column A: 2a+b^2 ? column B: a^2+2b

plug in the value a=b+1 in both column

2b+2+b^2 ? b^2+2b+1+2b

2 ? 1+2b [eliminating the common terms]

1 ? 2b [subtacting 1]

1/2 ? b [divide by 2]


case-1: if b= 2 then a=3 ab=6 [greater than 1/2]
case-2: if b=-3 then a=-2 ab=6 [less than 1/2]

b can be both positive and negative and still satisfies the condition 5<ab<8
therefore, D will be the answer.

We can solve this question using matching operations (my favorite Quantitative Comparison strategy)

Given:
Quantity A: \(2a+b^2\)
Quantity B: \(a^2+2b\)

Since we're told \(a = b + 1\), we can replace \(a\) with \(b + 1\) to get:
Quantity A: \(2(b + 1)+b^2\)
Quantity B: \((b + 1)^2+2b\)

Expand and simplify both quantities:
Quantity A: \(b^2+2b+2\)
Quantity B: \(b^2+4b+1\)

Subtract \(b^2\) from both quantities:
Quantity A: \(2b+2\)
Quantity B: \(4b+1\)

Subtract \(2b\) from both quantities:
Quantity A: \(2\)
Quantity B: \(2b+1\)

Subtract \(1\) from both quantities:
Quantity A: \(1\)
Quantity B: \(2b\)

Divide both quantities by \(2\):
Quantity A: \(0.5\)
Quantity B: \(b\)

We're told that 5 < ab < 8, however there are infinitely many values of a and b that satisfy this inequality.
For example, it could be the case that a = -2 and b = -3, in which case, Quantity A is greater
It could also be the case that a = 3 and b = 2, in which case, Quantity B is greater

Answer: D

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I am a bit confused with the solution posted. If a = 1 and b=6, wouldn't that not satisfy the requirement that a = b +1?

Good catch!
I've edited my response accordingly

Kudos for you!!!!
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