One can expect two to three questions from Statistics and Averages in the quant section of the GRE General Test. Questions from this topic can span the entire spectrum of easy to hard questions. So, having a good grasp of the following concepts in Statistics focusing on central tendancy is mandatory to get a good GRE score in the quant section.
The following concepts are also tested that measure extent of dispersion in a data set and distribution of data. Learn the basics, properties and these questions tend to be the medium to hard questions that appear in Statistics in the quant section of the GRE General Test.
Sample GRE practice questions from statistics & averages is given below. Attempt these questions and check whether you have got the correct answer. If you face difficulty with arriving at the answer to any question, go to the explanatory answer or the video explanations (provided for all questions) to learn how to crack the GRE sample question in this question bank.
Ideally, you should start by watching these two GRE Math lesson videos in Statistics and Averages to help you get better traction when solving the questions given below.
The maximum mark in an examination is 100 and the minimum is 0. The average mark of seven students such that no two of them have scored the same marks is 88. If the median score is 92 and all the marks are integers, what is the maximum possible difference between the highest and the least mark obtained by these seven students?
Step 1: Key to solving this question - the difference is maximized when the highest mark is maximized and the lowest mark is minimized.
Step 2: Maximizing the highest mark is easy. Complete that step
Step 3: Compute the sum of marks of all 7 students using information about the average marks of the 7 students.
Step 4: Key constraint - the median is 92. Therefore, marks of 3 students will be lesser than 92 and marks of 3 students will be more than 92.
Step 5: The lowest mark is minimized when the remaining marks are all maximized. Complete this step keeping constraint mentioned in step 4.
Step 6: Plug in the maximum values of all the marks other than the median and the lowest mark in sum of marks equation to compute the minimum value of the lowest marks.
Step 7: The difference between the maximum value of the highest mark and the minimum value of the lowest mark is the answer to this question.
The average (arithmetic mean) of two numbers x and y is 15. If x, y, and z are non-negative integers such that x < z < y, what is the minimum possible average of x, y, and z?
Step 1: Compute sum of x and y.
Step 2: The average of x, y, and z will be minimum if z is minimized.
Step 3: All numbers are non-negative integers. So, least value possible is 0.
Step 4: Because z > x, assign as low a value of x and z as possible.
Step 5: Compute y and subsequently the average of x, y, and z after assigning least values for x and z.
If the average (arithmetic mean) of 5 positive integers is 11, what is the maximum possible difference between the largest and the smallest of these 5 numbers?
Step 1: Compute sum of all 5 numbers from information about their average.
Step 2: The difference is maximized when the largest number is maximized and the smallest number is minimized.
Step 3: Because the numbers are positive integers, least possible value for the smallest number is 1.
Step 4: Note: the question does not mention that the numbers are distinct. So, keep that in mind and find what values of the remaining 3 numbers will maximize the largest number.
Step 5: Compute the difference after computing the maximum value of the largest number.
The average weight of the women in a group is 60 kg and that of the men is 72 kg. If the average weight of the group is 68 kg, what is the ratio of women to men in the group?
Two ways to solve this question. We will discuss the first method here. The alternative faster method is presented in the video solution. Step 1: Assign variables for the number of men and women in the group, say m and n respectively
Step 2: Compute the sum of weights of all men and that of all women in terms of m and n.
Step 3: Total number of people in the group = (m + n). Using this information and the overall average compute the sum of weights all men and women in the group.
Step 4: Sum of numbers computed in step 2 = number obtained in step 3
Step 5: Equate the two to compute the required ratio.
If the median of -10, 29, 6, 11, 31 and x is 20. What is the least possible average of the 6 numbers?
Step 1: Median of 6 numbers is the arithmetic mean of the 3rd and the 4th term. So, \\frac{\text{t_3 + t_4}}{2} = 20). Or t_{3} + t_{4} = 40
Step 2: The ascending order of the remaining 5 numbers is -10, 6, 11, 29, 31
Step 3: Iteratively, place x starting from a value less than 10 and check whether t_{3} + t_{4} = 40
Step 4: Because we are starting with the lowest possible value when we start with a value less than -10, the first value of x for which t_{3} + t_{4} = 40 is the least value of x
Step 5: Plug in the least value possible for x to compute the least average possible for the given 6 numbers.
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