GMAT-Arithmetic-3. Descriptive Statistics and Set theory

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Abner Shields
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1 GMAT-Arithmetic-3 Descriptive Statistics and Set theory Descriptive Statistics 1). If S is a set of consecutive integers, what is the standard deviation of S? (1) Set S contains 23 terms (2) The median of set S is 20 Statement I is Sufficient: From the question, we know that S is a set of consecutive integers. Statement 1 tells us that there are 23 terms in the set. Since, in any consecutive set with an odd number of terms, the middle value is the mean of the set, we can represent the set as 10 terms on either side of the middle term x: [x - 10, x - 9, x - 8, x - 7, x - 6, x - 5, x - 4, x - 3, x - 2, x - 1, x, x + 1, x + 2, x + 3, x + 4, x + 5, x + 6, x + 7, x + 8, x + 9, x + 10] Notice that the difference between the mean (x) and the first term in the set (x - 10) is 10. The difference between the mean (x) and the second term in the set (x - 9) is 9. As you can see, we can actually find the difference between each term in the set and the mean of the set without knowing the specific value of each term in the set! Statement II is not sufficient: Since the set is consecutive, we know that the median is equal to the mean. Thus, we know that the mean is 20. However, we do not know how big the set is so we cannot identify the difference between each term and the mean. Therefore, the correct answer is A. 2). A certain characteristic in a large population has a distribution that is symmetric about the mean m. If 68 percent of the distribution lies within one standard deviation d of the mean, what percent of the distribution is less than m + d? (A) 16% 1

2 (B) 32% (C) 48% (D) 84% (E) 92% Given that 68% lies between m-d and m+d, thus 32% lies out of this range. Now, since the distribution is symmetric about m, then half of the 32%, so 16%, lies to the right of m+d. Therefore, 16% lies to the right of m+d, and hence 84% lies to the left of m+d, which means that 84% is less than m+d. So the Answer is D. 3). Is c equal to the median of the integers a, b and c? I. a < b + c II. b = c Statement I is insufficient: a = 2, b = 4, c = 3 {c is the median} a = 2, b = 3, c = 4 {c is not the median} Statement II is sufficient: There are two possibilities: a>b and b>a in both the cases {a,c,c} and {c,c,a} will be the median. So the answer is B. 4). A set of 25 different integers has median of 30 and a range of 50. What is least possible integer that could be in this set? (A) -8 (B) -7 2

3 (C) 0 (D) 8 (E) Cannot be determined. Set: X X + 50 Since they are different integers let us make them consecutive integers and see if we strike a balance! Since the range is 50 and right now the range is = 24 which means we can still further reduce the lowest number by = 26 So the lowest possible number is = -8. So the answer is A. 5). If the average of five numbers, x, 7, 2, 16, and 11 = the median, what is x? I. 7 < x < 11 II. x is median of the five numbers Average of five numbers: (x + 36)/5 = Median Arranging the numbers = { 2, 7, 11, 16 and x} If x is between 7 and 11 then the median is x If x is less than 7 then the median is 7 If x is more than 7 then the median is 11 Statement I is sufficient: Since x is between 7 and 11 we know that median and the mean is x. (x + 36)/5 = x 3

4 X + 36 = 5x X = 9. Statement II is sufficient: (X + 36)/5 = x Again x = 9 Hence answer is D. 6). The arithmetic mean of 17 consecutive integers is an odd number. Which of the following must be true? I. Largest integer is even. II. Sum of all integers is odd. III. Difference between largest and smallest integer is even. (A) I (B) II (C) III (D) I, II (E) II, III Average of the 17 consecutive numbers is the middle number which is the 9th number. If 9th number is odd then 17th number is odd and the first number is also odd. Sum = There are 9 odd numbers and even numbers which means the sum of all will be odd I. Largest integer is even: Not true II. Sum of all integers is odd: True 4

5 III. Difference between largest and smallest integer is even. Odd Odd = Even (True) II and III are true. Answer option is E. 7). In a recent survey, twenty families reported their incomes for Was the range of the reported incomes for these families greater than $60,000? I. Thirteen of the reported incomes were between $20,000 and $35,000. II. Seven of the reported incomes were between $80,000 and $95,000. Is Range>60k? Statement I is insufficient: From this statement we will not know the highest and the lowest values which are required for calculating range. Similarly Statement II is also insufficient. Combining: Let us consider two scenarios: There could be a scenario where the lowest salary is and the highest be Range = <60k Also, consider a scenario where the lowest salary is and the highest is Range = >60k. So even after combining we cannot give a definite answer. Hence, the answer is E. 8).Peter, Paul, and Mary each received a passing score on his/her history midterm. The average (arithmetic mean) of the three scores was 78. What was the median of the three scores? (1) Peter scored a 73 on his exam. (2) Mary scored a 78 on her exam. 5

6 Statement I is not sufficient: Peter scored a 73 on his exam. If the other two scores are 73 and 78* (whatever it is) then the median is 73 but if the other two scores are 78 and 83 then the median is 78. Not sufficient. Statement II is sufficient: Mary scored a 78 on her exam. In order the average to be 78, one of the remaining scores (Peter or Paul) must be less than 78 and another must be more than 78 OR all three scores must be 78. In either case, the median is 78. Sufficient. So the Answer is B. 9). x is the sum of y consecutive integers. w is the sum of z consecutive integers. If y = 2z, and y and z are both positive integers, then each of the following could be true EXCEPT A. x = w B. x > w C. x/y is an integer D. w/z is an integer E. x/z is an integer If k is odd, the sum of k consecutive integers is always divisible by k. Given {9,10,11}, we have k=3 consecutive integers. The sum of =30, therefore, is divisible by 3. If k is even, the sum of k consecutive integers is never divisible by k. Given {9,10,11,12}, we have k=4 consecutive integers. The sum of =42, therefore, is not divisible by 4. Set theory: 1). Each of three charities in Sun Royalty Estates has 16 persons serving on its board of directors. If exactly 8 persons serve on 3 boards each and each pair of charities has 10 6

7 persons in common on their board of directors, then how many distinct persons serve on one or more boards of directors? (A) 16 (B) 26 (C) 32 (D) 48 (E) 54 At least one: A + B + C (AB + BC + CA) + ABC At least one: ( ) + 8 At least one: = 26. So the answer is B. 2). Of the members of a certain health club, 30 percent are women at least 40 years of age. What percent of the members are men under 40 years of age? (1) Of the female members of the health club, half are at least 40 years of age. (2) Of the members of the health club at least 40 years of age, half are men. Statement I is insufficient: 7

8 If half of the Women are at least 40 which mean that half of Women are under 40. We just know that there are 40 men and don t know the further classification hence this statement is insufficient. Statement II is insufficient: From this we would only know that 30 men are at least 40 and there are 40 people under the age of 40. We cannot break further and calculate the blank cells. Combining is sufficient: So the answer is C. 3). Of 300 members in an organization, each German speaker also speaks English, and 70 of members only speak Spanish. If no member can speak all 3 languages, how many of the members speak 2 of the 3 languages? (A) 60 members speak only English (B) 20 members do not speak any of the 3 languages Let us draw the diagram for the question: 8

9 The question asks us the value of a + b. a + b + c + d + 70 = 300 a + b + c + d = 230 Statement I is insufficient: Value of c is known to be 60 however we don t know the value of d or c. c + d = = 170 Statement II is insufficient: d = 20 a + b + c = = 210 Combining is sufficient: c = 60, d = 20 a + b = ).Of the 645 speckled trout in a certain fishery that contains only speckled and rainbow trout, the number of males is 45 more than twice the number of females. If the ratio of female speckled trout to male rainbow trout is 4:3 and the ratio of male rainbow trout to all trout is 3:20, how many female rainbow trout are there? A.192 B.195 9

10 C.200 D.205 E.208 Male Female Total Speckled Trout 645 Rainbow Trout Total Use M and F to represent the total number of males and females, respectively. We know that together they add up to the total number, so: M + F = 645 If the number of males is 45 more than twice the number of females, then: M = F Now we have a system of equations - 2 equations and 2 variables - so we can substitute for M: (45 + 2F) + F = 645 3F = 600 F = 200 M = 445 Male Female Total Speckled Trout Rainbow Trout Total 10

11 If the ratio of female speckled to male rainbow is 4/3, we know that the number of female speckled is 200, so: (200/x) = (4/3) 4x = 600 x = 150 Male Female Total Speckled Trout Rainbow Trout 150 Total There are 150 male rainbow trout, so put that in the the grid. If the ratio of male rainbow to all trout is 3/20, then: (150/y) = (3/20) 3Y = 3000 Y = 1000 Male Female Total Speckled Trout Rainbow Trout Total 1000 There are 1,000 total trout. Once we add this to our grid, we can easily solve for the number of female rainbow trout. There are 645 speckled trout and 1000 total trout, so: = 355. There are 355 total rainbow trout. Total rainbow trout - male rainbow trout = female rainbow trout, so: = 205. There are 205 female rainbow trout. So the answer is D. 11

12 5).Guests at a recent party ate a total of fifteen hamburgers. Each guest who was neither a student nor a vegetarian ate exactly one hamburger. No hamburger was eaten by any guest who was a student, a vegetarian, or both. If half of the guests were vegetarians, how many guests attended the party? (1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians. (2) 30% of the guests were vegetarian non-students. Let the total no of guests be X. Total no. of hamburgers = 15. From given data, half of guests are vegetarians (X/2). Hence, the remaining half of guests are non-vegetarians. And, only those guests who are neither students nor vegetarians ate hamburgers and NO OTHER GUEST ate hamburger. So, it is enough to find the no. of guests who are neither students nor vegetarians. With the given info, we have this table: Students Non-students Vegetarians A B x/2 Non-vegetarians C D = 15 x/2 x Statement I is sufficient: A/B = 2/3 => C/D = 4/3 C = 15 * 4/3 = 20 And, X/2 = C + D = 35 = > X = 70 Hence, (1) alone is sufficient. Statement II is insufficient: B = 3X/10 Nothing can be said about C from this. Hence, (2) alone is not sufficient. So the Answer is A 12

BRITISH COLUMBIA SECONDARY SCHOOL MATHEMATICS CONTEST, 2017 Junior Preliminary Problems & Solutions

BRITISH COLUMBIA SECONDARY SCHOOL MATHEMATICS CONTEST, 017 Junior Preliminary Problems & s 1. If x is a number larger than, which of the following expressions is the smallest? (A) /(x 1) (B) /x (C) /(x