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08 Sep 2022, 00:39
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Difficulty:
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Question Stats:
87% (01:21) correct
13%
(01:32)
wrong
based on 55
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If p and q are consecutive positive integers, is p greater than q ?
(1) p − 1 and q + 1 are consecutive positive integers.
(2) q is an odd integer.
(1) p − 1 and q + 1 are consecutive positive integers.
(2) q is an odd integer.
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08 Sep 2022, 02:06
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If p and q are consecutive positive integers, is p greater than q ?
We need to find whether \(p>q\)
By picking numbers
(1) \(p − 1\) and \(q + \)1 are consecutive positive integers.
\(p<q\)
Let \(p=3\) and \(q=4\)
then \(p-1=3-1=\)2
and \(q+1=4+1 = 5\)
So \(p-1\) and \(q+\)1 are not consecutive
\(p>q\)
Let \(p=6\) and \(q=\)5
then \(p-1=5\)
and \(q+1=6\)
So \(p-1\) and \(q+1\) are consecutive. Therefore \(p>q\).
(2) q is an odd integer.
Let say \(q=5\) , \(p\) can be 4 or even 6. So we cannot say whether \(p>q\).
So correct answer is A, Statement 1 is enough to answer whether \(p>q\).
Solving algebraically
(1) \(p − 1\) and \(q + 1\) are consecutive positive integers.
Case 1: \(p>q\), then if \(q=a\), \(p=a+1\) as p and q are consecutive integers
then according to Statement 1
\(p-1= (a+1)-1= a\)
\(q+1= (a)+1 = a+1 \)
So \(p-1\) and \(q+1\) end up as consecutive integers, as the difference between these two numbers is 1 \((a+1)-(a)=1\)
Case 2: \(p<q\), then if \(p=a\), \(q=a+1\) as p and q are consecutive integers
then according to Statement 1
\(p-1= (a)-1= a-1\)
\(q+1= (a+1)+1 = a+2 \)
So \(p-1\) and \(q+1\) doesn't end up as consecutive integers, as difference between these two numbers is 3 and hence more than 1 \({(a+2)-(a-1)=3}\)
(2) \(q\) is an odd integer.
It doesn't give enough information to answer whether \(p>q\). In any two consecutive integers an odd number can be the larger number or it can be a smaller number of the two numbers.
So correct answer is A, Statement 1 is enough to answer whether \(p>q\).
We need to find whether \(p>q\)
By picking numbers
(1) \(p − 1\) and \(q + \)1 are consecutive positive integers.
\(p<q\)
Let \(p=3\) and \(q=4\)
then \(p-1=3-1=\)2
and \(q+1=4+1 = 5\)
So \(p-1\) and \(q+\)1 are not consecutive
\(p>q\)
Let \(p=6\) and \(q=\)5
then \(p-1=5\)
and \(q+1=6\)
So \(p-1\) and \(q+1\) are consecutive. Therefore \(p>q\).
(2) q is an odd integer.
Let say \(q=5\) , \(p\) can be 4 or even 6. So we cannot say whether \(p>q\).
So correct answer is A, Statement 1 is enough to answer whether \(p>q\).
Solving algebraically
(1) \(p − 1\) and \(q + 1\) are consecutive positive integers.
Case 1: \(p>q\), then if \(q=a\), \(p=a+1\) as p and q are consecutive integers
then according to Statement 1
\(p-1= (a+1)-1= a\)
\(q+1= (a)+1 = a+1 \)
So \(p-1\) and \(q+1\) end up as consecutive integers, as the difference between these two numbers is 1 \((a+1)-(a)=1\)
Case 2: \(p<q\), then if \(p=a\), \(q=a+1\) as p and q are consecutive integers
then according to Statement 1
\(p-1= (a)-1= a-1\)
\(q+1= (a+1)+1 = a+2 \)
So \(p-1\) and \(q+1\) doesn't end up as consecutive integers, as difference between these two numbers is 3 and hence more than 1 \({(a+2)-(a-1)=3}\)
(2) \(q\) is an odd integer.
It doesn't give enough information to answer whether \(p>q\). In any two consecutive integers an odd number can be the larger number or it can be a smaller number of the two numbers.
So correct answer is A, Statement 1 is enough to answer whether \(p>q\).
11 Sep 2022, 22:58
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question stem
Since p,q are consecutive integers
either
p = q+1 , since they are +ve p < q
or
p = q-1 , , since they are +ve p > q
We need to find out if we can figure out which one is true
Statement 1 :
similar to the question stem ,
p-1 = q+1+1
=> p = q+3 , we can cancel this solution as p,q are not consecutive in this case.
or
p-1 = q +1 -1
p = q+1 , This makes p,q consecutive as well , this solution matches with our deduction in question stem .
Option A is sufficient
Option B,C,E is gone
We are left with A and D
Statement 2 :
We can take an example , q=5 , p can be 4 or 6 and make the question stem correct .We cannot deduct if p is > q .
Hence this is insufficient .
Option A is correct choice .
Quote:
Since p,q are consecutive integers
either
p = q+1 , since they are +ve p < q
or
p = q-1 , , since they are +ve p > q
We need to find out if we can figure out which one is true
Statement 1 :
Quote:
similar to the question stem ,
p-1 = q+1+1
=> p = q+3 , we can cancel this solution as p,q are not consecutive in this case.
or
p-1 = q +1 -1
p = q+1 , This makes p,q consecutive as well , this solution matches with our deduction in question stem .
Option A is sufficient
Option B,C,E is gone
We are left with A and D
Statement 2 :
Quote:
We can take an example , q=5 , p can be 4 or 6 and make the question stem correct .We cannot deduct if p is > q .
Hence this is insufficient .
Option A is correct choice .
