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BATCH #5 (10 QUESTIONS)
21. Statistics and Sets Problems
21.1. Nine training videos have an average length of 74 minutes and a median length of 80 minutes. If the range of the video lengths is 31 minutes, what is the maximum possible length, in minutes, of the longest video?
A. 90
B. 92
C. 94
D. 96
E. 98
Let the nine video lengths in ascending order be
a <= b <= c <= d <= e <= f <= g <= h <= i
Since the average is 74, the total length is
9 * 74 = 666
Since the median is 80, the fifth term is 80. So
e = 80
Since the range is 31,
i - a = 31
So
i = a + 31
We want to maximize i. Since i = a + 31, we need to maximize a.
To make a as large as possible, make all the other terms as small as the conditions allow.
The smallest possible values of b, c, and d are a.
The smallest possible values of f, g, and h are 80.
So,
a + a + a + a + 80 + 80 + 80 + 80 + (a + 31) = 666
5a + 351 = 666
5a = 315
a = 63
Therefore,
i = 63 + 31 = 94
Answer: C
Takeaway
For statistics questions such as this one, you should be comfortable working with average, median, range, and ordered lists. You should also know minimizing and maximizing strategies.
A useful general rule in problems where the total is fixed is this:
• to maximize one quantity, minimize the others;
• to minimize one quantity, maximize the others.
What This Question Tests
This question tests average, median, range, and ordered lists, along with minimizing and maximizing concepts.
21. Statistics and Sets Problems
21.1. Nine training videos have an average length of 74 minutes and a median length of 80 minutes. If the range of the video lengths is 31 minutes, what is the maximum possible length, in minutes, of the longest video?
A. 90
B. 92
C. 94
D. 96
E. 98
Let the nine video lengths in ascending order be
a <= b <= c <= d <= e <= f <= g <= h <= i
Since the average is 74, the total length is
9 * 74 = 666
Since the median is 80, the fifth term is 80. So
e = 80
Since the range is 31,
i - a = 31
So
i = a + 31
We want to maximize i. Since i = a + 31, we need to maximize a.
To make a as large as possible, make all the other terms as small as the conditions allow.
The smallest possible values of b, c, and d are a.
The smallest possible values of f, g, and h are 80.
So,
a + a + a + a + 80 + 80 + 80 + 80 + (a + 31) = 666
5a + 351 = 666
5a = 315
a = 63
Therefore,
i = 63 + 31 = 94
Answer: C
Takeaway
For statistics questions such as this one, you should be comfortable working with average, median, range, and ordered lists. You should also know minimizing and maximizing strategies.
A useful general rule in problems where the total is fixed is this:
• to maximize one quantity, minimize the others;
• to minimize one quantity, maximize the others.
What This Question Tests
This question tests average, median, range, and ordered lists, along with minimizing and maximizing concepts.
30 Apr 2026, 03:08
Kudos
Bookmarks
21. Statistics and Sets Problems
21.2. A data set consists of five consecutive multiples of the same prime number p: {a, b, c, d, e}. If the sum of the five numbers is 0, the median is m, and the range is r, which of the following data sets must have a standard deviation smaller than that of {a, b, c, d, e}?
I. {ma, mb, mc, md, me}
II. {a + r, b + r, c + r, d + r, e + r}
III. {a/p, b/p, c/p, d/p, e/p}
A. I only
B. II only
C. III only
D. I and III only
E. II and III only
The standard deviation of a list shows how much variation there is from the mean, that is, how spread out the numbers are. A low standard deviation means the values tend to be close to the mean, while a high standard deviation means the values are spread out over a wider range.
So, in a sense, standard deviation measures distance from the mean. Since distance cannot be negative, the standard deviation of any list must be greater than or equal to 0: SD >= 0. Also, the standard deviation of a list is 0 if and only if all the numbers in the list are identical. That is also why a list containing only one number has standard deviation 0.
Two more properties are especially useful here.
1. If we multiply or divide every number in a list by the same nonzero constant, the standard deviation is multiplied or divided by the absolute value of that constant.
For example, if the SD of {a, b, c} is s, then the SD of {10a, 10b, 10c} is 10s, and the SD of {-5a, -5b, -5c} is s * |-5| = 5s.
2. If we add or subtract the same constant to every number in a list, the standard deviation does not change.
For example, if the SD of {a, b, c} is s, then the SD of {a + 5, b + 5, c + 5} and the SD of {a - 2, b - 2, c - 2} are both still s.
Now back to the question.
Since the five numbers are consecutive multiples of the same prime number p, we can write them as
kp, (k+1)p, (k+2)p, (k+3)p, (k+4)p
Their sum is 0, so
kp + (k+1)p + (k+2)p + (k+3)p + (k+4)p = 0
p(5k + 10) = 0
Since p ≠ 0, we get
5k + 10 = 0
k = -2
So the original data set is
{-2p, -p, 0, p, 2p}
The median is the middle term, so m = 0.
The range is r = 2p - (-2p) = 4p
Also, since the numbers in {-2p, -p, 0, p, 2p} are not all equal, the original data set has a positive standard deviation.
Now check each statement.
I. {ma, mb, mc, md, me}
Since m = 0, this becomes
{0, 0, 0, 0, 0}
All values are identical, so the standard deviation is 0. That must be smaller than the standard deviation of the original data set.
Therefore, {ma, mb, mc, md, me} must have a standard deviation smaller than that of the original data set.
II. {a + r, b + r, c + r, d + r, e + r}
This adds the same constant r to every number in the set. Adding the same constant to every value does not change the standard deviation.
Therefore, {a + r, b + r, c + r, d + r, e + r} has the same standard deviation as the original data set.
III. {a/p, b/p, c/p, d/p, e/p}
This divides every number in the set by the positive number p. Dividing every value by the same nonzero constant divides the standard deviation by the absolute value of that constant.
Since p is a prime number, p > 1, so the new standard deviation must be smaller.
Therefore, {a/p, b/p, c/p, d/p, e/p} must have a standard deviation smaller than that of the original data set.
So the data sets that must have a smaller standard deviation are I and III only.
Answer: D.
Takeaway
For standard deviation questions, you should know when a transformation changes the spread and when it does not. Adding or subtracting the same constant keeps the standard deviation unchanged, while multiplying or dividing every value by the same nonzero constant changes the standard deviation by that factor.
What This Question Tests
This question tests standard deviation, especially the effect of shifting and scaling an entire data set. It also tests whether you can use the given structure of the set to determine the actual values, the median, and the range.
21.2. A data set consists of five consecutive multiples of the same prime number p: {a, b, c, d, e}. If the sum of the five numbers is 0, the median is m, and the range is r, which of the following data sets must have a standard deviation smaller than that of {a, b, c, d, e}?
I. {ma, mb, mc, md, me}
II. {a + r, b + r, c + r, d + r, e + r}
III. {a/p, b/p, c/p, d/p, e/p}
A. I only
B. II only
C. III only
D. I and III only
E. II and III only
The standard deviation of a list shows how much variation there is from the mean, that is, how spread out the numbers are. A low standard deviation means the values tend to be close to the mean, while a high standard deviation means the values are spread out over a wider range.
So, in a sense, standard deviation measures distance from the mean. Since distance cannot be negative, the standard deviation of any list must be greater than or equal to 0: SD >= 0. Also, the standard deviation of a list is 0 if and only if all the numbers in the list are identical. That is also why a list containing only one number has standard deviation 0.
Two more properties are especially useful here.
1. If we multiply or divide every number in a list by the same nonzero constant, the standard deviation is multiplied or divided by the absolute value of that constant.
For example, if the SD of {a, b, c} is s, then the SD of {10a, 10b, 10c} is 10s, and the SD of {-5a, -5b, -5c} is s * |-5| = 5s.
2. If we add or subtract the same constant to every number in a list, the standard deviation does not change.
For example, if the SD of {a, b, c} is s, then the SD of {a + 5, b + 5, c + 5} and the SD of {a - 2, b - 2, c - 2} are both still s.
Now back to the question.
Since the five numbers are consecutive multiples of the same prime number p, we can write them as
kp, (k+1)p, (k+2)p, (k+3)p, (k+4)p
Their sum is 0, so
kp + (k+1)p + (k+2)p + (k+3)p + (k+4)p = 0
p(5k + 10) = 0
Since p ≠ 0, we get
5k + 10 = 0
k = -2
So the original data set is
{-2p, -p, 0, p, 2p}
The median is the middle term, so m = 0.
The range is r = 2p - (-2p) = 4p
Also, since the numbers in {-2p, -p, 0, p, 2p} are not all equal, the original data set has a positive standard deviation.
Now check each statement.
I. {ma, mb, mc, md, me}
Since m = 0, this becomes
{0, 0, 0, 0, 0}
All values are identical, so the standard deviation is 0. That must be smaller than the standard deviation of the original data set.
Therefore, {ma, mb, mc, md, me} must have a standard deviation smaller than that of the original data set.
II. {a + r, b + r, c + r, d + r, e + r}
This adds the same constant r to every number in the set. Adding the same constant to every value does not change the standard deviation.
Therefore, {a + r, b + r, c + r, d + r, e + r} has the same standard deviation as the original data set.
III. {a/p, b/p, c/p, d/p, e/p}
This divides every number in the set by the positive number p. Dividing every value by the same nonzero constant divides the standard deviation by the absolute value of that constant.
Since p is a prime number, p > 1, so the new standard deviation must be smaller.
Therefore, {a/p, b/p, c/p, d/p, e/p} must have a standard deviation smaller than that of the original data set.
So the data sets that must have a smaller standard deviation are I and III only.
Answer: D.
Takeaway
For standard deviation questions, you should know when a transformation changes the spread and when it does not. Adding or subtracting the same constant keeps the standard deviation unchanged, while multiplying or dividing every value by the same nonzero constant changes the standard deviation by that factor.
What This Question Tests
This question tests standard deviation, especially the effect of shifting and scaling an entire data set. It also tests whether you can use the given structure of the set to determine the actual values, the median, and the range.
30 Apr 2026, 03:08
Kudos
Bookmarks
22. Word Problems
22.1. A bakery baked 3/4 as many loaves this week as it baked the previous week. However, its total spending on flour this week was 3/2 of what it had been the previous week. If each loaf uses the same amount of flour in both weeks, by what percent did the price of flour increase from last week to this week?
A. 50%
B. 75%
C. 100%
D. 125%
E. 150%
Assume that last week the bakery baked 100 loaves and spent 100 euros on flour. So last week, the flour cost per loaf was 100/100 = 1 euro per loaf.
This week, the bakery baked 3/4 as many loaves, so it baked 100 * 3/4 = 75 loaves.
Its flour spending was 3/2 of last week’s spending, so it spent 100 * 3/2 = 150 euros.
So this week, the flour cost per loaf was 150/75 = 2 euros per loaf.
So, the price went from 1 euro per loaf to 2 euros per loaf, which corresponds to a 100% increase.
Answer: C.
Takeaway
For word problems, you should be comfortable translating the setup into algebraic relationships. You should also know how to find percent increase or percent change once you identify the original value and the new value.
What This Question Tests
This question tests translating a word problem into algebraic relationships and then using those relationships to find a percent increase.
22.1. A bakery baked 3/4 as many loaves this week as it baked the previous week. However, its total spending on flour this week was 3/2 of what it had been the previous week. If each loaf uses the same amount of flour in both weeks, by what percent did the price of flour increase from last week to this week?
A. 50%
B. 75%
C. 100%
D. 125%
E. 150%
Assume that last week the bakery baked 100 loaves and spent 100 euros on flour. So last week, the flour cost per loaf was 100/100 = 1 euro per loaf.
This week, the bakery baked 3/4 as many loaves, so it baked 100 * 3/4 = 75 loaves.
Its flour spending was 3/2 of last week’s spending, so it spent 100 * 3/2 = 150 euros.
So this week, the flour cost per loaf was 150/75 = 2 euros per loaf.
So, the price went from 1 euro per loaf to 2 euros per loaf, which corresponds to a 100% increase.
Answer: C.
Takeaway
For word problems, you should be comfortable translating the setup into algebraic relationships. You should also know how to find percent increase or percent change once you identify the original value and the new value.
What This Question Tests
This question tests translating a word problem into algebraic relationships and then using those relationships to find a percent increase.
