GMAT Sets & Statistics

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Arithmetic mean: Alternate approach

For a set of n numbers x1, x2, x3, . . . xn, the arithmetic mean is calculated as x = xa + (((x1 − xa) + (x2 − xa) + (x3 − xa) + · · · + (xn − xa))/ n)

S is a set of integers. Is there a pair of integers that totals 102 in Set S? (1) The set S has 27 distinct positive odd integers. (2) The greatest integer in Set S is less than 100.

1 and 2 are insufficient by themselves because there is not enough information about the values.

Arithmetic mean:

For a set of n numbers x1, x2, x3, . . . xn, the arithmetic mean is calculated as: x = Sum of the n numbers / n

In a survey, it was found that 65 people keep dogs as pets, 70 people keep cats, 40 people keep birds, 20 people keep both dogs and cats, 10 people keep both cats and birds, 5 people keep both birds and dogs and 3 people keep all three. How many people participated in the survey if it is known that each person has at least pet?

A = 65, B = 40, C = 70 A ∩ B = 20, B ∩ C = 10, C ∩ A = 5 A ∩ B ∩ C = 3 The total number of people surveyed = A ∪ B ∪ C = A + B + C − (A ∩ B + B ∩ C + C ∩ A) + A ∩ B ∩ C = 65 + 40 + 70 − (20 + 10 + 5) + 3 = 143

Simple Set Breakdown

A = p + r, B = q + r, A ∩ B = r, A ∪ B = p + r + q A ∪ B = p + q + r = (p + r ) + (q + r ) − r => (A ∪ B)' = s A' = q + s, B' = p + s => A' ∩ B '= s

Three set breakdown

A = p + s + t + v B = q + s + u + v C = r + t + u + v A ∩ B = s + v, A ∩ C = t + v, B ∩ C = u + v A ∩ B ∩ C = v A ∪ B ∪ C = p + q + r + s + t + u + v A ∩ B + B ∩ C + C ∩ A = (s + v) + (t + v) + (u + v) = s + t + u + 3v

What is the median of 123, 134, 128, 139, 141 and 126 ?

After arranging data in ascending order, we have 123, 126, 128, 134, 139, 141

Complement:

Complement of set A, i.e. A−1 or A0 is a set that contains the elements outside set A. The universal set is the whole thing. For example: If A = {1, 2} and Universal Set U = {1, 2, 7, 8}, A0 = {7, 8}

Difference of Two Sets

Difference of two sets A and B, i.e. A − B is a set of elements present in set A but not in set B. For example: For A = {2, 4}, B = {2, 6, 5}, A − B = {4}, B − A = {5, 6}

What is the maximum percentage change in production of any item from 2013 to 2014 ?

Explanation: For this, we first need to identify which item has the maximum % change. Looking at the line graph, it is clear that the item may be mobiles or TVs, since each has a large difference in its two-year values. Percent increase in TVs = 180 − 120 /20 × 100% = 50% Percent increase in Mobiles = 300 − 200/200 × 100% = 50% Thus, the answer is 50%.

The average (arithmetic mean) of six numbers is 7.5. If one of the six numbers is multiplied by 4, the average of the numbers decreases by 0.5, which of the six numbers is multiplied by 4? (A) −3 (B) −2 (C) −1 (D) 2 (E) 3

Given: Average = 7.5 => Sum of 6 numbers = 6 × average = 6 × 7.5 = 45 Say the number multiplied by 4 is x, so instead of x, the number becomes 4x. Or the sum of the numbers increases by 3x. => New sum = 45 + 3x Given that new average = 7.5 − 0.5 = 7 => New sum = 6 × new average = 6 × 7 = 42 => 45 + 3x = 42 => x = −1

Which item has the smallest percentage increase in its production from 2013 to 2014? (Bar Graph)

Here we need to compare the percentage changes for each item and select the minimum value. For minimum % change, the initial base value should be high and the corresponding change should be low. We see that Laptops have a high initial value and the change is also small, so Laptops could be a possible answer. Among the others, DVDs have a decrease, so cannot be the answer, while the others have a relatively large change in their values.

Arithmetic mean w/ frequencies

If x1, x2, x3, . . . xn be n observations and f1, f2, f3, . . . fn, be their corresponding frequencies, their arithmetic mean is calculated as: x = (x1 × f1 + x2 × f2 + . . . xn × fn) / f1 + f2 + · · · + fn

Arithmetic mean w/ frequencies : Alternate Approach

If x1, x2, x3, . . . xn be n observations and f1, f2, f3, . . . fn, be their corresponding frequencies, their arithmetic mean is calculated as: x = xa + (((x1 − xa) × f1 + (x2 − xa) × f2 + . . . (xn − xa) × fn)) / (f1 + f2 + · · · + fn))

A survey is conducted on monthly salaries of households of a locality. For each household, the salary of the eldest member is considered. The data is then grouped according to different age-groups. From each age-group, the median salary is considered as shown in the table below. Age-group & Median Salary: Between 15 and 25 years $12,000 Between 25 and 35 years $24,000 Between 35 and 45 years $36,000 Between 45 and 55 years Between $38,000 55 and 65 years $28,000 If the data for the age groups 15-25 years and 55-65 years are ignored, which of the following statements would be correct? I. The mean of the above median salaries would change by less than $5,000. II. The median of the above median salaries would change by less than $5,000. III. The difference between the new mean and new median of the above median salaries is less than $5,000.

In the table given above, the median salary of each age-group is considered, i.e. the salary of one individual is considered as a representation of each age-group. Thus, the mean salary is simply the sum of the above 5 salaries, divided by 5. Similarly, the median salary is the ( 5 + 1)/ 2 th term, i.e. 3rd term, when the above salaries are arranged in order

Intersection

Intersection of two sets A and B, i.e. A ∩ B, that contains the elements common to both sets A and B. For example: For A = {2, 3}, B = {1, 3, 5}, A ∩ B = {3}

Median

Median refers to the middle value of all observations arranged in ascending or descending order. If n is odd: Median = Value of the (n + 1)/2 th observation. If n is even: Median = Average of (n/2) th and (n/2 +1)th observations

(1) What is the mean of 123, 134, 128, 139, 141, and 127?

Mean = (123 + 134 + 128 + 139 + 141 + 127) / 6 = 132

Is the standard deviation of Set A greater than the standard deviation of Set B? (1) Elements is set A are consecutive multiples of 5. (2) Elements is set B are consecutive multiples of 3.

Note that SDs of two sets would be same if they have equal number of elements, and the elements are equally placed. Set A would be like {25, 30, 35, 40, ....}, and Set B would be like {18, 21, 24, 27, ....}. Since the deviation among the elements of Set A (5) is greater than the deviation among the elements of Set B (3), we can conclude that the SD of set A is greater than the SD of set B, but the answer is E not C as Set B may have more number of elements than that in Set A, resulting in greater value of SD for itself. Note that we cannot assume that both the sets would have the same number of elements. Had the # of elements been the same, answer would have been C. The correct answer is option

Comparing two fractions:

Of two fractions, if their denominators are equal, the fraction with the greatest numerator is the greatest. • Of two fractions, if their numerators are equal, the fraction with the smallest denominator is the greatest. • Of two fractions, if one fraction has greater numerator and smaller denominator, it is the greater one. For example: Between 5/13 and 9/11, the latter is the greater fraction.

If one of P and Q is a positive number and the other is a negative number, which of the following could be the standard deviation of set {P, −8, −12, 4, 0, −4, 8, Q}? I. −2 II. −4 III. 0

Since SD is always positive, so option I & II are ruled out. SD for the given set CANNOT be '0' as there are deviations among elements. For SD to be '0', all the elements of a set must be equal. The correct answer is option E (none of them).

Range

Range refers to the difference between the values of maximum and minimum observations. For example, the range of the set of observations 11, 13, 23, 21, 13, 17, 19 is 23 − 11 = 12

Cricket/Football/Hockey Question part 2

Since C > F > H > (a + b + c)> d ≥ 1 , we assume d = 1, a + b + c = 2, and H = 3. This follows that F = 4 (minimum) (the answer) So we make the distribution in the diagram. we have C = 50 − 4 − 3 − 2 − 1 = 44, F = 4, H = 3, a = 1, b = 0, c = 1, and d = 1.

standard deviation

Standard deviation shows the spread of the observations. If the observations are far away from the mean, the standard deviation would be high; similarly, if the observations are close to the mean, the standard deviation would be low. For a set of n numbers x1, x2, x3, . . . xn, the Standard Deviation (s)

Cardinal Number of a Set:

The Cardinal number of a finite set A is the number of elements of the set, denoted by n(A). For example: For A = {1, 2, 3}, n

Disjoint Set:

Two sets are disjoint if they have no elements in common. For example: A = {1, 2, 3} and B = {4, 5, 6} are two disjoint sets.

What is the net difference in total production (in thousands) between 2013 and 2014?

We can simply add up the values of 2013 and 2014 and subtract them. Alternately, we can find the difference between each food item in each year and add up the differences (with sign):

In a class of 50 boys, some play cricket, some football and some hockey. The number of boys who play cricket is more than those who play football, which is more than those who play hockey, which is more than those who play only two games, which is more than those who play all three games. If each boy plays at least one game and there is at least one boy who plays all three games, what is the minimum number of boys who play football?

We know that the boys play cricket (C), football (F) or hockey (H). As the information, 50 = C + F + H − Only two games + All three games In order to minimize F, we need to minimize the number of students in regions a, b, c and d, as shown in the diagram below. 50 = C + F + H −(a + b + c) + d

If a constant number k is added to (or subtracted from) each term of the above set of observations, we have

◦ The Mean increases (or decreases) by k ◦ The Median increases (or decreases) by k ◦ The Range remains unchanged ◦ The Standard Deviation remains unchanged

If a new term, whose value is equal to the mean, is introduced in the above set of observations, we have

◦ The Mean remains unchanged ◦ Nothing can be concluded about Median; it depends on the distribution ◦ The Range remains unchanged ◦ The Standard Deviation decreases

For a set of fractions less than 1

if the difference between the numerator and its denominator is equal for all the fractions, the fraction with the greatest numerator is the greatest. ◦ For example: Among 2/3 , 5/6 , 9/10, and 11/12, the last fraction 11/12 is the greatest.

For a set of fractions greater than 1

if the difference between the numerator and its denominator is equal for all the fractions, the fraction with the least numerator is the greatest. ◦ For example: Among 3/2 , 6/5 , 10/9 , and 12/11, the first fraction 3/2 is the greatest.

If the terms of a set of observations arranged in ascending or descending order have a constant difference between consecutive terms

they form an Arithmetic Sequence, we have Mean = Median

Last way to compare fractions?

to cross-multiply the numerator of one fraction and the denominator of the other fraction and then compare the products. ◦ For example: For the fractions 13/23 and 7/13 We multiply 13 and 13 to get 169 and on the other hand multiply 7 and 23 to get 161. Since 169 > 161, we have (13 × 13) > (7 × 23) => 13/23 > 7/13.

What is the mean of the observations shown below? Observations Frequencies 11 - 14 7 - 8 15 - 6 13 - 12

=> Mean = 456 / 40 = 11.4

Union

The Union of two sets A and B, i.e. A ∪ B, is a set that contains all the elements contained in set A or set B. For example: If A = {2, 3}, B = {1, 3, 5} => A ∪ B = {1, 2, 3, 5}

How to compare fractions by making the denominator equal 1?

For example: For the fractions 5/13 and 9/20, we make the numerator 1 for both.

The average of five distinct integers is 65. If the largest integer is 75, what is the maximum possible value of the smallest integer? (A) 60 (B) 61 (C) 62 (D) 63 (E) 64

Since the average of 5 integers is 65; thus, the total = 65 × 5 = 325 - The largest integer is 75 Thus, sum of the remaining 4 integers = 325 − 75 = 250 - Thus, the average of the 4 integers = 250/4 = 62.5 - The smallest integer will be maximized if these 4 integers are as close to each other as possible. - Since the numbers are distinct integers, let us subtract 0.5 from the average (62.5) for one of the numbers and add 0.5 to the average (62.5) for the other number; giving us 62.5 - 0.5 = 62 and 62.5 + 0.5 = 63. -There are still two numbers left. Let us subtract 1.5 from the average (62.5) for one of them and add 1.5 to the average (62.5) for the other; giving us 62.5 - 1.5 = 61 and 62.5 + 1.5 = 64. -Thus, the 4 integers are 61, 62, 63 and 64. Thus, the maximum possible value of the smallest integer = 61.

John, a student of the ninth grade, scored 75 in an exam in which the mean score of all students was 70 and standard deviation of the scores was 2. Bob, a student of the tenth grade, scored 72 in an exam in which the mean score of all students was 65 and standard deviation of the scores was 3. Which of the following statements must be correct? I. John's performance was better than Bob's performance in their respective exams. II. There was at least one student who scored more than John. III. There was at least one student who scored less than Bob.

Statement II: Since John's score is higher than the mean score, there may or may not be a student scoring higher than John. - Statement II may be correct but must not be correct

If a constant number k is multiplied (divided) to each term of the above set of observations, we have

The Mean is multiplied (divided) by k ◦ The Median is multiplied (divided) by k ◦ The Range is multiplied (divided) by k ◦ The Standard Deviation is multiplied (divided) by k


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