How to add negative numbers - PSAT Math
Card 0 of 14
a, b, c are integers.
abc < 0
ab > 0
bc > 0
Which of the following must be true?
a, b, c are integers.
abc < 0
ab > 0
bc > 0
Which of the following must be true?
Tap to see back →
Let's reductively consider what this data tells us.
Consider each group (a,b,c) as a group of signs.
From abc < 0, we know that the following are possible:
(–, +, +), (+, –, +), (+, +, –), (–, –, –)
From ab > 0, we know that we must eliminate (–, +, +) and (+, –, +)
From bc > 0, we know that we must eliminate (+, +, –)
Therefore, any of our answers must hold for (–, –, –)
This eliminates immediately a > 0, b > 0
Likewise, it eliminates a – b > 0 because we do not know the relative sizes of a and b. This could therefore be positive or negative.
Finally, ac is a product of negatives and is therefore positive. Hence ac < 0 does not hold.
We are left with a + b < 0, which is true, for two negatives added must be negative.
Let's reductively consider what this data tells us.
Consider each group (a,b,c) as a group of signs.
From abc < 0, we know that the following are possible:
(–, +, +), (+, –, +), (+, +, –), (–, –, –)
From ab > 0, we know that we must eliminate (–, +, +) and (+, –, +)
From bc > 0, we know that we must eliminate (+, +, –)
Therefore, any of our answers must hold for (–, –, –)
This eliminates immediately a > 0, b > 0
Likewise, it eliminates a – b > 0 because we do not know the relative sizes of a and b. This could therefore be positive or negative.
Finally, ac is a product of negatives and is therefore positive. Hence ac < 0 does not hold.
We are left with a + b < 0, which is true, for two negatives added must be negative.
a, b, c are integers.
abc < 0
ab > 0
bc > 0
Which of the following must be true?
a, b, c are integers.
abc < 0
ab > 0
bc > 0
Which of the following must be true?
Tap to see back →
Let's reductively consider what this data tells us.
Consider each group (a,b,c) as a group of signs.
From abc < 0, we know that the following are possible:
(–, +, +), (+, –, +), (+, +, –), (–, –, –)
From ab > 0, we know that we must eliminate (–, +, +) and (+, –, +)
From bc > 0, we know that we must eliminate (+, +, –)
Therefore, any of our answers must hold for (–, –, –)
This eliminates immediately a > 0, b > 0
Likewise, it eliminates a – b > 0 because we do not know the relative sizes of a and b. This could therefore be positive or negative.
Finally, ac is a product of negatives and is therefore positive. Hence ac < 0 does not hold.
We are left with a + b < 0, which is true, for two negatives added must be negative.
Let's reductively consider what this data tells us.
Consider each group (a,b,c) as a group of signs.
From abc < 0, we know that the following are possible:
(–, +, +), (+, –, +), (+, +, –), (–, –, –)
From ab > 0, we know that we must eliminate (–, +, +) and (+, –, +)
From bc > 0, we know that we must eliminate (+, +, –)
Therefore, any of our answers must hold for (–, –, –)
This eliminates immediately a > 0, b > 0
Likewise, it eliminates a – b > 0 because we do not know the relative sizes of a and b. This could therefore be positive or negative.
Finally, ac is a product of negatives and is therefore positive. Hence ac < 0 does not hold.
We are left with a + b < 0, which is true, for two negatives added must be negative.
a, b, c are integers.
abc < 0
ab > 0
bc > 0
Which of the following must be true?
a, b, c are integers.
abc < 0
ab > 0
bc > 0
Which of the following must be true?
Tap to see back →
Let's reductively consider what this data tells us.
Consider each group (a,b,c) as a group of signs.
From abc < 0, we know that the following are possible:
(–, +, +), (+, –, +), (+, +, –), (–, –, –)
From ab > 0, we know that we must eliminate (–, +, +) and (+, –, +)
From bc > 0, we know that we must eliminate (+, +, –)
Therefore, any of our answers must hold for (–, –, –)
This eliminates immediately a > 0, b > 0
Likewise, it eliminates a – b > 0 because we do not know the relative sizes of a and b. This could therefore be positive or negative.
Finally, ac is a product of negatives and is therefore positive. Hence ac < 0 does not hold.
We are left with a + b < 0, which is true, for two negatives added must be negative.
Let's reductively consider what this data tells us.
Consider each group (a,b,c) as a group of signs.
From abc < 0, we know that the following are possible:
(–, +, +), (+, –, +), (+, +, –), (–, –, –)
From ab > 0, we know that we must eliminate (–, +, +) and (+, –, +)
From bc > 0, we know that we must eliminate (+, +, –)
Therefore, any of our answers must hold for (–, –, –)
This eliminates immediately a > 0, b > 0
Likewise, it eliminates a – b > 0 because we do not know the relative sizes of a and b. This could therefore be positive or negative.
Finally, ac is a product of negatives and is therefore positive. Hence ac < 0 does not hold.
We are left with a + b < 0, which is true, for two negatives added must be negative.
a, b, c are integers.
abc < 0
ab > 0
bc > 0
Which of the following must be true?
a, b, c are integers.
abc < 0
ab > 0
bc > 0
Which of the following must be true?
Tap to see back →
Let's reductively consider what this data tells us.
Consider each group (a,b,c) as a group of signs.
From abc < 0, we know that the following are possible:
(–, +, +), (+, –, +), (+, +, –), (–, –, –)
From ab > 0, we know that we must eliminate (–, +, +) and (+, –, +)
From bc > 0, we know that we must eliminate (+, +, –)
Therefore, any of our answers must hold for (–, –, –)
This eliminates immediately a > 0, b > 0
Likewise, it eliminates a – b > 0 because we do not know the relative sizes of a and b. This could therefore be positive or negative.
Finally, ac is a product of negatives and is therefore positive. Hence ac < 0 does not hold.
We are left with a + b < 0, which is true, for two negatives added must be negative.
Let's reductively consider what this data tells us.
Consider each group (a,b,c) as a group of signs.
From abc < 0, we know that the following are possible:
(–, +, +), (+, –, +), (+, +, –), (–, –, –)
From ab > 0, we know that we must eliminate (–, +, +) and (+, –, +)
From bc > 0, we know that we must eliminate (+, +, –)
Therefore, any of our answers must hold for (–, –, –)
This eliminates immediately a > 0, b > 0
Likewise, it eliminates a – b > 0 because we do not know the relative sizes of a and b. This could therefore be positive or negative.
Finally, ac is a product of negatives and is therefore positive. Hence ac < 0 does not hold.
We are left with a + b < 0, which is true, for two negatives added must be negative.
a, b, c are integers.
abc < 0
ab > 0
bc > 0
Which of the following must be true?
a, b, c are integers.
abc < 0
ab > 0
bc > 0
Which of the following must be true?
Tap to see back →
Let's reductively consider what this data tells us.
Consider each group (a,b,c) as a group of signs.
From abc < 0, we know that the following are possible:
(–, +, +), (+, –, +), (+, +, –), (–, –, –)
From ab > 0, we know that we must eliminate (–, +, +) and (+, –, +)
From bc > 0, we know that we must eliminate (+, +, –)
Therefore, any of our answers must hold for (–, –, –)
This eliminates immediately a > 0, b > 0
Likewise, it eliminates a – b > 0 because we do not know the relative sizes of a and b. This could therefore be positive or negative.
Finally, ac is a product of negatives and is therefore positive. Hence ac < 0 does not hold.
We are left with a + b < 0, which is true, for two negatives added must be negative.
Let's reductively consider what this data tells us.
Consider each group (a,b,c) as a group of signs.
From abc < 0, we know that the following are possible:
(–, +, +), (+, –, +), (+, +, –), (–, –, –)
From ab > 0, we know that we must eliminate (–, +, +) and (+, –, +)
From bc > 0, we know that we must eliminate (+, +, –)
Therefore, any of our answers must hold for (–, –, –)
This eliminates immediately a > 0, b > 0
Likewise, it eliminates a – b > 0 because we do not know the relative sizes of a and b. This could therefore be positive or negative.
Finally, ac is a product of negatives and is therefore positive. Hence ac < 0 does not hold.
We are left with a + b < 0, which is true, for two negatives added must be negative.
a, b, c are integers.
abc < 0
ab > 0
bc > 0
Which of the following must be true?
a, b, c are integers.
abc < 0
ab > 0
bc > 0
Which of the following must be true?
Tap to see back →
Let's reductively consider what this data tells us.
Consider each group (a,b,c) as a group of signs.
From abc < 0, we know that the following are possible:
(–, +, +), (+, –, +), (+, +, –), (–, –, –)
From ab > 0, we know that we must eliminate (–, +, +) and (+, –, +)
From bc > 0, we know that we must eliminate (+, +, –)
Therefore, any of our answers must hold for (–, –, –)
This eliminates immediately a > 0, b > 0
Likewise, it eliminates a – b > 0 because we do not know the relative sizes of a and b. This could therefore be positive or negative.
Finally, ac is a product of negatives and is therefore positive. Hence ac < 0 does not hold.
We are left with a + b < 0, which is true, for two negatives added must be negative.
Let's reductively consider what this data tells us.
Consider each group (a,b,c) as a group of signs.
From abc < 0, we know that the following are possible:
(–, +, +), (+, –, +), (+, +, –), (–, –, –)
From ab > 0, we know that we must eliminate (–, +, +) and (+, –, +)
From bc > 0, we know that we must eliminate (+, +, –)
Therefore, any of our answers must hold for (–, –, –)
This eliminates immediately a > 0, b > 0
Likewise, it eliminates a – b > 0 because we do not know the relative sizes of a and b. This could therefore be positive or negative.
Finally, ac is a product of negatives and is therefore positive. Hence ac < 0 does not hold.
We are left with a + b < 0, which is true, for two negatives added must be negative.
a, b, c are integers.
abc < 0
ab > 0
bc > 0
Which of the following must be true?
a, b, c are integers.
abc < 0
ab > 0
bc > 0
Which of the following must be true?
Tap to see back →
Let's reductively consider what this data tells us.
Consider each group (a,b,c) as a group of signs.
From abc < 0, we know that the following are possible:
(–, +, +), (+, –, +), (+, +, –), (–, –, –)
From ab > 0, we know that we must eliminate (–, +, +) and (+, –, +)
From bc > 0, we know that we must eliminate (+, +, –)
Therefore, any of our answers must hold for (–, –, –)
This eliminates immediately a > 0, b > 0
Likewise, it eliminates a – b > 0 because we do not know the relative sizes of a and b. This could therefore be positive or negative.
Finally, ac is a product of negatives and is therefore positive. Hence ac < 0 does not hold.
We are left with a + b < 0, which is true, for two negatives added must be negative.
Let's reductively consider what this data tells us.
Consider each group (a,b,c) as a group of signs.
From abc < 0, we know that the following are possible:
(–, +, +), (+, –, +), (+, +, –), (–, –, –)
From ab > 0, we know that we must eliminate (–, +, +) and (+, –, +)
From bc > 0, we know that we must eliminate (+, +, –)
Therefore, any of our answers must hold for (–, –, –)
This eliminates immediately a > 0, b > 0
Likewise, it eliminates a – b > 0 because we do not know the relative sizes of a and b. This could therefore be positive or negative.
Finally, ac is a product of negatives and is therefore positive. Hence ac < 0 does not hold.
We are left with a + b < 0, which is true, for two negatives added must be negative.
How many elements of the set 
are less than 
?
How many elements of the set
Tap to see back →
The absolute value of a negative number can be calculated by simply removing the negative symbol. Therefore,
All four (negative) numbers in the set 
are less than this positive number.
The absolute value of a negative number can be calculated by simply removing the negative symbol. Therefore,
All four (negative) numbers in the set
How many elements of the set are less than ?
How many elements of the set
Tap to see back →
The absolute value of a negative number can be calculated by simply removing the negative symbol. Therefore,
All four (negative) numbers in the set are less than this positive number.
The absolute value of a negative number can be calculated by simply removing the negative symbol. Therefore,
All four (negative) numbers in the set
How many elements of the set are less than ?
How many elements of the set
Tap to see back →
The absolute value of a negative number can be calculated by simply removing the negative symbol. Therefore,
All four (negative) numbers in the set are less than this positive number.
The absolute value of a negative number can be calculated by simply removing the negative symbol. Therefore,
All four (negative) numbers in the set
How many elements of the set are less than ?
How many elements of the set
Tap to see back →
The absolute value of a negative number can be calculated by simply removing the negative symbol. Therefore,
All four (negative) numbers in the set are less than this positive number.
The absolute value of a negative number can be calculated by simply removing the negative symbol. Therefore,
All four (negative) numbers in the set
How many elements of the set are less than ?
How many elements of the set
Tap to see back →
The absolute value of a negative number can be calculated by simply removing the negative symbol. Therefore,
All four (negative) numbers in the set are less than this positive number.
The absolute value of a negative number can be calculated by simply removing the negative symbol. Therefore,
All four (negative) numbers in the set
How many elements of the set are less than ?
How many elements of the set
Tap to see back →
The absolute value of a negative number can be calculated by simply removing the negative symbol. Therefore,
All four (negative) numbers in the set are less than this positive number.
The absolute value of a negative number can be calculated by simply removing the negative symbol. Therefore,
All four (negative) numbers in the set
How many elements of the set are less than ?
How many elements of the set
Tap to see back →
The absolute value of a negative number can be calculated by simply removing the negative symbol. Therefore,
All four (negative) numbers in the set are less than this positive number.
The absolute value of a negative number can be calculated by simply removing the negative symbol. Therefore,
All four (negative) numbers in the set