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Bunuel
Is \(q^r + 1\) an odd number?

(1) q + r is even
(2) q is even

(1) q + r is even

Let q=2 & r= 0..........\(2^0 + 1\)=2..........Answer No

Let q=2 & r= 2..........\(2^2 + 1\)=5..........Answer Yes

Insufficient

(2) q is even

Use same examples above

Insufficient

Combine 1 & 2

Use same examples.............No clear answer

Insufficient

Answer: E
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Bunuel
Is \(q^r + 1\) an odd number?

(1) q + r is even
(2) q is even


we are asked to determine whether \(q^r\) + 1 = odd or not.

Statement 1 :

q + r = even . This fact can be proved by 2 ways.

1. even + even = even . 2 +2 = 4
2. odd + odd = even . 3 + 1 = 4

If both q and r are even , we will always get odd number but the scenario is different when q and r are odd.

NOT sufficient.

Statement 2 :

This statement is a trap. anyone can say that if q is even \(q^r\) + 1 will be odd. But we don't sufficient information about r. Thus this statement becomes fruitless. r could be anything. so, NOT sufficient .

combining both statements:

if q is even r has to be even if we consider statement 1. even or odd can't be fraction .

So, both statements together are sufficient.

Thus the best answer is C.
If you take r=0 then q^r+1=even.so together also it is insufficient to ans.
Hence E is right choice.

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