Integers - GRE Quantitative Reasoning
Card 0 of 1224
x = the total number of positive, odd 2 digit numbers less than 100
Column A
x
Column B
45
x = the total number of positive, odd 2 digit numbers less than 100
Column A
x
Column B
45
There are 50 positive, odd numbers less than 100, and 45 of them are 2 digit numbers.
There are 50 positive, odd numbers less than 100, and 45 of them are 2 digit numbers.
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What is the sum of all of the four-digit integers that can be created with the digits 1, 2, 3, and 4?
What is the sum of all of the four-digit integers that can be created with the digits 1, 2, 3, and 4?
First we need to find out how many possible numbers there are. The number of possible four-digit numbers with four different digits is simply 4 * 4 * 4 * 4 = 256.
To find the sum, the formula we must remember is sum = average * number of values. The last piece that's missing in this formula is the average. To find this, we can average the first and last number, since the numbers are consecutive. The smallest number that can be created from 1, 2, 3, and 4 is 1111, and the largest number possible is 4444. Then the average is (1111 + 4444)/2.
So sum = 5555/2 * 256 = 711,040.
First we need to find out how many possible numbers there are. The number of possible four-digit numbers with four different digits is simply 4 * 4 * 4 * 4 = 256.
To find the sum, the formula we must remember is sum = average * number of values. The last piece that's missing in this formula is the average. To find this, we can average the first and last number, since the numbers are consecutive. The smallest number that can be created from 1, 2, 3, and 4 is 1111, and the largest number possible is 4444. Then the average is (1111 + 4444)/2.
So sum = 5555/2 * 256 = 711,040.
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On a real number line, x1 = -4 and x2 = 14. What is the distance between these two points?
On a real number line, x1 = -4 and x2 = 14. What is the distance between these two points?
The distance between two points is always positive. We calculate lx2 - x1l, which will give us the distance between the points.
|14- (-4)| = |14+4| = |18| = 18
The distance between two points is always positive. We calculate lx2 - x1l, which will give us the distance between the points.
|14- (-4)| = |14+4| = |18| = 18
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If 
is even, and 
is odd. Which of the following must be odd?
If
To solve, pick numbers to represent 
and 
. Let 
and 
. Now try each of the equations given:
.
Only 
works and is thus our answer.
To solve, pick numbers to represent
Only
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x = the total number of positive, odd 2 digit numbers less than 100
Column A
x
Column B
45
x = the total number of positive, odd 2 digit numbers less than 100
Column A
x
Column B
45
There are 50 positive, odd numbers less than 100, and 45 of them are 2 digit numbers.
There are 50 positive, odd numbers less than 100, and 45 of them are 2 digit numbers.
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The sum of seven consecutive even integers is 0.
Column A: The product of the seven integers
Column B: 2
The sum of seven consecutive even integers is 0.
Column A: The product of the seven integers
Column B: 2
For the sum of 7 consecutive even integers to be zero, the only sequence possible is –6, –4, –2, 0, 2, 4, 6. This can be determined algebraically by assigning the lowest number in the sequence to be “y” and adding 2 for each consecutive even integer, and then setting this equal to zero.
y, y + 2, y + 4, y + 6 . . .
The product of any number and 0 is 0.
Therefore, column B must be greater.
For the sum of 7 consecutive even integers to be zero, the only sequence possible is –6, –4, –2, 0, 2, 4, 6. This can be determined algebraically by assigning the lowest number in the sequence to be “y” and adding 2 for each consecutive even integer, and then setting this equal to zero.
y, y + 2, y + 4, y + 6 . . .
The product of any number and 0 is 0.
Therefore, column B must be greater.
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and 
are both even whole numbers.
What is a possible solution for 
?
What is a possible solution for
If 
and 
are both even whole numbers, then their addition must be an even whole number as well. Although 
is an even number, it is not a whole number and could therefore not be a solution. This means the only possible solution would be 
.
If
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If John buys twenty-two apples on Monday, and thirty-four bananas on Tuesday, how many pieces of fruit does John have?
If John buys twenty-two apples on Monday, and thirty-four bananas on Tuesday, how many pieces of fruit does John have?
Plugging in the values given we arrive at the total fruit John has:
A good note about adding even numbers--any even numbers--is that if you add two even numbers, their sum will ALWAYS be an even number.
Plugging in the values given we arrive at the total fruit John has:
A good note about adding even numbers--any even numbers--is that if you add two even numbers, their sum will ALWAYS be an even number.
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Choose the answer below which best solves the following problem:
Choose the answer below which best solves the following problem:
To deal with a problem with this many digits, often the best strategy is to line up one number over the other, then add the places one at a time. Don't forget to carry a one every time the addition goes over ten. Also, note that any time you add two even numbers, their sum will ALWAYS be an even number.
To deal with a problem with this many digits, often the best strategy is to line up one number over the other, then add the places one at a time. Don't forget to carry a one every time the addition goes over ten. Also, note that any time you add two even numbers, their sum will ALWAYS be an even number.
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The sum of seven consecutive even integers is 0.
Column A: The product of the seven integers
Column B: 2
The sum of seven consecutive even integers is 0.
Column A: The product of the seven integers
Column B: 2
For the sum of 7 consecutive even integers to be zero, the only sequence possible is –6, –4, –2, 0, 2, 4, 6. This can be determined algebraically by assigning the lowest number in the sequence to be “y” and adding 2 for each consecutive even integer, and then setting this equal to zero.
y, y + 2, y + 4, y + 6 . . .
The product of any number and 0 is 0.
Therefore, column B must be greater.
For the sum of 7 consecutive even integers to be zero, the only sequence possible is –6, –4, –2, 0, 2, 4, 6. This can be determined algebraically by assigning the lowest number in the sequence to be “y” and adding 2 for each consecutive even integer, and then setting this equal to zero.
y, y + 2, y + 4, y + 6 . . .
The product of any number and 0 is 0.
Therefore, column B must be greater.
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and are both even whole numbers.
What is a possible solution for ?
What is a possible solution for
If and are both even whole numbers, then their addition must be an even whole number as well. Although is an even number, it is not a whole number and could therefore not be a solution. This means the only possible solution would be .
If
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If John buys twenty-two apples on Monday, and thirty-four bananas on Tuesday, how many pieces of fruit does John have?
If John buys twenty-two apples on Monday, and thirty-four bananas on Tuesday, how many pieces of fruit does John have?
Plugging in the values given we arrive at the total fruit John has:
A good note about adding even numbers--any even numbers--is that if you add two even numbers, their sum will ALWAYS be an even number.
Plugging in the values given we arrive at the total fruit John has:
A good note about adding even numbers--any even numbers--is that if you add two even numbers, their sum will ALWAYS be an even number.
Compare your answer with the correct one above
Choose the answer below which best solves the following problem:
Choose the answer below which best solves the following problem:
To deal with a problem with this many digits, often the best strategy is to line up one number over the other, then add the places one at a time. Don't forget to carry a one every time the addition goes over ten. Also, note that any time you add two even numbers, their sum will ALWAYS be an even number.
To deal with a problem with this many digits, often the best strategy is to line up one number over the other, then add the places one at a time. Don't forget to carry a one every time the addition goes over ten. Also, note that any time you add two even numbers, their sum will ALWAYS be an even number.
Compare your answer with the correct one above
The sum of seven consecutive even integers is 0.
Column A: The product of the seven integers
Column B: 2
The sum of seven consecutive even integers is 0.
Column A: The product of the seven integers
Column B: 2
For the sum of 7 consecutive even integers to be zero, the only sequence possible is –6, –4, –2, 0, 2, 4, 6. This can be determined algebraically by assigning the lowest number in the sequence to be “y” and adding 2 for each consecutive even integer, and then setting this equal to zero.
y, y + 2, y + 4, y + 6 . . .
The product of any number and 0 is 0.
Therefore, column B must be greater.
For the sum of 7 consecutive even integers to be zero, the only sequence possible is –6, –4, –2, 0, 2, 4, 6. This can be determined algebraically by assigning the lowest number in the sequence to be “y” and adding 2 for each consecutive even integer, and then setting this equal to zero.
y, y + 2, y + 4, y + 6 . . .
The product of any number and 0 is 0.
Therefore, column B must be greater.
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and are both even whole numbers.
What is a possible solution for ?
What is a possible solution for
If and are both even whole numbers, then their addition must be an even whole number as well. Although is an even number, it is not a whole number and could therefore not be a solution. This means the only possible solution would be .
If
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If John buys twenty-two apples on Monday, and thirty-four bananas on Tuesday, how many pieces of fruit does John have?
If John buys twenty-two apples on Monday, and thirty-four bananas on Tuesday, how many pieces of fruit does John have?
Plugging in the values given we arrive at the total fruit John has:
A good note about adding even numbers--any even numbers--is that if you add two even numbers, their sum will ALWAYS be an even number.
Plugging in the values given we arrive at the total fruit John has:
A good note about adding even numbers--any even numbers--is that if you add two even numbers, their sum will ALWAYS be an even number.
Compare your answer with the correct one above
Choose the answer below which best solves the following problem:
Choose the answer below which best solves the following problem:
To deal with a problem with this many digits, often the best strategy is to line up one number over the other, then add the places one at a time. Don't forget to carry a one every time the addition goes over ten. Also, note that any time you add two even numbers, their sum will ALWAYS be an even number.
To deal with a problem with this many digits, often the best strategy is to line up one number over the other, then add the places one at a time. Don't forget to carry a one every time the addition goes over ten. Also, note that any time you add two even numbers, their sum will ALWAYS be an even number.
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Of a group of 335 graduating high school atheletes, 106 played basketball, 137 ran track and field, and 51 participated in swimming. What is the maximum number of students that did both track and field and swimming upon graduation?
Of a group of 335 graduating high school atheletes, 106 played basketball, 137 ran track and field, and 51 participated in swimming. What is the maximum number of students that did both track and field and swimming upon graduation?
Simply recognize that logically, the participation of either sport is non-exclusive, that is, just because people took track and field does not necessarily mean they did not take swimming as well. As such, those 51 who took swimming could have all potentially done track and field, which means all 51 students.
Simply recognize that logically, the participation of either sport is non-exclusive, that is, just because people took track and field does not necessarily mean they did not take swimming as well. As such, those 51 who took swimming could have all potentially done track and field, which means all 51 students.
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Of 300 students, 120 are enrolled in math club, 150 are enrolled in chess club, and 100 are enrolled in both. How many students are not members of either club?
Of 300 students, 120 are enrolled in math club, 150 are enrolled in chess club, and 100 are enrolled in both. How many students are not members of either club?
There are 120 students in the math club and 150 students in the chess club, for a total membership of 270. However, 100 students are in both clubs, which means they are counted twice. You simply subtract 100 from 270, which will give you a total of 170 different students participating in both clubs. This means that the remaining 130 students do not participate in either club.
There are 120 students in the math club and 150 students in the chess club, for a total membership of 270. However, 100 students are in both clubs, which means they are counted twice. You simply subtract 100 from 270, which will give you a total of 170 different students participating in both clubs. This means that the remaining 130 students do not participate in either club.
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Choose the answer which best solves the following equation:
Choose the answer which best solves the following equation:
When adding integers, one needs to pay close attention to the sign. When you add a negative integer, it's the same thing as subtracting that integer. Therefore:
When adding integers, one needs to pay close attention to the sign. When you add a negative integer, it's the same thing as subtracting that integer. Therefore:
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