How to find the number of integers between two other integers - GRE Quantitative Reasoning
Card 1 of 48
is a positive integer between 200 and 500
Quantity A: The number of possible values of 
with a units digit of 5
Quantity B: 31
Quantity A: The number of possible values of
Quantity B: 31
Tap to reveal answer
An integer with a units digit of 5 occurs once every 10 consecutive integers. There are 300 integers between 200 and 500, so there must be 30 values with a units digit of 5.
An integer with a units digit of 5 occurs once every 10 consecutive integers. There are 300 integers between 200 and 500, so there must be 30 values with a units digit of 5.
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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
Tap to reveal answer
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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0 < x < y < z < 10
x, y, and z are integers.
Quantity A: –7
Quantity B: x + y – z
0 < x < y < z < 10
x, y, and z are integers.
Quantity A: –7
Quantity B: x + y – z
Tap to reveal answer
Since:
- There are only 9 integers between 0 and 10
- x, y, and z must all be unique
- They must be specifically ordered such that x < y < z
There are actually not too many ways in which these numbers can be chosen. So what we can do is find a range of answers for Quantity B, and see if 6 falls a) below b) above or c) in between the range.
For the maximum:
Note that the term (x + y) is maximized when x and y are maximum. The (–z) term is maximized when z is minimized. However, there are 2 terms in (x + y) and one term in (–z); thus intuitively it seems we should prioritize (x+y). To make x and y maximum:
0 < 7 < 8 < 9 < 10 since x, y, and z must be unique.
Thus maximum: (x + y – z) = 7 + 8 – 9 = 6
For the minimum:
Note that (x + y) is minimum when (x) and (y) are minimum, and (–z) is minimum when (z) itself is maximimized. However since there are 2 terms in (x+y) and1 of (–z) , again intuititively you should prioritize (x+y) over (-z). Then in order to make this the least number possible, x and y would be:
min(x + y – z) = 1 + 2 – 9 = –6
Thus, the range of possible answers is:
(x + y – z): \[–6, 6\]
and –7 is always less than this amount.
Since:
- There are only 9 integers between 0 and 10
- x, y, and z must all be unique
- They must be specifically ordered such that x < y < z
There are actually not too many ways in which these numbers can be chosen. So what we can do is find a range of answers for Quantity B, and see if 6 falls a) below b) above or c) in between the range.
For the maximum:
Note that the term (x + y) is maximized when x and y are maximum. The (–z) term is maximized when z is minimized. However, there are 2 terms in (x + y) and one term in (–z); thus intuitively it seems we should prioritize (x+y). To make x and y maximum:
0 < 7 < 8 < 9 < 10 since x, y, and z must be unique.
Thus maximum: (x + y – z) = 7 + 8 – 9 = 6
For the minimum:
Note that (x + y) is minimum when (x) and (y) are minimum, and (–z) is minimum when (z) itself is maximimized. However since there are 2 terms in (x+y) and1 of (–z) , again intuititively you should prioritize (x+y) over (-z). Then in order to make this the least number possible, x and y would be:
min(x + y – z) = 1 + 2 – 9 = –6
Thus, the range of possible answers is:
(x + y – z): \[–6, 6\]
and –7 is always less than this amount.
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Quantity A: The number of positive even integers less than 1000
Quantity B: The number of positive odd integers less than 1000
Quantity A: The number of positive even integers less than 1000
Quantity B: The number of positive odd integers less than 1000
Tap to reveal answer
The question asks for the number of positive even and odd integers less than 1000. Because 1000 is not included, the numbers to consider are 1 through 999. Every positive odd integer will have a corresponding even integer (1 and 2, 3 and 4, 5 and 6, etc.) until you get to 999. This gives the positive odd integers one more number than the number of positive even integers.
The question asks for the number of positive even and odd integers less than 1000. Because 1000 is not included, the numbers to consider are 1 through 999. Every positive odd integer will have a corresponding even integer (1 and 2, 3 and 4, 5 and 6, etc.) until you get to 999. This gives the positive odd integers one more number than the number of positive even integers.
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Miles is 3 years older than Ashley. Ashley is 5 years younger than Bill. How old is Ashley if together the three of their ages sum to 44?
Miles is 3 years older than Ashley. Ashley is 5 years younger than Bill. How old is Ashley if together the three of their ages sum to 44?
Tap to reveal answer
Miles is 3 years older than Ashley, so M = A + 3. Also, Bill is 5 years older than Ashley, so B = A + 5. Together the three of their ages sum to 44, thus:
A + A + 3 + A + 5 = 44
3_A_ + 8 = 44
3_A_ = 36
A = 12
Miles is 3 years older than Ashley, so M = A + 3. Also, Bill is 5 years older than Ashley, so B = A + 5. Together the three of their ages sum to 44, thus:
A + A + 3 + A + 5 = 44
3_A_ + 8 = 44
3_A_ = 36
A = 12
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In a certain game, integers are called magic numbers if they are multiples of either 
or 
.
How many magic numbers are there in the game between 
and 
?
In a certain game, integers are called magic numbers if they are multiples of either
How many magic numbers are there in the game between
Tap to reveal answer
There are 13 "magic" numbers: 3,4,6,8,9,12, their negative counterparts, and 0.
There are 13 "magic" numbers: 3,4,6,8,9,12, their negative counterparts, and 0.
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is a positive integer between 200 and 500
Quantity A: The number of possible values of with a units digit of 5
Quantity B: 31
Quantity A: The number of possible values of
Quantity B: 31
Tap to reveal answer
An integer with a units digit of 5 occurs once every 10 consecutive integers. There are 300 integers between 200 and 500, so there must be 30 values with a units digit of 5.
An integer with a units digit of 5 occurs once every 10 consecutive integers. There are 300 integers between 200 and 500, so there must be 30 values with a units digit of 5.
← Didn't Know|Knew It →
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
Tap to reveal answer
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.
← Didn't Know|Knew It →
0 < x < y < z < 10
x, y, and z are integers.
Quantity A: –7
Quantity B: x + y – z
0 < x < y < z < 10
x, y, and z are integers.
Quantity A: –7
Quantity B: x + y – z
Tap to reveal answer
Since:
- There are only 9 integers between 0 and 10
- x, y, and z must all be unique
- They must be specifically ordered such that x < y < z
There are actually not too many ways in which these numbers can be chosen. So what we can do is find a range of answers for Quantity B, and see if 6 falls a) below b) above or c) in between the range.
For the maximum:
Note that the term (x + y) is maximized when x and y are maximum. The (–z) term is maximized when z is minimized. However, there are 2 terms in (x + y) and one term in (–z); thus intuitively it seems we should prioritize (x+y). To make x and y maximum:
0 < 7 < 8 < 9 < 10 since x, y, and z must be unique.
Thus maximum: (x + y – z) = 7 + 8 – 9 = 6
For the minimum:
Note that (x + y) is minimum when (x) and (y) are minimum, and (–z) is minimum when (z) itself is maximimized. However since there are 2 terms in (x+y) and1 of (–z) , again intuititively you should prioritize (x+y) over (-z). Then in order to make this the least number possible, x and y would be:
min(x + y – z) = 1 + 2 – 9 = –6
Thus, the range of possible answers is:
(x + y – z): \[–6, 6\]
and –7 is always less than this amount.
Since:
- There are only 9 integers between 0 and 10
- x, y, and z must all be unique
- They must be specifically ordered such that x < y < z
There are actually not too many ways in which these numbers can be chosen. So what we can do is find a range of answers for Quantity B, and see if 6 falls a) below b) above or c) in between the range.
For the maximum:
Note that the term (x + y) is maximized when x and y are maximum. The (–z) term is maximized when z is minimized. However, there are 2 terms in (x + y) and one term in (–z); thus intuitively it seems we should prioritize (x+y). To make x and y maximum:
0 < 7 < 8 < 9 < 10 since x, y, and z must be unique.
Thus maximum: (x + y – z) = 7 + 8 – 9 = 6
For the minimum:
Note that (x + y) is minimum when (x) and (y) are minimum, and (–z) is minimum when (z) itself is maximimized. However since there are 2 terms in (x+y) and1 of (–z) , again intuititively you should prioritize (x+y) over (-z). Then in order to make this the least number possible, x and y would be:
min(x + y – z) = 1 + 2 – 9 = –6
Thus, the range of possible answers is:
(x + y – z): \[–6, 6\]
and –7 is always less than this amount.
← Didn't Know|Knew It →
Quantity A: The number of positive even integers less than 1000
Quantity B: The number of positive odd integers less than 1000
Quantity A: The number of positive even integers less than 1000
Quantity B: The number of positive odd integers less than 1000
Tap to reveal answer
The question asks for the number of positive even and odd integers less than 1000. Because 1000 is not included, the numbers to consider are 1 through 999. Every positive odd integer will have a corresponding even integer (1 and 2, 3 and 4, 5 and 6, etc.) until you get to 999. This gives the positive odd integers one more number than the number of positive even integers.
The question asks for the number of positive even and odd integers less than 1000. Because 1000 is not included, the numbers to consider are 1 through 999. Every positive odd integer will have a corresponding even integer (1 and 2, 3 and 4, 5 and 6, etc.) until you get to 999. This gives the positive odd integers one more number than the number of positive even integers.
← Didn't Know|Knew It →
Miles is 3 years older than Ashley. Ashley is 5 years younger than Bill. How old is Ashley if together the three of their ages sum to 44?
Miles is 3 years older than Ashley. Ashley is 5 years younger than Bill. How old is Ashley if together the three of their ages sum to 44?
Tap to reveal answer
Miles is 3 years older than Ashley, so M = A + 3. Also, Bill is 5 years older than Ashley, so B = A + 5. Together the three of their ages sum to 44, thus:
A + A + 3 + A + 5 = 44
3_A_ + 8 = 44
3_A_ = 36
A = 12
Miles is 3 years older than Ashley, so M = A + 3. Also, Bill is 5 years older than Ashley, so B = A + 5. Together the three of their ages sum to 44, thus:
A + A + 3 + A + 5 = 44
3_A_ + 8 = 44
3_A_ = 36
A = 12
← Didn't Know|Knew It →
In a certain game, integers are called magic numbers if they are multiples of either or .
How many magic numbers are there in the game between and ?
In a certain game, integers are called magic numbers if they are multiples of either
How many magic numbers are there in the game between
Tap to reveal answer
There are 13 "magic" numbers: 3,4,6,8,9,12, their negative counterparts, and 0.
There are 13 "magic" numbers: 3,4,6,8,9,12, their negative counterparts, and 0.
← Didn't Know|Knew It →
is a positive integer between 200 and 500
Quantity A: The number of possible values of with a units digit of 5
Quantity B: 31
Quantity A: The number of possible values of
Quantity B: 31
Tap to reveal answer
An integer with a units digit of 5 occurs once every 10 consecutive integers. There are 300 integers between 200 and 500, so there must be 30 values with a units digit of 5.
An integer with a units digit of 5 occurs once every 10 consecutive integers. There are 300 integers between 200 and 500, so there must be 30 values with a units digit of 5.
← Didn't Know|Knew It →
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
Tap to reveal answer
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.
← Didn't Know|Knew It →
0 < x < y < z < 10
x, y, and z are integers.
Quantity A: –7
Quantity B: x + y – z
0 < x < y < z < 10
x, y, and z are integers.
Quantity A: –7
Quantity B: x + y – z
Tap to reveal answer
Since:
- There are only 9 integers between 0 and 10
- x, y, and z must all be unique
- They must be specifically ordered such that x < y < z
There are actually not too many ways in which these numbers can be chosen. So what we can do is find a range of answers for Quantity B, and see if 6 falls a) below b) above or c) in between the range.
For the maximum:
Note that the term (x + y) is maximized when x and y are maximum. The (–z) term is maximized when z is minimized. However, there are 2 terms in (x + y) and one term in (–z); thus intuitively it seems we should prioritize (x+y). To make x and y maximum:
0 < 7 < 8 < 9 < 10 since x, y, and z must be unique.
Thus maximum: (x + y – z) = 7 + 8 – 9 = 6
For the minimum:
Note that (x + y) is minimum when (x) and (y) are minimum, and (–z) is minimum when (z) itself is maximimized. However since there are 2 terms in (x+y) and1 of (–z) , again intuititively you should prioritize (x+y) over (-z). Then in order to make this the least number possible, x and y would be:
min(x + y – z) = 1 + 2 – 9 = –6
Thus, the range of possible answers is:
(x + y – z): \[–6, 6\]
and –7 is always less than this amount.
Since:
- There are only 9 integers between 0 and 10
- x, y, and z must all be unique
- They must be specifically ordered such that x < y < z
There are actually not too many ways in which these numbers can be chosen. So what we can do is find a range of answers for Quantity B, and see if 6 falls a) below b) above or c) in between the range.
For the maximum:
Note that the term (x + y) is maximized when x and y are maximum. The (–z) term is maximized when z is minimized. However, there are 2 terms in (x + y) and one term in (–z); thus intuitively it seems we should prioritize (x+y). To make x and y maximum:
0 < 7 < 8 < 9 < 10 since x, y, and z must be unique.
Thus maximum: (x + y – z) = 7 + 8 – 9 = 6
For the minimum:
Note that (x + y) is minimum when (x) and (y) are minimum, and (–z) is minimum when (z) itself is maximimized. However since there are 2 terms in (x+y) and1 of (–z) , again intuititively you should prioritize (x+y) over (-z). Then in order to make this the least number possible, x and y would be:
min(x + y – z) = 1 + 2 – 9 = –6
Thus, the range of possible answers is:
(x + y – z): \[–6, 6\]
and –7 is always less than this amount.
← Didn't Know|Knew It →
Quantity A: The number of positive even integers less than 1000
Quantity B: The number of positive odd integers less than 1000
Quantity A: The number of positive even integers less than 1000
Quantity B: The number of positive odd integers less than 1000
Tap to reveal answer
The question asks for the number of positive even and odd integers less than 1000. Because 1000 is not included, the numbers to consider are 1 through 999. Every positive odd integer will have a corresponding even integer (1 and 2, 3 and 4, 5 and 6, etc.) until you get to 999. This gives the positive odd integers one more number than the number of positive even integers.
The question asks for the number of positive even and odd integers less than 1000. Because 1000 is not included, the numbers to consider are 1 through 999. Every positive odd integer will have a corresponding even integer (1 and 2, 3 and 4, 5 and 6, etc.) until you get to 999. This gives the positive odd integers one more number than the number of positive even integers.
← Didn't Know|Knew It →
Miles is 3 years older than Ashley. Ashley is 5 years younger than Bill. How old is Ashley if together the three of their ages sum to 44?
Miles is 3 years older than Ashley. Ashley is 5 years younger than Bill. How old is Ashley if together the three of their ages sum to 44?
Tap to reveal answer
Miles is 3 years older than Ashley, so M = A + 3. Also, Bill is 5 years older than Ashley, so B = A + 5. Together the three of their ages sum to 44, thus:
A + A + 3 + A + 5 = 44
3_A_ + 8 = 44
3_A_ = 36
A = 12
Miles is 3 years older than Ashley, so M = A + 3. Also, Bill is 5 years older than Ashley, so B = A + 5. Together the three of their ages sum to 44, thus:
A + A + 3 + A + 5 = 44
3_A_ + 8 = 44
3_A_ = 36
A = 12
← Didn't Know|Knew It →
In a certain game, integers are called magic numbers if they are multiples of either or .
How many magic numbers are there in the game between and ?
In a certain game, integers are called magic numbers if they are multiples of either
How many magic numbers are there in the game between
Tap to reveal answer
There are 13 "magic" numbers: 3,4,6,8,9,12, their negative counterparts, and 0.
There are 13 "magic" numbers: 3,4,6,8,9,12, their negative counterparts, and 0.
← Didn't Know|Knew It →
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
Tap to reveal answer
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.
← Didn't Know|Knew It →
0 < x < y < z < 10
x, y, and z are integers.
Quantity A: –7
Quantity B: x + y – z
0 < x < y < z < 10
x, y, and z are integers.
Quantity A: –7
Quantity B: x + y – z
Tap to reveal answer
Since:
- There are only 9 integers between 0 and 10
- x, y, and z must all be unique
- They must be specifically ordered such that x < y < z
There are actually not too many ways in which these numbers can be chosen. So what we can do is find a range of answers for Quantity B, and see if 6 falls a) below b) above or c) in between the range.
For the maximum:
Note that the term (x + y) is maximized when x and y are maximum. The (–z) term is maximized when z is minimized. However, there are 2 terms in (x + y) and one term in (–z); thus intuitively it seems we should prioritize (x+y). To make x and y maximum:
0 < 7 < 8 < 9 < 10 since x, y, and z must be unique.
Thus maximum: (x + y – z) = 7 + 8 – 9 = 6
For the minimum:
Note that (x + y) is minimum when (x) and (y) are minimum, and (–z) is minimum when (z) itself is maximimized. However since there are 2 terms in (x+y) and1 of (–z) , again intuititively you should prioritize (x+y) over (-z). Then in order to make this the least number possible, x and y would be:
min(x + y – z) = 1 + 2 – 9 = –6
Thus, the range of possible answers is:
(x + y – z): \[–6, 6\]
and –7 is always less than this amount.
Since:
- There are only 9 integers between 0 and 10
- x, y, and z must all be unique
- They must be specifically ordered such that x < y < z
There are actually not too many ways in which these numbers can be chosen. So what we can do is find a range of answers for Quantity B, and see if 6 falls a) below b) above or c) in between the range.
For the maximum:
Note that the term (x + y) is maximized when x and y are maximum. The (–z) term is maximized when z is minimized. However, there are 2 terms in (x + y) and one term in (–z); thus intuitively it seems we should prioritize (x+y). To make x and y maximum:
0 < 7 < 8 < 9 < 10 since x, y, and z must be unique.
Thus maximum: (x + y – z) = 7 + 8 – 9 = 6
For the minimum:
Note that (x + y) is minimum when (x) and (y) are minimum, and (–z) is minimum when (z) itself is maximimized. However since there are 2 terms in (x+y) and1 of (–z) , again intuititively you should prioritize (x+y) over (-z). Then in order to make this the least number possible, x and y would be:
min(x + y – z) = 1 + 2 – 9 = –6
Thus, the range of possible answers is:
(x + y – z): \[–6, 6\]
and –7 is always less than this amount.
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