) = (x+p)(12-x-p), where x and p are integers When p = 3, we get: 6x = 27 => x = 9/2 which is not an integer, therefore p != 3 => ANSWER C

Answer 9

Answer would be (D) (A)X ^2 = Y^2 ---> |X| = |Y| once you get that the question can be answered (B) X+ Y=0 --> X= -Y ---> |X| = |Y| again main question can be answered.

Answer 10

chosters, Official Answer is right. Total people interviewd=N of people said "yes" to question 1 = N/4 In the question it states that 1/3 of people who answered "yes" to question 1 also answered "yes" question 2. What it means is people answered "yes" to question 1 and 2 = 1/3*N/4=N/12 So the number of people interviewed who did not answer "yes" to both questions = Total people interviwed- number of people answered "yes" to both questions=N-N/12 =11N/12 Hence E. Let me know if you have any questions.

Answer 11

Good post? | Answer 50% 80 = 75 +55 - p(a int b)

Answer 12

Total of 64 classes Total of 37 teachers A teacher has to teach atleast 1 class If each teacher were to teach 2 classes, then there would need to be 37*2 = 74 classes but there are only 64 so least number of teachers who teach 3 classes = 0 Now there are 64-37 = 27 more classes than there are teachers. The greatest number of teachers who can teach 3 classes = 26/2 = 13 (and one teacher will teach 2 classes) So solution set {greatest, least} = {13,0}

Answer 13

You are trying to find the number of ways that each team can play with each other. This is about combinations not permutations. Order does not matter. ``` So out of 9 teams, you want to pick 2, such that the order does not matter. That's 9nCr2 That's 9! / (2! * (9-2)!) = 9! / (2! *7!) = 8*9 / 2 = 72/2= 36 ```

Answer 14

Originally Posted by choked I'm having a lot of trouble making my answer look like the Official Answer. Can someone help? Here's the problem: In how many different ways can 12 people be seated at two tables with one seating 7 and the other seating 5? Official Answer to follow after some discussion It depends on whether or the each seat at a table is unique. What I mean is that if we consider the tables to have chair1, chair2, chair3, etc... we get a different answer than if we just think of the tables as having 5 or 7 chairs. In other words, does seating the people in the same order but shifting everyone over to the right or the left count as the same arrangement or not? If we're only concerned about which people are next to each other and not which particular chairs they occupy, then we have 12c5 (or 12c7) ways to divide them into groups of 7 and 5. Then we have 6! ways to arrange the people at the larger table and 4! ways to arrange the people at the smaller table. This gives us (12c5)*6!*4!, which will be a huge number. If we care about which chairs they occupy, then the 6! and 4! become 5! and 7!. They would cancel out with the denominator in 12c5, so we would just get 12!. These calculations seem right to me for these 2 different cases. What do you think?

Answer 15

Originally Posted by vd17 Hi, Could anyone of you please help me explain how can we solve and visualize the below two scenarios: 1) How to arrange 7 people in 5 chairs. 2) How to arrange 5 people in 7 chairs. Thanks. A lot of people prefer to just apply the permutation formula here, but the truth of the matter is that you don't need to know the permutation formula for the GMAT. In fact, there are very few true permutation questions on the GMAT. Instead, you should have a solid grasp of the Fundamental Counting Principle (FCP). ``` 1) How to arrange 7 people in 5 chairs Take the task of sitting people and break it into stages Stage 1: seat someone in chair #1 Stage 2: seat someone in chair #2 Stage 3: seat someone in chair #3 Stage 4: seat someone in chair #4 Stage 5: seat someone in chair #5 ``` Stage 1 can be accomplished in 7 ways (7 people to choose from) Stage 2 can be accomplished in 6 ways (6 people to choose from, once the first person was seated) Stage 3 can be accomplished in 5 ways Stage 4 can be accomplished in 4 ways Stage 5 can be accomplished in 3 ways Applying the FCP, the total number of ways to complete all 5 stages, and occupy all 5 chairs = 7x6x5x4x3 2) How to arrange 5 people in 7 chairs. We need to take a different approach here, since some chairs will not be occupied. However, every person will get a seat. ``` So, take the task of sitting people and break it into stages Stage 1: seat person #1 in a chair Stage 2: seat person #2 in a chair Stage 3: seat person #3 in a chair Stage 4: seat person #4 in a chair Stage 5: seat person #5 in a chair ``` Stage 1 can be accomplished in 7 ways (7 chairs to choose from) Stage 2 can be accomplished in 6 ways (6 chairs to choose from, once the first chair was occupied) Stage 3 can be accomplished in 5 ways Stage 4 can be accomplished in 4 ways Stage 5 can be accomplished in 3 ways Applying the FCP, the total number of ways to complete all 5 stages, and seat all 5 people = 7x6x5x4x3 Cheers, Brent

Answer 16

Good post? | Hi Dave, First of all, you got the Combination and Permutation mixed up - but I guess that was the whole point of your post I came up with this mnemonic device to keep these straight: P=Prizes (permutations) C=Committee (combinations) Here's the full story: [/SIZE]Permutations ("Prizes")[/SIZE] Let's say you have 5 people who are running a race, and you want to know in how many ways three prizes (gold, silver, and bronze medal) could be awarded. Any of the 5 could win the gold, any of the remaining 4 could win silver, and any of the other 3 could win bronze. So you get 5 x 4 x 3 possibilities. You can get this result using the permutation formula: P(5,3) = 5! / (5-3)! = (5 x 4 x 3 x 2 x 1) / (2 x 1) = 60 Notice that the effect of the denominator in the formula is just to cancel out the last two terms from the numerator. That's why I prefer to think of it as simply an incomplete factorial where you multiply out only as many terms as you have selections. In this case, instead of the complete factorial of 5 x 4 x 3 x 2 x 1, you only have three prizes to be awarded, so you stop after the third term. [/SIZE]Combinations ("Committee")[/SIZE] Now think of the same 5 people, but your task this time is to form a committee of 3 (with no special roles, just equal members). You could do your selection in the same way as above, but you would find that some of these permutations give you the same committee. For example, it doesn't matter whether your selection is person A, then C, then D or whether it is C, then D, then A. So the straightforward selection process we used to award prizes for the race needs to be adapted a bit. We need to divide by the number of possible ways in which a particular committee could have been picked. This is simply the factorial of the number of committee members, in this case 3! For example, consider these 6 permutations: ACD ADC CAD CDA DAC DCA These selections all result in the same committee, so they are all equivalent to a single combination. That's why the combination formula includes the division by n!, so C(5,3) = P(5,3) / 3! = (5 x 4 x 3) / (3 x 2) = 10 Hope that clears it up. I completely agree with you that understanding the concept behind it is a much more solid approach than just memorizing the formulas. Once you understand the concept, you can reconstruct the formulas very quickly, even if you forget them under pressure. Also, if you can "map" the problem to one of the two scenarios above, you should have no problem picking the correct formula to use.

Answer 17

For a 4 digit num to be div by 4, the last two digits shuld be divisible by 4. we shall have 14 such combinations. The remaining digits are 6. (i) For the combinations which include 0 (there wuld be 4 combinations - 20, 40, 60, 04)-> there are 6 ways to choose the first digit and 5 ways to choose the second digit to make it 4 digit num. (ii) For the remaining combinations (14-4 = 10)-> there are only 5 ways to choose the first digit (coz 0 would make it 3 digit num) and 5 ways to choose the second digit to make it 4 digit num. so, (i) 30*4 = 120 (ii) 25*10 = 250 total = 370 ans . (2) It took some time to get this answer... is there a way to solve this in 3 to 4 steps..!!??

Answer 18

A DERANGEMENT is a permutation in which NO element is in the correct position. Given n elements (where n>1): Number of derangements = n! (1/2! - 1/3! + 1/4! + ... + ((-1)^n)/n!) Given 5 letters A, B, C, D and E: Total number of derangements = 5! (1/2! - 1/3! + 1/4! - 1/5!) = 60-20+5-1 = 44. Total possible arrangements = 5! = 120. P(no letter is in the correct position) = 44/120 = 11/30. 2nd Part A DERANGEMENT is a permutation in which NO element is in the correct position. Given n elements (where n>1): Number of derangements = n! (1/2! - 1/3! + 1/4! + ... + ((-1)^n)/n!) Let the correct ordering of the 5 letters be A-B-C-D-E. P(A is correctly placed): P(A is in the correct position) = 1/5. (Of the 5 positions, only 1 is correct.) P(B, C, D and E are all incorrectly placed): Total number of derangements = 4! (1/2! - 1/3! + 1/4!) = 12-4+1 = 9. Total possible arrangements = 4! = 24. P(no letter is in the correct position) = 9/24 = 3/8. Since we want both events to happen, we multiply the probabilities: 1/5 * 3/8. Since the correctly placed letter could be A, B, C, D or E -- yielding 5 options for the correctly placed letter -- the result above must be multiplied by 5: 5 * 1/5 * 3/8 = 3/8.

Answer 19

Hi, I would approach this problem this way: if there are 6 people to be arranged in 6 places, then number of ways they can be arranged would be 6! ways. Since Frankie has to be always behind Joe not necessarily immediately behind him: So let us fix Joe's position & count the number of options for frankie & remaining people: A : if Joe at 1st position then Frankie has 5 slots behind him. B C D E F Say for eg: he occupies 3rd slot then remaining 4 persons can be arranged in total 4! ways. This means no of ways this option will have is a) 1(for Joe) X 5 (for Frankin) X 4!(for remaining 4) ``` Second possibility: A B Joe's position C D E F ``` Now Joe is at second position so franlin has 4 slots (if he has to remain behind him) b) This means no of ways this option will have is 1(for Joe) X 4 (for Frankin) X 4!(for remaining 4) Similarly c) 1(for Joe) X 3 (for Frankin) X 4!(for remaining 4) d) 1(for Joe) X 2 (for Frankin) X 4!(for remaining 4) e) 1(for Joe) X 1 (for Frankin) X 4!(for remaining 4) Add all the options from a to e: You will get 360. Hope this helps. Cheers Kundan

Answer 20

Good post? | Originally Posted by targetsep08 yes Official Answer is (2). can someone explain. Prize one [1st, 2nd and 3rd] can be distribute among three boy in 3P3=6 ways. Prize two, three and four also can be distribute in 6 ways So total way is =6*6*6*6=36*36

Answer 21

Good post? | There are 8C3 combination. Out of these there are 6 combination where both the inexperienced engineers will be together. 8C3 - 6 = 50

Answer 22

I was doing the IMS GMAT guide for Perm & Comb and came across a couple of confusing ones. someone por favor, el señor shed some light Problem Solving 1. There are 4 copies of 5 different books. In how many ways can they be arranged on a shelf? (a) 20!/4! (b) 20!/5(4!) (c) 20!/(4!)5 (d) 20! (e) 5! Official Answer : (C) 20!/(4!)5. Can someone explain in detail? Data Sufficiency 1. How many ways m digit numbers can be formed using n distinct digits? I. m=3 and n =5 II. n=2m Official Answer: A 2. what is the value of nCr if the value of nC3=56? I. n - r = 3 II. r = 3 Official Answer: D. Can Some explain in detail? Will post the Official Answer later

Answer 23

Good post? | solution: D this is the repeated question Answer is (10 * 8 * 6) / 3! = 80

Answer 24

i agree with mauni it should be (4! * 4! * 2) - (4 * 3! * 3! *2). can some body plaese explain? thanks.

Answer 25

Originally Posted by mbafan I disagree with the Official Answer. However, my answer does not fit any of those above. If 0 is chosen for the first number, that's OK > then the number automatically defaults to a 4-digit number. The lowest 4-digit number is 01345 or 1345. You cannot do that. For example, choose 0 as your first digit. Then you only have 1 option for that. Then you choose the other 4 digits from 4*3*2*1 [notice when you do this, you have effectively eliminated the possibility of a 4 digit>1000 that includes a 0. So you've under-counted. And then you do the 5 digits by excluding the numbers that started with 0. So 4*4*3*2*1--All these just tells you that you have counted 4 digit numbers greater than 1000 that do no include 0, and all 5 digit numbers. [add those two pieces, 96+24=120, which is what you have] [The only 4 digit numbers you can create by 0 being the first digit are those that do not include 0's. And if the number starts with anything other than a 0, then that number is a 5 digit number] comments please.

Answer 26

Originally Posted by ACETARGET How many different combinations of outcomes can you make by rolling three standard dice if the order of the dice doesnot matter. A.24 B.30 C.56 D.120 E.216 The key here is WITHOUT ORDERING, as I had mentioned in my previous post.. such problems can be classified in 4 categories! This one falls in WITH REPETITION, W/O ORDER i.e. (1, 1, 1) is allowed (1, 2, 3) is allowed but not (1, 3, 2) Thus answer is: C(6+3-1, 3) = C(8,3) = 56

Answer 27

Originally Posted by shootout A talent contest has 8 contestants. Judges must award prizes for 1st, Second and third places . If there are no ties a) How many different ways can the prizes be awarded, and b) how many different groups of 3 people can get prizes? These problems can be classified into 4 categories: ``` R - Repetition O - Order 1) With R, With O 2) With R, Without O 3) Without R, With O 4) Without R, Without O ``` a) Falls in without repetition, with order: ex: 123, 132 are countable but not 122, 222 - (assuming same person cannot get more than one prizes) Then the applied rule is permutation: 8*7*6 (8P3) ``` b) Falls in Without R, Without O ex: 123, 134 allowed, but not 123, 132 (w/o order) Then the applied rule is combination (8C3) = 8!/(3!*5!) = 42 ``` c) If a person can get more than one prize, then it falls into with R, with O. Then the rule applied is: (8^3) = 8 * 8 * 8 i dont remember the 4th one, but can derive it if required

Answer 28

Good post? | Hi, 8 Men and 2 Women can be seated around the table in (10-1)! ways. No of ways you can arrange 2 women around a table sitting opposite to each other = 10 No of ways we can arrange 8 men = 8! totoal no of ways you can arrange 8 men and 2 women where the 2 women sitting opposite to each other = 10 * 8! total no of ways you can arrange 8 men and 2 women around a table, where the 2 women are not sitting opposite to each other = 9! - 10 * 8! Thanks, Sharath Reply Reply With Quote

Answer 29

Good post? | Wherever the host sits, he has to be surrounded by two female guests, and so the gender arrangement gets completely defined. Because this is a rectangular table, he only has 2 options of where to sit (had this been a circular table, he'd have 10). So wherever he sits, there are only 4 places where female guests can sit and 4 places where male guests can sit. Since order is important, we use factorials: 2*4!*4!=1152 as cesar82 said. The other possibility is if the couples have to sit together, but the problem didn't specify that explicitly so it seems you have to go with the solution above.

Answer 30

``` Good post? | 1. B [ 3/9 * 2/8 = 1/12 ] 2. C [ 4/11 ] 3. ? [ 1-2/8 * 1/7 = 27/28] 4. C [ 1/2 * 2/6 = 1/6 ] 5. D [ 1-3/10*2/9 = 14/15 ] 6. B [ 3/6 = 1/2 ] 7. B [ 2/8 = 1/4 ] 8. E [ 2/6 * 2/5 = 2/15 ] 9. E [ 4/36 = 1/9 ] 10. D [ 1/2*1/2*1/2*1/2 = 1/16 ] 11. B [ 1-5/9 = 4/9 ] 12. D [ 1-0.2-0.3-0.1 = 0.4 ] 13. E [ 3/9*1/6 = 1/18 ] 14. E [ 3/6 = 1/2 ] 15. B only wine=200-50=150 only sailing = 400-50-150=200 sailing=200+50=250 probability=250/400=5/8 16. E [ 3 * 2/6 * 1/5 = 1/5 ] 17. A [ 3/5 * 2/4 = 3/10 ] 18. D [ 1- 3/7 * 2/6 * 1/5 = 34/35 ] 19. D [ 6/12 * 5/11 = 5/22 ] ```

Answer 31

35 Good post? | 1. For a quadrilateral, we need 4 vertices out of the 7 of a heptagon. Hence, 7C4 = 35 quadrilaterals can be formed.

Answer 32

816 2. Once can select to answer the first 2 and 2 from the last three in 3 ways. These can be answered in 3*3*4*4 = 144 ways, yielding a total of 3*144 = 432 ways. Or one may choose to answer the last 3 and 1 of the first two. This can be done in 2 ways. Each way would result in 4*4*4*3 = 192 ways, giving a total of 384 ways. Adding the two together would give a total of 816 ways.

Answer 33

14!*13 3. First, we arrange the remaining 13 students in 13! ways. This gives 14 places where the 2 children can be placed. The first child can be put in any of the 14 places and the second child can be put in any of the 13 places. This gives a total of 13! * 14 * 13 = 14! * 13 ways.

Answer 34

1500 4. I agree that the answer should be 2000, as you have explained.

Answer 35

120 5. Similar to first question. The answer is 10C7 = 120 ways.

Answer 36

1000 6. I think the answer should be 10*10*10*10 = 10,000 ways as it does not mention anything about repetition not being allowed.

Answer 37

3/4x + 5 = 7/8x 1/8x = 5 x = 40 (B)

Answer 38

f(v) = f(m)/3^2 => 3^y / 3^2 => difference between m and v is y-2. Since y is the tens digit, the difference will be 20. Therefore (d) is the answer.

Answer 39

To determine # of apples and oranges in first selection: (40(x) + 60 (10-x)) / 10 = 56 This leads to x = 2, therefore 2 apples and 8 oranges. To determine how many to return: (80 + 60(x)) / (2+x) = 52 This leads to x = 3...Therefore she must return 8-3 = 5 oranges... E

Answer 40

If f(v) = f(432) = (2^4)(3^3)(5^2), f(m)= 3^2((2^4)(3^3)(5^2)) Therefore f(m) = (2^4)(3^5)(5^2) Therefore m=452 and v=432...452-432=20

Answer 41

Answer IMO B 7/8x-3/4x = 1/8x = 5 Therefore x = 8*5 = 4

Answer 42

Answer : IMO D Let m = 100A + 10B +C V = 100D + 10E + F ``` Then f(m) = 2a3b5c f(v) = 2d3e5f ``` ``` multiplying f(v) by 9 = 9f(v) = 32*2d*3e*5f comparing powers of f(m) we get a=d, b=e+2 and c=f ``` putting these in the original equation we get m-v = 100A +10(e+2) +C – 100A -10e –C = 20 Hence answer is D

Answer 43

IMO B (0.4x+0.6y)/10 =5.6 where x = number of apples and y = number of oranges Also x+y = 10. from the two equation we can calculate that x = 2 and y = 8 Now the average price after A oranges are put back (i.e A = number of oranges to be put back) (0.4x+0.6y-0.6A)/(10-A) = 0.52 Solving for A we get A = 2

Answer 44

IMO E Statement I is tells us that the lowest limit is same but the upper limit can be different therefore the range (i.e maximum – minimum can be different). Hence it is insufficient. Statement II gives us the mean, does not tell us about the maximum or minimum limits therefore it is also insufficient Together we cannot calculate the upper limit therefore the combined equations are also insufficient.

Answer 45

IMO A Statement I: 1,125 = 53*32 therefore x = 2 and y = 3 as bases are same powers must be same too. Hence equation I is sufficient. Statement II: value of y is given. We cannot calculate x from this equation therefore the statement is insufficient.

Answer 46

IMO D Since multiple of d is equal to the absolute value of an number therefore it must be positive. Rule out A and B. Plug in values of C, D and E. Only D satisfies the equation therefore D must be the answer

Answer 47

IMO E | 11C2*9C2 = 1980

Answer 48

IMO B Let x be the length of each side of the square. Draw two lines as shown in the figure below. Let us look at the green line first. The length of the green line = root 2*x (property of an isosles triangle…) therefore from the figure we can infer that the width of the rectangle = root 2*x. Next look at the blue line. Use same principle but this time we have two squares so length of rectangle = 2*root 2*x.x Now the perimeter of the rectangle is 2L+2W= 18 * root 2. Put in values of Length and width. We get x = 3. Thereofre perimeter of each square = 4*3 = 12 Hence answer is B.

Answer 49

IMO C Statement I : since the sum of all the angles of a quadrilateral is 360 and we know that two of the angles are 90 degrees therefore the sum of the other two angles would be 180. (i.e 360-2*90= A + B) where A and B are the other two angles. Therefore there would be an angle that is atleat greater than 60 degrees. (i.e if one angle is 10 degrees then the other angle would have to be 170 degrees). However it is uncertain whether the angle would be equal to 60 or not. Therefore statement I is insufficient. Statement II : A = 2 B. On its own the equation is insufficient. Combined Statement I and II : 360-2*90=A+B Therefore 3B =180 hence B = 60 Therefore the one of the angle is equal to 60 hence C should be the answer

Answer 50

IMO D Both equations satisfy as D(t) = 11,000 tells us the amount the deposit will grow in three years. I am assuming that there is a misprint as t = 3. Also in statement II r is given there we can calculate how much will the deposit grow in three years by pluggin in the values in the equation.

Answer 51

IMO C | Statement 1 tells us that a is even. Since a is even in statement 2 therefore b must also be even.

Answer 52

IMO D Machine K took 3 hours to do ¼ of the job. Therefore it would take it 4*3 = 12 hours to do the complete job on its own Similarly both Machine K and M took 6 hours to do ¾ of the job therefore it would take 6*4/3 = 8 hours to do the complete job. Use the formula 1/k + 1/m = 1/T = where k = hours that would be taken by machine K if it had operated alone. m= hours that would be taken by machine M if it had operated alone. T = hours that would be taken by both machines if they operated together 1/12+1/m = 1/8 solve for m . M = 24

Answer 53

IMO B Let price list price be 100. Then regular price = 80 85% of regular price = 68 100-68 = 32 Hence B is the answer

Answer 54

E ``` 5*5 = 25 4*4 = 16 ``` Unit digit of 5 minus unit digit of 6 should result in a unit digit of 9. Option e is the only one with unit digit of 9.

Answer 55

A ``` 60/d=25/100 d=6000/125 d=48 60-d= 12 Hence A ```

Answer 56

B A=1/100 B=1/5 C= 1/10 Hence a is the least as the denominator is the highest followed by c and then b

Answer 57

B If a^x = b then a = b^(1/x) similarly b= c^(1/y) and c=a^(1/z) Put value of c in equation 2 b=[a^(1/z)]^(1/y) Therefore b= a^(1/(ya)) But b= a^x then A^x=a^(1/(ya)) Taking root x on both sides A=a^(1/(xyz)) Since bases are same powers would have to be same 1/(xyz) = 1 Xyz=1 Hence B is the correct answer

Answer 58

D For question 1), you can solve as following : After expanding the two equation given in the problem, you get the following : x^2 - 4y^2 = 5 .....(i) 4x^2 - y^2 = 35.....(ii) Add the two equations and you get following : 5x^2 - 5y^2 = 40 Therefore, x^2 - y^2 = 8.... (a) Now subtract (i) from (ii) 3x^2 + 3y^2 = 30 Therefore, x^2 + y^2 = 10.....(b) Divide (a) by (b) and you get = 8/10 = 4/5 Answer is (d)

Answer 59

C A number will have to be a multiple of 2,3 and 5 5*2*3= 30 Hence 30, 60 and 90 lie between 1 to 100 therefore 3 numbers

Answer 60

B For question 2) L/m+n = k therefore, L/k = m+n Similarly, m/k = n+L n/k = L+m Add all three equations : 1/k (m+n+L) = 2(m+n+L) Therefore k = 1/2 Answer = B

Answer 61

This was my way of doing it. I took just more than 2 minutes but this was due to the final step which required lengthy calculations. The dealer plans to cell the cameras for USD 250 which is 20% more than dealer's intial cost. Let the initial cost be "a" Therefore, a + 0.2a = 250 Solve for "a" , a = 208. Therefore initial cost = 208 * 60 = $ 12480. The dealer sold 54 cameras at $ 250 = $ 13500 The dealer sold 6 cameras at $ 104 (50% refund on initial cost) = $ 624 Total selling price = $ 14124. Profit = (14124 - 12480)/ 12480 = 13 % profit.

Answer 62

B they will have gone 4 and 6 miles, respectively. Just do the pythagorean theorem : 4^2 + 6 ^ 2 = 52, The square root of this is closest to 7. _________________ If you find my posts helpful. Please take the time to click the "thank" icon at the top of the page and/or follow me! Full Length GMAT Exam | GMAT classes | GMAT Prep|GMAT Tutoring|MBA Admissions Consulting|Stratus Prep Reviews

Answer 63

E x^2 = (x +30)(x-20) x^2 = x^2 - 10x - 600 when you solve this you will get -60 -> the magnitude of this answer matches and honestly the sign does not matter for this question. Really the wording should be flipped though.

Answer 64

C Simple interest over 5 periods would be 20% (4 x 5). You know you are going to be more than this because of compounding but not almost double or double as is in D and E.

Answer 65

E The previous explanation is correct. I will try to elaborate in more detail, and hopefully that will help. Let's break it down piece by piece..... h(100) = multiply every even number from 2 to 100 h(100) =2x4x6x8....etc Can we simplify this? Yes. h(100) = 2 * (2*2) * (2*3) * (2*4)....all the way up to (2*50) I think the first "trick" of this question is that they are trying to get you to misapply the distributive property of addition. If the question asked 2+4+6+8, you could solve by figuring out (2*1)+(2*2)+.... and you could pull out all your 2's, so you have (2*(1+2+3+4)....) - and then find the sum of all the numbers from 1 to 50). However, this is not that kind of question. It is asking you to MULTIPLY every even number from 2 to 100. So we can't distribute. It is literally (2*1)*(2*2)*(2*3)......so we are multiplying the number 2 times every single number between 1 and 50 in a composite way. Why is it not between 1 and 100, you ask........because its only the even numbers between 1 and 100, so there are only 50 numbers. This is the same as saying multiply 2, 50 times, and then multiply all the numbers from 1 to 50. Translated into math, this is 2^50 * 50!. So, h(100) = 2^50 * 50! Ok - now we have a probem. We have simplified our equation, but we still are working with an incalculable number for GMAT purposes. So we can't use our standard divisibility rules in any predicable way. This is where the solution using prime numbers comes into play. Let's play around with a smaller version of h(100) for practice. Let's try h(6). h(6) = 2x4x6 = 48 h(6) + 1 = 49 What is the SMALLEST PRIME NUMBER THAT DIVIDES h(6)? Well, 49 = 7x7 - so 7 is the smallest prime number that divides h(6) Let's try h(4). h(4) = 2x4 = 8 8+1 = 9 What's the smallest prime number that divides 9? Well, it's 3. Hopefully you are noticing a pattern. The smallest prime number that divides h(n)+1 is always larger than the largest prime number in h(n). So, before we even get to the formal explanation, we see that rule. If we extend that rule to h(100) +1 - then the largest prime number that divides h(100)+1 is larger than the largest prime in the set H(100). We can easily see from above that 47 is the largest prime number in h(100). So our answer has to be greater than 47.....so we know that it must be E. Now, a more formal "proof" for this question is along the lines of what someone explained before. We know h(100) = 2^50 * 50!, as explained earlier. We know from that if a number is divisible by a prime number, and then you add 1 to the number in question, it CANNOT be divisible by that same prime. For example. if 10 is divisible by 5, can 11 be divisible by 5? Absolutely not. Take this logic, and apply it to the problem. We are looking for the largest PRIME divisor. So by the above logic, if h(100) is divisible by a certain prime divisor, h(100) + 1 CANNOT be divisible by that same prime divisor. We know that h(100) = 50! * 2^50, and 50! contains every prime from 1 to 50. Therefore, h(100) must be divisible by every prime from 1 to 50. (If you don't get the part about being divisible by primes, then you need to study this for the GMAT. While this type of question in particular will not come up often, you need to know prime factorization and its applications to do well on GMAT Math). Since h(100) is divisible by every prime from to 1 to 50, H(100) + 1 CANNOT be divisible by ANY of those primes. Therefore, our prime in question must be greater than 50.

Answer 66

A GMAT Kolaveri wrote: In an election, candidate Smith won 52% of the total vote in Counties A and B. He won 61% of the vote in County A. If the ratio of people who voted in County A to County B is 3:1, what percent of the vote did candidate Smith win in County B ? ``` 25% 27% 34% 43% 49% ``` Need to know alternate methods of solving this problem in less than 2 min!! source: https://www.freequestionaday.com/gmat Let the number of voters in A = 3. Let the number of voters in B = 1. The total number of voters in A and B = 3+1 = 4. Treat the percentages as averages. Sum in A and B = (number)(average) = 4*52 = 208. Sum in A = (number)(average) = 3*61 = 183. Sum in B = 208-183 = 25. Average in B = sum/number = 25/1 = 25. The correct answer is A.

Answer 67

D ``` avada wrote: Working alone, Printers X, Y, and Z can do a certain printing job, in 12, 15, and 18 hours respectively. What is the ratio of the time it takes printer x to do the job tot he time it takes printers Y and Z working together at their individual rates? 4/11 1/2 15/22 22/15 11/4 The time ratio for X:Y:Z = 12:15:18. To make the math easier, divide all of the times by 3: X:Y:Z = 4:5:6. ``` Let the job = the LCM of 4, 5, and 6 = 60. X's rate = w/t = 60/4 = 15 pages per hour. Y's rate = w/t = 60/5 = 12 pages per hour. Z's rate = w/t = 60/6 = 10 pages per hour. Combined rate for Y+Z = 12+10 = 22 pages per hour. Rate and time are RECIPROCALS. Since (rate for X alone) : (rate for Y+Z) = 15:22, (time for X alone) : (time for Y+Z) = 22:15. The correct answer is D.

Answer 68

etters and Words A common GMAT P&C problem type is letter problems. You would be asked to take letters from an existing word to form new words. Example: Four letters are taken at random from the word AUSTRALIA. What is the probability to have two vowels and two consonants? Solution: Vowels: A, U, A, I, A. Consonants: S, T, R, L This is a case of "without replacement". Total outcomes: C(9,4)=9*8*7*6/4!=126 Outcomes with two vowels and two consonants: C(5,2)*C(4,2)=60 Probability=60/126=10/21 Example: From the word AUSTRALIA, a letter is taken at random and put back. Then a second letter is taken and put back. This process is repeated for four letters. What would be the possibility that two vowels and two consonants have been chosen? This is a case of "with replacement": Total outcomes: 9^4 Outcomes with two vowels and two consonants: C(4,2)*5^2*4^2 You can calculate the probability from here. Note that there is a special thing with letter problems, ie. the repeating letters. In the above examples the repeating letters don't matter. However they would matter if the question is asked differently. Example: There is a word AUSTRALIA. Four letters are taken at random. How many different words can be formed that have two vowels and two consonants? Solution: Vowels: A, U, A, I, A. Distinguish vowles: A, U, I. Consonants: S, T, R, L Outcomes with two vowels and two consonants: 1) The two vowels are different: C(3,2)*C(4,2) 2) The two vowels are the same: C(1,1)*C(4,2) After you've got the four letters you need to order them to get different outcomes. So the first case (with different vowels) would be C(3,2)*C(4,2)*P(4,4)=432, and the second case (with same vowels) would be C(1,1)*C(4,2)*P(4,4)/2=72. The final outcome would be 504.