If a and b are consecutive negative integers, is b greater than a?

Question

(1) a + 1 and b – 1 are consecutive negative integers.
(2) b is an odd number.

Kudos for a correct  solution .

chetan2u · Answer

(1) a + 1 and b – 1 are consecutive negative integers....
only possiblity if b>a... so suff
2) b is odd.... nothing about a...insuff
ans A

thembaseeker · Answer

IF a=-5 and b= -4 then a<b-----(i)
IF a=-4 and b=-5 the a>b-------(ii)

From statement (1)

a+1 and b-1 are consecutive -ve inetegers
therefore 
from (i) a+1=-4 and b-1=-5 consecutive therefore a<b
from (ii) a+1=-3 and b-1 =-6 not consecutive doesn't hold.

Hence (1) is sufficient.

From Statement(2) -- Clearly not sufficient

Ans:- A

ramajha · Answer

a=-4,b=-3 a+1=-3,b-1=-4 and b>a so yes statement 1 is sufficient, 
Statement 2. suppose b=-3 or -5 but a is still unknown so can't answer. Answer A

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION:

Question Type: Yes/No  This question asks whether b > a.

Given information from the question stem: a and b are consecutive negative integers.

On this question you can start with Statement 2 first since that statement is easier to evaluate.

Statement 2 : b is an odd number. It should not matter if b is odd when determining if “b > a.” You can quickly Play Devil’s Advocate with a couple of consecutive negative integers. b could = -5 and a could equal -6. In this case b > a. Of course b could also equal -5 and a could equal -4, in which case b is not greater than a. Since you can get both a “yes” and a “no” for this statement it is not sufficient. Eliminate choices B and D (since you did Statement 2 first.)

Statement 1 : a + 1 and b – 1 are consecutive negative integers. You already know from the facts that a and b are consecutive negative integers. Now when you add one to “a” and subtract one from “b” you also have consecutive negative integers. Playing Devil’s Advocate is a good strategy here to take the abstraction of variables and make it concrete with numbers. You can use the numbers from above: Let b = -5 and a = -6. Adding 1 to a and subtracting 1 from b changes it to b = -6 and a = -5, meaning that these are still consecutive negative numbers. So b can be greater (less negative) than a.

Can the opposite be true? Can a be larger than b? Try using the reverse numbers. Make a = -5 and b = -6. Now when you add one to a and subtract one from b the result is a = -4 and b = - 7, so this will not work with Statement 2. a cannot be larger than b; b must be the larger number. The statement is sufficient.  The correct answer is A .

Nevernevergiveup · Answer

If a and b are consecutive negative integers, is b greater than a?

(1)  a + 1  and  b – 1  are consecutive negative integers.
(2) b is an odd number.

Consecutive numbers are numbers that are next to each other in a number line.
So either  a=b+1  or  b=a+1  i.e.,  a>b  or  b>a

Statement 1:   a+1 and b-1 are consecutive numbers
i,e.,  a+1  must be have a distance of +1 or -1 of  b-1  on numerical line of integers.
Thus  (a+1)=(b-1)+1  or  (a+1)=(b-1)-1 
 a+1=b  or  a=b-3 (here a,b have a difference of 3 i.e., these two cannot be consecutive numbers thus failing the given information in statement 1.)

Thus for  a+1=b  given that  a, b  are negative integers b is always greater than a.

Statement 2:   b is an odd number.
b can be any odd negative number and there is no info regarding number a except that it is negative as mentioned.
So it we cannot determine whether b>a or not!
Not sufficient.

Answer choice A

bumpbot · Answer

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If a and b are consecutive negative integers, is b greater than a?

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