saubaer

Since there is an even number of students, the median score is the average of 10th and 11th score = 77.

1. Let the original amount of production be a and original selling prize be b. So revenues in year X = ab. Revenues in year Y = (1.25a)(0.75b) = 0.9375ab = decrease of approximately 6 percent. 2. We have to distinguish between the average number of vehicles and the total number of commuters. It is given that the average number of vehicles transporting commuters to county W is 30.000. Now by using the information provided in the graph on the left (the average number of persons per vehicle) we can infer that the total number of commuters to county W = 30.000*1.2 = 36.000. It is also given that the number of commuters to county Z is half the number of county W = 18.000. Since the average occupancy rate for Z is 1.5 we get the average number of vehicles driving to county Z by dividing 18.000 by 1.5, which yields 12.000.

Explanation for 1st question: Let the digits of the five digit number be a ,b, c, d and e. Number = 10.000a + 1.000b + 100c + 10d +e. Reflection = 10.000e + 1.000d + 100c +10b +a Difference = ( 10.000a + 1.000b + 100c + 10d +e) - (10.000e + 1.000d + 100c +10b +a) = 9.999a + 990b + 0c - 990d - 9.999e = 99(101a + 10b -10d -101e). So difference would be divisible by 3, 9, 11 and 33, but only 9 is among the answer choices.

I have some questions about the difference between 'either or', 'at least one' and 'only one'. 1. Every one who passes the test will be awarded a degree. The probability that Tom passes the test is 0.3 and the probability that John passes the test is 0.4. What is the probability that at least one of them gets the degree? A 0.7, B. 0.58, C. 0.46, D 0.42, E 0.8 2. The probability that Tom will win the Booker prize is 0.5 and the probability that John will win the prize is 0.4. There is only one prize to win. What is the probability that at least one of them wins the prize? A 0.2, B 0.4, C 0.7, D 0.8, E 0.9 –––––––––––––––––––––––––––––––––––––––––––––––– P (only one) = P(A and not B) + P(B and not A) = P(A\B) + P(B\A) P (A u B) = P(A) + P(B) – P(A and B) = P(A\B) + P(B\A) + P(A and B) P (at least one) = P(A\B) + P(B\A) + P(A and B) = P(A) + P(B) – P(A and B) P(A\B) + P(B\A) + P(A and B) + P(neither A nor B) = 1 Is this correct? I don't see a difference between 'either or' and 'at least one'. The first question (which is not mutually exclusive) can be solved in 3 ways: P (at least one) = 1 – P(neither) = 1 – (0.7*0.6) = 0.58 P(A\B) + P(B\A) + P(A and B) = 0.3*0.6 + 0.4*0.7 + 0.3*0.4 = 0.58 P(A) + P(B) – P(A and B) = 0.3+ 0.4 – 0.3*0.4 = 0.58 But for the second question this does not work: Since there is only one prize to win we have two mutually exclusive events, right? So P(A and B) = 0 Let's try to solve this like the first question: 1 - P (neither) = 1 - 0.5*0.6 = 0.7 P(A\B) + P(B\A) + P (A and B) = 0.5 + 0.4 - 0 = 0.9 P(A) + P(B) - P(A and B) = 0.5 + 0.4 -0 = 0.9 Can someone please explain what is wrong with my reasoning. One more question: 3. 3 high school classes have 4 soccer players each. If a team of 9 is to be formed with an equal number of players from each class, in how many ways can the selections be done? O.A. is 4C3 + 4C3 + 4C3 = 12, but isn't it 4C3 * 4C3 * 4C3 = 64

1. Let the side of the smaller square be x: x2 = s2 + t2, x2 = area of shaded region, so quantities are equal. 2. r = s3t6 = s3t3 t3 = (st)3t 3. Since st = 10: r = 103t3 –> t3 = r/1000

2. Distance from center (0,a) to point (d,e) = d2 + (e-a)2 Distance from center to point (b,c) = b2 + (c-a)2 Since they are equal we can set up an equation: d2 + (e-a)2 = b2 + (c-a)2 d2 + e2 -2ae + a2 = b2 + c2 - 2ac + a2 (we can cancel out a2) (d2 + e2) - (b2 + c2) = 2ae - 2ac = 2a(e-c) Now a is positive (above x-coord.) and e > c --> the difference is positive and therefore d2 + e2 > b2 + c2

@ jomalon: I don't know if there is an actual site of the quant database, I only know about the forum: Dr Raju's Institute of Graduate Studies Abroad :: View Forum - Quantitative

1. If you draw a line connecting points Q and T you will get 4 congruent right triangles with hypotenuse √5. Now let x be the side of square. x2 + (x/2)2 = 5 –> x = 2 Area of square = 4. Region PQST is composed of 2 of the 4 equal right triangles –> area = 4/2 = 2, which is grater than 3/2 so answer is A.

Should be 3/8. Monthly salary = x. Car payment = x*(1/6) and rent = x*(1/6)*(5/4) = 5x/24, car + rent = 4x/24 + 5x/24 = 9x/24 = 3x/8 --> (3x/8)/x = 3/8

1.B Let the 2 unknown numbers be x and y, set of the 5 numbers in ascending order: x, y, 32, 35, 35 Sum of all numbers = 5*30 = 150 and 32+35+35 = 102 --> x+y = 150-102 = 48. Least possible value of x if y = 32, so x = 48-32 = 16 2. 14 For number 1 we have 1 term,for 2 - 2terms, 3 - 3 terms and so on, so number of terms = sum of first natural numers= 1+2+3+4+5+6.... = (n(n+1))/2. If we plug in 13 we get (13*14)/2 = 91. After that we have 14 times the number 14 so 100th term is 14. 3. C Number of at least one digit repeating = total number of possible arrangements minus number of possible arrangements where no digit is repeating. (Since this is a telephone number I think that we can use zero as leftmost digit) Total number of ways = 105 no digit repeating = 10*9*8*7*6 = 30240 at least one digit repeating: 100.000 - 30240 = 69760

I meant with repetition allowed, forgot to mention it.

If this is not the case, can we only solve this by systematically listing all possible permutations? Are there any other constraints? It seems not to work if repetition is allowed. E.g: In how many ways can the letters ABA be arranged? In this case the above conditions are fulfilled. But the right answer is obviously not 33/2! = 13.5 (in fact, there are 8 possible permutations) How can one solve this without writing down all options? Another question: How do we handle indistinguishable objects in combination problems? E.g: How many different 3 letter combinations are there of the set ABBBC (rep. not allowed)? Possible combinations are ABC,ABB,BBB,BBC = 4 and not 5C3/2! = 5 One more question (from Dr. Raju's april quant update): Given the two names JULLIE and LILLY, if one letter is picked from both simultaneously at random, what is the probability that it is the same letter? Can we solve it like this: Probability of picking a L from both names + prob. of picking I from both names = (2/6)*(3/5) + (1/6)*(1/5) = 7/30 How would you solve this by using the combination formula? Do we have to be concerned about the indistinguishable letters?

I solved it like this (there is probably a better way): A day has 24 hours so the number of hours in x days = 24x and since x is an integer the number must be divisible by 24 or 8 and 3 (8*3=24). Since 2 digits are blotted out we won't get the answer by checking for divisibility by 3, so let's see if any of the numbers is divisible by 8. For that matter we have to be concerned about the last 3 digits only. If the hundreds digit is even, the number is divisible by 8 if the last 2 digits are. (e.g: 264 and 616). But this is not the case for any of the answer choices (6,26,2,50 and 20 all not div. by 8) and therefore the hundreds digit has to be odd. If the hundreds digit is odd, the number is divisible if the number formed by the last 2 digits +/ - 4 is div. by 8 (e.g: 320 - 4 ->16 or 504 + 4 ->8). This is only true for 65_,_20 since 20 +/-4 is divisible by 8. Possible numbers would be 652320 or 651120 (the sum of the digits must be div. by 3 and the number formed by the last 3 digits by 8)

I. For any numbers a and b, a⋅b=a+b−ab. If a⋅b=0, which of the following CANNOT be a value of b? (A) 2 (B) 1 © 0 (D) −1 (E) −3/2 IIa. A fair, six-sided die is rolled. What is the probability of obtaining 3 or an odd number? IIb. A fair,six sided die is rolled. What is the probability of obtaining a multiple of 3 or an even number? III. S is a series of natural numbers. Sn represents the 'n' th number of the series. Sn (for n > 2) is equal to the sum of Sn-1 and Sn-2. What is the difference between the square of S100 and the product of S99 and S101 if S1 = S2 = 5? IV. The calorie content of five fruits is measured. The sum of calorie content of three fruits taken at a time are 150, 160, 120, 180, 140, 150, 170, 130, 160 and 140. What is the average calorie content of a fruit? V. A pentagonal prism is to be painted in such a way that no two adjacent faces are painted. In how many ways can the prism be painted? VI. Of the science books in a certain supply room, 50 are on botany, 65 are on zoology, 90 are on physics, 50 are on geology, and 110 are on chemistry. If science books are removed randomly from the supply room, how many must be removed to ensure that 80 of the books removed are on the same science? VII. The front wheels of a toy truck are 4 inches in circumference. The back wheels are 7 inches in circumference. If the truck travels in a straight line without slippage, how many inches will the truck have traveled when the front wheels have made 12 more revolutions than the back wheels? VIII. In performing a sequence of experiments, a scientist made 20 measurements. The average (arithmetic mean) of these measurements was 34. For security reasons the scientist coded the data by multiplying each of the measurements by 10 and then adding 40 to each product. What is the average of the coded measurements? IX. A straight fence is to be constructed from posts 6 inches wide and separated by lengths of chains 5 feet long. If a certain fence begins and ends with a post, which of the following could not be the length of the fence in feet? (A) 17 (B) 28 © 35 (D) 39 (E) 50

1) Col. A: xy Col. B: yx (clearly, they are equal) 2) Col. A: xy/(x+y) Col. B: (x+y)/xy for x = y = 2 the quantities are equal, but for any other value > 1 they are not, so answer is D.