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msignatius
20 Dec 2025, 05:20
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Some pre-thinking helps here.
A, B and C are not zero, but any other integer. The standard deviation will be determined by looking at the middle term and the distance on the number line between that term and the ones smaller and larger than it.
We are told the standard deviation of A, B, C is greater than that of |A|, |B|, and |C|. That's not hard to figure out - clearly, it is the existence of a negative and positive value simultaneously in the A, B, C set that makes this happen. The values will also not have to be too far apart to make this happen. One option is -3, 2, 4 (here, the -3 is 5 away from 2, the median); which in absolute terms would be 3, 2, 4, where the SD is much smaller. Also, the range of A, B, C is 4.
Know, let's see which can be true:
I'll start with the product and A, B, and C is a prime number.
3 is a small prime number, so let's see if this works with 3:
-3, 1, -1: Here, the SD is naturally larger than 3, 1, 1; plus: -3*1*-1 = 3, is possible.
The mode of A, B, C, is one. Well, look at the example above: -3, 1, -1 = here |1| = 1, is the mode. So yes, possible.
The median of A, B, C is -2. This makes sense if you consider another example:
-3, -2, 1 (higher SD than 3, 2, 1); plus same range of 4. Here, the -2 is the MEDIAN.
Hence, all 3 can be true: E
Here, we get -2, wh
A, B and C are not zero, but any other integer. The standard deviation will be determined by looking at the middle term and the distance on the number line between that term and the ones smaller and larger than it.
We are told the standard deviation of A, B, C is greater than that of |A|, |B|, and |C|. That's not hard to figure out - clearly, it is the existence of a negative and positive value simultaneously in the A, B, C set that makes this happen. The values will also not have to be too far apart to make this happen. One option is -3, 2, 4 (here, the -3 is 5 away from 2, the median); which in absolute terms would be 3, 2, 4, where the SD is much smaller. Also, the range of A, B, C is 4.
Know, let's see which can be true:
I'll start with the product and A, B, and C is a prime number.
3 is a small prime number, so let's see if this works with 3:
-3, 1, -1: Here, the SD is naturally larger than 3, 1, 1; plus: -3*1*-1 = 3, is possible.
The mode of A, B, C, is one. Well, look at the example above: -3, 1, -1 = here |1| = 1, is the mode. So yes, possible.
The median of A, B, C is -2. This makes sense if you consider another example:
-3, -2, 1 (higher SD than 3, 2, 1); plus same range of 4. Here, the -2 is the MEDIAN.
Hence, all 3 can be true: E
Here, we get -2, wh
Bunuel
20 Dec 2025, 05:27
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SD for {a, b, c} is greater than {|a|, |b|, |c|}
This implies a larger range for {a, b, c} than {|a|, |b|, |c|}
Range of the set is 4 and a or b or c != 0
This is possible when we have either one or two negative numbers in the set
Possible sets could be { -3, -1, 1}, {-3, -2, 1}, {-2, 1, 2}, {-2, -1, 2}, {-1, 1, 3}, {-1, 2, 3}
Median of set is -2 can be satisfied by {-3, -2, 1}
Product of a, b, c is a prime number is satisfied by {-3, -1, 1}
Mode of {|a|, |b|, |c|} is 1 is satisfied by {-3, -1, 1}
All three values are possible with different combinations of {a, b, c}
Option E
This implies a larger range for {a, b, c} than {|a|, |b|, |c|}
Range of the set is 4 and a or b or c != 0
This is possible when we have either one or two negative numbers in the set
Possible sets could be { -3, -1, 1}, {-3, -2, 1}, {-2, 1, 2}, {-2, -1, 2}, {-1, 1, 3}, {-1, 2, 3}
Median of set is -2 can be satisfied by {-3, -2, 1}
Product of a, b, c is a prime number is satisfied by {-3, -1, 1}
Mode of {|a|, |b|, |c|} is 1 is satisfied by {-3, -1, 1}
All three values are possible with different combinations of {a, b, c}
Option E
Bunuel
20 Dec 2025, 06:09
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a, b, and c are distinct non-zero integers, and the standard deviation of {a, b, c} is greater than the standard deviation of {|a|, |b|, |c|}. If the range of {a, b, c} is 4, which of the following could be true?
We need to follow the above constraints and find which options could be true.
The SD decreases after taking absolute values when the original set has mixed signs.
I. The median of {a, b, c} is -2
Lets take (-3,-2,1), they are all distinct, range is 4 and median is -2. And original SD > SD (3,2,1)
It is possible.
II. The product of a, b, and c is a prime number
Lets take (-1,1,3), they are all distinct, product= -3 and range=4
Original SD> SD(1,1,3)
It is possible
III. The mode of {|a|, |b|, |c|} is 1
Lets take (-1,1,3), they are all distinct and non zero. Their absolute values are (1,1,3) and mode is 1.
It is possible
E. I,II, and III
We need to follow the above constraints and find which options could be true.
The SD decreases after taking absolute values when the original set has mixed signs.
I. The median of {a, b, c} is -2
Lets take (-3,-2,1), they are all distinct, range is 4 and median is -2. And original SD > SD (3,2,1)
It is possible.
II. The product of a, b, and c is a prime number
Lets take (-1,1,3), they are all distinct, product= -3 and range=4
Original SD> SD(1,1,3)
It is possible
III. The mode of {|a|, |b|, |c|} is 1
Lets take (-1,1,3), they are all distinct and non zero. Their absolute values are (1,1,3) and mode is 1.
It is possible
E. I,II, and III
